Explanation:
It is given that,
Mass of a car is 900 kg
Radius of curve is 600 m
Speed of the car in the curve is 25 m/s
We need to find the centripetal force exerted on a car. The formula used to find the centripetal force is given by :
[tex]F=\dfrac{mv^2}{r}\\\\F=\dfrac{900\times (25)^2}{600}\\\\F=937.5\ N[/tex]
So, the centripetal force exerted on a car is 937.5 N.
Static friction prevents the car from slipping. It means that the magnitude of centripetal force is balanced by the frictional force. So, the frictional force of 937.5 N is acting on the car.
The froghopper, a tiny insect, is a remarkable jumper. Suppose a colony of the little critters is raised on Rhea, a moon of Saturn, where the acceleration due to gravity is only 0.264 m/s2 , whereas gravity on Earth is =9.81 m/s2 . If on Earth a froghopper's maximum jump height is ℎ and its maximum horizontal jump range is R, what would its maximum jump height and range be on Rhea in terms of ℎ and R? Assume the froghopper's takeoff velocity is the same on Rhea and Earth.
Answer:
Maximum height of jump on Rhea is 37.16 times of that on Earth, i.e 37.16h
Maximum range of jump on Rhea is 37.16 of times that on Earth, i.e 37.16R
Explanation:
The acceleration due to gravity on Rhea = 0.264 m/s^2
Acceleration due to gravity on earth here = 9.81 m/s^2
this means that the acceleration due to gravity g on earth is 9.81/0.264 = 37.16 times that on Rhea.
maximum height that can be achieved by the froghopper is given by the equation;
h = [tex]\frac{u^{2}sin^{2} \alpha}{2g}[/tex]
let us put all the numerator of the equation as k, since the velocity of take off is the same for Earth and Rhea. The equation is simplified to
h = [tex]\frac{k}{2g}[/tex]
for earth,
h = [tex]\frac{k}{2*9.81}[/tex] = [tex]\frac{k}{19.62}[/tex]
for Rhea,
h = [tex]\frac{k}{2*0.264}[/tex] = [tex]\frac{k}{0.528}[/tex]
therefore,
h on Rhea is [tex]\frac{k}{0.528}[/tex] ÷ [tex]\frac{k}{19.62}[/tex] = 37.16 times of that on Earth, i.e 37.16h
Equation for range R is given as
R = [tex]\frac{u^{2}sin 2\alpha}{g}[/tex]
following the same approach as before,
R on Rhea will be [tex]\frac{k}{0.264}[/tex] ÷ [tex]\frac{k}{9.81}[/tex] = 37.16 of times that on Earth, i.e 37.16R
Lebron James and Stephen Curry are playing an intense game of minigolf. The final(18th) hole is 8.2 m away from the tee box (starting location) at an angle of 20◦ east of north. Lebron’s first shot lands 8.6 m away at an angle of 35.2◦ east of north and Steph’s first shot lands 6.1 m away at an angle of 20◦ east of north. Assume that the minigolf course is flat.
(A) Which ball lands closer to the hole?
(B) Each player sunk the ball on the second shot. At what angle did each player hit their ball to reach the hole?
Answer:
A. we will see that the notion [tex]\mathbf{|r ^ \to - r_2 ^\to| = 2.10006 \ m}[/tex] which denotes Stephen Curry illustrates that Stephen Curry minigolf ball shot is closer
B. Lebron James hits at an angle of 17.48° North -East.
The direction of Stephen is = 20° due to East of North
Explanation:
Let [tex]r ^ {\to[/tex] represent the position vector of the hole;
Also; using the origin as starting point. Let the east direction be along the positive x axis and the North direction be + y axis
Thus:
[tex]r ^ {\to[/tex] = [tex]8.2 \ sin 20^0 \hat i + 8.2 \ cos 20 \hat j[/tex]
[tex]r ^ {\to[/tex] = [tex](2.8046 \hat i + 7.7055 \hat j ) m[/tex]
Let [tex]r_1 ^ \to[/tex] be the position vector for Lebron James's first shot
So;
[tex]r_1 ^ \to[/tex] = [tex](8.6 \ sin \ 35.2 )^0 \hat i + 8.6 \ cos \ ( 35.2)^0 \hat j[/tex]
[tex]r^ \to = (4.9573 \hat i + 7.02745 \hat j) m[/tex]
Let [tex]r_2 ^ \to[/tex] be the position vector for Stephen Curry's shot
[tex]r_2 ^ \to[/tex] [tex]=6.1 \ sin 20^0 \hat i + 6.1 \ cos \ 20 \hat j[/tex]
[tex]r_2 ^ \to[/tex] = [tex](2.0863 \hat i + 5.7321 \hat j )m[/tex]
However;
[tex]r ^ \to - r_1 ^\to = (-2.1527 \hat i + 0.67805 \hat j) m[/tex]
[tex]\mathbf{|r ^ \to - r_1 ^\to| =2.25696 \ m }[/tex]
Also;
[tex]r ^ \to - r_2 ^\to = (0.71013 \hat i - 1.9734 \hat j) m[/tex]
[tex]\mathbf{|r ^ \to - r_2 ^\to| = 2.10006 \ m}[/tex]
Thus; from above ; we will see that the notion [tex]\mathbf{|r ^ \to - r_2 ^\to| = 2.10006 \ m}[/tex] which denotes Stephen Curry illustrates that Stephen Curry minigolf ball shot is closer
B .
