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1 year ago

Rearrange the equation = KE -1 2 - my? to solve for v. Show your work.

Physics

1 answer:

vesna_86 [32]1 year ago

3 0

Answer:

See below

Explanation:

KE = 1/2 m v^2 multiply both sides by 2

2 (KE) = mv^2 divide both sides by m

2(KE) / m = v^2 sqrt both sides

√ [(2KE)/m ] = v

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If a 100-N net force acts on a 50-kg car, what will the acceleration of the car be?

Tema [17]

Newton's 2nd law of motion:

   Force = (mass) x (acceleration)

Divide each side by (mass):

   Acceleration = (force) / (mass)

   = (100 N) / (50 kg)

   = 2 m/s²

5 0

2 years ago

The ms = 0.5645 and mk=0.1113 for a 30 N box on level ground. What level force is needed to start the box moving

nevsk [136]

Answer:

16.935 N

Explanation:

In order to make the box start moving, the level force applied on the box (F) must be greater than the force of static friction that keeps the box at rest, which is equal to

$F_f = \mu_s (mg)$

where

$\mu_s = 0.5645$ is the coefficient of static friction

(mg) = 30 N is the weight of the box

Therefore, the condition for F must be:

$F \geq F_f\\F \geq \mu_k (mg)=(0.5645)(30 N)=16.935 N$

So, the applied force must be greater than this value.

3 0

3 years ago

Consider the speed of light in another universe to be only 100 m/s. Two cars are traveling along a highway in opposite direction

olganol [36]

Answer:

-62.45m/s and +62.45m/s

Explanation:

The formula for relativistic speed

This is the speed of A with respect to B

$V_{AB}=\frac{V_{A}-V_{B}}{1-\frac{V_{A}V_{B}}{C^2} }$

where

$V_{A}$ will be the velocity of person 1: 39m/s

$V_{B}$ will be the velocity of person 2: -31m/s (negative because is travelling in opposite direction)

and $C$ the velocity of light: 100m/s

The velocity of person 1 measured by person 2 is:

$V_{AB}=\frac{39m/s-(-31m/s)}{1-\frac{(39m/s)(-31m/s)}{(100m/s)^2}}=62.45m/s$

The velocity of person 2 measured by person 1 is:

$V_{BA}=\frac{V_{B}-V_{A}}{1-\frac{V_{B}V_{A}}{C^2} }$

$V_{BA}=\frac{-31m/s-39m/s}{1-\frac{(-31m/s)(39m/s)}{(100)^2} }=-62.45m/s$

8 0

2 years ago

What was the direction of the ball’s velocity

Tju [1.3M]

Part of the question is missing. Here it is:

<em>A 72 g autographed baseball slides off of a 1.3 m high table and strikes the floor a horizontal distance of 0.7m away from the table. The acceleration of gravity is 9.81 m/s2. What was the direction of the ball’s velocity just before it hit the floor? </em>

Answer:

$\theta=-75.7^{\circ}$

Explanation:

The motion of the ball is a projectile motion, which consists of two separate motions:

- A horizontal motion at constant velocity

- A vertical motion at constant acceleration (free fall)

We start by analyzing the vertical motion, to find the time of flight of the ball. This can be done by using the suvat equation

$s=ut+\frac{1}{2}at^2$

where, choosing downward as positive direction:

s =1.3 m is the vertical displacement of the ball

u = 0 is the initial vertical velocity

$a=g=9.8 m/s^2$ is the acceleration of gravity

t is the time

Solving for t,

$t=\sqrt{\frac{2s}{a}}=\sqrt{\frac{2(1.3)}{9.8}}=0.52 s$

Now we can find the final vertical velocity of the ball, using:

$v_y=u+at$

And susbtituting t = 0.52 s, we find

$v_y = 0 +(9.8)(0.52)=5.1 m/s$

It is important to keep in mind that the direction of this velocity is downward, since we chose downward as positive direction.

The horizontal velocity of the ball instead is constant; we know that the ball covers a horizontal distance of

d = 0.7 m

In a time of

t = 0.52 s

So, the horizontal velocity is

$v_x = \frac{0.7}{0.52}=1.3 m/s$

So now we can find the direction of the ball's velocity using:

$\theta=tan^{-1}(\frac{v_y}{v_x})=tan^{-1}(\frac{5.1}{1.3})=75.7^{\circ}$

And since the vertical direction is downward, this means that this velocity is below the horizontal, so the answer is

$\theta=-75.7^{\circ}$

8 0

2 years ago

A 0.150-kg ball traveling horizontally on a frictionless surface approaches a very massive stone at 20.0 m/s perpendicular to wa

Hitman42 [59]

Answer:

The magnitude of the change in momentum of the stone is 5.51kg*m/s.

Explanation:

the final kinetic energy = 1/2(0.15)v^2

1/2(0.15)v^2 = 70%*1/2(0.15)(20)^2

v^2 = 21/0.075

v^2 = 280

v = 16.73 m.s

if u is the initial speed and v is the final speed, then:

u = 20 m/s and v = - 16.73m/s

change in momentum = m(v-u)

= 0.15(- 16.73-20)

= -5.51 kg*m.s

Therefore, The magnitude of the change in momentum of the stone is 5.51kg*m/s.

3 0

3 years ago