Answer:
The average velocity of Jackson is 18.056 m/s South
The average velocity of Hunter is 10.65 m/s East
Explanation:
initial velocity of Jackson, u = 25 km/h east = 6.944 m/s east
time for this motion, [tex]t_i[/tex] = 20 minutes = 1200 seconds
⇒initial displacement of Jackson, [tex]x_i[/tex] = (6.944 m/s) x (1200 s) = 8332.8 m
Final velocity of Jackson, v = 45 km/h South = 12.5 m/s South
time at Jackson's final position, [tex]t_f[/tex] = 20 minutes + [tex]t_i[/tex] = 20 minutes + 20 minutes
time at Jackson's final position, [tex]t_f[/tex] = 40 minutes = 2400 s
⇒Final displacement of Jackson,[tex]x_f[/tex] = (12.5 m/s) x (2400 s) = 30,000m
Average velocity of Jackson;
[tex]= \frac{x_f-x_i}{t_f-t_i} \\\\= \frac{30,000-8332.8}{2400-1200} \\\\= 18.056 \ m/s \ South[/tex]
initial velocity of Hunter, u = 45 km/h South = 12.5 m/s South
time for this motion, [tex]t_i[/tex] = 10 minutes = 600 seconds
⇒initial displacement of Hunter, [tex]x_i[/tex] = (12.5 m/s) x (600 s) = 7500 m
Final velocity of Hunter, v = 40 km/h east = 11.11 m/s east
time at Hunter's final position, [tex]t_f[/tex] = 30 minutes + [tex]t_i[/tex] = 30 minutes + 10 minutes
time at Hunter's final position, [tex]t_f[/tex] = 40 minutes = 2400 s
⇒Final displacement of Hunter,[tex]x_f[/tex] = (11.11 m/s) x (2400 s) = 26,664m
Average velocity of Hunter;
[tex]= \frac{x_f-x_i}{t_f-t_o} \\\\= \frac{26,664-7500}{2400-600} \\\\= 10.65 \ m/s \ east[/tex]
The starships of the Solar Federation are marked with the symbol of the Federation, a circle, whereas starships of the Denebian Empire are marked with the Empire's symbol, an ellipse whose major axis is n times its minor axis (a=nb in the figure ).
How fast, relative to an observer, does an Empire ship have to travel for its markings to be confused with those of a Federation ship? Use c for the speed of light in a vacuum.
Express your answer in terms of n and c.
Complete question
The complete question is shown on the first uploaded image
Answer:
The velocity is [tex]v = c* \sqrt{1 - \frac{1}{n^2} }[/tex]
Explanation:
From the question we are told that
a = nb
The length of the minor axis of the symbol of the Federation, a circle, seen by the observer at velocity v must be equal to the minor axis(b) of the Empire's symbol, (an ellipse)
Now this length seen by the observer can be mathematically represented as
[tex]h = t \sqrt{1 - \frac{v^2}{c^2} }[/tex]
Here t is the actual length of the major axis of of the Empire's symbol, (an ellipse)
So t = a = nb
and b is the length of the minor axis of the symbol of the Federation, (a circle) when seen by an observer at velocity v which from the question must be the length of the minor axis of the of the Empire's symbol, (an ellipse)
i.e h = b
So
[tex]b = nb [\sqrt{1 - \frac{v^2}{c^2} } ][/tex]
[tex][\frac{1}{n} ]^2 = 1 - \frac{v^2}{c^2}[/tex]
[tex]v^2 =c^2 [1- \frac{1}{n^2} ][/tex]
[tex]v^2 =c^2 [\frac{n^2 -1}{n^2} ][/tex]
[tex]v = c* \sqrt{1 - \frac{1}{n^2} }[/tex]
During a football game, a receiver has just caught a pass and is standing still. Before he can move, a tackler, running at a velocity of 2.60 m/s, grabs and holds onto him so that they move off together with a velocity of 1.30 m/s. If the mass of the tackler is 122 kg, determine the mass of the receiver. Assume momentum is conserved.
