Ask question

Ask question

All categories

2 years ago

12

Your friend says an appliance uses energy. How would you correct his statement?

1 answer:

madreJ [45]2 years ago

3 0

Answer:

Not all appliances run on energy. Some of them run on gas. Some both. It just depends on the age of the appliance, the make of the appliance, and the company who made it.

You might be interested in

PLEASE HELP! Can you do the write about it! Will thank and try to mark brainiest!

goblinko [34]

Move the objects faster to get more friction.

8 0

2 years ago

Read 2 more answers

A 0.125 kg hockey puck moving at 20.0 m/s is caught and held by an 85.0 kg goalkeeper at rest. After catching the puck, with wha

jenyasd209 [6]

Answer:

a

Explanation:cuz i saiad so

4 0

3 years ago

Which do you think changes the land more frozen water or flowing water?

Maksim231197 [3]

I believe flowing water changes the land because eventually frozen water has to melt to flowing water, right?

8 0

3 years ago

Read 2 more answers

In Anchorage, collisions of a vehicle with a moose are so common that they are referred to with the abbreviationMVC. Suppose a 1

lara31 [8.8K]

Answer:

Part a)

$f = \frac{8}{9}$

Part b)

$f = \frac{120}{169}$

Part c)

So from above discussion we have the result that energy loss will be more if the collision occurs with animal with more mass

Explanation:

Part a)

Let say the collision between Moose and the car is elastic collision

So here we can use momentum conservation

$m_1v_{1i} = m_1v_{1f} + m_2v_{2f}$

$1000 v_o = 1000 v_{1f} + 500 v_{2f}$

also by elastic collision condition we know that

$v_{2f} - v_{1f} = v_o$

now we have

$2v_o = 2v_{1f} + v_o + v_{1f}$

now we have

$v_{1f} = \frac{v_o}{3}$

Now loss in kinetic energy of the car is given as

$\Delta K = \frac{1}{2}m(v_o^2 - v_{1f}^2)$

$\Delta K = \frac{1}{2}m(v_o^2 - \frac{v_o^2}{9})$

so fractional loss in energy is given as

$f = \frac{\Delta K}{K}$

$f = \frac{\frac{4}{9}mv_o^2}{\frac{1}{2}mv_o^2}$

$f = \frac{8}{9}$

Part b)

Let say the collision between Camel and the car is elastic collision

So here we can use momentum conservation

$m_1v_{1i} = m_1v_{1f} + m_2v_{2f}$

$1000 v_o = 1000 v_{1f} + 300 v_{2f}$

also by elastic collision condition we know that

$v_{2f} - v_{1f} = v_o$

now we have

$10v_o = 10v_{1f} + 3(v_o + v_{1f})$

now we have

$v_{1f} = \frac{7v_o}{13}$

Now loss in kinetic energy of the car is given as

$\Delta K = \frac{1}{2}m(v_o^2 - v_{1f}^2)$

$\Delta K = \frac{1}{2}m(v_o^2 - \frac{49v_o^2}{169})$

so fractional loss in energy is given as

$f = \frac{\Delta K}{K}$

$f = \frac{\frac{60}{169}mv_o^2}{\frac{1}{2}mv_o^2}$

$f = \frac{120}{169}$

Part c)

So from above discussion we have the result that energy loss will be more if the collision occurs with animal with more mass

8 0

2 years ago

A uniform disk with a mass of 5.0 kg and diameter 30 cm rotates on a frictionless fixed axis through its center and perpendicula

Marina86 [1]

Answer: $\alpha =10.66 rad/s^2$

Explanation:

Given

mass of disk $m=5 kg$

diameter of disc $d=30 cm$

Force applied $F=4 N$

Now this force will Produce a torque of magnitude

$T=F\cdot r$

$T=4\dot 0.15$

$T=0.6 N-m$

And Torque is given Product of moment of inertia and angular acceleration $(\alpha )$

$T=I\cdot \alpha$

Moment of inertia for Disc $I= \frac{Mr^2}{2}$

$I=0.05625 kg-m^2$

$0.6=0.05625\cdot \alpha$

$\alpha =10.66 rad/s^2$

3 0

2 years ago