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11Alexandr11 [23.1K]
3 years ago
10

Our galaxy, the Milky Way, has a diameter of about 100,000 light years. How many years would it take a spacecraft to cross the g

alaxy if it could travel at 99% the speed of light?
Physics
1 answer:
Marat540 [252]3 years ago
3 0

Answer:

It takes to a spacecraft 100,837.13 years to cross the galaxy if could travels at 99% the speed of light.

Explanation:

d= 9.461 *10²⁰m

V= 297 *10⁶ m/s

t= d/V

t= 3.18 * 10¹² seconds = 100837.13 years

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Plz help!!!
shtirl [24]

Answer:

I think the answer to your question is true

3 0
3 years ago
At the instant the traffic light turns green, a car that has been waiting at an intersection starts with a constant acceleration
Setler79 [48]

Answer:

(a) 300 ft

(b) 60 ft/s

Explanation:

distance d=0.5at^{2} where a is acceleration and t is time

d=0.5\times 6\times t^{2}=3t^{2}

Also, d=vt where v is the velocity

d=30t

Therefore

30t=3t^{2} hence t=10 s

Substituting t is either formula

d=30t=30*10=300 ft

Also

v=at hence v=6\times 10=60 ft/s

4 0
3 years ago
A rock dropped from a 20 meter bridge falls into the river below. How long did it take to reach the water? What velocity did it
Brut [27]

Answer:

2 s, -20 m/s

Explanation:

Given:

y₀ = 20 m

y = 0 m

v₀ = 0 m/s

a = -9.8 m/s²

Find: t and v

y = y₀ + v₀ t + ½ at²

0 = 20 + 0 + ½ (-9.8) t²

0 = 20 − 4.9 t²

t ≈ 2 s

v² = v₀² + 2a(y − y₀)

v² = 0 + 2(-9.8)(0 − 20)

v ≈ ±20 m/s

Since the rock is falling, v = -20 m/s.

6 0
4 years ago
An object in free fall is at heights y1, y2, and y3 at times t1, t2, and t3 respectively.
Dvinal [7]

Answer:

  a = (v₃₂ - v₂₁) / (t₃₂ -t₂₁)

Explanation:

This is an exercise of average speed, which is defined with the variation of the distance in the unit of time

         v = (y₃ - y₂) / (t₃-t₂)

the midpoint of a magnitude is the sum of the magnitude between 2

         t_mid = (t₂ + t₃) / 2

the same reasoning is used for the mean acceleration

         a = (v_f - v₀) / (t_f - t₀)

   

in our case

        a = (v₃₂ - v₂₁) / (t₃₂ -t₂₁)

5 0
3 years ago
An 1120 kg car traveling at 17.2 m/s is brought to a stop while skidding 40m. Calculate the work done on the car by the friction
Nesterboy [21]

Answer:

Work = 165670.4 J = 165.67 KJ

Explanation:

First, we will find the deceleration of the car, using the 3rd equation of motion:

2as = v_{f}^2 - v_{i}^2\\

where,

a = deceleration = ?

s = skid distance = 40 m

vf = final speed = 0 m/s

vi = initial speed = 17.2 m/s

Therefore,

2a(40\ m) = (0\ m/s)^2 - (17.2\ m/s)^2\\a = - 3.698\ m/s^2

the negative sign indicates deceleration here.

Now, we will calculate the braking force applied by the brakes on the car:

F = ma\\F = (1120\ kg)(-3.698\ m/s^2)\\F = - 4141.76\ N

the negative sign indicates braking force.

Now, we will calculate the work done using the magnitude of this force:

Work = |F|s\\Work = (4141.76\ N)(40\ m)\\

<u>Work = 165670.4 J = 165.67 KJ</u>

8 0
3 years ago
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