Kyle has a mass of 54kg and is moving at 3 m/s what is his kinetic energy

Suppose you are on a cart, initially at rest, which rides on a frictionless horizontal track. You throw a ball at a vertical sur

Len [333]

Answer:

$F_c t_ c = -F_b t_b$

And the forces are equal but in the opposite direction. So then we can write by general rule:

$m_c \Delta V_{c} = -m_b \Delta V_b$

Or equivalently:

$m_c \Delta V_{c} +m_b \Delta V_b =0$

Where: $V_c$ represent the speed of the car and $V_b$ the speed of the ball

$m_c$ represent the mass of the car

$m_b$ represent the mass of the ball

Since the ball is moving to the left and we assume that the total momentum not changes then the car need to move to the right in order to satisfy the equation and satisfy the balance.

By conservation of the momentum the car will move to the right since the ball is moves to the left.

So then the correct option for this case is :

A.Yes, and it moves to the right.

Explanation:

If we assume that we have the situation in the figure attached.

For this case we assume that the momentum changes are equal in magnitude and opposite in direction, so then we satisfy this:

$F_c t_ c = -F_b t_b$

And the forces are equal but in the opposite direction. So then we can write by general rule:

$m_c \Delta V_{c} = -m_b \Delta V_b$

Or equivalently:

$m_c \Delta V_{c} +m_b \Delta V_b =0$

Where: $V_c$ represent the speed of the car and $V_b$ the speed of the ball

$m_c$ represent the mass of the car

$m_b$ represent the mass of the ball

Since the ball is moving to the left and we assume that the total momentum not changes then the car need to move to the right in order to satisfy the equation and satisfy the balance.

By conservation of the momentum the car will move to the right since the ball is moves to the left.

So then the correct option for this case is :

A.Yes, and it moves to the right.

3 0

3 years ago

This object is located 13.0 cm to the left of the lens and the image forms at 20.8 cm to the right of the lens.

Alina [70]

Answer:

0.125 cm

Explanation:

1/f = 1/d¡ + 1/d。

Find the focal point

(13.0^-1 + 20.8^-1) = 0.125 m

Focal point = 0.125 m

6 0

2 years ago

Read 2 more answers

Tony drove to the mountains last weekend. There was heavy traffic on the way there, and the trip took hours. When Tony drove hom

polet [3.4K]

Answer:

Question not completed, so I analysed the question first

Tony drove to the mountains last weekend. there was heavy traffic on the way there, and the trip took 6 hours. when tony drove home, there was no traffic and the trip only took 4 hours. if his average rate was 22 miles per hour faster on the trip home, how far away does tony live from the mountains?

Explanation:

Let use variables to solve the problems

Let the first trip to be mountain take x hours

Let the trip back home take y hours

Let the speed to while going to the mountain be a miles/hour

Then, while going home it was b miles/hour faster than while going to the mountain.

Then, speed going home is (a+b)miles / hour

The formula for speed is given as

Speed=distance/time

The constant through out the journey is distance, the two journey has the same distance.

Then,

Distance =speed×time

For first journey going to the mountain

Distance = a×x=ax miles

For the second journey going home

Distance =y×(a+b)

Distance Mountain= distance home

ax=y(a+b)

Make a subject of the formula

ax=ya+yb

ax-ya=yb

a(x-y)=yb

a=yb/(x-y)

Therefore, distance from mountain is

Distance=speed ×time

Distance= a×x=ax

Now, applying the questions

So from the questions

x=6hours, y=4hours

Also, b=22miles/hour

Then,

a=yb/(x-y)

a=4×22/(6-4)

a=88/2

a=44miles/hour

Then, the house distance from the mountain is

Distance=ax

Distance =44×6

Distance =264miles

4 0

3 years ago

Read 2 more answers

A bus decreases its speed

eduard

Answer:

-6 km/h per minute

Explanation:

7 0

3 years ago

A 2 kg ball is moving 3 m/s when it starts rolling up a hill.

AURORKA [14]

Answer:

the height reached is = 0.458 [m]

Explanation:

We need to make a sketch of the ball and see the location of the reference point where the potential energy is zero. But the kinetic energy will be defined by the following expression:

$Ek=\frac{1}{2} *m*v^{2} \\where:Ek= kinetic energy [J]\\m = mass of the ball [kg]\\v = velocity of the ball [m/s]$

Replacing the values on the equation we have:

$Ek=\frac{1}{2}*(2)*(3^{2} )\\ Ek=9[J]\\$

This kinetic energy will be transformed in potential energy in the moment when the ball starts to rolling up. Therefore the maximum height reached by the ball depends of the initial velocity given to the ball.

$Ek=Ep\\where\\Ep=potential energy [J]\\Ep=m*g*h\\where\\g=gravity = 9.81[m/s^2]\\h=height reached [m]\\$

Now we have:

$h=\frac{Ep}{m*g} \\h=\frac{9}{2*9.81} \\\\h=0.45 [m]$

In that moment when the ball reach the 0.45 [m] the potencial energy will be maximum and equal to the kinetic energy when the ball has a velocity of 3[m/s]

6 0

3 years ago