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erik [133]
2 years ago
6

Surveyors frequently use trig functions. Suppose that you measure the angle from your position to the top of a mountain to be 2.

50∘. (Figure 6) If the mountain is 1.00 km higher in elevation than your position, how far away is the mountain?
Physics
1 answer:
Darina [25.2K]2 years ago
8 0

Answer:

22.9km

Explanation:

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Something that can not be used up or depleted​
Jet001 [13]

Answer:

oxygen

Explanation:

as it is in the air it can't be depleted or used up

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What is thermal conductivity?
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Thermal conductivity is the ability of a material to conduct heat.
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3 0
3 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.

7 0
3 years ago
A train is moving at 15ms-1. It slows down to 10.5ms over a time of 4s. Calculate the distance it travels before stopping if acc
Nostrana [21]

Answer:

3WGGRGWERGRG

Explanation:

GERAGAETGAERR GHERUERHGRRGF;SBDF;JKSRDMFNSDFLDGGD;GDVF

4 0
2 years ago
A ball has an electric charge of +1.5 × 10-9 coulombs. At what distance from the ball's center is the electric field strength eq
dimaraw [331]
E=Fe/q                                                      
5.8x10^5N/C=Fe/(1.5x10^-9C)             
Fe=(5.8x10^5N/C)(1.5x10^-9C)
Fe=8.7x10^-4N

Fe=kq1q2/r²
8.7X10^-4N=(8.99x10^9N·m²/C²)(1.5x10^-9C)(1.5x10-9C)/r²
r²=(8.99x10^9N·m²/C²)(1.5x10^-9C)(1.5x10-9C)/(8.7x10^-4N)
√r²=√0.00002325
The final answer is r=4.8x10-3m

5 0
3 years ago
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