Answer:
122kgExplanation:
Using the law of conservation of momentum which states that 'the sum of momentum of bodies before collision is equal to their sum after collision. The bodies will move together with a common velocity after collision.
Momentum = Mass * Velocity
Before collision;
Momentum of receiver m1u1= 0 kgm/s (since the receiver is standing still)
Momentum of the tackler
m2u2 = 2.60*122 = 317.2 kgm/s
where m2 and u2 are the mass and velocity of the tacker respectively.
Sum of momentum before collision = 0+317.2 = 317.2 kgm/s
After collision
Momentum of the bodies = (m1+m2)v
v = their common velocity
m1 = mass of the receiver
Momentum of the bodies = (122+m1)(1.30)
Momentum of the bodies = 158.6+1.30m1
According to the law above;
317.2 = 158.6+1.30m1
317.2-158.6 = 1.30m1
158.6 = 1.30m1
m1 = 158.6/1.30
m1 = 122kg
The mas of the receiver is 122kg
A water slide is constructed so that swimmers, starting from rest at the top of the slide, leave the end of the slide traveling horizontally. One person hits the water 5.00 m from the end of the slide in a time of 0.504 s after leaving the slide. Ignore friction and air resistance. Find the height H.
Answer:
4.93 m
Explanation:
According to the question, the computation of the height is shown below:
But before that first we need to find out the speed which is shown below:
As we know that
[tex]Speed = \frac{Distance}{Time}[/tex]
[tex]Speed = \frac{5}{0.504}[/tex]
= 9.92 m/s
Now
[tex]v^2 - u^2 = 2\times g\times h[/tex]
[tex]9.92^2 = 2\times 9.98 \times h[/tex]
98.4064 = 19.96 × height
So, the height is 4.93 m
We simply applied the above formulas so that the height i.e H could arrive
We say that the displacement of a particle is a vector quantity. Our best justification for this assertion is: A. a displacement is obviously not a scalar. B. displacement can be specified by a magnitude and a direction. C. operating with displacements according to the rules for manipulating vectors leads to results in agreement with experiments. D. displacement can be specified by three numbers. E. displacement is associated by motion.
Answer:
Option B - displacement can be specified by a magnitude and a direction.
Explanation:
A Vector quantity is defined as a physical quantity characterized by the presence of both magnitude as well as direction. Examples include displacement, force, torque, momentum, acceleration, velocity e.t.c
Whereas a scalar quantity is defined as a physical quantity which is specified with the magnitude or size alone. Examples include length, speed, work, mass, density, etc.
Displacement is the difference between the initial position and the final position of a body. Displacement is a vector quantity and not a scalar quantity because it can be described by using both magnitude as well as direction.
Looking at the options, the only one that truly justifies this definition is option B.
Which of the following best describes the current age of the Sun?
A.) It is near the end of its lifespan.
B.) It is about halfway through its lifespan.
C.) It is early in its lifespan.
D.) We do not have a good understanding of the Sun's age.
Answer: Its b, The only problem with this is is there supposed to be a picture?
Explanation: NASA has used there fancy gadgets to figure this out but if there was a picture, this answer could be different.
a steel ball is dropped from a diving platform use the approximate value of g as 10 m/s^2 to solve the following problem what is the velocity of the ball 0.9 seconds after its released
Answer:
The final speed of the ball is 9 m/s.
Explanation:
We have,
A steel ball is dropped from a diving platform. It is required to find the velocity of the ball 0.9 seconds after its released. It will move under the action of gravity. Using equation of motion to find it as :
[tex]v=u+at[/tex]
u = 0 (at rest), a = g
[tex]v=gt\\\\v=10\times 0.9\\\\v=9\ m/s[/tex]
So, the final speed of the ball is 9 m/s.
On April 13, 2029 (Friday the 13th!), the asteroid 99942 mi Apophis will pass within 18600 mi of the earth-about 1/13 the distance to the moon! It has a density of 2600 kg/m^3, can be modeled as a sphere 320 m in diameter, and will be traveling at 12.6 km/s.
1)If, due to a small disturbance in its orbit, the asteroid were to hit the earth, how much kinetic energy would it deliver?