For Lebron James ;
The angle can be determine using the trigonometric function:
[tex]tan \theta = ( \dfrac{0.67805}{-2.1527}) \\ \\ tan \theta = -0.131498 \\ \\ \theta = tan ^{-1} ( -0.31498) \\ \\ \mathbf{\theta = -17.48^0}[/tex]
Thus Lebron James hits at an angle of 17.48° North -East.
For Stephen Curry;
[tex]tan \theta = ( \dfrac{-1.9734}{0.7183}) \\ \\ tan \theta = -2.74732 \\ \\ \theta = tan ^{-1} ( -2.74732) \\ \\ \mathbf{\theta = -70.0^0}[/tex]
The direction of Stephen is = 90° - 70° = 20° due to East of North
3. A ray of light incident on one face of an equilateral glass prism is refracted in such a way that it emerges from the opposite surface at an angle of 900 to the normal. Calculate the i. angle of incidence. ii. minimum deviation of the ray of light passing through the prism [n_glass=1.52]
Answer:
i) angle of incidence;i = 29.43°
ii) δm = 38.92°
Explanation:
Prism is equilateral so angle of prism (A) = 60°
Refractive index of glass; n_glass = 1.52
A) Let's assume the incident angle = i and Critical angle = θc
We know that, sin θc = 1/n
Thus;
sin θc = 1/n_glass
θc = sin^(-1) (1/n_glass)
θc = sin^(-1) (1/1.52)
θc = 41.14°
Now, the angle of prism will be the sum of external angle that is critical angle and reflected angle.
Thus;
A = r + θc
r = A - θc
So;
r = 60° - 41. 14°
r = 18.86°
From, Snell's law. If we apply it to this question, we will have;
(sin i)/(sin r) = n_glass
Where;
i is angle of incidence and r is angle of reflection.
Let's make i the subject;
i = sin^(-1) (n_glass × sin r)
i = sin^(-1) (1.52 × sin 18.86)
i = sin^(-1) 0.4914
i = 29.43°
B) The formula to calculate minimum deviation would be from;
μ = [sin ((A + δm)/2)]/(sin A/2)
Where;
μ is Refractive index
δm is minimum angle of deviation
A is angle of prism
Now Refractive index is given by a formula; μ = (sin i)/(sin r)
So; μ = (sin 29.43)/(sin 18.86)
μ = 1.52
Thus;
1.52 = [sin ((60 + δm)/2)]/(sin 60/2)
1.52 * sin 30 = sin ((60 + δm)/2)
0.76 = sin ((60 + δm)/2)
sin^(-1) 0.76 = ((60 + δm)/2)
49.46 × 2 = (60 + δm)
98.92 - 60 = δm
δm = 38.92°
A car moving in a straight line starts at X=0 at t=0. It passesthe point x=25.0 m with a speed of 11.0 m/s at t=3.0 s. It passes the point x=385 with a speed of 45.0 m/s at t=20.0 s. Find the average velocity and the average acceleration between t=3.0 s and 20.0 s.
Answer:
Average velocity v = 21.18 m/s
Average acceleration a = 2 m/s^2
Explanation:
Average speed equals the total distance travelled divided by the total time taken.
Average speed v = ∆x/∆t = (x2-x1)/(t2-t1)
Average acceleration equals the change in velocity divided by change in time.
Average acceleration a = ∆v/∆t = (v2-v1)/(t2-t1)
Where;
v1 and v2 are velocities at time t1 and t2 respectively.
And x1 and x2 are positions at time t1 and t2 respectively.
Given;
t1 = 3.0s
t2 = 20.0s
v1 = 11 m/s
v2 = 45 m/s
x1 = 25 m
x2 = 385 m
Substituting the values;
Average speed v = ∆x/∆t = (x2-x1)/(t2-t1)
v = (385-25)/(20-3)
v = 21.18 m/s
Average acceleration a = ∆v/∆t = (v2-v1)/(t2-t1)
a = (45-11)/(20-3)
a = 2 m/s^2
A sphere of diameter 6.0cm is moulded into a thin uniform wire of diameter 0.2mm. Calculate the length of the wire in metres (Take π = 22/7) *
Answer:
2025m
Explanation:
Since all materials of the sphere is made to a cylindrical wire, it implies the volume of the sphere material is same as that of the cylinder. This is expressed mathematically thus.
Volume of Sphere= volume of cylinder
4/3 ×π×R^3= π× r2× L
4/3 ×R^3= r^2×L
Hence
L = 3/4 × R^3/ r^2
But R = 6.0/2 = 3.0cm{ Diameter is twice raduis}
r= 0.2/2 = 0.1mm=>0.01cm{ Diameter is twice raduis and unit converted by dividing by 10 since 10mm = 1cm}
Substituting R and r into the expression for L, we have :
L = 3/4 × 3^3/ 0.01^2= 0.75 ×27/0.0001 = 202500cm
202500/100= 2025m{ we divide by 100 because 100cm=1m}
an aluminium bar 600mm long, with diameter 40mm, has a hole drilled in the center of the bar. the hole is 30mm in diameter and is 30mm and is 100mm long. if modulus of elasticity for the aluminium is 85GN/m2, calculate the total contraction on the bar due to a compressive load of 180KN
Answer:
ΔL = 1.011 mm
Explanation:
Let's begin by listing out the given information:
Length (L) = 600 mm = 0.6 m,
Diameter (D) = 40 mm = 0.04 m ⇒ Radius (r) = 20 mm = 0.2 m,
Area (cross sectional) = πr² = 3.14 x .02² = 0.001256 m²,
Modulus of Elasticity (E) = 85 GN/m²,
Compressive load (F) = 180 KN
Using the formula, Stress = Load ÷ Area
Mathematically,
σ = F ÷ A = 180 x 10³ ÷ 0.001256
σ = 143312.1 KN/m²
Modulus of elasticity = stress ÷ strain
E = σ ÷ ε
ε = ΔL/L
85 x 10⁹ = 143312.1 x 10³ ÷ (ΔL/L)
ΔL = 143312.1 x 10³ ÷ 85 X 10⁹ = 1686.02 * 10⁻⁶
ΔL = L x 1686.02 * 10⁻⁶
ΔL = 0.6 * 1686.02 * 10⁻⁶ = 1011.61 x 10⁻⁶
ΔL = 1.011 x 10⁻³ m
ΔL = 1.011 mm
∴The bar contracts by 1.011 mm
Two parallel plates having charges of equal magnitude but opposite sign are separated by 21.0 cm. Each plate has a surface charge density of 39.0 nC/m2. A proton is released from rest at the positive plate. (a) Determine the magnitude of the electric field between the plates from the charge density.
