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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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2. What is the mass of an object if it accelerates at 3 m/s2 and has a force of 1 N?

Dvinal [7]

Answer:

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2 years ago

An 85-kg man plans to tow a 109 000-kg airplane along a runway by pulling horizontally on a cable attached to it. Suppose that h

Lelu [443]

Answer:

The greatest acceleration the man can give the airplane is 0.0059 m/s².

Explanation:

Given that,

Mass of man = 85 kg

Mass of airplane = 109000 kg

Distance = 9.08

Coefficient of static friction = 0.77

We need to calculate the greatest friction force

Using formula of friction

$F=\mu mg$

Where, m = mass of man

g = acceleration due to gravity

Put the value into the formula

$F = 0.77\times85\times9.8$

$F= 641.41\ N$

We need to calculate the acceleration

Using formula of newton's second law

$F = ma$

$a=\dfrac{F}{m}$

Put the value into the formula

$a=\dfrac{ 641.41}{109000}$

$a=0.0059\ m/s^2$

Hence, The greatest acceleration the man can give the airplane is 0.0059 m/s².

3 0

3 years ago

A school bus has a mass (including the driver and passengers) of 1.64 times 10^4 kg and is driving north at a speed of 15.2 km/h

Butoxors [25]

Answer:

Explanation:

Given

mass of bus along with travelers travelling in North direction is $m_1=1.6\times 10^4 kg$

speed of bus towards North $v_1=15.2 km/h\approx 4.22\ m/s$

mass of bus travelling in South direction is $m_2=1.578\times 10^4 kg$

speed of bus $v_2=12.2 km/h\approx 3.38\ m/s$

mass of each Passenger in south moving bus $m_0=64.8 kg$

Momentum of North moving bus

$P_1=m_1\times v_1$

$P_1=1.6\times 10^4\times 4.22$

$P_1=6.768\times 10^4 kg-m/s$

Momentum with south moving bus

$P_2=m_2\times v_2+n\cdot m_0\times v_2$

$P_2=(1.578\times 10^4+n\cdot 64.8 )\cdot 3.38$

For total momentum to be towards south

$P_2-P_1$ should be greater than 0

thus for least value of n

$P_2=P_1$

$(1.578\times 10^4+n\cdot 64.8 )\cdot 3.38=6.768\times 10^4$

$1.578\times 10^4+n\cdot 64.8=2.0023\times 10^4$

$n=\frac{4243.6686}{64.8}=65.48\approx 66$

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3 years ago

A block weighing 10 newtons is resting on a plane inclined 30° to the horizontal. What is the magnitude of the normal force acti

aleksley [76]

Magnitude of normal force acting on the block is 7 N

Explanation:

10N = 1.02kg

Mass of the block = m = 1.02 kg

Angle of incline Θ = 30°

Normal force acting on the block = N

From the free body diagram,

N = mgCos Θ

N = (1.02)(9.81)Cos(30)

N = 8.66 N

Rounding off to nearest whole number,

N = 7 N

Magnitude of normal force acting on the block = 7 N

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3 years ago

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SOVA2 [1]

Answer:

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Explanation:

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3 years ago