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3 years ago

13

After two tetionic plates shift deep underground, which event is most likely to occur

1 answer:

Andrew [12]3 years ago

7 0

An earthquake will occur

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On a frictionless horizontal surface, a 1.50 kg mass traveling at 3.50 m/s collides with and sticks to a 3.00 kg mass that is in

My name is Ann [436]

Answer:

√ $\frac{3}{5} m$

Explanation:

Hope it helps!.

4 0

3 years ago

Consult Interactive Solution 7.10 for a review of problem-solving skills that are involved in this problem. A stream of water st

AlekseyPX

Explanation:

Initial speed of the incident water stream, u = 16 m/s

Final speed of the exiting water stream, v = -16 m/s

The mass of water per second that strikes the blade is 48.0 kg/s.

We need to find the magnitude of the average force exerted on the water by the blade. The force acting on an object is given by :

$F=\dfrac{m(v-u)}{t}$

Here, $\dfrac{m}{t}=48\ kg/s$

$F=48\times (-16-16)\ N\\\\F=-1536\ N\\\\|F|=1536\ N$

So, the magnitude of the average force exerted on the water by the blade is 1536 N.

5 0

3 years ago

David is mowing his lawn. He has been riding on the mower at 50m/hr for 2.5 hours? What is the distance of the area David has mo

DaniilM [7]

Answer:

125 metres

Explanation:

Distance = speed × time

Here , speed = 50m/hr

time = 2.5 hr

So,

Distance = 50 × 2.5

=125 m

6 0

3 years ago

Read 2 more answers

A 100-kg block being released from rest from a height of 1.0 m. It then takes it 1.40 s to reach the floor. What is the mass m o

dimulka [17.4K]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The mass of the other block is $m_1 = 81.14 \ kg$

Explanation:

From the question we are told that

Mass of the first block is $m_1 = 100 \ kg$

The height is $s = 1.0 \ m$

The time it takes it is $t = 1.40 \ s$

Generally from kinematic equation

$s = ut + \frac{1}{2} at^2$

Here u is the initial velocity which zero given that it was at rest initially

So

$s = 0 * t + \frac{1}{2} at^2$

=> $s = \frac{1}{2} at^2$

=> $1 = \frac{1}{2}* a * (1.40 )^2$

=> $a = 1.0204 \ m/s^2$

Generally from the diagram the resultant force due to the weight of the first object and the tension on the string is mathematically represented as

$mg - T = ma$

=> $T = m g - ma$

=> $T = m(g - a)$

=> $T = 877.96 \ N$

Generally from the diagram the resultant force due to the weight of the second object and the tension on the string is mathematically represented as

$T - m_1g = m_1 a$

=> $877.96 = m_1 (a + g)$

=> $877.96 = m_1 (1.0204 + 9.8 )$

=> $m_1 = 81.14 \ kg$

5 0

3 years ago

A river 500 ft wide flows with a speed of 8 ft/s with respect to the earth. A woman swims with a speed of 4 ft/s with respect to

White raven [17]

Answer:

1) $\Delta s=1000\ ft$

2) $\Delta s'=998.11\ ft.s^{-1}$

3) $t\approx125\ s$

$t'\approx463.733\ s$

Explanation:

Given:

width of river, $w=500\ ft$

speed of stream with respect to the ground, $v_s=8\ ft.s^{-1}$

speed of the swimmer with respect to water, $v=4\ ft.s^{-1}$

<u>Now the resultant of the two velocities perpendicular to each other:</u>

$v_r=\sqrt{v^2+v_s^2}$

$v_r=\sqrt{4^2+8^2}$

$v_r=8.9442\ ft.s^{-1}$

<u>Now the angle of the resultant velocity form the vertical:</u>

$\tan\beta=\frac{v_s}{v}$

$\tan\beta=\frac{8}{4}$

$\beta=63.43^{\circ}$

Now the distance swam by the swimmer in this direction be d.

so,

$d.\cos\beta=w$

$d\times \cos\ 63.43=500$

$d=1118.034\ ft$

Now the distance swept downward:

$\Delta s=\sqrt{d^2-w^2}$

$\Delta s=\sqrt{1118.034^2-500^2}$

$\Delta s=1000\ ft$

2)

On swimming 37° upstream:

<u>The velocity component of stream cancelled by the swimmer:</u>

$v'=v.\cos37$

$v'=4\times \cos37$

$v'=3.1945\ ft.s^{-1}$

<u>Now the net effective speed of stream sweeping the swimmer:</u>

$v_n=v_s-v'$

$v_n=8-3.1945$

$v_n=4.8055\ ft.s^{-1}$

<u>The component of swimmer's velocity heading directly towards the opposite bank:</u>

$v'_r=v.\sin37$

$v'_r=4\sin37$

$v'_r=2.4073\ ft.s^{-1}$

<u>Now the angle of the resultant velocity of the swimmer from the normal to the stream</u>:

$\tan\phi=\frac{v_n}{v'_r}$

$\tan\phi=\frac{4.8055}{2.4073}$

$\phi=63.39^{\circ}$

Now let the distance swam in this direction be d'.

$d'\times \cos\phi=w$

$d'=\frac{500}{\cos63.39}$

$d'=1116.344\ ft$

<u>Now the distance swept downstream:</u>

$\Delta s'=\sqrt{d'^2-w^2}$

$\Delta s'=\sqrt{1116.344^2-500^2}$

$\Delta s'=998.11\ ft.s^{-1}$

3)

Time taken in crossing the rive in case 1:

$t=\frac{d}{v_r}$

$t=\frac{1118.034}{8.9442}$

$t\approx125\ s$

Time taken in crossing the rive in case 2:

$t'=\frac{d'}{v'_r}$

$t'=\frac{1116.344}{2.4073}$

$t'\approx463.733\ s$

7 0

4 years ago