Answer:
E = 3.45*10^-19 N/C
Explanation:
a) The electric field between two parallel plates id given by the following formula:
[tex]E=\frac{\sigma}{\epsilon_o}[/tex] (1)
where:
σ: surface charge density of the plates = 39.0nC/m^2
εo: dielectric permittivity of vacuum = 8.85*10^-12 C/Nm^2
You replace these values in the equation (1):
[tex]E=\frac{39.0*10^{-9}C/m^2}{8.85*10^{-12}C^2/Nm^2}\\\\E=3.45*10^{-19}\frac{N}{C}[/tex]
The electric field in between the parallel plates is 3.45*10^-19 N/C
A surveyor measures the distance across a river that flows straight north by the following method. Starting directly across from a tree on the opposite bank, the surveyor walks distance, D = 130 m along the river to establish a baseline. She then sights across to the tree and reads that the angle from the baseline to the tree is an angle θ = 25°. How wide is the river?
Answer:
The width of the river is [tex]z = 60.62 \ m[/tex]
Explanation:
From the question we are told that
The distance of the base line is D = 130 m
The angle is [tex]\theta = 25^o[/tex]
A diagram illustration the question is shown on the first uploaded image
Applying Trigonometric Rules for Right-angled Triangle,
[tex]tan 25 = \frac{z}{130}[/tex]
Now making z the subject
[tex]z = 130 * tan (25)[/tex]
[tex]z = 60.62 \ m[/tex]
Question 9 of 10
2 Powie
You are riding a bicycle. You apply a forward force of 100 N, and you and the
bicycle have a combined mass of 80 kg. What is the acceleration of the
bicycle?
A. 125 m/s
B. 1.5 m/s2
c. 1.8 m/s?
D. 0.8 m/s
Answer:
1.25 m/s^2
Explanation:
F = m*a ...... force = mass * acceleration
force = 100 N, mass = 80 kg
100 = 80 * a
100/80 = a = 1.25 m/s^2
Answer:
The acceleration is 1.25m/s².
Explanation:
You have to apply Newton's Second Law which is F = m×a where F represents force, m is mass and a is acceleratipn. Then you have to substitute the following values into the formula :
[tex]f = m \times a[/tex]
Let F = 100,
Let m = 80,
[tex]100 = 80 \times a[/tex]
[tex]100 = 80a[/tex]
[tex]a = 100 \div 80[/tex]
[tex]a = 10 \div 8[/tex]
[tex]a = 1.25[/tex]
An organism has 20 chromosomes after fertilization.after meiosis, how many chromosomes would each sex cell have?
Answer:
EACH SEX CELL WILL HAVE 10 CHROMOSOMES BECAUSE n+n=2n
means haploid parent cells join or fuse to form diploid zygote
Answer:
10
Explanation:
A 5.50-kg bowling ball moving at 9.00 m/s collides with a 0.850-kg bowling pin, which is scattered at an angle of 85.0 0 to the initial direction of the bowling ball and with a speed of 15.0 m/s. (a) Calculate the final velocity (magnitude and direction) of the bowling ball.
Answer:
9.05 m/s , -14.72° (respect to x axis)
Explanation:
To find the final velocity of the bowling ball you take into account the conservation of the momentum for both x and y component of the total momentum. Then, you have:
[tex]p_{xi}=p_{xf}\\\\p_{yi}=p_{yf}\\\\[/tex]
[tex]m_1v_{1xi}+m_2v_{2xi}=m_1v_1cos\theta+m_2v_{2}cos\phi\\\\0=m_1v_1sin\theta-m_2v_2sin\phi[/tex]
m1: mass of the bowling ball = 5.50 kg
m2: mass of the bowling pin = 0.850 kg
v1xi: initial velocity of the bowling ball = 9.0 m/s
v2xi: initial velocity of bowling pin = 0m/s
v1: final velocity of bowling ball = ?
v2: final velocity of bowling pin = 15.0 m/s
θ: angle of the scattered bowling pin = ?
Φ: angle of the scattered bowling ball = 85.0°
Where you have used that before the bowling ball hits the pin, the y component of the total momentum is zero.
