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serg [7]
3 years ago
15

If you are seeing a solar eclipse but you are not located in the umbra of the moon, what might you see in the sky?

Physics
1 answer:
damaskus [11]3 years ago
3 0
That is pretty tricky, I'd say you would see the moon over the sun, but not all the way.



I Hope this helps
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What physics concept is used when designing a motorcycle helmet to protect the head from injury
fomenos

Answer:

Explanation:

From the 1st Law of linear motion which states that when a body goes into motion, it will continue doing so until it is stopped by force. So the body of the rider keeps moving until it is been stopped and the stopping could be as a result of an impact of any part of the body which includes the head it an object. Also the 3rd Law of Motion will also be applicable because for every action, there will be equal and opposite reaction. The magnitude of the impact will be as a result of the force with which the crash took place as well.

When crashes take place the rider does not always experience a head impact square on with a solid obstruction. During a bike crash, your head comes in contact with the ground. The ground exerts a force that causes your head to stop moving. Often impact will be at an angle and may not be head first. It may be your shoulder will hit first, then your side, and then your head will receive a glancing blow against the ground as you slide.

The ground exerts so much force that it can stop our forward motion within seconds. Without your motorcycle helmet, your head experiences a huge amount of concentrated force during a crash.

3 0
2 years ago
If A is the area of a circle with radius r and the circle expands as time passes, find dA/dt in terms of dr/dt. dA dt = dr dt (b
Lorico [155]

Answer:157.1 m^2/s

Explanation:

Given

A is the area of circle with radius r

A=\pi r^2

Differentiate w.r.t time

\frac{\mathrm{d} A}{\mathrm{d} t}=\pi \times 2\times \frac{\mathrm{d} r}{\mathrm{d} t}

Also \frac{\mathrm{d} r}{\mathrm{d} t}=1 m/s at r=25 m

\frac{\mathrm{d} A}{\mathrm{d} t}=\pi \times 2\times 25\times 1=157.1 m^2/s

5 0
3 years ago
Read 2 more answers
A planet orbits a star, in a year of length 2.35 x 107 s, in a nearly circular orbit of radius 3.49 x 1011 m. With respect to th
Naya [18.7K]

Answer:

a)   w = 9.599 10⁴ rad / s , b)   v = 3.35 10¹⁶ m / s , c)    a = 3.22  10²¹ m / s²

Explanation:

For this exercise we must use the relation of angular kinematics

a) angular velocity, the distance remembered in orbit between time (period)

         w = 2π r / T

         w = 2 π 3.59 10¹¹ / 2.35 10⁷

         w = 9.599 10⁴ rad / s

b) linear and angular velocity are related by the equation

          v = w r

          v = 9,599 10⁴ 3.49 10¹¹

          v = 3.35 10¹⁶ m / s

c) the centripetal acceleration is

            a = v² / r = w² r

            a = (9,599 10⁴)²   3.49 10¹¹

            a = 3.22  10²¹ m / s²

7 0
3 years ago
Which of the following quantities provide enough information to calculate the tension in a string of mass per unit length μ that
Bad White [126]

Answer:

A. the wave speed v and Wavelength

Explanation:

Given that

Mass density per unit length=μ

Frequency = f

The velocity V given as

\mu=\dfrac{T}{V^2}\ kg/m

V=\sqrt{\dfrac{T}{\mu}}

T=Tension

V=Velocity

V= f λ

λ=Wavelength

Therefore to find the tension ,only wavelength and speed is required.

The answer is A.

8 0
3 years ago
A​ blimp, suspended in the air at a height of 600 ​feet, lies directly over a line from a sports stadium to a planetarium. If th
MAXImum [283]

Answer:

1946 ft

Explanation:

The distance is the addition of the distance gotten from both triangles.

d = x + y

d = 1178 ft + 768 ft

d = 1946 ft

Attached is a picture showing how I arrived at the answer

3 0
2 years ago
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