It is recommended that you said at least how many inches away from the airbag ?

An observer on Earth sees rocket 1 leave Earth and travel toward Planet X at 0.3c. The observer on Earth also sees that Planet X

Verizon [17]

Answer:

0.625 c

Explanation:

Relative speed of a body may be defined as the speed of one body with respect to some other or the speed of one body in comparison to the speed of second body.

In the context,

The relative speed of body 2 with respect to body 1 can be expressed as :

$u'=\frac{u-v}{1-\frac{uv}{c^2}}$

Speed of rocket 1 with respect to rocket 2 :

$u' = \frac{0.4 c- (-0.3 c)}{1-\frac{(0.4 c)(-0.3 c)}{c^2}}$

$u' = \frac{0.7 c}{1.12}$

$u'=0.625 c$

Therefore, the speed of rocket 1 according to an observer on rocket 2 is 0.625 c

5 0

3 years ago

What is transferred by a radio wave?

Alex

Answer: B. energy

Explanation:

4 0

3 years ago

Which type of curved mirror uses its inside as the reflecting surface? A. a convex mirror B. a plane mirror C. a virtual mirror

ser-zykov [4K]

D. a concave mirror

6 0

3 years ago

Read 2 more answers

WILL GIVE BRAINLIEST

Diano4ka-milaya [45]

Answer:

nice to help you

Explanation:

If for any reason the Sun shrank smaller than the Earth, this shrunken Sun wouldn't have the mass to create fusion and would burn out completely. Our solar system would lose its only star. ... Earth's mass would be at least 333,000 times bigger than it is now. Imagine the gravity that planet would have.

5 0

3 years ago

At time t=0 a 2150 kg rocket in outer space fires an engine that exerts an increasing force on it in the +x−direction. This forc

amm1812

Explanation:

Given that,

Mass of the rocket, m = 2150 kg

At time t=0 a rocket in outer space fires an engine that exerts an increasing force on it in the +x−direction. The force is given by equation :

$F=At^2$

Here F = 888.93 N when t = 1.25 s

(c) We can find the value of A first as :

$F=At^2\\\\A=\dfrac{F}{t^2}\\\\A=\dfrac{888.93}{(1.25)^2}\\\\A=568.91\ N/s^2$

The value of A is $568.91\ N/s^2$ .

(a) Let J is the impulse does the engine exert on the rocket during the 4.0 s interval starting 2.00 s after the engine is fired. It is given in terms of force as :

$J=\int\limits {F{\cdot} dt}$

Limits will be from 2 s to 2+ 4 = 6 s

It implies :

$J=\int\limits^6_2 {At^2{\cdot} dt}\\\\J=A\int\limits^6_2 {t^2{\cdot} dt}\\\\J=A\dfrac{t^3}{3}|_2^6\\\\J=568.91\times \dfrac{1}{3}\times (6^3-2^3)\\\\J=39444.42\ Ns$

(b) Impulse is also equal to the change in momentum as :

$J=m\Delta v\\\\\Delta v=\dfrac{J}{m}\\\\\Delta v=\dfrac{39444.42}{2150}\\\\\Delta v=18.34\ m/s$

Hence, this is the required solution.

5 0

3 years ago