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umka21 [38]
3 years ago
8

Two blocks are sliding to the right across a horizontal surface, as the drawing shows. In Case A the mass of each block is 3.0 k

g. In Case B the mass of block 1 (the block behind) is 6.0 kg, and the mass of block 2 is 3.0 kg. No frictional force acts on block 1 in either Case A or Case B. However, a kinetic frictional force of 5.8 N does act on block 2 in both cases and opposes the motion. For both Case A and Case B determine (a) the magnitude of the forces with which the blocks push against each other and (b) the magnitude of the acceleration of the blocks.
Physics
1 answer:
evablogger [386]3 years ago
7 0

Answer:

Explanation:

According to a free body diagram the forces in the horizontal direction on body 1 would be:

F₁ = a₁*m₁ = -N

and on body 2:

F₂ = a₂*m₂ = N - F

N: normal force between the two blocks

F: frictional force on block 2

Since the two blocks are moving together, they need to have the same acceleration:

a₁ = a₂

This gives two equations with two unknown. Solving for a and N gives:

a = - \frac{F}{m_1+m_2}

N = -am_1

case A:

|a| = 0.96 m/s²

|N| = 2.9 N

case B:

|a| = 0.644 m/s²

|N| = 3.86 N

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Alex73 [517]

Answer:

The boat moves away from the dock at 0.5 m/s.

Explanation:

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Since no external forces are acting on the system boy-boat at the moment at which the boy lands on the boat, the momentum of the system is conserved (i.e. it remains constant).

The momentum of the system is calculated as the sum of the momentum of the boy plus the momentum of the boat. Before the boy lands on the boat, the momentum of the system is given by the momentum of the boy.

momentum of the system before the boy lands on the boat:

momentum of the boy + momentum of the boat

m1 · v1 + m2 · v2 = momentum of the system

Where:

m1 and v1: mass and velocity of the boy.

m2 and v2: mass and velocity of the boat.

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50 kg · 2.0 m/s + 150 kg · 0 m/s = momentum of the system

momentum of the system = 100 kg m/s

After the boy lands on the boat, the momentum of the system will be equal to the momentum of the boat moving with the boy on it:

momentum of the system = (m1 + m2) · v (where v is the velocity of the boat).

100 kg m/s = (50 kg + 150 kg) · v

100 kg m/s / 200 kg = v

v = 0.5 m/s

The boat moves away from the dock at 0.5 m/s.

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You create a ramp using two text books and a 0.50m board. Using a timer you determine that a cart can roll down the ramp in 0.55
ahrayia [7]

Answer:

The velocity of the cart at the bottom of the ramp is 1.81m/s, and the acceleration would be 3.30m/s^2.

Explanation:

Assuming the initial velocity to be zero, we can obtain the velocity at the bottom of the ramp using the kinematics equations:

v=v_0+at\\\\v^2=v_0^2+2ad

Dividing the second equation by the first one, we obtain:

v=\frac{v_0^2+2ad}{v_0+at}

And, since v_0=0, then:

v=\frac{2ad}{at}\\\\v=\frac{2d}{t}\\\\v=\frac{2(0.50m)}{0.55s}\\\\v=1.81m/s

It means that the velocity at the bottom of the ramp is 1.81m/s.

We could use this data, plus any of the two initial equations, to determine the acceleration:

v=v_0+at\\\\\implies a=\frac{v}{t}\\\\a=\frac{1.81m/s}{0.55s}\\\\a=3.30m/s^2

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Answer:

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a cylindrical jar is 10cm long and has a cross sectional area of 36cm. if it is completely filled with a fluid of relative densi
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Answer:

The mass of the fluid is 72 g.

Explanation:

The following data were obtained from the question:

Height (h) = 10 cm

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Next, we shall determine the volume of the cylinder. This can be achieved by doing the following:

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Next, we shall determine the density of the liquid.

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Relative density = density of substance/density of water.

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Cross multiply

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Cross multiply

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