2)The largest nuclear bomb ever tested by the United States was the "Castle/Bravo" bomb, having a yield of 15 megatons of TNT. (A megaton of TNT releases 4.184x10^15 J of energy.) How many Castle/Bravo bombs would be equivalent to the energy of Apophis?
Answer:
Explanation:
Volume of asteroid = 4/3 x π x 160³
= 17.15 x 10⁶
mass = volume x density
= 17.15 x 10⁶ x 2600
= 445.9 x 10⁸ kg
kinetic energy
= 1/2 x 445.9 x 10⁸ x( 12.6 )² x 10⁶
= 35.4 x 10¹⁷ J .
2 )
energy of 15 megaton
= 4.184 x 10¹⁵ x 15 J
= 62.76 x 10¹⁵ J
No of bombs required
= 35.4 x 10¹⁷ / 62.76 x 10¹⁵
= 56.4 Bombs .
When you "crack" a knuckle, you suddenly widen the knuckle cavity, allowing more volume for the synovial fluid inside it and causing a gas bubble suddenly to appear in the fluid. The sudden production of the bubble, called "cavitation", produces a sound pulse---the cracking sound. Assume that the sound is transmitted uniformly in all directions and that it fully passes from the knuckle interior to the outside, at a distance of 0.29 m from your ear. If the pulse has a sound level of 61 dB at your ear, what is the rate at which energy is produced by the cavitation
Answer:If a wave y(x, t) = (6.0 mm) sin(kx + (600 rad/s)t + Φ) travels along a string, how much time does any given point on the string take to move between displacements y = +2.0 mm and y = -2.0 mm?
Explanation:
A cheetah bites into its prey. One tooth exerts a force of 320 N. The area of the point of the tooth is 0.5 cm². The pressure of the tooth on the prey, in N/cm², is
a) 0.0013 N/cm²
b) 128 N/cm²
c) 320 N/cm²
d) 640 N/cm²
Answer:
640N/cm^2Answer D is correct
Explanation:
[tex]pressure = \frac{force}{area} \\ = \frac{320}{0.5} \\ = 640[/tex]
hope this helps
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Parallel light rays with a wavelength of 610nm fall on a single slit. On a screen 3.10m away, the distance between the first dark fringes on either side of the central maximum is 4.00mm.
What is the width of the slit?
Answer:
The width of the slit will be ".946 mm".
Explanation:
The given values are:
Wavelength = 610 × 10⁻⁹
Length, L = 3 m
As we know,
⇒ [tex]\frac{y}{L} = \frac{m(wavelength)}{a}[/tex]
On putting the estimated values, we get
⇒ [tex]\frac{2\times 10^{-3}}{3.1} = \frac{(1)(610 X 10^{-9})}{a}[/tex]
On applying cross-multiplication, we get
⇒ [tex]a=9.46\times 10^{-4}[/tex]
⇒ [tex]a = .946 mm[/tex]
Q) A particle in simple harmonic motion starts its motion from its mean position. If T be the time period, calculate the ratio of kinetic energy and potential energy of the particle at the instant when t = T/12.
t\12 and the parties are spreading ever
Explanation:
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A rifle fires a 2.05 x 10-2-kg pellet straight upward, because the pellet rests on a compressed spring that is released when the trigger is pulled. The spring has a negligible mass and is compressed by 8.01 x 10-2 m from its unstrained length. The pellet rises to a maximum height of 4.46 m above its position on the compressed spring. Ignoring air resistance, determine the spring constant.
Answer:
Spring Constant = 279.58 N/m
Explanation:
We are given;
Mass; m = 2.05 x 10^(-2) kg = 0.0205 kg
Distance of compression; x = 8.01 × 10^(-2) m = 0.0801 m
Maximum height; h = 4.46 m
The formula for the energy in the spring is given by;
E = ½kx²
where:
k is the spring constant
x is the distance the spring is compressed.
