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alexdok [17]
2 years ago
10

A ______ employs a weight and a wire to detect horizontal movement.

Physics
2 answers:
Finger [1]2 years ago
7 0

A CREEP METER employs a weight and a wire to detect horizontal movement.

Explanation:

A creep meter is an apparatus utilized to measure ground above an existing fault line has gradually shifted in a horizontal orientation. A creep meter operates by utilizing a wire extended across a fault line. As the ground shifts in a horizontal orientation, the creep meter measures the difference.

eimsori [14]2 years ago
6 0
The answer is creep meter
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When we look at a star
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4 0
3 years ago
If a 54 kg sprinter can accelerate from a standing start to a speed of 10 m/s in 3 s, what average power is generated?
lora16 [44]

Answer:

Power, P = 600 watts

Explanation:

It is given that,

Mass of sprinter, m = 54 kg

Speed, v = 10 m/s

Time taken, t = 3 s

We need to find the average power generated. The work done divided by time taken is called power generated by the sprinter i.e.

P=\dfrac{W}{t}

Work done is equal to the change in kinetic energy of the sprinter.

P=\dfrac{\dfrac{1}{2}mv^2}{t}

P=\dfrac{\dfrac{1}{2}\times 54\ kg\times (10\ m/s)^2}{3\ s}

P = 900 watts

So, the average power generated by the sprinter is 900 watts. Hence, this is the required solution.

3 0
3 years ago
A man pushes a stalled car with a force of 100 N. If the acceleration of the car is 0.75 m/s 2 , what is the mass of the car?
shutvik [7]
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4 0
3 years ago
The plates of a parallel-plate capacitor are 2.50 mm apart, and each carries a charge of magnitude 77.0 nC. The plates are in va
WINSTONCH [101]

Answer:

Part A: 7500 V

Part B: 2.899×10⁻³ m²

Part C: 10.27 pF or 10.27×10⁻¹² F

Explanation:

Part A:

Applying,

E = V/d................ Equation 1

Where E = electric field intensity between the plates, V = potential difference between the plates, d = distance of separation between the plates

make V the subject of the equation above,

V = Ed............. Equation 2

Given: E = 3.0×10⁶ V/m, d = 2.5 mm = 2.5×10⁻³ m

Substitute into equation 2

V =  3.0×10⁶ (2.5×10⁻³ )

V = 7.5×10³ V

V = 7500 V

Part B:

Using,

E = Q/(e₀A).................... Equation 3

Where Q = Charge on each plate of the capacitor, A = Area of each plate, e₀ = constant = dielectric = permitivity of free space

make A the subject of the equation,

A = Q/(e₀E).............. Equation 4

Given: Q = 77 nC = 77×10⁻⁹ C, E = 3.0×10⁶ V/m

Constant: e₀ = 8.854×10⁻¹² F/m

Substitute into equation 4

A = 77×10⁻⁹/(8.854×10⁻¹²× 3.0×10⁶)

A = 77×10⁻⁹/(26.562×10⁻⁶)

A = 2.899×10⁻³ m²

A = 2.899×10⁻³ m².

Part C:

Using,

Q = CV.................. Equation 5

Where C = Capacitance of the capacitor

make C the subject of the equation

C = Q/V.............. Equation 6

Given: Q = 77 nC = 77×10⁻⁹ C, V = 7500 V

Substitute into equation 6

C = 77×10⁻⁹/7500

C = 10.27×10⁻¹² F

C = 10.27 pF

5 0
3 years ago
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