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lana [24]
3 years ago
8

State what is meant by graviration potential at a point in an orbit 6.5×10^7​

Physics
1 answer:
laila [671]3 years ago
5 0

Explanation:

The gravitational potential at a point in a gravitational field is the work done per unit mass that would have to be done by some externally applied force to bring a massive object to that point from some defined position of zero potential,

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A centrifuge in a forensics laboratory rotates at an angular speed of 3,700 rev/min. When switched off, it rotates 46.0 times be
ra1l [238]

Answer:

Explanation:

Given,

initial angular speed, ω = 3,700 rev/min

                                      = 3700\times \dfrac{2\pi}{60}=387.27\ rad/s

final angular speed = 0 rad/s

Number of time it rotates= 46 times

angular displacement, θ = 2π x 46 = 92 π

Angular acceleration

\alpha = \dfrac{\omega_f^2 - \omega^2}{2\theta}

\alpha = \dfrac{0 - 387.27^2}{2\times 92\ pi}

\alpha = -259.28 rad/s^2

3 0
3 years ago
How does the state of matter of the medium through which the sound waves and light waves travel theough affect their velocity
yanalaym [24]

Answer:

idk

Explanation:

idk

5 0
3 years ago
On May 26, 1934, a streamlined, stainless steel diesel train called the Zephyr set the world’s nonstop long-distance speed recor
Andre45 [30]

Answer:

124.88 km/h

34.69 m/s

Explanation:

1633.8 km = 1633800 m

13 hours 4 minutes 58 seconds = 13 + 4/60 + 58/3600 = 13.083 hours

13 hours 4 minutes 58 seconds = 13*3600 + 4*60 + 58 = 47098 seconds

So the average speed in km/h is

1633.8 / 13.083 = 124.88 km/h

The average speed in m/s is

1633800 / 47098 = 34.69 m/s

8 0
3 years ago
Why is it important for earth’s magnetic field that the inner core is hotter then the outer core?
Shkiper50 [21]

A. The inner core heat produce stuff

8 0
2 years ago
An object is propelled straight up from ground level with an initial velocity of 48 feet per second. Its height at time t is mod
Ket [755]

Answer:

Explanation:

For a. its max height and when it occurs. First the max height. That's a y-dimension thing, and in the y-dimension we have this info:

v₀ = 48 ft/s

a = -32 ft/s/s

v = 0 (the max height of an object occurs when the final velocity of the object is 0). Use the following equation for this part of the problem:

v² = v₀² + 2aΔx and filling in:

0=48^2+2(-32)Δx and

0 = 2300 - 64Δx and

-2300 = -64Δs so

Δx = 36 feet.

Now for the time it takes to get to this max height. Final velocity is still 0 here, but the equation is a different one for this part of the problem. Use:

v = v₀ + at and filling in:

0 = 48 - 32t and

-48 = -32t so

t = 1.5 sec.  That's part a. Onto part b:

The object hits the ground when its displacement, Δx, is 0. Use this equation for this problem:

Δx = v₀t + \frac{1}{2}at^2 and filling in:

0=48t+\frac{1}{2}(-32)t^2 and

0=48t-16t^2 and

0 = 16t(3 - t) so

t = 0 and t = 3.  t = 0 is before the object is propelled, so it makes sense that at 0 seconds, the object was still on the ground, right? Then at 3 seconds, it's back on the ground. (Isn't math just perfectly, beautifully sensible!?) Now onto part c:

We are looking for the time interval when the object is >32 feet. So we use the same equation we just used, but with an inequality instead of an equals sign:

48t+\frac{1}{2}(-32)t^2 >32 and get everything on one side and factor it again:

-16t^2+48t-32>0 and we find that

1 < t < 2 so the time interval is between 1 and 2 seconds that the object is over 32 feet in the air.

8 0
3 years ago
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