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aalyn [17]
3 years ago
11

Boxes A and B are in contact on a horizontal, frictionless surface. Box A has mass 21.0 kg and box B has mass 8.0 kg. A horizont

al force of 100N is exerted on box A. What is the magnitude of the force that box A exerts on box B?
Physics
1 answer:
vfiekz [6]3 years ago
8 0

Answer:

2.75 N

Explanation:

Given that,

Box A has a mass 21.0 kg and box B has a mass 8.0 kg.

A horizontal force of 100N is exerted on box A.

Let a be the acceleration of the system. Using second law of motion,

F=(m_A+m_B)a\\\\a=\dfrac{F}{(m_A+m_B)}\\\\a=\dfrac{10}{(21+8)}\\\\a=0.344\ m/s^2

Now applying Newton's second law to box B. So,

F_A=m_Ba\\\\=8\times 0.344\\\\=2.75\ N

So, 2.75 N is the force that box A exerts on box B.

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Compare the ozone layer to a sunscreen?
sergij07 [2.7K]

Answer:

The ozone layer is like the Earth's sunscreen, as it absorbs some of the sun's radiation hitting our planet

Explanation:

6 0
2 years ago
A projectile is shot vertically upward with a given initial velocity. It reaches a maximum height of 72.0 m. On a second shot, t
exis [7]

Answer:

Explanation:

Given

maximum height reached h=72\ m

suppose Projectile is launched vertically with initial velocity u

applying equation of motion

v^2-u^2=2 as

where v=final velocity

u=initial velocity

a=acceleration

s=displacement

Here final velocity is zero so

s=frac{u^2}{2g}

for twice initial velocity

s'=\frac{(2u)^2}{2g}

s'=4\times \frac{u^2}{2g}

s'=4\times 72

s'=288\ m

5 0
3 years ago
2)
ludmilkaskok [199]
B).  A <span>car that rounds a curve at a constant speed is accelerating.

</span><span>D).  A car that is set to a constant speed of 60 miles per hour is
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3 0
3 years ago
Describe at least two ways that you can reduce the friction between two solid surfaces
Firlakuza [10]
I will try to name as much as I can :

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• Butter
• Petroleum
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8 0
4 years ago
Like the filters falling through the air, a car on the freeway represents an object with a high Reynolds number traveling throug
Goshia [24]

Answer:

ΔF=125.22 %

Explanation:

We know that drag force on the car given as

F_D=\dfrac{1}{2}\rho C_DA v^2

C_D=Drag coefficient

A=Projected area

v=Velocity

ρ=Density

All other quantity are constant so we can say that drag force and velocity can be given as

\dfrac{F_D_1}{F_D_2}=\dfrac{v_1^2}{v_2^2}

Now by putting the values

\dfrac{F_D_1}{F_D_2}=\dfrac{v_1^2}{v_2^2}

\dfrac{F_D_1}{F_D_2}=\dfrac{50^2}{75^2}

\dfrac{F_D_1}{F_D_2}=0.444

Percentage Change in the drag force

\Delta F=\dfrac{F_D_2-F_D_1}{F_D_1}\times 100

\Delta F=\dfrac{F_D_2-0.444F_D_2}{0.444F_D_2}\times 100

\Delta F=\dfrac{1-0.444}{0.444}\times 100

ΔF=125.22 %

Therefore force will increase by 125.22  %.

3 0
3 years ago
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