Answer:
A , D , E
Explanation:
Solution:-
- Consider the two identical objects with mass ( m ).
- The stiffness of the springs are ( k1 and k2 ).
- Both the spring store 55.0 J of potential energy.
- We will apply the principle of energy conservation on both the systems. In both cases the spring stores 55.0 Joules of energy. Once released, the objects gain kinetic energy with a consequent loss of potential energy in either spring.
- The maximum speed ( v ) is attained when all the potential energy is converted to kinetic energy.
- Apply Energy conservation for spring with stiffness ( k1 ).
ΔU = ΔEk
55.0 = 0.5*m*v^2
v = √ ( 110 / m )
- Apply Energy conservation for spring with stiffness ( k2 ).
ΔU = ΔEk
55.0 = 0.5*m*v^2
v = √ ( 110 / m )
Answer: Both objects will have the same maximum speed ( A )
- We are told that one spring is more stiff as compared to the other one. The measure of stiffness is proportionally quantified by the spring constant. To mathematically express we can write it as:
k1 > k2
Where,
k1: The stiff spring
k2: The flexible spring
Answer: The stiff spring has a larger spring constant than the flexible spring. ( D )
- We will assume that the spring with constant ( k1 ) undergoes a displacement ( x1 ) and the spring with constant ( k2 ) undergoes a displacement ( x2 ). The potential energy stored in both spring is the same. Hence,
U1 = U2
0.5*( k1 ) * ( x1 )^2 = 0.5*( k2 ) * ( x2 )^2
[ k1 / k2 ] = [ x2 / x1 ]^2
Since,
k1 > k2 , then [ k1 / k2 ] > 1
Then,
[ x2 / x1 ]^2 > 1
[ x2 / x1 ] > 1
x2 > x1
Answer: The flexible spring ( x2 ) was compressed more than the stiff spring ( x1 ). ( E )
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]
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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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
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 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]
A parallel-plate capacitor in air has a plate separation of 1.30 cm and a plate area of 25.0 cm2. The plates are charged to a potential difference of 255 V and dis-connected from the source. The capacitor is then immersed in distilled water. Determine a) the charge on the plates before and after immersion.b) the capacitance and potential difference after immersion.c) the change in energy of the capacitor.
Answer:
Explanation:
capacitance of air capacitor
C = ε₀ A / d
ε₀ is permittivity of medium , A is plate area , d is distance between plate .
C = 8.85 x 10⁻¹² x 25 x 10⁻⁴ / 1.3 x 10⁻²
= 170.19 x 10⁻¹⁴ F .
charge on the capacitor when it is charged to potential of 255 V
= CV , C is capacitance and V is potential
= 170.19 x 10⁻¹⁴ x 255
= 4.34 x 10⁻¹⁰ C .
After it is disconnected from the source , and it is immersed in water , charge on it remains the same .
So its charge when immersed in water will be constant at 4.34 x 10⁻¹⁰ C.
b )
When it is immersed in water its capacity increases k times where k is dielectric constant of water which is 80 .
capacitance of capacitor in water = 80 x 170.19 x 10⁻¹⁴ F
= 13615.2 x 10⁻¹⁴ F .
= 1.36 x 10⁻¹⁰ F
potential difference = charge / capacitance
= 4.34 x 10⁻¹⁰ / 1.36 x 10⁻¹⁰
= 3.2 V
c )
Energy of capacitor = 1/2 C V²
Initial energy = 1/2 x 170.19 x 255² x 10⁻¹⁴
= 55.33 x 10⁻⁹ J
Final energy = 1/2 x 1.36 x 10⁻¹⁰ x 3.2²
= .7 x 10⁻⁹ J .
decrease of energy = 54.63 x 10⁻⁹ J .
commune time to work ( physics) i need help pls :(
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
During a baseball game, a baseball is struck at ground level by a batter. The ball leaves the baseball bat with an initial speed v0 = 38 m/s at an angle θ = 35° above horizontal. Let the origin of the Cartesian coordinate system be the ball's position the instant it leaves the bat. Air resistance may be ignored throughout this problem.
