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3 years ago

When a rocket lifts off what provides the momentum match of the upper lift

1 answer:

5 0

The momentum of the mass expelled in the opposite direction ... the rocket-engine exhaust, or the ionized matter expelled from an ion drive.

THAT's why every propulsion engine has outlet nozzles designed with super-high-intensity math, to achieve the highest possible velocity for the mass that gets shot out the back ... so that it will carry the maximum possible momentum.

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Joe drove at the speed of 45 miles per hour for a certain distance. He then drove at the speed of 55 miles per hour for the same

Snowcat [4.5K]

Answer:

$v_{avg} = 49.5 mph$

Explanation:

Let the distance moved by Joe is "d"

so the time taken by him to drove it by speed 45 mph is given as

$t_1 = \frac{d}{v_1}$

$t_1 = \frac{d}{45}$

now the same distance is traveled by him with speed 55 mph

so the time taken by him

$t_2 = \frac{d}{55}$

so total time taken by him for complete distance 2d

$t = t_1 + t_2$

$t = \frac{d}{45} + \frac{d}{55}$

$t = 0.0404 d$

now the average speed is given as

$v_{avg} = \frac{2d}{t}$

$v_{avg} = \frac{2d}{0.0404d}$

$v_{avg} = 49.5 mph$

5 0

2 years ago

Read 2 more answers

Compared to the weak nuclear force the electromagnetic force

Crazy boy [7]

Explanation:

Weak nuclear force:

The interaction between the subatomic particles is called weak nuclear force.

The weak nuclear force is one of the four fundamental forces.

The weak nuclear force is effective at very short distance.

The range and relative strength of weak nuclear force is 10⁻¹⁸ m and 10²⁵ with respect to gravitational force respectively

Deuterium is formed due to the fusion of protons and neutrons under the action the weak force.

Example : Beta decay

Electromagnetic force:

The interaction between the charged particles is called electromagnetic force.

The electromagnetic force is one of the four fundamental forces.

The electromagnetic force is effective at long range distance.

The range and relative strength of electromagnetic force is infinity and 10³⁶ with respect to gravitational force respectively

Example : light

8 0

2 years ago

Read 2 more answers

Describe the difference between contact forces and field forces.

Salsk061 [2.6K]

Contact forces has to be touching for it to be an actual force. A field force does not have to be touching but it does have to be acting on particles at different positions in a space.

8 0

3 years ago

Explain why astronomers use the term "blueshifted" for objects moving toward us and "redshifted" for objects moving away from us

ValentinkaMS [17]

<span>Two of them are "redshift" and "blueshift", which are used to describe an object'smotion toward or away from us in space. Redshift indicates that an object is moving away from us. "Blueshift" is a term that astronomers use to describe an object that is moving toward another object or toward us.</span>

8 0

2 years ago

A projectile is thrown with velocity v at an angle θ with horizontal. When the projectile is at a height equal to half of the ma

raketka [301]

Let, the maximum height covered by projectile be $\sf{H_m}$

$\purple{ \longrightarrow \bf{h_m = \dfrac{ {v}^{2} \: {sin}^{2} \theta }{2g} }}$

Projectile is thrown with a velocity = v
Angle of projection = θ

Velocity of projectile at a height half of the maximum height covered be $\sf{v_0}$

$\qquad$ ______________________________

Then –

$\qquad$ $\pink{ \longrightarrow \bf{ \dfrac{h_m}{2} = \dfrac{ {v_0}^{2} \: {sin}^{2} \theta }{2g} }}$

$\qquad$ $\longrightarrow \sf{ \dfrac{ {v}^{2} \: {sin}^{2} \theta }{2g} \times \dfrac{1}{2} = \dfrac{ {v_0}^{2} \: {sin}^{2} \theta }{2g} }$

$\qquad$ $\longrightarrow \sf{ \dfrac{ {v}^{2} \: {sin}^{2} \theta }{4g} = \dfrac{ {v_0}^{2} \: {sin}^{2} \theta }{2g} }$

$\qquad$ $\longrightarrow \sf{ \dfrac{ {v}^{2} \: {sin}^{2} \theta }{2} = {v_0}^{2} \: {sin}^{2} \theta }$

$\qquad$ $\longrightarrow \sf{ \dfrac{ {v}^{2} }{2} = {v_0}^{2} }$

$\qquad$ $\longrightarrow \bf{v_0 = \sqrt{ \dfrac{ {v}^{2} }{2} } = \dfrac{v}{ \sqrt{2} } }$

Now, the vertical component of velocity of projectile at the height half of $\sf{h_m}$ will be –

$\qquad$ $\longrightarrow \bf{v_{(y)}=v_0 \: sin \theta }$

$\qquad$ $\longrightarrow \bf{v_{(y)} = \dfrac{v}{ \sqrt{2} } \: sin \theta = \dfrac{v \: sin \: \theta}{ \sqrt{2} } }$

Therefore, the vertical component of velocity of projectile at this height will be–

☀️ $\qquad$ $\pink {\bf{ \dfrac{v \: sin \: \theta}{ \sqrt{2} }} }$

6 0

2 years ago

Read 2 more answers