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3 years ago

8

In general, what happens to temperature as depth below earths surface increases

1 answer:

lesantik [10]3 years ago

8 0

Temperature increases as depth increases.

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What would changing the frequency of a wave do to the wave?

nata0808 [166]

The data convincingly show that wave frequency does not affect wave speed. An increase in wave frequency caused a decrease in wavelength while the wave speed remained constant. The last three trials involved the same procedure with a different rope tension.

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3 years ago

A spring requires 10 N of force to hold it 1 m past its natural length. How much work is required to stretch it from its natural

Ulleksa [173]

Answer:

5 J

Explanation:

F = spring force required to stretch the spring = 10 N

x = stretch in the spring = 1 m

k = spring constant of the spring

Spring force is given as

F = k x

Inserting the values

10 = k (1)

k = 10 N/m

W = work done in stretching the spring

Work done is given as

W = (0.5)k x²

Inserting the values

W = (0.5) (10) (1)²

W = 5 J

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4 years ago

Temperature can be described as ________

Agata [3.3K]

Temperature can be described as an indirect measure of heat energy
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3 years ago

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Nonamiya [84]

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2 years ago

The area of the piston to the master cylinder in a hydraulic braking system of a car is 0.4 square inches. If a force of 6.4 lb

Anit [1.1K]

Answer:

The force applied on one wheel during braking = 6.8 lb

Explanation:

Area of the piston (A) = 0.4 $in^{2}$

Force applied on the piston(F) = 6.4 lb

Pressure on the piston (P) = $\frac{F}{A}$

⇒ P = $\frac{6.4}{0.4}$

⇒ P = 16 $\frac{lb}{in^{2} }$

This is the pressure inside the cylinder.

Let force applied on the brake pad = $F_{1}$

Area of the brake pad ( $A_{1}$ )= 1.7 $in^{2}$

Thus the pressure on the brake pad ( $P_{1}$ ) = $\frac{F_{1} }{A_{1} }$

When brake is applied on the vehicle the pressure on the piston is equal to pressure on the brake pad.

⇒ P = $P_{1}$

⇒ 16 = $\frac{F_{1} }{A_{1} }$

⇒ $F_{1}$ = 16 × $A_{1}$

Put the value of $A_{1}$ we get

⇒ $F_{1}$ = 16 × 1.7

⇒ $F_{1}$ = 27.2 lb

This the total force applied during braking.

The force applied on one wheel = $\frac{F_{1} }{4}$ = $\frac{27.2}{4}$ = 6.8 lb

⇒ The force applied on one wheel during braking.

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3 years ago