Now, this energy of the spring will be equal to the energy of the pellet at its highest point. Energy of pallet = mgh So;
½kx² = mgh
Plugging in the relevant values, we have;
½ * k * 0.0801² = 0.0205 * 9.81 * 4.46
0.003208005k = 0.8969
k = 0.8969/0.003208005
k = 279.58 N/m
PLEASE HELP !
Complete the following sentence. Choose the right answer from the given ones. The internal energy of the body can be changed A / B / C. A. only when the body is warmed or cooled B. when work is done on the body or heat flow C. only when the body does work
B
HOPE IT HELPS LET ME KNOW IF U NEED EXPLANATION
I really need help with this question someone plz help !
Answer:
The answer is option 2.
Explanation:
Both sides are pulling the rope with equal force where the rope doesn't move. So they have a balanced forces.
Use Hooke's Law to determine the work done by the variable force in the spring problem. Nine joules of work is required to stretch a spring 0.5 meter from its natural length. Find the work required to stretch the spring an additional 0.40 meter.
Answer:
29.16 J
Explanation:
From Hook's law,
W = 1/2(ke²)..................... Equation 1
Where W = work done, k = Spring constant, e = extension.
Given: W = 9 J, e = 0.5 m.
Substitute into equation 1
9 = 1/2(k×0.5²)
Solve for k
k = 18/0.5²
k = 72 N/m.
The work done required to stretch the spring by additional 0.4 m is
W = 1/2(72)(0.4+0.5)²
W = 36(0.9²)
W = 29.16 J.
You measure the power delivered by a battery to be 1.15 W when it is connected in series with two equal resistors. How much power will the same battery deliver if the resistors are now connected in parallel across it
Answer:
The power is [tex]P_p = 4.6 \ W[/tex]
Explanation:
From the question we are told that
The power delivered is [tex]P_{s} = 1.15 \ W[/tex]
Let it resistance be denoted as R
The resistors are connected in series so the equivalent resistance is
[tex]R_{eqv} = R+ R = 2 R[/tex]
Considering when it is connected in series
Generally power is mathematically represented as
[tex]P_s = V * I[/tex]
Here I is the current which is mathematically represented as
[tex]I = \frac{V}{2R}[/tex]
The power becomes
[tex]P_s = V * \frac{V}{2R}[/tex]
[tex]P_s = \frac{V^2}{2R}[/tex]
substituting value
[tex]1.15 = \frac{V^2}{2R}[/tex]
Considering when resistance is connected in parallel
The equivalent resistance becomes
[tex]R_{eqv} = \frac{R}{2}[/tex]
So The current becomes
[tex]I = \frac{V}{\frac{R}{2} } = \frac{2V}{R}[/tex]
And the power becomes
[tex]P_p = V * \frac{2V}{R} = \frac{2V^2}{R} = \frac{4 V^2}{2 R} = 4 * P_s[/tex]
substituting values
[tex]P_p = 4 * 1.15[/tex]
[tex]P_p = 4.6 \ W[/tex]
During last year’s diving competition, the divers always pull their limbs in and curl up their bodies when they do flips. Just before entering the water, they fully extend their limbs to enter straight down as shown. Explain the effect of both actions on their angular velocities and kinetic energy (support your answer with working). Also explain the effect on their angular momentum.
Answer:
the angular speed of the person increases, being able to make more turns and faster.
K₂ = K₁ I₁ / I₂
Explanation:
When the divers are turning the system is isolated, so all the forces are internal and therefore also the torque, therefore the angular momentum is conserved
initial, joint when starting to turn
L₀ = I₁ w₁
final. When you shrink your arms and legs
Lf = I₂ w₂
L₀ = Lf
I₁ w₁ = I₂ w₂
when you shrink your arms and legs the distance to the turning point decreases and since the moment of inertia depends on the distance squared, the moment of inertia also decreases
I₂ <I₁
w₂ = I₁ / I₂ w₁
therefore the angular speed of the person increases, being able to make more turns and faster.
When it goes into the water it straightens the arm and leg, so the moment of inertia increases
I₁> I₂
w₁ = I₂ / I₁ w₂
therefore we see that the angular velocity decreases, therefore the person trains the water like a stone and can go deeper faster.
