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3 years ago

Define in own words.

Physics

1 answer:

professor190 [17]3 years ago

8 0

A recurring illness or an illness that lasts for a long time. Chronic illnesses are typically very hard to cure.

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What do it mean by the value of gravitational constant is 6.67×10^-11Nm^2/kg^2

luda_lava [24]

Answer:

gravitational constant value means it was never change in any particular area of the Earth

8 0

1 year ago

A ball filled with an unknown material starts from rest at the top of a 2 m high incline that makes a 28o with respect to the ho

Lady_Fox [76]

Answer:

Searching in google I found the total mass and the radius of the ball (m = 1.5 kg and r = 10 cm) which are needed to solve the problem!

The ball rotates 6.78 revolutions.

Explanation:

Searching in google I found the total mass and the radius of the ball (m = 1.5 kg and r = 10 cm) which are needed to solve the problem!

At the bottom the ball has the following angular speed:

$\omega_{f} = \frac{v_{f}}{r} = \frac{4.9 m/s}{0.10 m} = 49 rad/s$

Now, we need to find the distance traveled by the ball (L) by using θ=28° and h(height) = 2 m:

$sin(\theta) = \frac{h}{L} \rightarrow L = \frac{h}{sin(\theta)} = \frac{2 m}{sin(28)} = 4.26 m$

To find the revolutions we need the time, which can be found using the following equation:

$v_{f} = v_{0} + at$

$t = \frac{v_{f} - v_{0}}{a}$ (1)

So first, we need to find the acceleration:

$v_{f}^{2} = v_{0}^{2} + 2aL \rightarrow a = \frac{v_{f}^{2} - v_{0}^{2}}{2L}$ (2)

By entering equation (2) into (1) we have:

$t = \frac{v_{f} - v_{0}}{\frac{v_{f}^{2} - v_{0}^{2}}{2L}}$

Since it starts from rest (v₀ = 0):

$t = \frac{2L}{v_{f}} = \frac{2*4.26 m}{4.9 m/s} = 1.74 s$

Finally, we can find the revolutions:

$\theta_{f} = \frac{1}{2} \omega_{f}*t = \frac{1}{2}*49 rad/s*1.74 s = 42.63 rad*\frac{1 rev}{2\pi rad} = 6.78 rev$

Therefore, the ball rotates 6.78 revolutions.

I hope it helps you!

3 0

2 years ago

If the Earth and distant stars were stationary (motionless) in space, what would we observe about the wavelength from these star

torisob [31]

There's no such thing as "stationary in space". But if the distance
between the Earth and some stars is not changing, then (A) wavelengths
measured here would match the actual wavelengths emitted from these
stars. 

If a star is moving toward us in space, then (A) Wavelengths measured
would be shorter than the actual wavelengths emitted from that star.

In order to decide what's actually happening, and how that star is moving,
the trick is: How do we know the actual wavelengths the star emitted ?

7 0

3 years ago

How can you describe the motion of an object in a race?

Hitman42 [59]

You can describe the motion of an object by its position, speed, direction, and acceleration

4 0

2 years ago

Gravity does not actually "pull" objects at

Radda [10]

I believe it’s (D. Any object)

5 0

3 years ago