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Alexus [3.1K]
3 years ago
15

A rocket powered sled is accelerating along a straight, level track with a constant acceleration of magnitude 20 m/s2. By how mu

ch will the speed of the sled change in 3.00 seconds
Physics
1 answer:
Dafna1 [17]3 years ago
7 0

Answer:

the change in speed of the sled after 3.0 s is 60 m/s.

Explanation:

Given;

constant acceleration of the sled, a = 20 m/s²

change in time of motion, dt = 3.0 s

The change in speed of the sled after 3.0 s is calculated as;

a = \frac{dv}{dt} \\\\dv = a \times dt\\\\dv = 20 \ \times \ 3\\\\dv = 60 \ m/s

Therefore, the change in speed of the sled after 3.0 s is 60 m/s.

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The downward pull of an object due to gravity is the objects _______?
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Read 2 more answers
A space shuttle sits on the launch pad for 2.0 minutes, and then goes from rest to 4600 m/s in 8.0 minutes. Treat its motion as
SpyIntel [72]

Answer:

a.) a = 0 ms⁻²

b.) a = 9.58 ms⁻²

c.) a = 7.67 ms⁻²

Explanation:

a.)

    Acceleration (a) is defined as the time rate of change of velocity

                       a = \frac{v_{2} - v_{1} } {t}  

Given data

 Final velocity = v₂ = 0 m/s

 Initial velocity = v ₁ = 0 m/s

  As the space shuttle remain at rest for the first 2 minutes i.e there is no change in velocity so,

                 a = 0 ms⁻²

b.)

     Given data

As the space shuttle start from rest, So initial velocity is zero

    Initial velocity = v₁ = 0 ms⁻¹

    Final velocity  = v₂ = 4600 ms⁻¹

     Time = t = 8 min = 480 s

By the definition of Acceleration (a)

             a = \frac{v_{2} - v_{1} } {t}  

             a = \frac{4600 - 0 } {480}

                     a = 9.58 ms⁻²

c.)

    Given data

As the space shuttle is at rest for first 2 min then start moving, So initial velocity is zero

    Initial velocity = v₁ = 0 ms⁻¹

    Final velocity  = v₂ = 4600 ms⁻¹

     Time = t = 10 min = 600 s

By the definition of Acceleration (a)

             a = \frac{v_{2} - v_{1} } {t}  

             a = \frac{4600 - 0 } {600}

                     a = 7.67 ms⁻²

8 0
3 years ago
Star A and Star B have different apparent brightnesses but identical luminosities. Star A is 10 light years away from earth and
STALIN [3.7K]

intensity of a star is inversely depends on the square of the distance from the star

we can say it is given as

\frac{I_1}{I_2} = \frac{r_2^2}{r_1^2}

here we know that

\frac{I_1}{I_2} = 36 times

also we know that

r_1 = 10 Ly

now we will have

\frac{I_1}{I_2} = \frac{r_2^2}{r_1^2}

36 = \frac{r_2^2}{10^2}

r_2 = 60 Ly

so other star is at distance 60 Light years

8 0
3 years ago
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