First you solve for v1cosθ in the equation for the x component of the momentum:
[tex]v_1cos\theta=\frac{m_1v_{1xi}-m_2v_2cos\phi}{m_1}\\\\v_1cos\theta=\frac{(5.50kg)(9.0m/s)-(0.850kg)(15.0m/s)cos85.0\°}{5.50kg}\\\\v_1cos\theta=8.79m/s[/tex]
and also you solve for v1sinθ in the equation for the y component of the momentum:
[tex]v_1sin\theta=\frac{(0.850kg)(15.0m/s)sin(85.0\°)}{5.50kg}\\\\v_1sin\theta=2.3m/s[/tex]
Next, you divide v1cosθ and v1sinθ:
[tex]\frac{v_1sin\theta}{v_1cos\theta}=tan\theta=\frac{2.3}{8.79}=0.26\\\\\theta=tan^{-1}(0.26)=14.72[/tex]
the direction of the bawling ball is -14.72° respect to the x axis
The final velocity of the bawling ball is:
[tex]v_1=\frac{2.3m/s}{sin\theta}=\frac{2.3}{sin(14.72\°)}=9.05\frac{m}{s}[/tex]
hence, the final velocity of the bawling ball is 9.05 m/s
An alpha particle has a charge of +2e and a mass of 6.64 x 10-27 kg. It is accelerated from rest through a potential difference of 1.2 x 106 V and then enters a uniform magnetic field whose strength is 2.2 T. The alpha particle moves perpendicular to the field. Calculate (a) the speed of the alpha particle, (b) the magnitude of the magnetic force exerted on it, and (c) the radius of its circular path.
Answer:
a) v = 1.075*10^7 m/s
b) FB = 7.57*10^-12 N
c) r = 10.1 cm
Explanation:
(a) To find the speed of the alpha particle you use the following formula for the kinetic energy:
[tex]K=qV[/tex] (1)
q: charge of the particle = 2e = 2(1.6*10^-19 C) = 3.2*10^-19 C
V: potential difference = 1.2*10^6 V
You replace the values of the parameters in the equation (1):
[tex]K=(3.2*10^{-19}C)(1.2*10^6V)=3.84*10^{-13}J[/tex]
The kinetic energy of the particle is also:
[tex]K=\frac{1}{2}mv^2[/tex] (2)
m: mass of the particle = 6.64*10^⁻27 kg
You solve the last equation for v:
[tex]v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2(3.84*10^{-13}J)}{6.64*10^{-27}kg}}\\\\v=1.075*10^7\frac{m}{s}[/tex]
the sped of the alpha particle is 1.075*10^6 m/s
b) The magnetic force on the particle is given by:
[tex]|F_B|=qvBsin(\theta)[/tex]
B: magnitude of the magnetic field = 2.2 T
The direction of the motion of the particle is perpendicular to the direction of the magnetic field. Then sinθ = 1
[tex]|F_B|=(3.2*10^{-19}C)(1.075*10^6m/s)(2.2T)=7.57*10^{-12}N[/tex]
the force exerted by the magnetic field on the particle is 7.57*10^-12 N
c) The particle describes a circumference with a radius given by:
[tex]r=\frac{mv}{qB}=\frac{(6.64*10^{-27}kg)(1.075*10^7m/s)}{(3.2*10^{-19}C)(2.2T)}\\\\r=0.101m=10.1cm[/tex]
the radius of the trajectory of the electron is 10.1 cm
The speed, magnetic force and radius are respectively; 10.75 * 10⁶ m/s; 7.57 * 10⁻¹² N; 0.101 m
What is the Magnetic force?
A) We know that the formula for kinetic energy can be expressed as;
K = qV
where;
q is charge of the particle = 2e = 2(1.6 × 10⁻¹⁹ C) = 3.2 × 10⁻¹⁹ C
V is potential difference = 1.2 × 10⁶ V
K = 3.2 × 10⁻¹⁹ * 1.2 × 10⁶
K = 3.84 × 10⁻¹³ J
Also, formula for kinetic energy is;
K = ¹/₂mv²
where v is speed
Thus;
v = √(2K/m)
v = √(2 * 3.84 × 10⁻¹³)/(6.64 * 10⁻²⁷)
v = 10.75 * 10⁶ m/s
B) The magnetic force is given by the formula;
F_b = qvB
F_b = (3.2 × 10⁻¹⁹ * 10.75 * 10⁶ * 2.2)
F_b = 7.57 * 10⁻¹² N
C) The formula to find the radius is;
r = mv/qB
r = (6.64 * 10⁻²⁷ * 10.75 * 10⁶)/(1.6 × 10⁻¹⁹ * 2.2)
r = 0.101 m
Read more about magnetic field at; https://brainly.com/question/7802337
commune time to work ( physics) i need help pls :(
Scientists studying an anomalous magnetic field find that it is inducing a circular electric field in a plane perpendicular to the magnetic field. The electric field strength 1.5 m from the center of the circle is 3.5 mV/m. At what rate is the magnetic field changing?