In both cases the kinetic energy is
K = ½ I w²
the initial kinetic energy is
K₁ = ½ I₁ w₁²
the final kinetic energy is
K₂ = ½ I₂ w₂²
we substitute
K₂ = ½ I₂ (I₁ / I₂ w1² 2
K₂ = ½ I₁² / I₂ w₁² = (½ I₁ w₁²) I₁ / I₂
K₂ = K₁ I₁ / I₂
therefore we see that the kinetic energy increases by factor I₁/I₂
In which situation is chemical energy being converted to another form of energy?
Answer:
A burning candle. (chemical energy into energy of heat and light, i.e. thermal and wave)
Explanation:
A solid sphere has a temperature of 556 K. The sphere is melted down and recast into a cube that has the same emissivity and emits the same radiant power as the sphere. What is the cube's temperature in kelvins
Answer:
Cube temperature = 526.83 K
Explanation:
Volume of the cube and sphere will be the same.
Now, volume of cube = a³
And ,volume of sphere = (4/3)πr³
Thus;
a³ = (4/3)πr³
a³ = 4.1187r³
Taking cube root of both sides gives;
a = 1.6119r
Formula for surface area of sphere is;
As = 4πr²
Also,formula for surface area of cube is; Ac = 6a²
Thus, since a = 1.6119r,
Then, Ac = 6(1.6119r)²
Ac = 15.5893r²
The formula for radiant power is;
Q' = eσT⁴A
Where;
e is emissivity
σ is Stefan boltzman constant = 5.67 x 10^(-8) W/m²k
T is temperate in kelvin
A is Area
So, for the cube;
(Qc)' = eσ(Tc)⁴(Ac)
For the sphere;
(Qs)' = eσ(Ts)⁴(As)
We are told (Qc)' = (Qs)'
Thus;
eσ(Tc)⁴(Ac) = eσ(Ts)⁴(As)
eσ will cancel out to give;
(Tc)⁴(Ac) = (Ts)⁴(As)
Since we want to find the cube's temperature Tc,
(Tc)⁴ = [(Ts)⁴(As)]/Ac
Plugging in relevant figures, we have;
(Tc)⁴ = [556⁴ × 4πr²]/15.5893r²
r² will cancel out to give;
(Tc)⁴ = [556⁴ × 4π]/15.5893
Tc = ∜([556⁴ × 4π]/15.5893)
Tc = 526.83 K
510 g squirrel with a surface area of 935 cm2 falls from a 4.8-m tree to the ground. Estimate its terminal velocity. (Use the drag coefficient for a horizontal skydiver. Assume that the squirrel can be approximated as a rectanglar prism with cross-sectional area of width 11.6 cm and length 23.2 cm. Note, the squirrel may not reach terminal velocity by the time it hits the gr
Answer:
The terminal velocity is [tex]v_t =17.5 \ m/s[/tex]
Explanation:
From the question we are told that
The mass of the squirrel is [tex]m_s = 50\ g = \frac{50}{1000} = 0.05 \ kg[/tex]
The surface area is [tex]A_s = 935 cm^2 = \frac{935}{10000} = 0.0935 \ m^2[/tex]
The height of fall is h =4.8 m
The length of the prism is [tex]l = 23.2 = 0.232 \ m[/tex]
The width of the prism is [tex]w = 11.6 = 0.116 \ m[/tex]
The terminal velocity is mathematically represented as
[tex]v_t = \sqrt{\frac{2 * m_s * g }{\dho_s * C * A } }[/tex]
Where [tex]\rho[/tex] is the density of a rectangular prism with a constant values of [tex]\rho = 1.21 \ kg/m^3[/tex]
[tex]C[/tex] is the drag coefficient for a horizontal skydiver with a value = 1
A is the area of the prism the squirrel is assumed to be which is mathematically represented as
[tex]A = 0.116 * 0.232[/tex]
[tex]A = 0.026912 \ m^2[/tex]
substituting values
[tex]v_t = \sqrt{\frac{2 * 0.510 * 9.8 }{1.21 * 1 * 0.026912 } }[/tex]
[tex]v_t =17.5 \ m/s[/tex]
Crystalline germanium (Z=32, rho=5.323 g/cm3) has a band gap of 0.66 eV. Assume the Fermi energy is half way between the valence and conduction bands. Estimate the ratio of electrons in the conduction band to those in the valence band at T = 300 K. (See eq. 10-11) Assume the width of the valence band is ΔΕV ~ 10 eV.
