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lisov135 [29]
1 year ago
8

A space craft is moving relative to the earth , an observer on the earth finds that, between 1pm and 2pm according to her clock,

3601 seconds elapse on the space craft clock . What is the space craft speed relative to the earth?c=2.998×10^8ms
Physics
1 answer:
madam [21]1 year ago
4 0

The speed of the space craft relative to the earth is given as: 0.024c. This is solved using the the equation for time dilation.

<h3>What is time dilation?</h3>

Time dilation is the "slowing down" of a clock as determined by an observer in relative motion with regard to that clock under the theory of special relativity.

The formula is given as :

Δt = [Δr]/ √ 1 - (v²/c²)

Thus,

v = c√1 - (Δr/Δt)²

= c √(1 - (3600/3601)²

v = 0.024c

Learn more about time dilation at:
brainly.com/question/1933572
#SPJ1

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A nonzero net force acts on a particle and does work. Which one of the following statements is true?
arlik [135]

Answer:

option (D) is correct.

Explanation:

According to the work energy theorem, the work done by all forces is equal to the change in kinetic energy of the body.

the kinetic energy of a body is directly proportional to the square of the speed of the body.

As the kinetic energy change, the speed of the body also change.

Option (D) is correct.

5 0
3 years ago
3.00 m^3 of water is at 20.0°C.
krok68 [10]

Answer:

9m^3

Explanation:

Given data

volume  v1=  3m^3

volume  v2=  ???

Temperature T1= 20.0°C.

Temperature T2= 60.0°C.

Applying the relation for temperature and volume

V1/T1= V2/T2

substitute

3/20= V2/60

3*60= V2*20

180= 20*V2

180/20= V2

V2= 9m^3

Hence the final volume is 9m^3

6 0
2 years ago
Two bodies of specific heats S1 and S2 having the same heat capacities are combined to form a single composite body. What is the
Dafna11 [192]

\qquad\qquad\huge\underline{{\sf Answer}}♨

Heat capacity of body 1 :

\qquad \sf  \dashrightarrow \:m_1s_1

Heat capacity of body 2 :

\qquad \sf  \dashrightarrow \:m_2s_2

it's given that, the the head capacities of both the objects are equal. I.e

\qquad \sf  \dashrightarrow \:m_1s_1 = m_2s_2

\qquad \sf  \dashrightarrow \:m_1 =  \dfrac{m_2s_2}{s_1}

Now, consider specific heat of composite body be s'

According to given relation :

\qquad \sf  \dashrightarrow \:(m_1 + m_2) s' = m_1s_1 + m_2s_2

\qquad \sf  \dashrightarrow \:s' = \dfrac{ m_1s_1 + m_2s_2}{m_1 + m_2}

\qquad \sf  \dashrightarrow \:s' = \dfrac{ m_2s_2+ m_2s_2}{ \frac{m_2s_2}{s_1} + m_2 }

[ since, m_2s_2 = m_1s_1 ]

\qquad \sf  \dashrightarrow \:s' = \dfrac{ 2m_2s_2}{ m_2(\frac{s_2}{s_1} + 1)}

\qquad \sf  \dashrightarrow \:s' = \dfrac{ 2 \cancel{m_2}s_2}{  \cancel{m_2}(\frac{s_2}{s_1} + 1)}

\qquad \sf  \dashrightarrow \:s' = \dfrac{ 2 s_2}{  (\frac{s_2 + s_1}{s_1} )}

\qquad \sf  \dashrightarrow \: s' =  \dfrac{2s_1s_2}{s_1 + s_2}

➖➖➖➖➖➖➖➖➖➖➖➖➖➖➖➖➖

6 0
2 years ago
Read 2 more answers
At what distance will the weight of a body be halved , Earth's radius=6.4×10^6
Sholpan [36]

A body of mass m has weight

F = GMm/r²

on the surface of the Earth, where G is the universal gravitational constant, M is the mass of the Earth, and r is it's radius.

If the weight is to be halved, then we have

1/2 F = 1/2 GMm/r² = (1/√2)² GMm/r² = GMm/(√2 r²)

so the distance between the body and the planet's center needs to be

√2 × 6.4 × 10⁶ m ≈ 9.1 × 10⁶ m

5 0
2 years ago
Although planets orbit the Sun in ellipses, all the planetary orbits are fairly close to circular and not very eccentric.
Nutka1998 [239]

Answer:

False

Explanation:

The Sun rotates in this same, right-hand-rule direction. All planetary orbits lie in nearly the same plane. All planetary orbits are nearly circular (eccentricity near zero).

Rate as Brainliest please

8 0
2 years ago
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