Part (a) Express the magnitude of the ball's initial horizontal velocity Or in terms of vo and 20%
Part (b) Express the magnitude of the ball's initial vertical velocity vOy in terms of vo and 0. 20%
Part (c) Find the ball's maximum vertical height Amat in meters above the ground.
Part (d) Create an expression in terms of vo-e, and g for the time-ur İt takes te ball to travel to its maximum vertical height.
Part (e) Calculate the horizontal distance in meters the ball has traveled when it returns to ground level.
Answer:
a) v₀ₓ = v₀ cos θ , b) v_{oy} = v₀ sin θ , c) y = v_{oy}² / 2g, y = 24.25 m
e) R = 138.46 m
Explanation:
This is a projectile launch exercise
a) let's use trigonometry to find the components of the initial velocity
cos θ = v₀ₓ / v₀
v₀ₓ = v₀ cos θ
v₀ₓ = 38 cos 35
v₀ₓ = 31.13 m / s
b) sin θ = [tex]v_{oy}[/tex] / v₀
v_{oy} = v₀ sin θ
v_{oy} = 38 sint 35
v_{oy} = 21 80 m / s
c, d) to find the maximum height, the vertical speed is zero
v_{y}² = v_{oy}² - 2 g y
0 = [tex]v_{oy}[/tex]² - 2 gy
y = v_{oy}² / 2g
let's calculate
y = 21.80 2 / (2 9.8)
y = 24.25 m
e) They ask to find the horizontal distance
for this we can use the expression of reaches
R = v₀² sin 2θ / g
let's calculate
R = 38² sin (2 35) / 9.8
R = 138.46 m
BEST ANSWER GETS BRAINLIEST!
At what distance from a 70.0 Watt speaker is the intensity 0.0195 W/m^2
(Treat the speaker as point of the source)
(Unit=meters)
PLEASE HELP ME!
Answer:
Distance = 16.9 m
Explanation:
We are given;
Power; P = 70 W
Intensity; I = 0.0195 W/m²
Now, for a spherical sound wave, the intensity in the radial direction is expressed as a function of distance r from the center of the sphere and is given by the expression;
I = Power/Unit area = P/(4πr²)
where;
P is the sound power
r is the distance.
Thus;
Making r the subject, we have;
r² = P/4πI
r = √(P/4πI)
r = √(70/(4π*0.0195))
r = √285.6627
r = 16.9 m
Answer:
16.9 m
Explanation:
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 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}
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 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]
What do you call a group of sea turtles?
Answer:
a bale
Explanation:
a bale is a group of turtles
Answer:
A bale or nest
Explanation:
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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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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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°
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
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 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
Light bulb 1 operates with a filament temperature of 3000 K, whereas light bulb 2 has a filament temperature of 2000 K. Both filaments have the same emissivity, and both bulbs radiate the same power. Find the ratio A1/A2 of the filament areas of the bulbs.
Answer:
A₁/A₂ = 0.44
Explanation:
The emissive power of the bulb is given by the formula:
P = σεAT⁴
where,
P = Emissive Power
σ = Stefan-Boltzman constant
ε = Emissivity
A = Surface Area
T = Absolute Temperature of Surface
FOR BULB 1:
Since, emissivity and emissive power are constant.
Therefore,
P = σεA₁T₁⁴ ----------- equation 1
where,
A₁ = Surface Area of Bulb 1
T₁ = Temperature of Bulb 1 = 3000 k
FOR BULB 2:
Since, emissivity and emissive power are constant.
Therefore,
P = σεA₂T₂⁴ ----------- equation 2
where,
A₂ = Surface Area of Bulb 2
T₂ = Temperature of Bulb 1 = 2000 k
Dividing equation 1 by equation 2, we get:
P/P = σεA₁T₁⁴/σεA₂T₂⁴
1 = A₁(3000)²/A₂(2000)²
A₁/A₂ = (2000)²/(3000)²
A₁/A₂ = 0.44
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]
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
In each pair, select the body with more internal energy.
Answer:
rt
Explanation:
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
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
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
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
