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ki77a [65]
3 years ago
13

Given: Each block has masses m1 = m2 = m3 = m and the coefficient of kinetic friction is µ. The magnitude of F equals twice the

total frictional force. Apply a horizontal force F in pushing an array of three identical blocks in the horizontal plane (see sketch). m1 m2 m3 F µ Find the force F23 with which the second block is pushing the third block.

Physics
1 answer:
MariettaO [177]3 years ago
3 0

Answer:

2umg

Explanation:

See it in the pic

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What force is the total force felt by an object?
skelet666 [1.2K]

Answer:

net force

Explanation:

Net force felt by an object.

6 0
3 years ago
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11. You want to calculate the displacement of an object thrown over a bridge. Using -10 m/s2 for acceleration due to gravity, wh
IrinaVladis [17]

Answer:

The displacement was 320 meters.

Explanation:

Assuming projectile motion and zero initial speed (i.e., the object was dropped, not thrown down), you can calculate the displacement using the kinematic equation:

d = \frac{1}{2}gt^2=\frac{1}{2}10\frac{m}{s^2}\cdot 8^2 s^2=320 m

The displacement was 320 meters.

7 0
3 years ago
A 1.0-kg ball is attached to the end of a 2.5-m string to form a pendulum. This pendulum is released from rest with the string h
hram777 [196]

Answer:

v_{2}=3.5 m/s

Explanation:

Using the conservation of energy we have:

\frac{1}{2}mv^{2}=mgh

Let's solve it for v:

v=\sqrt{2gh}

So the speed at the lowest point is v=7 m/s

Now, using the conservation of momentum we have:

m_{1}v_{1}=m_{2}v_{2}

v_{2}=\frac{1*7}{2}

Therefore the speed of the block after the collision is v_{2}=3.5 m/s

I hope it helps you!

       

8 0
3 years ago
A 4.00 m long, massless beam rests horizontally on a support 3.00 m from the left
Rus_ich [418]

If the beam is in static equilibrium, meaning the Net Torque on it about the support is zero, the value of x₁ is 2.46m

Given the data in the question;

  • Length of the massless beam;L = 4.00m
  • Distance of support from the left end; x = 3.00m
  • First mass; m1 = 31.3 kg
  • Distance of beam from  the left end( m₁ is attached to ); x_1 = ?
  • Second mass; m_2 = 61.7 kg
  • Distance of beam from  the right of the support( m₂ is attached to ); x_1 = 0.273m

Now, since it is mentioned that the beam is in static equilibrium, the Net Torque on it about the support must be zero.

Hence, m_1g( x-x_1) = m_2gx_2

we divide both sides by g

m_1( x-x_1) = m_2x_2

Next, we make x_1, the subject of the formula

x_1 = x - [ \frac{m_2x_2}{m_1} ]

We substitute in our given values

x_1 = 3.00m - [ \frac{61.7kg\ * \ 0.273m}{31.3kg} ]

x_1 = 3.00m - 0.538m

x_1 = 2.46m

Therefore, If the beam is in static equilibrium, meaning the Net Torque on it about the support is zero, the value of x₁ is 2.46m

Learn more; brainly.com/question/3882839

6 0
3 years ago
An electric field from a charge has a magnitude of 1.5 × 104 N/C at a certain location that points inward. If another charge wit
zysi [14]

Answer:

-0.045 N, they will attract each other

Explanation:

The strength of the electrostatic force exerted on a charge is given by

F=qE

where

q is the magnitude of the charge

E is the electric field magnitude

In this problem,

q=3.0\cdot 10^{-6}C

E=-1.5\cdot 10^4 N/C (negative because inward)

So the strength of the electrostatic force is

F=(-3.0\cdot 10^{-6}C)(1.5\cdot 10^4 N/C)=-0.045 N

Moreover, the charge will be attracted towards the source of the electric field. In fact, the text says that the electric field points inward: this means that the source charge is negative, so the other charge (which is positive) is attracted towards it.

6 0
4 years ago
Read 2 more answers
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