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3 years ago

What occurs when someone places an ice pack on an injured ankle?

Physics

2 answers:

erica [24]3 years ago

7 0

The answer is:

Heat from the ankle is transferred to the ice pack.

The explanation:

-when the ice pack a lower temp than the ankle , it is very cool.

- and the injured ankle has a high temp or heat

and we know that Q transfer from the object with high heat to the object with the low temp So,

Heat will transfer from the ankle to the ice pack.

Dimas [21]3 years ago

5 0

Answer: 4

Explanation:

Heat from the ankle is transferred to the ice pack.

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Which best describes how energy changes form in a car's engine?

andrezito [222]

Answer:

Multiple transformations occur because the chemical energy of the fuel is changed to several forms of energy

Explanation:

In a car engine, multiple energy transformation takes place. The chemical energy storef in fuel is transformed into mechanical energy which helps move the wheels of the vehicle.

The mechanical energy can also be transformed into electrical energy through a sort of dynamo system in vehicles. Stereo players use the electrical energy to produce sound.

We see that multiple energy conversions are common in a motor car.

6 0

3 years ago

Read 2 more answers

How is energy conserved in a transformation?

irina [24]

As the water plunges, its velocity increases. Its potential energy becomes kineticenergy. The law of conservation of energy states that when one form of energy istransformed to another, no energy is destroyed in the process. ... So the total amount of energy is the same before and after any transformation.

hope it helps

5 0

3 years ago

WHO WANts TO HAVE SOME GLIZZY ACTION

Lilit [14]

Bruh huh.............

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2 years ago

A parachutist falls 50.0 m without friction. When the parachute opens, he slows down at a rate of 67 m/s*2. If he reaches the gr

KIM [24]

Answer:

3.49 seconds

3.75 seconds

-43200 ft/s²

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

$s=ut+\frac{1}{2}at^2\\\Rightarrow 50=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{50\times 2}{9.81}}\\\Rightarrow t=3.19\ s$

Time the parachutist falls without friction is 3.19 seconds

$v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 9.81\times 50+0^2}\\\Rightarrow v=31.32\ m/s$

Speed of the parachutist when he opens the parachute 31.32 m/s. Now, this will be considered as the initial velocity

$v=u+at\\\Rightarrow 11=31.32+9.81t\\\Rightarrow t=\frac{11-31.32}{-67}=0.3\ s$

So, time the parachutist stayed in the air was 3.19+0.3 = 3.49 seconds

$s=ut+\frac{1}{2}at^2\\\Rightarrow \frac{s}{2}=0t+\frac{1}{2}\times a\times t^2\\\Rightarrow \frac{s}{2}=\frac{1}{2}at^2$

$s=ut+\frac{1}{2}at^2\\\Rightarrow \frac{s}{2}=u1.1+\frac{1}{2}\times a\times 1.1^2$

Now the initial velocity of the last half height will be the final velocity of the first half height.

$v=u+at\\\Rightarrow v=at$

Since the height are equal

$\frac{1}{2}at^2=u1.1+\frac{1}{2}\times a\times 1.1^2\\\Rightarrow \frac{1}{2}at^2=at1.1+\frac{1}{2}\times a\times 1.1^2\\\Rightarrow 0.5t^2-1.1t-0.605=0\\\Rightarrow 500t^2-1100t-605=0$

$t=\frac{11\left(1+\sqrt{2}\right)}{10},\:t=\frac{11\left(1-\sqrt{2}\right)}{10}\\\Rightarrow t=2.65, -0.45$

Time taken to fall the first half is 2.65 seconds

Total time taken to fall is 2.65+1.1 = 3.75 seconds.

When an object is thrown with a velocity upwards then the velocity of the object at the point to where it was thrown becomes equal to the initial velocity.

$v^2-u^2=2as\\\Rightarrow a=\frac{v^2-u^2}{2s}\\\Rightarrow a=\frac{0^2-240^2}{2\times \frac{8}{12}}\\\Rightarrow a=-43200\ ft/s^2$

Magnitude of acceleration is -43200 ft/s²

5 0

3 years ago

Please help this is physics!!!

larisa [96]

I’m pretty sure u have it right

8 0

3 years ago