Would time move slower or faster approaching black hole?

a large sphere is on a horizontal field on a sunny day. at a certain time the shadow reaches out a distance of 10 m from the poi

Mars2501 [29]

The radius of the sphere in meters is ,r = $10\sqrt{5} -20$

Think about the angle the ground and the shadow make. Since the sun's beams are parallel, the angle created by the stick's shadow is also equal. Since the stick is 1 m high and its shadow is 2 m long, we know that the stick's angle is arctan 1/2. Therefore, by thinking of a right-angled triangle,

r/10 = tan [arctan(1/2)] = tan (1/2)

Since, tan (θ/2) = 1-cos(θ) / sin(θ)

we find that,

r/10 = $\sqrt{5} -2$

Hence, r = $10\sqrt{5} -20$

So, the radius of the sphere in meters is ,r = $10\sqrt{5} -20$

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5 0

1 year ago

6th grade science !!!!!!!!!!!!!!!

posledela

Answer:

71

Explanation:

5 0

3 years ago

A jogger moves from a position x =

vlada-n [284]

Answer:

3 m/s

Explanation:

We'll begin by calculating the change in displacement of the jogger. This can be obtained as follow:

Initial displacement (d₁) = 4 m

Final displacement (d₂) = 16 m

Change in displacement (Δd) =?

Δd = d₂ – d₁

Δd = 16 – 4

Δd = 12 m

Finally, we shall determine the determine the average velocity. This can be obtained as follow:

Change in displacement (Δd) = 12 m

Time (t) = 4 s

Velocity (v) =?

v = Δd / t

v = 12 / 4

v = 3 m/s

Thus, the average velocity of the jogger is 3 m/s

5 0

3 years ago

A proton, starting from rest, accelerates through a potential difference of 1.0 kV and then moves into a magnetic field of 0.040

Minchanka [31]

Answer:

r = 0.11 m

Explanation:

The radius of the proton's resulting orbit can be calculated equaling the force centripetal (Fc) with the Lorentz force ( $F_{B}$ ), as follows: