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Nikolay [14]
3 years ago
11

A 10.0-kg box starts at rest on a level floor. An external, horizontal force of 2.00 × 102 N is applied to the box for a distanc

e of 4.00 m. If the coefficient of kinetic friction between the box and the floor is 0.440, how fast is the block moving at the end of the 4.00 m? Assume it starts from rest.
Physics
1 answer:
Harman [31]3 years ago
7 0

Answer:

vf = 11.2 m/s

Explanation:

m = 10 Kg

F = 2*10² N

x = 4.00 m

μ = 0.44

vi = 0 m/s

vf = ?

We can apply Newton's 2nd Law

∑ Fx = m*a   (→)

F - Ffriction = m*a  ⇒  F - (μ*N) = F - (μ*m*g) = m*a   ⇒  a = (F - μ*m*g)/m

⇒    a = (2*10² N - 0.44*10 Kg*9.81 m/s²)/10 Kg = 15.6836 m/s²

then , we use the equation

vf² = vi² + 2*a*x    ⇒    vf = √(vi² + 2*a*x)

⇒   vf = √((0)² + 2*(15.6836 m/s²)*(4.00m)) = 11.2 m/s

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lions [1.4K]

Answer:

7229 N

Explanation:

The gravitational force between the Death Star and the Millenium Falcon is given by:

F=G\frac{mM}{R^2}

where

G=6.67\cdot 10^{-11} m^3 kg^{-1} s^{-2} is the gravitational constant

M=5.1\cdot 10^{17} kg is the mass of the Death Star

m=1.36\cdot 10^6 kg is the mass of the Millennium Falcon

R=\frac{160,000 m}{2}=80,000 m is the radius of the Death Star

Substituting numbers into the equation, we find the force

F=(6.67\cdot 10^{-11})\frac{(1.36\cdot 10^6 kg)(5.1\cdot 10^{17} kg)}{(80,000 m)^2}=7229 N

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3 years ago
Because so little can be found of the first rocks to form on
fgiga [73]

Answer:

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You punt a ball straight up at 20 m/s. What is the balls hangtime
Natalija [7]

The hang time of the ball is 4.08 s

Explanation:

The ball is in free fall motion: this means that it is acted upon gravity only, so its acceleration is the acceleration of gravity,

a=g=-9.8 m/s^2

downward (the negative sign refers to the downward direction).

Since this is a uniformly accelerated motion, we can solve the problem by using the following suvat equation:

v=u+at

where

v is the final velocity

u is the initial velocity

a is the acceleration

t is the time

First we calculate the time it takes for the ball to reach the maximum height, where the velocity is zero:

v = 0

Substituting:

u = +20 m/s

a=-9.8 m/s^2

we find t

t=\frac{v-u}{a}=\frac{0-20}{-9.8}=2.04 s

The motion of the ball is symmetrical, so the total time of flight is just twice the time needed to reach the maximum height, therefore:

T=2t=2(2.04)=4.08 s

Learn more about free fall:

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4 0
3 years ago
An object is originally moving at a constant velocity of 8 m/s in the -x direction. It moves at this constant velocity for 3 sec
aivan3 [116]

Answer:

244.64m

Explanation:

First, we find the distance traveled with constant velocity. It's simply multiplying velocity time the time that elapsed:

x = V*t = -8\frac{m}{s} *3s = -24m

After this, the ball will start traveling with a constant acceleration motion. Due to the fact that the acceleration is the opposite direction to the initial velocity, this motion will have 2 phases:

1. The velocity will start to decrease untill it reaches 0m/s.

2. Then, the velocity will start to increase at the rate of the acceleration.

The distance that the ball travels in the first phase can be found with the following expression:

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Where v is the final velocity (0m/s), v_0 is the initial velocity (-8m/s) and a is the acceleration (+9m/s^2). We solve for d:

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Now, before finding the distance traveled in the second phase, we need to find the time that took for the velocity to reach 0:

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Then, the time of the second phase will be:

t_2 = 9s - t_1 = 9s - 1.143s = 7.857s

Using this, we using the equations for constant acceleration motion in order to calculate the distance traveled in the second phase:

x = \frac{1}{2}a*t^2 + v_0*t + x_0

V_0, the initial velocity of the second phase, will be 0 as previously mentioned. X_0, the initial position, will be 0, for simplicity:

x = \frac{1}{2}*7\frac{m}{s^2}*t^2 + 0m/s*t + 0m = 216.07m

So, the total distance covered by this object in meters will be the sum of all the distances we found:

x_total = 24m + 4.57m + 216.07m = 244.64m

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3 years ago
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