Answer
The rate at which the magnetic field is changing is [tex][\frac{dB}{dt} ] = 0.000467 T/s[/tex]
Explanation
From the question we are told that
The electric field strength is [tex]E = 3.5mV/m = 3.5 *10^{-3} \ V/m[/tex]
The radius is [tex]r = 1.5 \ m[/tex]
The rate of change of the magnetic field is mathematically represented as
[tex]\frac{d \phi }{dt} = \int\limits^{} {E \cdot dl}[/tex]
Where [tex]dl[/tex] is change of a unit length
[tex]\frac{d \phi}{dt} = A * \frac{dB}{dt}[/tex]
Where A is the area which is mathematically represented as
[tex]A = \pi r^2[/tex]
So
[tex]E \int\limits^{} { dl} = ( \pi r^2) (\frac{dB}{dt} )[/tex]
[tex]E L = ( \pi r^2) (\frac{dB}{dt} )[/tex]
where L is the circumference of the circle which is mathematically represented as
[tex]L = 2 \pi r[/tex]
So
[tex]E (2 \pi r ) = (\pi r^2 ) [\frac{dB}{dt} ][/tex]
[tex]E = \frac{r}{2} [\frac{dB}{dt} ][/tex]
[tex][\frac{dB}{dt} ] = \frac{E}{ \frac{r}{2} }[/tex]
substituting values
[tex][\frac{dB}{dt} ] = \frac{3.5 *10^{-3}}{ \frac{15}{2} }[/tex]
[tex][\frac{dB}{dt} ] = 0.000467 T/s[/tex]
The rate of heat conduction out of a window on a winter day is rapid enough to chill the air next to it. To see just how rapidly windows conduct heat, calculate the rate of conduction in watts through a 2.82 m2 window that is 0.675 cm thick if the temperatures of the inner and outer surfaces are 5.00°C and −10.0°C, respectively. This rapid rate will not be maintained — the inner surface will cool, and frost may even form. The thermal conductivity of glass is 0.84 J/(s · m · °C).
Answer:
Q = - 5264 W = - 5.26 KW
Here, negative sign indicates the outflow of heat
Explanation:
Fourier's Law of heat conduction, gives the following formula:
Q = - KAΔT/t
where,
Q = Rate of Heat Conduction out of window = ?
K = Thermal Conductivity of Glass = 0.84 W/m.°C
A =Surface Area of window = 2.82 m²
ΔT = Difference in Temperature of both sides of surface
ΔT = Inner Surface Temperature - Outer Surface Temperature= 5°C - (- 10°C)
ΔT = 15°C
t = thickness of window = 0.675 cm = 0.00675 m
Therefore,
Q = - (0.84 W/m.°C)(2.82 m²)(15°C)/0.00675 m
Q = - 5264 W = - 5.26 KW
Here, negative sign indicates the outflow of heat.
A cantilever beam with a width b=100 mm and depth h=150 mm has a length L=2 m and is subjected to a point load P =500 N at B. Calculate the state of plane stress at point C located 50 mm below the top of the beam and 0.5 m to the right of point A. Also find the principal stresses and the maximum shear stress at C. Neglect the weight of the beam.
Answer:
Explanation:
Given that:
width b=100mm
depth h=150 mm
length L=2 m =200mm
point load P =500 N
Calculate moment of inertia
[tex]I=\frac{bh^3}{12} \\\\=\frac{100 \times 150^3}{12} \\\\=28125000\ m m^4[/tex]
Point C is subjected to bending moment
Calculate the bending moment of point C
M = P x 1.5
= 500 x 1.5
= 750 N.m
M = 750 × 10³ N.mm
Calculate bending stress at point C
[tex]\sigma=\frac{M.y}{I} \\\\=\frac{(750\times10^3)(25)}{28125000} \\\\=0.0667 \ MPa\\\\ \sigma =666.67\ kPa[/tex]
Calculate the first moment of area below point C
[tex]Q=A \bar y\\\\=(50 \times 100)(25 +\frac{50}{2} )\\\\Q=250000\ mm[/tex]
Now calculate shear stress at point C
[tex]=\frac{FQ}{It}[/tex]
[tex]=\frac{500*250000}{28125000*100} \\\\=0.0444\ MPa\\\\=44.4\ KPa[/tex]
Calculate the principal stress at point C
[tex]\sigma_{1,2}=\frac{\sigma_x+\sigma_y}{2} \pm\sqrt{(\frac{\sigma_x-\sigma_y}{2} ) + (\tau)^2} \\\\=\frac{666.67+0}{2} \pm\sqrt{(\frac{666.67-0}{2} )^2 \pm(44.44)^2} \ [ \sigma_y=0]\\\\=333.33\pm336.28\\\\ \sigma_1=333.33+336.28\\=669.61KPa\\\\\sigma_2=333.33-336.28\\=-2.95KPa[/tex]
Calculate the maximum shear stress at piont C
[tex]\tau=\frac{\sigma_1-\sigma_2}{2}\\\\=\frac{669.61-(-2.95)}{2} \\\\=336.28KPa[/tex]
The velocity of an object is given by the expression v (t) = 3.00 m / s + (2.00 m / s ^ 3) t ^ 2. Determine the position of the object as a function of time if it is located at x = 1.00 m at time t = 0.00 s.
Answer: [tex]x=\frac{2}{3}t^3+3t+1[/tex]
Explanation:
Given
velocity of object is given by
[tex]v(t)=3+2t^2[/tex]
and we know change of position w.r.t time is velocity
[tex]\Rightarrow \dfrac{dx}{dt}=v[/tex]
[tex]\Rightarrow \dfrac{dx}{dt}=3+2t^2[/tex]
[tex]\Rightarrow dx=(3+2t^2)dt[/tex]
Integrating both sides we get
[tex]\Rightarrow \int_{1}^{x}dx=\int_{0}^{t}(3+2t^2)dt[/tex]
[tex]\Rightarrow x\mid _{1}^{x}=(3t+\frac{2}{3}t^3)\mid _{0}^{t}[/tex]
[tex]\Rightarrow x-1=3(t-0)+\frac{2}{3}(t^3-0)[/tex]
[tex]\Rightarrow x=\frac{2}{3}t^3+3t+1[/tex]
A block of mass 15.0 kg slides down a ramp inclined at 28.0∘ above the horizontal. As it slides, a kinetic friction force of 30.0 N parallel to the ramp acts on it. If the block slides for 5.50 m along the ramp, find the work done on the block by friction.
