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Fofino [41]
3 years ago
11

List the copositional layers in order of most dense to least dense

Physics
1 answer:
Dominik [7]3 years ago
4 0

Answer:

core, mantle, crust is the most dense to least dense

Explanation:

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Name 2 things that have gears in them
Anvisha [2.4K]

BICYCLE

MOTOR BIKES

CARS

THEY ARE THE VEHICLES WHICH HAS GEARS IN THEM.

6 0
3 years ago
The magnitude of each force is 208 N the force on the right is applied at an angle 36° and the mass of the block is 17 kg the co
djyliett [7]

Answer:

<em>11.06m/s²</em>

Explanation:

According to Newtons second law of motion

\sm F_x = ma_x\\F_m - F_f = ma_x\\mgsin \theta - \mu R mgcos \theta = ma_x\\

Given

Mass m = 17kg

Fm = 208N

theta = 36 degrees

g = 9.8m/s²

a is the acceleration

Substitute

208 - 0.148(17)(9.8)cos 36 = 17a

208 - 24.6568cos36 = 17a

208 - 19.9478 = 17a

188.05 = 17a

a = 188.05/17

a = 11.06m/s²

<em>Hence the  the magnitude of the resulting acceleration is 11.06m/s²</em>

6 0
2 years ago
Write down the boiling point and freezing point of mercury an alcohol​
nekit [7.7K]
Mercury has a high boiling point of 357 degrees C.
Mercury has a freezing point of −39 degrees C.
8 0
3 years ago
A star is moving away from an observer. Toward which end of the spectrum does its visible light shift?.
scZoUnD [109]

Answer:

Redshift, or lower power

Explanation:

doppler effect

waves get stretched when you are moving away from something, and squished when you are moving towards it. Imagine you have a long, bent wire. if you stretch out the wire, the wavelength becomes longer. This also applies to sound.

4 0
2 years ago
One end of a steel rod of radius R = 10 mm and length L = 80cm is held in a vise. A force of magnitude F = 60kN is then applied
Alexandra [31]

Answer: 1.91*10^8 N/m²

Explanation:

Given

Radius of the steel, R = 10 mm = 0.01 m

Length of the steel, L = 80 cm = 0.8 m

Force applied on the steel, F = 60 kN

Stress on the rod, = ?

Area of the rod, A = πr²

A = 3.142 * 0.01²

A = 0.0003142

Stress = Force applied on the steel/Area of the steel

Stress = F/A

Stress = 60*10^3 / 0.0003142

Stress = 1.91*10^8 N/m²

From the calculations above, we can therefore say, the stress on the rod is 1.91*10^8 N/m²

8 0
3 years ago
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