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Vesna [10]
3 years ago
14

There was a major collision of an asteroid with the Moon in medieval times. It was described by monks at Canterbury Cathedral in

England as a red glow on and around the Moon. How long (in s) after the asteroid hit the Moon, which is 3.58 ✕ 105 km away, would the light first arrive on Earth?
Physics
1 answer:
soldi70 [24.7K]3 years ago
4 0

Answer:

1.28 s

Explanation:

The expression to find the distance traveled by light is,

d = ct

Here is distance is d,  v is the speed and t is the time.

Rearrange the expression for the time,

t = \frac{d}{c}

Substitute 3.58 x 10^{8} m for d and 3 .00 x 10^{8}  m/s  for c in the expression of time,

t = \frac{3.58 * 10^{8 } m  }{3.00 * 10^{8} m/s} = 1.28 s

Hence light will arrive at earth after.

1.28 s

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A sled is slowing down at the bottom of a snowy hill.
Kitty [74]

Answer:

A sled and its rider are moving at a speed of along a horizontal stretch of snow, as Figure 4.24a illustrates. The snow exerts a kinetic frictional force on the runners of the sled, so the sled slows down and eventually comes to a stop. The coefficient of kinetic friction is 0.050. What is the displacement x of the sled?

5 0
3 years ago
Physical values in the real world have two components: magnitude and
Nadusha1986 [10]

Answer:

Dimension.

Explanation:

6 0
3 years ago
A 0.150 kg stone rests on a frictionless, horizontal surface. A bullet of mass 9.50 g, traveling horizontally at 380 m/s, strike
Anvisha [2.4K]

Answer:

(a)Magnitude=28.81 m/s

Direction=33.3 degree below the horizontal

(b) No, it is not perfectly elastic collision

Explanation:

We are given that

Mass of stone, M=0.150 kg

Mass of bullet, m=9.50 g=9.50\times 10^{3} kg

Initial speed of bullet, u=380 m/s

Initial speed of stone, U=0

Final speed of bullet, v=250m/s

a. We have to find the magnitude and direction of the velocity of the stone after it is struck.

Using conservation of momentum

mu+ MU=mv+ MV

Substitute the values

9.5\times 10^{-3}\times 380 i+0.150(0)=9.5\times 10^{-3} (250)j+0.150V

3.61i=2.375j+0.150V

3.61 i-2.375j=0.150V

V=\frac{1}{0.150}(3.61 i-2.375j)

V=24.07i-15.83j

Magnitude of velocity of stone

=\sqrt{(24.07)^2+(-15.83)^2}

|V|=28.81 m/s

Hence, the magnitude and direction of the velocity of the stone after it is struck, |V|=28.81 m/s

Direction

\theta=tan^{-1}(\frac{y}{x})

=tan^{-1}(\frac{-15.83}{24.07})

\theta=tan^{-1}(-0.657)

=33.3 degree below the horizontal

(b)

Initial kinetic energy

K_i=\frac{1}{2}mu^2+0=\frac{1}{2}(9.5\times 10^{-3})(380)^2

K_i=685.9 J

Final kinetic energy

K_f=\frac{1}{2}mv^2+\frac{1}{2}MV^2

=\frac{1}{2}(9.5\times 10^{-3})(250)^2+\frac{1}{2}(0.150)(28.81)^2

K_f=359.12 J

Initial kinetic energy is not equal to final kinetic energy. Hence, the collision is not perfectly elastic collision.

5 0
3 years ago
You lift a 10-N physics book up in the air a distance of 1.0 m, at a constant velocity
Anarel [89]

The book is lifted upward, but gravity points down, so the work done by gravity must be negative (so you can eliminate options 1 and 3).

The force exerted on the book by gravity has magnitude

<em>F</em> = <em>mg</em> = (10 N) (9.80 m/s^2) = 9.8 N ≈ 10 N

You raise the book 1.0 m in the opposite direction, so the work done is

<em>W</em> = (10 N) (-1.0 m) = -10 J

5 0
3 years ago
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