Answer:
Work is done by friction = -165 J
Explanation:
Given:
Mass of block (m) = 15 kg
Ramp inclined = 28°
Friction force (f) = 30 N
Distance (d) = 5.5 m
Find:
Work is done by friction.
Computation:
Work is done by friction = -Fd
Work is done by friction = -(30)(5.5)
Work is done by friction = -165 J
Block 1, of mass m1 = 2.50 kg , moves along a frictionless air track with speed v1 = 27.0 m/s. It collides with block 2, of mass m2 = 33.0 kg , which was initially at rest. The blocks stick together after the collision.A. Find the magnitude pi of the total initial momentum of the two-block system. Express your answer numerically.B. Find vf, the magnitude of the final velocity of the two-block system. Express your answer numerically.C. what is the change deltaK= Kfinal- K initial in the two block systems kinetic energy due to the collision ? Express your answer numerically in joules.
Answer:
a
The total initial momentum of the two-block system is [tex]p_t = 67.5 \ kg \cdot m/s^2[/tex]
b
The magnitude of the final velocity of the two-block system [tex]v_f = 1.9014 \ m/s[/tex]
c
the change ΔK=Kfinal−Kinitial in the two-block system's kinetic energy due to the collision is
[tex]\Delta KE =- 847.08 \ J[/tex]
Explanation:
From the question we are told that
The mass of first block is [tex]m_1 = 2.50 \ kg[/tex]
The initial velocity of first block is [tex]u_1 = 27.0 \ m/s[/tex]
The mass of second block is [tex]m_2 = 33.0\ kg[/tex]
initial velocity of second block is [tex]u_2 = 0 \ m/s[/tex]
The magnitude of the of the total initial momentum of the two-block system is mathematically repented as
[tex]p_i = (m_1 * u_1 ) + (m_2 * u_2)[/tex]
substituting values
[tex]p_i = (2.50* 27 ) + (33 * 0)[/tex]
[tex]p_t = 67.5 \ kg \cdot m/s^2[/tex]
According to the law of linear momentum conservation
[tex]p_i = p_f[/tex]
Where [tex]p_f[/tex] is the total final momentum of the system which is mathematically represented as
[tex]p_f = (m_+m_2) * v_f[/tex]
Where [tex]v_f[/tex] is the final velocity of the system
[tex]p_i = (m_1 +m_2 ) v_f[/tex]
substituting values
[tex]67.5 = (2.50+33 ) v_f[/tex]
[tex]v_f = 1.9014 \ m/s[/tex]
The change in kinetic energy is mathematically represented as
[tex]\Delta KE = KE_f -KE_i[/tex]
Where [tex]KE_f[/tex] is the final kinetic energy of the two-body system which is mathematically represented as
[tex]KE_f = \frac{1}{2} (m_1 +m_2) * v_f^2[/tex]
substituting values
[tex]KE_f = \frac{1}{2} (2.50 +33) * (1.9014)^2[/tex]
[tex]KE_f =64.17 J[/tex]
While [tex]KE_i[/tex] is the initial kinetic energy of the two-body system
[tex]KE_i = \frac{1}{2} * m_1 * u_1^2[/tex]
substituting values
[tex]KE_i = \frac{1}{2} * 2.5 * 27^2[/tex]
[tex]KE_i = 911.25 \ J[/tex]
So
[tex]\Delta KE = 64.17 -911.25[/tex]
[tex]\Delta KE =- 847.08 \ J[/tex]
For the RC circuit and the RL circuit, assume that the period of the source square wave is much larger than the time constant for each. Make a sketch of vR(t) as a function of t for each of the circuits?
Answer with Explanation:
Concepts and reason
The concept to solve this problem is that if a capacitor is connected in a RC circuit then it allows the flow of charge through circuit only till it gets fully charged. Once the capacitor is charged it will not allow any charge or current to flow.
Opposite is the case with inductor in the RL circuit. According to Faraday's law an inductor develops an emf to oppose the voltage applied but once the flux change stops then the inductor behaves just like a normal wire as if no inductor is there.
In attached figure, resistor is connected in series to the capacitor.
As we considered [tex]V_{C}[/tex] the voltage across the capacitor and [tex]V_{s}[/tex] the voltage across the source.
Voltage across a resistor In RC circuit.
[tex]V_{R}=V_S\left ( e^{-\frac{t}{RC}} \right )[/tex]
Voltage across a resistor In RL circuit.
[tex]V_{R}=V_S\left (1- e^{-\frac{Rt}{L}} \right )[/tex]
The sketch of [tex]\mathbf{v_R(t)}[/tex] as a function of t for each of the circuits can be seen in the diagram attached below.
For the Pre-Laboratory exercise, based on the assumption that the RC circuit has a capacitor and a sensing resistor while the RL circuit has a sensing resistor and an inductor.
The input voltage for both circuits is regarded as the square wave and if the square wave is much larger than the time constant for each.
Therefore, we can conclude that the below diagram shows an appropriate sketch of [tex]\mathbf{v_R(t)}[/tex] as a function of t for each of the circuits.
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What is the power of a child that has
done work of 50J in 10 seconds.
(a)50W (b)20W (c)30W (d)5W
_____________________________
Solution,
Work=50 Joule
Time=10 seconds
Power=?
Now,
Power=Work/time
= 50/10
= 5 Watt.
