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Marizza181 [45]

3 years ago

6

SHORT ANSWER TYPE QUESTIONS1 1. Why school bags have broad shoulder straps ?

1 answer:

ANTONII [103]3 years ago

7 0

Broad shoulder straps provide larger surface area. So, the pressure exerted by the bag on the shoulder is less and carrying the bad is easier.

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The velocity selector in in a mass spectrometer consists of a uniform magnetic field oriented at 90 degrees to a uniform electri

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Sandra takes a class field trip to a geology museum. She sees an interesting sample of igneous rock. She wonders if the material

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Yes

Explanation:

It is possible for sedimentary rocks to be converted to igneous rocks. Under conditions of high temperature and pressure, sedimentary rocks can be broken down into igneous rock by melting this rock type.

When the rock is broken down, it forms melt which when cooled and solidifies will form igneous rocks.

Sedimentary rocks are formed from the breaking down of pre-existing rocks through the action of weathering, erosion and sediment transportation. Within a basin, the sediments are compacted and lithified.

When this is subjected to intense pressure and temperature, the rock hardens and might further break down to melt.

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2 years ago

Plot StartRoot 1.5 EndRoot and StartRoot 1.9 EndRoot on the number line to find which inequalities are true. Check all that appl

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A, B, C, E

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2 years ago

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3 years ago

In a game of pool, the cue ball strikes another ball of the same mass and initially at rest. After the collision, the cue ball m

ikadub [295]

(a) $-39.4^{\circ}$

Let's take the initial direction (before the collision) of the cue ball has positive x-direction.

Along the y-direction, the total initial momentum is zero:

$p_y =0$

Therefore, since the total momentum must be conserved, it must be zero also after the collision. So we write:

$0 = m v_1 sin \phi_1 + m v_2 sin \phi_2 \\0 = m(4.60) sin (28^{\circ}) + m(3.40) sin \phi_2$

where

m is the mass of each ball

$v_1= 4.60 m/s$ is the velocity of the cue ball after the collision

$v_2 = 3.40 m/s$ is the velocity of the second ball after the collision

$\phi_1=28.0^{\circ}$ is the angle of the cue ball with the x-axis

$\phi_2$ is the angle of the second ball

Solving for $\phi_2$ , we find the angle between the direction of motion of the second ball and the original direction of motion:

$sin \phi_2 = -\frac{4.60 sin 28}{3.40}=-0.635\\\phi_2 = -39.4^{\circ}$

(b) 6.69 m/s

To find the original speed of the cue ball, we analyze the situation along the horizontal direction.

First, we calculate the total momentum along the x-direction after the collision, which is:

$p_x = m v_1 cos \phi_1 + m v_2 cos \phi_2 \\0 = m(4.60) cos (28^{\circ}) + m(3.40) cos (-39.4^{\circ})=6.69 m$

The initial total momentum along the x-direction as

$p_x = m u$

where

m is the mass of the cue ball

$u$ is the initial velocity of the cue ball

The momentum along this direction must be conserved, so we can equate the two expressions and find the value of u:

$mu = 6.69 m\\u = 6.69 m/s$

7 0

3 years ago