All categories

3 years ago

What is the kinetic energy of a 1 kg pie if it is thrown at 10 m/s?

Physics

1 answer:

nignag [31]3 years ago

7 0

Given

mass (m) = 1 kg

velocity (v) = 10 m/s

kinetic energy ( ke) = ?

,we know

K.E =1/2 m v²

= 1/2 * 1 * 10²

= 100/2

=50 joule

hope it helps :)

You might be interested in

do y’all think a toxic relationship is good ? like me and him are on and off one minute i say something wrong and BOOM he says b

ziro4ka [17]

Answer:

A toxic relationship is the worst thing someone could have,because people need to learn to chill out.

Explanation:

5 0

3 years ago

You are a member of an alpine rescue team and must get a box of supplies, with mass 2.20 kg , up an incline of constant slope an

a_sh-v [17]

Answer:

The minimum speed of the box bottom of the incline so that it will reach the skier is 8.19 m/s.

Explanation:

It is given that,

Mass of the box, m = 2.2 kg

The box is inclined at an angle of 30 degrees

Vertical distance, d = 3.1 m

The coefficient of friction, $\mu=6\times 10^{-2}$

Using the work energy theorem, the loss of kinetic energy is equal to the sum of gain in potential energy and the work done against friction.

$KE=PE+W$

$\dfrac{1}{2}mv^2=mgh+W$

W is the work done by the friction.

$W=f\times d$

$f=\mu mg\ cos\theta$

$W=\mu mg\ cos\theta\times \dfrac{d}{sin\theta}$

$W=\dfrac{\mu mgh}{tan\theta}$

$\dfrac{1}{2}mv^2=mgh+\dfrac{\mu mgh}{tan\theta}$

$\dfrac{1}{2}v^2=gh+\dfrac{\mu gh}{tan\theta}$

$\dfrac{1}{2}v^2=9.8\times 3.1+\dfrac{6\times 10^{-2}\times 9.8\times 3.1}{tan(30)}$

v = 8.19 m/s

So, the speed of the box is 8.19 m/s. Hence, this is the required solution.

8 0

3 years ago

A 3kg ball moving at 8 m/s strikes a 2kg ball at rest,if the collision is elastic,what is the speed of the lighter ball if the h

zloy xaker [14]

Answer:

Speed of lighter ball is 4 m/s.

Explanation:

Applying the principle of conservation of linear momentum,

momentum before collision = momentum after collision.

$m_{1}$ $u_{1}$ + $m_{2}$ $u_{2}$ = $m_{1}$ $v_{1}$ - $m_{2}$ $v_{2}$

$m_{1}$ = 3 kg, $u_{1}$ = 8 m/s, $m_{2}$ = 2 kg, $u_{2}$ = 0 m/s ( since it is at rest), $v_{1}$ = 2 m/s, $v_{2}$ = ?

(3 x 8) + (2 x 0) = (8 x 2) - (2 x $v_{2}$ )

24 + 0 = 16 - 2 $v_{2}$

2 $v_{2}$ = 16 - 24

2 $v_{2}$ = -8

$v_{2}$ = $\frac{-8}{2}$

= -4 m/s

This implies that the light ball moves at the speed of 4 m/s in the opposite direction of the heavier ball after collision.

7 0

2 years ago

What is the resistance of a circuit with a voltage of 10 volts (V) and a current of 5 amps (A)? Use Ohm’s law to calculate the r

iogann1982 [59]

Answer: 2Ω

Explanation:

V = IR

R = V/I

R = 10/5 = 2Ω

Hope it helped!

5 0

3 years ago

Read 2 more answers

What are two ways friction can be useful

Mnenie [13.5K]

Gravity is an example of friction. If we didn't have gravity, we would be flying all over the place. Also, friction keeps us from sliding on the ground and falling.

4 0

3 years ago