Answer:
= 8.2*10⁻¹²
Explanation:
Probability of finding an electron to occupy a state of energy, can be expressed by using Boltzmann distribution function
[tex]f(E) = exp(-\frac{E-E_f}{K_BT} )[/tex]
From the given data, fermi energy lies half way between valence and conduction bands, that is half of band gap energy
[tex]E_f = \frac{E_g}{2}[/tex]
Therefore,
[tex]f(E) = exp(-\frac{E-\frac{E_g}{2} }{K_BT} )[/tex]
Using boltzman distribution function to calculate the ratio of number of electrons in the conduction bands of those electrons in the valence bond is
[tex]\frac{n_{con}}{n_{val}} =\frac{exp(-\frac{[E_c-E_g/2]}{K_BT} )}{exp(-\frac{[E_v-E_fg/2}{K_BT} )}[/tex]
[tex]= exp(\frac{-(E_c-E_v}{K_BT} )\\\\=exp(\frac{-(0.66eV)}{(8.617\times10^-^5eV/K)(300K)} )\\\\=8.166\times10^-^1^2\approx8.2\times10^{-12}[/tex]
Q.1- Find the distance travelled by a particle moving in a straight line with uniform acceleration, in the 10th unit of time.
Answer:
If the acceleration is constant, the movements equations are:
a(t) = A.
for the velocity we can integrate over time:
v(t) = A*t + v0
where v0 is a constant of integration (the initial velocity), for the distance traveled between t = 0 units and t = 10 units, we can solve the integral:
[tex]\int\limits^{10}_0 {A*t + v0} \, dt = ((A/2)10^2 + v0*10) = (A*50 + v0*10)[/tex]
Where to obtain the actual distance you can replace the constant acceleration A and the initial velocity v0.
If you were to drop a rock from a tall building, assuming that it had not yet hit the ground, and neglecting air resistance. What is its vertical displacement (in m) after 4 s? (g = 10 m/s2)
Answer:
d = 80 m
its vertical displacement (in m) after 4 s is 80 m
Explanation:
From the equation of motion;
d = vt + 0.5at^2 ......1
Where;
d = displacement
v = initial velocity = 0 (dropped with no initial speed)
t = time of flight = 4s
a = g = acceleration due to gravity = 10 m/s^2
Substituting the given values into equation 1;
d = 0(4) + 0.5(10 × 4^2)
d = 0.5(10×16)
d = 80 m
its vertical displacement (in m) after 4 s is 80 m
Plaskett's binary system consists of two stars that revolve In a circular orbit about a center of mass midway between them. This statement implies that the masses of the two stars are equal . Assume the orbital speed of each star is |v | = 240 km/s and the orbital period of each is 12.5 days. Find the mass M of each star. (For comparison, the mass of our Sun is 1.99 times 1030 kg Your answer cannot be understood or graded.
