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dsp73
3 years ago
9

Three blocks are arranged in a stack on a frictionless horizontal surface. The bottom block has a mass of 37.0 kg. A block of ma

ss 18.0 kg sits on top of it and a 16.0 kg block sits on top of the middle block. A downward vertical force of 170 N is applied to the top block. What is the magnitude of the normal force exerted by the bottom block on the middle block?
Physics
1 answer:
Alex777 [14]3 years ago
7 0

Answer:

N₂ = 503.8 N

Explanation:

given,

mass of bottom block = 37 Kg

mass of middle block = 18 Kg

mass of the top block = 16 Kg

force acting on the top block = 170 N

force on the block at top

N₁ be the normal force from block at middle

now,

N₁ = 170 + m g

N₁ = 170 + 16 x 9.8

now, force on block at middle

N₂ be the normal force exerted by the bottom block

N₂ = N₁ + m₂ g

N₂ = 326.8 + 18 x 9.8

N₂ = 503.8 N

hence, normal force by bottom block is equal to N₂ = 503.8 N

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Answer:

he fall movement we see that both the force is different from zero, and the torque is different from zero.

When analyzing the statements the d is true

Explanation:

Let's pose the solution of this problem, to be able to analyze the firm affirmations.

When the person is falling, the weight acts on them all the time, initially the rope has no force, but at the moment it begins to lash it exerts a force towards the top that is proportional to the lengthening of the rope.

The equation for this part is

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As the axis of rotation is located at the top where they jump, there is a torque.

What is it

                Fe y - W y = I α

angular and linear acceleration are related

       a = α r

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In the fall movement we see that both the force is different from zero, and the torque is different from zero.

When analyzing the statements the d is true

4 0
3 years ago
Example 2.1
tatiyna

Answer:

5 min 20 km --------

Explanation:

5 min 20 km --------

8 0
3 years ago
If a current is two amps and the resistance is 3 ohms, how much voltage was needed?
Hunter-Best [27]

Answer:

6 V

Explanation:

We can solve the problem by using Ohm's law:

V=RI

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7 0
3 years ago
At t=0 bullet A is fired vertically with an initial (muzzle) velocity of 450 m/s. When 3s. bullet B is fired upward with a muzzl
Debora [2.8K]

Answer:

At time 10.28 s after A is fired bullet B passes A.

Passing of B occurs at 4108.31 height.

Explanation:

Let h be the height at which this occurs and t be the time after second bullet fires.

Distance traveled by first bullet can be calculated using equation of motion

s=ut+0.5at^2 \\

Here s = h,u = 450m/s a = -g and t = t+3

Substituting

h=450(t+3)-0.5\times 9.81\times (t+3)^2=450t+1350-4.9t^2-29.4t-44.1\\\\h=420.6t-4.9t^2+1305.9

Distance traveled by second bullet

Here s = h,u = 600m/s a = -g and t = t

Substituting

h=600t-0.5\times 9.81\times t^2=600t-4.9t^2\\\\h=600t-4.9t^2 \\

Solving both equations

600t-4.9t^2=420.6t-4.9t^2+1305.9\\\\179.4t=1305.9\\\\t=7.28s \\

So at time 10.28 s after A is fired bullet B passes A.

Height at t = 7.28 s

h=600\times 7.28-4.9\times 7.28^2\\\\h=4108.31m \\

Passing of B occurs at 4108.31 height.

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3 years ago
For an isolated system, the sum of the kinetic and potential energies
Elenna [48]
I am pretty sure the answer to your question is B
4 0
3 years ago
Read 2 more answers
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