Bem ²4.23=1.4m/s

If the sun suddenly ceased to shine, how long would it take earth to become dark? you will have to look up the speed of light in

sleet_krkn [62]

We are 8 light minutes from the sun. That means two things, we see the sun as it was 8 minutes ago, and we WOULD continue to see the sun for 8 minutes after it disappeared.

3 0

3 years ago

jet is flying at 500 mph east relative to the ground. A Cessna is flying at 150 mph 60° north of west relative to the ground. Wh

Greeley [361]

Answer:

C. 590 mph

$\vert v_{cj}\vert=589.49\ mph$

Explanation:

Given:

velocity of jet, $v_j=500\ mph$
direction of velocity of jet, east relative to the ground
velocity of Cessna, $v_c=150\ mph$
direction of velocity of Cessna, 60° north of west

Taking the x-axis alignment towards east and hence we have the velocity vector of the jet as reference.

Refer the attached schematic.

So,

$\vec v_j=500\ \hat i\ mph$

&

$\vec v_c=150\times (\cos120\ \hat i+\sin120\ \hat j)$

$\vec v_c=-75\ \hat i+75\sqrt{3}\ \hat j\ mph$

Now the vector of relative velocity of Cessna with respect to jet:

$\vec v_{cj}=\vec v_j-\vec v_c$

$\vec v_{cj}=500\ \hat i-(-75\ \hat i+75\sqrt{3}\ \hat j )$

$\vec v_{cj}=575\ \hat i-75\sqrt{3}\ \hat j\ mph$

Now the magnitude of this velocity:

$\vert v_{cj}\vert=\sqrt{(575)^2+(75\sqrt{3} )^2}$

$\vert v_{cj}\vert=589.49\ mph$ is the relative velocity of Cessna with respect to the jet.

8 0

3 years ago

If two stars in the sky are separated by twice the angle projected by your index and pinky fingers, then how many degrees are th

quester [9]

Answer:

The answer is 30 degrees.

Explanation:

The angle between the index and pinky fingers is 15 degrees.

Twice this angle is 2×15=30degrees.

7 0

3 years ago

A 600kg car is moving at 5 m/s to the right and elastically collided with a stationary 900 kg car. What is the velocity of the 9

11111nata11111 [884]

Answer:

$\mathrm{v}_{2} \text { velocity after the collision is } 3.3 \mathrm{m} / \mathrm{s}$

Explanation:

It says “Momentum before the collision is equal to momentum after the collision.” Elastic Collision formula is applied to calculate the mass or velocity of the elastic bodies.

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

$\mathrm{m}_{1} \text { and } \mathrm{m}_{2} \text { are masses of the object }$

$\mathrm{v}_{1} \text { velocity before the collision }$

$\mathrm{v}_{2} \text { velocity after the collision }$

$\mathrm{m}_{1}=600 \mathrm{kg}$

$\mathrm{m}_{2}=900 \mathrm{kg}$

$\text { Velocity before the collision } v_{1}=5 \mathrm{m} / \mathrm{s}$

$600 \times 5=900 \times v_{2}$

$3000=900 \times v_{2}$

$\mathrm{v}_{2}=\frac{3000}{900}$

$\mathrm{v}_{2}=3.3 \mathrm{m} / \mathrm{s}$

$\mathrm{v}_{2} \text { velocity after the collision is } 3.3 \mathrm{m} / \mathrm{s}$

5 0

4 years ago

A block slides on a table pulled by a string attached to a hanging weight. In case 1 the block slides without friction and in ca

ycow [4]

The tension in the string with friction would be the biggest because of the involvement of the force of gravity. This would result in that the friction force that is acting on the system. There is no friction in the frictionless system, and only the force of gravity is relevant.

7 0

3 years ago

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