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tangare [24]
3 years ago
10

How long does it take to go 3000 miles at the speed of light?

Physics
1 answer:
Crank3 years ago
4 0

3000 miles (1609 meters/1 mile)

= 4827000 meters

Light travels at 2.998 X 10^8 m/s

4827000 meters (1 s/2.998 X 10^8 m/s)

= 0.016 seconds

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Hatshy [7]
The last one is correct (D)
8 0
3 years ago
A stone is thrown at an angle of 30 degrees above the horizontal from the top edge of a cliff with an initial speed of 12 m/s. A
natta225 [31]

v = initial velocity of launch of the stone = 12 m/s

θ = angle of the velocity from the horizontal = 30

Consider the motion of the stone along the vertical direction taking upward direction as positive and down direction as negative.

v₀ = initial velocity along vertical direction = v Sinθ = 12 Sin30 = 6 m/s

a = acceleration of the stone = - 9.8 m/s²

t = time of travel = 4.8 s

Y = vertical displacement of stone = vertical height of the cliff = ?

using the kinematics equation

Y = v₀ t + (0.5) a t²

inserting the values

Y = 6 (4.8) + (0.5) (- 9.8) (4.8)²

Y = - 84.1 m

hence the height of the cliff comes out to be 84.1 m

5 0
3 years ago
A wave with a high frequency generally has a _____.
il63 [147K]
High frequency = D, short wavelength
4 0
3 years ago
what are the speeds of (a) a proton that is accelerated from rest through a potential difference of −1000 v−1000 v and (b) an el
Evgen [1.6K]

Answer:

This is the answer: The speed of a proton is about 5.0 × 10⁵ m/s

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5 0
1 year ago
A 1400 kg car starts from rest on a horizontal road and gains a speed of 61 km/h in 19 s. (a) what is its kinetic energy at the
lana [24]
(a) Let's convert the final speed of the car in m/s:
v_f = 61 km/h = 16.9 m/s
The kinetic energy of the car at t=19 s is
K= \frac{1}{2}mv_f^2= \frac{1}{2}(1400 kg)(16.9 m/s)^2=2.00 \cdot 10^5 J

(b) The average power delivered by the engine of the car during the 19 s is equal to the work done by the engine divided by the time interval:
P= \frac{W}{\Delta t}
But the work done is equal to the increase in kinetic energy of the car, and since its initial kinetic energy is zero (because the car starts from rest), this translates into
P= \frac{K}{\Delta t}= \frac{2.00 \cdot 10^5 J}{19 s}=1.05 \cdot 10^4 W

(c) The instantaneous power is given by
P_i = Fv_f
where F is the force exerted by the engine, equal to F=ma.

So we need to find the acceleration first:
a= \frac{v_f-v_i}{\Delta t}=  \frac{16.9 m/s}{19 s}=0.89 m/s^2
And the problem says this acceleration is constant during the motion, so now we can calculate the instantaneous power at t=19 s:
P_i = Fv=(ma)v=(1400 kg)(0.89 m/s^2)(16.9 m/s)=2.11 \cdot 10^4 W
5 0
3 years ago
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