So the right answer is 5 W
Hope it helps..
Good luck on your assignment
__________________________
A flat coil of wire is used with an LC-tuned circuit as a receiving antenna. The coil has a radius of 0.30 m and consists of 420 turns. The transmitted radio wave has a frequency of 1.3 MHz. The magnetic field of the wave is parallel to the normal of the coil and has a maximum value of 1.7 x 10-13 T. Using Faraday's Law of electromagnetic induction and the fact that the magnetic field changes from zero to its maximum value in one-quarter of a wave period, find the magnitude of the average emf induced in the antenna in this time.
Answer:
The average emf induce is [tex]V = 2.625 * 10^{-5} \ V[/tex]
Explanation:
From the question we are told that
The radius of the coil is [tex]r = 0.30 \ m[/tex]
The number of turns is [tex]N = 420 \ turns[/tex]
The frequency of the transition radio wave is [tex]f = 1.3\ MHz = 1.3 *10^{6} Hz[/tex]
The magnetic field is [tex]B_,{max} = 1.7 * 10^{-13} \ T[/tex]
The time taken for the magnetic field to go from zero to maximum is [tex]\Delta T = \frac{T}{4}[/tex]
The period of the transmitted radio wave is [tex]T = \frac{1}{f}[/tex]
So
[tex]\Delta T = \frac{T}{4} = \frac{1}{4 f}[/tex]
The potential difference can be mathematically represented as
[tex]V = NA (\frac{\Delta B}{\Delta T} )[/tex]
[tex]V = NA ([B_{max} - B_{min} ] * 4f)[/tex]
Where [tex]B_{min} = 0T[/tex]
substituting values
[tex]V = 420 * (\pi *(0.30)^2) * (1.7 *10^{-13} * 4 * 1.3 *10^{6})[/tex]
[tex]V = 2.625 * 10^{-5} \ V[/tex]
A 1200 kg car reaches the top of a 100 m high hill at A with a speed vA. What is the value of vA that will allow the car to coast in neutral so as to just reach the top of the 150 m high hill at B with vB
Complete question is:
A 1200 kg car reaches the top of a 100 m high hill at A with a speed vA. What is the value of vA that will allow the car to coast in neutral so as to just reach the top of the 150 m high hill at B with vB = 0 m/s. Neglect friction.
Answer:
(V_A) = 31.32 m/s
Explanation:
We are given;
car's mass, m = 1200 kg
h_A = 100 m
h_B = 150 m
v_B = 0 m/s
From law of conservation of energy,
the distance from point A to B is;
h = 150m - 100 m = 50 m
From Newton's equations of motion;
v² = u² + 2gh
Thus;
(V_B)² = (V_A)² + (-2gh)
(negative next to g because it's going against gravity)
Thus;
(V_B)² = (V_A)² - (2gh)
Plugging in the relevant values;
0² = (V_A)² - 2(9.81 × 50)
(V_A) = √981
(V_A) = 31.32 m/s
Mr. Patel is photocopying lab sheets for his first period class. A particle of toner carrying a charge of 4.0 * 10^9 C in the copying machine experiences an electric field of 1.2 * 10^6 N/C as it’s pulled toward the paper. What is the electric force acting on the toner particle?
Answer:
4.8 × 10^15 N
Explanation:
Electric Field is defined as Force per unit Charge.
This is expressed mathematically as;
E= F/Q
Where E- Electric Field
F- Force
Q- charge
From the expression above by change of subject of formula for F, we have;
F=E×Q
= 1.2 * 10^6 ×4.0 * 10^9
= 4.8 × 10^15 N
Your lab instructor has asked you to measure a spring constant using a dynamic method—letting it oscillate—rather than a static method of stretching it. You and your lab partner suspend the spring from a hook, hang different masses, m, on the lower end, and start them oscillating. One of you uses a meter stick to measure the amplitude, A, and the other uses a stopwatch to time 10 oscillations, t. Your data are as follows:Mass, m(g) Amplitude, A(cm) Time, T(s) 100 6.5 7.8150 5.5 9.8200 6.0 10.9250 3.5 12.4Use the best-fit line of an appropriate graph to determine the spring constant.
Answer:
k = 6,547 N / m
Explanation:
This laboratory experiment is a simple harmonic motion experiment, where the angular velocity of the oscillation is
w = √ (k / m)
angular velocity and rel period are related
w = 2π / T
substitution
T = 2π √(m / K)
in Experimental measurements give us the following data
m (g) A (cm) t (s) T (s)
100 6.5 7.8 0.78
150 5.5 9.8 0.98
200 6.0 10.9 1.09
250 3.5 12.4 1.24
we look for the period that is the time it takes to give a series of oscillations, the results are in the last column
T = t / 10
To find the spring constant we linearize the equation
T² = (4π²/K) m
therefore we see that if we make a graph of T² against the mass, we obtain a line, whose slope is
m ’= 4π² / k
where m’ is the slope
k = 4π² / m'
the equation of the line of the attached graph is
T² = 0.00603 m + 0.0183
therefore the slope
m ’= 0.00603 s²/g
we calculate
k = 4 π² / 0.00603
k = 6547 g / s²
we reduce the mass to the SI system
k = 6547 g / s² (1kg / 1000 g)
k = 6,547 kg / s² =
k = 6,547 N / m
let's reduce the uniqueness
[N / m] = [(kg m / s²) m] = [kg / s²]
The spring-mass system forms a linear graph between the time period and mass. And the value of spring-constant from the given data is 6.46 N/m.