Complete Question
The complete question is shown on the first uploaded image
Answer:
The mass is [tex]M =1.43 *10^{32} \ kg[/tex]
Explanation:
From the question we are told that
The mass of the stars are [tex]m_1 = m_2 =M[/tex]
The orbital speed of each star is [tex]v_s = 240 \ km/s =240000 \ m/s[/tex]
The orbital period is [tex]T = 12.5 \ days = 12.5 * 2 4 * 60 *60 = 1080000\ s[/tex]
The centripetal force acting on these stars is mathematically represented as
[tex]F_c = \frac{Mv^2}{r}[/tex]
The gravitational force acting on these stars is mathematically represented as
[tex]F_g = \frac{GM^2 }{d^2}[/tex]
So [tex]F_c = F_g[/tex]
=> [tex]\frac{mv^2}{r} = \frac{Gm_1 * m_2 }{d^2}[/tex]
=> [tex]\frac{v^2}{r} = \frac{GM}{(2r)^2}[/tex]
=> [tex]\frac{v^2}{r} = \frac{GM}{4r^2}[/tex]
=> [tex]M = \frac{v^2*4r}{G}[/tex]
The distance traveled by each sun in one cycle is mathematically represented as
[tex]D = v * T[/tex]
[tex]D = 240000 * 1080000[/tex]
[tex]D = 2.592*10^{11} \ m[/tex]
Now this can also be represented as
[tex]D = 2 \pi r[/tex]
Therefore
[tex]2 \pi r= 2.592*10^{11} \ m[/tex]
=> [tex]r= \frac{2.592*10^{11}}{2 \pi }[/tex]
=> [tex]r= 4.124 *10^{10} \ m[/tex]
So
[tex]M = \frac{v^2*4r}{G}[/tex]
=> [tex]M = \frac{(240000)^2*4*(4.124*10^{10})}{6.67*10^{-11}}[/tex]
=> [tex]M =1.43 *10^{32} \ kg[/tex]
Which is the best description of the scientific theory
Explanation:
a scientific theory is a well substantiated explanation of some aspect of the nature world, based on a body of facts that have been repeatedly confirmed through observation and experiment. search fact-supported theories are not "guesses" but reliable account of the real world .
A 72.0 kg swimmer jumps into the old swimming hole from a tree limb that is 3.90 m above the water.
A. Use energy conservation to find his speed just as he hits the water if he just holds his nose and drops in.
b) Use energy conservation to find his speed just he hits the water if he bravely jumps straight up (but just beyond the board!) at 2.90 m/s .
c) Use energy conservation to find his speed just he hits the water if he manages to jump downward at 2.90 m/s .
Answer:
Explanation:
The Law of Energy Conservation states that K1 + U1 = K2 + U2
m= 72.0 kg
h= 3.90 m
a)
K1 + U1 = K2 + U2
0 + mgh = 1/2mvf^2 + 0
mass cancels out so gh=1/2vf^2
(9.8 m/s^2)(3.9 m)=(.5)(vf^2)
vf= 8.74 m/s
b)
1/2mv^2 + mgh = 1/2mv^2 + 0
mass cancels again
(.5)(2.9^2 m/s) + (9.8 m/s^2)(3.9 m) = (.5)(vf^2)
vf= 9.21 m/s
c)
This would be the same as the past problem as the velocity gets squared so direction along the axis doesn't matter. Thus, vf= 9.21 m/s
Volume of an block is 5 cm3. If the density of the block is 250 g/cm3, what is the mass of the block ?
Answer:
The mass of the block is 1250g.
Explanation:
Given that the formula for density is ρ = mass/volume. Then you have to substitute the values into the formula :
[tex]ρ = \frac{mass}{volume} [/tex]
Let density = 250,
Let volume = 5,
[tex]250 = \frac{m}{5} [/tex]
[tex]m = 250 \times 5[/tex]
[tex]m = 1250g[/tex]
A 3 kg mass object is pushed 0.6 m into a spring with spring constant 210 N/m on a frictionless horizontal surface. Upon release, the object moves across the surface until it encounters a rough incline. The object moves UP the incline and stops a height of 1.5 m above the horizontal surface.
(a) How much work must be done to compress the spring initially?
(b) Compute the speed of the mass at the base of the incline.
(c) How much work was done by friction on the incline?
Answer with Explanation:
We are given that
Mass of spring,m=3 kg
Distance moved by object,d=0.6 m
Spring constant,k=210N/m
Height,h=1.5 m
a.Work done to compress the spring initially=[tex]\frac{1}{2}kx^2=\frac{1}{2}(210)(0.6)^2=37.8J[/tex]
b.
