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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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Answer:

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sergiy2304 [10]

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Astronomers suspect that a galaxy’s type can be affected both by the conditions in the protogalactic cloud from which it forms (
Katen [24]

Answer:

The items here are describing either a condition in a later interacton or a protogalactic cloud.  The results matching with spiral and elliptical galaxy are:

For spiral galaxy are options 6,3,2 and 5.

and for elliptical galaxy are options 4 and 1.

Explanation:

Here it is given that astrnomers suspect that types of galaxy can be affected both by the conditions which occurs due to protogalactic cloud and then from it forms the initial conditions and then by the later interactions with the other galaxies.

so, both types of galaxies are matched with their respective items given:

A. Spiral galaxy:

    2. A galaxy collision results tostripping of gas.

    3. The protogalactic cloud rotates in a very slow motion.

    5. The density of protogalactic cloud is very high.

    6. when the protogalactic cloud shrinks cloud forms very rapidly.

B. Elliptical galaxy:

    1. The protogalactic cloud has high angular momentum.

    4. Most of the protogalactic gases settles down into a disk.

6 0
3 years ago
A 150 kg line backer sacks the 120 kg quarterback. With what force is the quarterback sacked if the line backer has an accelerat
Gekata [30.6K]

Answer:

The force required to move the quarterback with linebacker is <u>1215 N</u>

Explanation:

\text { Mass of linebacker } \mathrm{m}_{2}=150 \mathrm{kg}

\text { Mass of quarterback } \mathrm{m}_{2}=120 \mathrm{kg}

\text { Moved at an acceleration }(a)=4.5 \mathrm{m} / \mathrm{s}^{2}

Using Newton's second law, it is established that  F = Ma

Where F is net force acting on the system, a is the acceleration and M is mass of the two object \left(m_{1}+m_{2}\right)

Now consider both \mathrm{m}_{1} \text { and } \mathrm{m}_{2}as a system, so net force acting on the system is \text { Force }=\left(m_{1}+m_{2}\right) a

Substitute the given values in the above formula,

\text { Force }=(150+120) \mathrm{kg} \times 4.5 \mathrm{m} / \mathrm{s}^{2}

\text { Force }=270 \mathrm{kg} \times 4.5 \mathrm{m} / \mathrm{s}^{2}

Force = 1215 N

<u>1215 N </u>is the force required to move the quarterback with linebacker.

5 0
3 years ago
A hot-air balloon is rising upward with a constant speed of 2.51 m/s. When the balloon is 3.16 m above the ground, the balloonis
icang [17]

Answer:

  t = 1.099 s

Explanation:

given,

constant speed = 2.51 m/s

height of balloon above ground = 3.16 m

time elapsed before it hit the ground = ?

Applying equation of motion to the compass

y = u t + \dfrac{1}{2}at^2

-3.16 = 2.51 t + \dfrac{1}{2}\times (-9.8)t^2

4.9 t^2 - 2.51 t - 3.16 = 0

using quadratic formula to solve the equation

t = \dfrac{-b\pm \sqrt{b^2-4ac}}{2a}

t = \dfrac{-(-2.51)\pm \sqrt{2.51^2-4(4.9)(-3.16)}}{2\times 4.9}

  t = 1.099 s, -0.586 s

hence, the time elapses before the compass hit the ground is equal to 1.099 s.

8 0
3 years ago
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