Given data:
Mass suspended by spring is, [tex]m=100 \;\rm g =0.1 \;\rm kg[/tex].
Number of oscillations is, [tex]n =10\;\rm oscillations[/tex].
Time period of oscillation is, [tex]T=7.8 \;\rm s[/tex].
The expression for the angular frequency of spring-mass system is,
[tex]\omega =\drac \sqrt{\dfrac{k}{m} }[/tex] ......................................................(1)
Here, k is the spring constant.
Angular frequency is also expressed as,
[tex]\omega = 2 \pi f[/tex] .........................................................(2)
here, f is the linear frequency of spring-mass system.
And linear frequency is,
[tex]f=\dfrac{n}{T}\\f=\dfrac{10}{7.81}\\f=1.28 \;\rm cycles/sec[/tex]
Then substitute equation (2) in equation (1) as,
[tex]2 \pi f=\drac \sqrt{\dfrac{k}{m} }\\2 \pi \times 1.28=\drac \sqrt{\dfrac{k}{0.1} }\\(2 \pi \times 1.28)^{2}= \dfrac{k}{0.1}\\k = 6.46 \;\rm N/m[/tex]
Thus, the value of spring constant is 6.46 N/m. And the suitable graph for the spring-mass system is given below.
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A population _____ follows a period of
Answer:
a population increase
Explanation:
During the 20th century, the world population increased from 1.65 billion to 6 billion. In 1970, the world's population was half that of today. In less than 15 years, 47% of the population will live in areas already under heavy water stress. In Africa, between 75 and 250 million people will face growing shortages in 2020 due to climate change. The scarcity of some arid and semi-arid regions will have a decisive impact on migration.
How much work must be done on a 10 kg snowboard to increase its speed from 4 m/s to 6 m/s?
Answer:
100 J
Explanation:
Work = change in energy
W = ΔKE
W = ½ mv² − ½ mv₀²
W = ½ m (v² − v₀²)
W = ½ (10 kg) ((6 m/s)² − (4 m/s)²)
W = 100 J
An automobile moving along a straight track changes its velocity from 40 m/s to 80 m/s in a distance of 200 m. What is the (constant) acceleration of the vehicle during this time? Group of answer choices
Answer:
Dear Kaleb
Answer to your query is provided below
Acceleration of the vehicle is 12m/s^2
Explanation:
Explanation for the same is attached in image
A proton moving along the x axis has an initial velocity of 4.0 × 106 m/s and a constant acceleration of 6.0 × 1012 m/s2. What is the velocity of the proton after it has traveled a distance of 80 cm? Group of answer choices
Answer:
5.06*10^6 m/s
Explanation:
Given that
Initial velocity, u = 4*10^6 m/s
Acceleration, a = 6*10^12 m/s²
Distance traveled, s = 80 cm
Final velocity, v = ?
We can find the final velocity by using one of the equations of motion.
v² = u² + 2as
On substituting the values, we have
v² = (4*10^6)² + 2 * 6*10^12 * 0.8
v² = 2.56*10^13
v = √2.56*10^13
v = 5.06*10^6 m/s
Therefore, the final velocity of the proton is adjudged to be 5.06*10^6 m/s
The final velocity of the proton over the given distance is [tex]5.06 \times 10^6 \ m/s[/tex].
The given parameters;
initial velocity of the proton, u = 4 x 10⁶ m/sacceleration of the proton, a = 6 x 10¹² m/s²distance traveled by the proton, s = 80 cm = 0.8 mThe final velocity of the proton over the given distance is calculated as follows;
[tex]v^2 = u^2 + 2as\\\\v^2 = (4\times 10^6)^2 \ + \ 2(6.0 \times 10^{12})(0.8)\\\\v^2 = 2.56 \times 10^{13} \\\\v = \sqrt{2.56 \times 10^{13} } \\\\v = 5.06 \times 10^6 \ m/s[/tex]
Thus, the final velocity of the proton over the given distance is [tex]5.06 \times 10^6 \ m/s[/tex]
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A Ferris wheel has radius 5.0 m and makes one revolution every 8.0 s with uniform rotation. A person who normally weighs 670 N is sitting on one of the benches attached at the rim of the wheel. What is the apparent weight (the normal force exerted on her by the bench) of the person as she passes through the highest point of her motion? ( type in your answer with no units in form xx0)
Answer:
The apparent weight of the person as she pass the highest point is [tex]N = 458.8 \ N[/tex]
Explanation:
From the question we are told that
The radius of the Ferris wheel is [tex]r = 5.0 \ m[/tex]
The period of revolution is [tex]T = 8.0 \ s[/tex]
The weight of the person is [tex]W = 670 \ N[/tex]
Generally the speed of the wheel is mathematically represented as
[tex]v = \frac{2 \pi r}{T }[/tex]
substituting values
[tex]v = \frac{2 * 3.142 * 5}{8 }[/tex]
[tex]v = 3.9 3 \ m/s[/tex]
The apparent weight (the normal force exerted on her by the bench) at the highest point is mathematically evaluated as
[tex]N = mg - \frac{mv^2}{r}[/tex]
Where m is the mass of the person which is mathematically evaluated as
[tex]m = \frac{W}{g}[/tex]
substituting values
[tex]m = \frac{670}{9.8}[/tex]
[tex]m = 68.37 \ kg[/tex]
So
[tex]N = 68.37 * 9.8 - \frac{68.37 * {3.93}^2}{5}[/tex]
[tex]N = 458.8 \ N[/tex]
