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alexdok [17]
2 years ago
15

As light from a star spreads out and weakens, do gaps form between the photons?

Physics
2 answers:
Sati [7]2 years ago
8 0

Answer:

There are no gaps in space between the photons as they travel.

Yes, you can form the shadow of a fire

Explanation:

gtnhenbr [62]2 years ago
5 0
Yes you can form the shadow of a fire
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7 0
3 years ago
19) A child on a sled starts from rest at the top of a 15.0° slope. If the trip to the bottom takes 15.2 s,
sveticcg [70]

Answer: 288.8 m

Explanation:

We have the following data:

t=15.2 s is the time it takes to the child to reach the bottom of the slope

V_{o}=0 is the initial velocity (the child started from rest)

\theta=15\° is the angle of the slope

d is the length of the slope

Now, the Force exerted on the sled along the ramp is:

F=ma (1)

Where m is the mass of the sled and a its acceleration

In addition, if we draw a free body diagram of this sled, the force along the ramp will be:

F=mg sin \theta (2)

Where g=9.8 m/s^{2} is the acceleration due gravity

Then:

ma=mg sin \theta (3)

Finding a:

a=g sin \theta (4)

a=9.8 m/s^{2} sin(15\°) (5)

a=2.5 m/s^{2} (6)

Now, we will use the following kinematic equations to find d:

V=V_{o}+at (7)

V^{2}=V_{o}^{2}+2ad (8)

Where V is the final velocity

Finding V from (7):

V=at=(2.5 m/s^{2})(15.2 s) (9)

V=38 m/s (10)

Substituting (10) in (8):

(38 m/s)^{2}=2(2.5 m/s^{2})d (11)

Finding d:

d=288.8 m

6 0
3 years ago
An amateur player is about to throw a dart with an initial velocity of 15 meters/second onto a dartboard that is at a distance o
Minchanka [31]

Answer:

B. 0.16 m

Explanation:

The vertical distance by which the player will miss the target is equal to the vertical distance covered by the dart during its motion.

Since the dart is thrown horizontally, the initial vertical velocity is zero:

v_y = 0

While the horizontal velocity is

v_x = 15 m/s

The horizontal distance covered is

d_x = 2.7 m

Since the dart moves by uniform motion along the horizontal direction, the time it takes for covering this distance is

t=\frac{d_x}{v_x}=\frac{2.7 m}{15 m/s}=0.18 s

along the vertical direction, the motion is a uniformly accelerated motion with constant downward acceleration g=9.8 m/s^2, so the vertical distance covered is given by

d_y = \frac{1}{2}gt^2=\frac{1}{2}(9.8 m/s^)(0.18 s)^2=0.16 m

8 0
4 years ago
Please help with this question !!!!!
lesantik [10]

Answer:

te puedo ayudar pero corazón y estrella

y ando buscando novia por si quieres ser mia

7 0
3 years ago
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