By conservation law of energy
Initial energy of spring=Kinetic energy of object
[tex]37.8=\frac{1}{2}(3)v^2[/tex]
[tex]v^2=\frac{37.8\times 2}{3}[/tex]
[tex]v=\sqrt{\frac{37.8\times 2}{3}}[/tex]
v=5.02 m/s
c.Work done by friction on the incline,[tex]w_{friction}=P.E-spring \;energy[/tex]
[tex]W_{friction}=3\times 9.8\times 1.5-37.8=6.3 J[/tex]
Davina accelerates a box across a smooth frictionless horizontal surface over a displacement of 18.0 m with a constant 25.0 N force angled at 23.0° below the horizontal. How much work does she do on the box? A. 176 J B. 414 J C. 450 J D. 511 J Group of answer choices
Answer:
W = 414 J, correct is B
Explanation:
Work is defined by
W = ∫ F .dx
where F is the force, x is the displacement and the point represents the dot product
this expression can also be written with the explicit scalar product
W = ∫ F dx cos θ
where is the angle between force and displacement
for this case as the force is constant
W = F x cos θ
calculate
W = 25.0 18.0 cos (-23)
W = 414 J
the correct answer is B
A machinist turns the power on to a grinding wheel, which is at rest at time t = 0.00 s. The wheel accelerates uniformly for 10 s and reaches the operating angular velocity of 25 rad/s. The wheel is run at that angular velocity for 37 s and then power is shut off. The wheel decelerates uniformly at 1.5 rad/s2 until the wheel stops. In this situation, the time interval of angular deceleration (slowing down) is closest to: A machinist turns the power on to a grinding wheel, which is at rest at time t = 0.00 s. The wheel accelerates uniformly for 10 s and reaches the operating angular velocity of 25 rad/s. The wheel is run at that angular velocity for 37 s and then power is shut off. The wheel decelerates uniformly at 1.5 rad/s2 until the wheel stops. In this situation, the time interval of angular deceleration (slowing down) is closest to:__________.
a) 19 s
b) 17 s
c) 21 s
d) 23 s
e) 15 s
Starting from rest, the wheel attains an angular velocity of 25 rad/s in a matter of 10 s, which means the angular acceleration [tex]\alpha[/tex] is
[tex]25\dfrac{\rm rad}{\rm s}=\alpha(10\,\mathrm s)\implies\alpha=2.5\dfrac{\rm rad}{\mathrm s^2}[/tex]
For the next 37 s, the wheel maintains a constant angular velocity of 25 rad/s, meaning the angular acceleration is zero for the duration. After this time, the wheel undergoes an angular acceleration of -1.5 rad/s/s until it stops, which would take time [tex]t[/tex],
[tex]0\dfrac{\rm rad}{\rm s}=25\dfrac{\rm rad}{\rm s}+\left(-1.5\dfrac{\rm rad}{\mathrm s^2}\right)t\implies t=16.666\ldots\,\mathrm s[/tex]
which makes B, approximately 17 s, the correct answer.
The time interval of angular deceleration is 16.667 seconds, whose closest integer is 17 seconds. (B. 17 s.)
Let suppose that the grinding wheel has uniform Acceleration and Deceleration. In this question we need to need to calculate the time taken by the grinding wheel to stop, which is found by means of the following Kinematic formula:
[tex]t = \frac{\omega - \omega_{o}}{\alpha}[/tex] (1)
Where:
[tex]\omega_{o}[/tex] - Initial angular velocity, in radians per second.
[tex]\omega[/tex] - Final angular velocity, in radians per second.
[tex]\alpha[/tex] - Angular acceleration, in radians per square second.
[tex]t[/tex] - Time, in seconds.
If we know that [tex]\omega = 0\,\frac{rad}{s}[/tex], [tex]\omega_{o} = 25\,\frac{rad}{s}[/tex] and [tex]\alpha = -1.5\,\frac{rad}{s^{2}}[/tex], then the time taken by the grinding wheel to stop:
[tex]t = \frac{0\,\frac{rad}{s}-25\,\frac{rad}{s}}{-1.5\,\frac{rad}{s^{2}} }[/tex]
[tex]t = 16.667\,s[/tex]
The time interval of angular deceleration is 16.667 seconds. (Answer: B)
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