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FinnZ [79.3K]
1 year ago
12

Alex throws a 0.15-kg rubber ball down onto the floor. The ball’s speed just before impact is 6.5 m/s, and just after is 3.5 m/s

. If the ball is in contact with the floor for 0.025 s, what is the magnitude of the average force applied by the floor on the ball?
Physics
1 answer:
Jet001 [13]1 year ago
7 0

Answer: Change in ball's momentum is 1.5 kg-m/s.

Explanation: It is given that,

Mass of the ball, m = 0.15 kg

Speed before the impact, u = 6.5 m/s

Speed after the impact, v = -3.5 m/s (as it will rebound)

We need to find the change in the magnitude of the ball's momentum. It is given by :

So, the change in the ball's momentum is 1.5 kg-m/s. Hence, this is the required solution.

Read more on Brainly.com - brainly.com/question/12946012#readmore

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The wheels of the locomotive push back on the tracks with a constant net force of 7.50 × 105 N, so the tracks push forward on th
Rasek [7]

Answer:

The freight train would take 542.265 second to increase the speed of the train from rest to 80.0 kilometers per hour.

Explanation:

Statement is incomplete. Complete description is presented below:

<em>A freight train has a mass of </em>1.83\times 10^{7}\,kg<em>. The wheels of the locomotive push back on the tracks with a constant net force of </em>7.50\times 10^{5}\,N<em>, so the tracks push forward on the locomotive with a force of the same magnitude. Ignore aerodynamics and friction on the other wheels of the train. How long, in seconds, would it take to increase the speed of the train from rest to 80.0 kilometers per hour?</em>

If locomotive have a constant net force (F), measured in newtons, then acceleration (a), measured in meters per square second, must be constant and can be found by the following expression:

a = \frac{F}{m} (1)

Where m is the mass of the freight train, measured in kilograms.

If we know that F = 7.50\times 10^{5}\,N and m = 1.83\times 10^{7}\,kg, then the acceleration experimented by the train is:

a = \frac{7.50\times 10^{5}\,N}{1.83\times 10^{7}\,kg}

a = 4.098\times 10^{-2}\,\frac{m}{s^{2}}

Now, the time taken to accelerate the freight train from rest (t), measured in seconds, is determined by the following formula:

t = \frac{v-v_{o}}{a} (2)

Where:

v - Final speed of the train, measured in meters per second.

v_{o} - Initial speed of the train, measured in meters per second.

If we know that a = 4.098\times 10^{-2}\,\frac{m}{s^{2}}, v_{o} = 0\,\frac{m}{s} and v = 22.222\,\frac{m}{s}, the time taken by the freight train is:

t = \frac{22.222\,\frac{m}{s}-0\,\frac{m}{s}  }{4.098\times 10^{-2}\,\frac{m}{s^{2}} }

t = 542.265\,s

The freight train would take 542.265 second to increase the speed of the train from rest to 80.0 kilometers per hour.

6 0
1 year ago
Now that we have a feel for the state of the circuit in its steady state, let us obtain the expression for the current in the ci
vesna_86 [32]

Answer:

i(t) = (E/R)[1 - exp(-Rt/L)]

Explanation:

E−vR−vL=0

E− iR− Ldi/dt = 0

E− iR = Ldi/dt

Separating te variables,

dt/L = di/(E - iR)

Let x = E - iR, so dx = -Rdi and di = -dx/R substituting for x and di we have

dt/L = -dx/Rx

-Rdt/L = dx/x

interating both sides, we have

∫-Rdt/L = ∫dx/x

-Rt/L + C = ㏑x

x = exp(-Rt/L + C)

x = exp(-Rt/L)exp(C)     A = exp(C) we have

x = Aexp(-Rt/L) Substituting x = E - iR we have

E - iR = Aexp(-Rt/L) when t = 0, i(0) = 0. So

E - i(0)R = Aexp(-R×0/L)

E - 0 = Aexp(0) = A × 1

E = A

So,

E - i(t)R = Eexp(-Rt/L)

i(t)R = E - Eexp(-Rt/L)

i(t)R = E(1 - exp(-Rt/L))

i(t) = (E/R)(1 - exp(-Rt/L))

5 0
2 years ago
A civil engineer wishes to redesign the curved roadway in the example What is the Maximum Speed of the Car? in such a way that a
vlabodo [156]

Answer:

24.3 degrees

Explanation:

A car traveling in circular motion at linear speed v = 12.8 m/s around a circle of radius r = 37 m is subjected to a centripetal acceleration:

a_c = \frac{v^2}{r} = \frac{12.8^2}{37} = 4.43 m/s^2

Let α be the banked angle, as α > 0, the outward centripetal acceleration vector is split into 2 components, 1 parallel and the other perpendicular to the road. The one that is parallel has a magnitude of 4.43cosα and is the one that would make the car slip.

Similarly, gravitational acceleration g is split into 2 component, one parallel and the other perpendicular to the road surface. The one that is parallel has a magnitude of gsinα and is the one that keeps the car from slipping outward.

So gsin\alpha = 4.43cos\alpha

\frac{sin\alpha}{cos\alpha} = \frac{4.43}{g} = \frac{4.43}{9.81} = 0.451

tan\alpha = 0.451

\alpha = tan^{-1}0.451 = 0.424 rad = 0.424*180/\pi \approx 24.3^0

3 0
2 years ago
The inductor in a radio receiver carries a current of amplitude 200 mA when a voltage of amplitude 2.40 V is across it at a freq
White raven [17]

Answer:92

Explanation:

3 0
2 years ago
Astronomers have discovered a new planet called "Xandar" beyond the orbit of Pluto (No, not really but I need a fake planet for
Burka [1]

Answer:

m = 1.82E+23 kg

Explanation:

G = universal gravitational constant = 6.67E-11 N·m²/kg²

r = radius of orbit = 72,600 km = 7.26E+07 m

C = circumference of orbit = 2πr = 4.56E+08 m

P = period of orbit = 12.9 d = 1,114,560 s

v = orbital velocity of satellite Jim = C/P = 409 m/s

m = mass of Xandar = to be determined

v = √(Gm/r)

v² = [√(Gm/r)]²

v² = Gm/r

rv² = Gm

rv²/G = m

m = rv²/G

mG = universal gravitational constant = 6.67E-11 N·m²/kg²

r = radius of orbit = 72,600 km = 7.26E+07 m

C = circumference of orbit = 2πr = 4.56E+08 m

P = period of orbit = 12.9 d = 1,114,560 s

v = orbital velocity of satellite Jim = C/P = 409 m/s

m = mass of Xandar = to be determined

v = √(Gm/r)

v² = [√(Gm/r)]²

v² = Gm/r

rv² = Gm

rv²/G = m

m = rv²/G

m = 1.82E+23 kg

3 0
2 years ago
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