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RSB [31]
2 years ago
14

Suppose a 1300 kg car is traveling around a circular curve in a road at a constant

Physics
1 answer:
Scrat [10]2 years ago
4 0

Answer:

F = 4212 N

Explanation:

Given that,

Mass of a car, m = 1300 kg

Speed of car on the road is 9 m/s

Radius of curve, r = 25 m

We need to find the magnitude of the unbalanced force that steers the car out of its natural straight-  line path. The force is called centripetal force. It can be given by :

F=\dfrac{mv^2}{r}\\\\F=\dfrac{1300\times 9^2}{25}\\\\F=4212\ N

So, the force has a magnitude of 4212 N

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A soccer ball with a mass of 0.45 kg is kicked and is moving at 8.9 m/s. Find the kinetic
strojnjashka [21]

Answer:

17.82J

Explanation:

Kinetic energy = 1/2 mv^2

Given

Mass M = 0.45kg

Velocity v = 8.9m/s

Therefore,

K.E. = 1/2 x 0.45 x (8.9)^2

= 1/2 x 0.45 x (8.9 x 8.9)

= 1/2 x 0.45 x 79.21

Multiply through

= 35.6445/2

= 17.82J

The kinetic energy of the ball is 17.82J

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2 years ago
When calculating the power bill, power companies use _____.
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I think the correct answer from the choices listed above is option B. When calculating the power bill, power companies use kilowatt-hours. This unit is a derived unit of energy equal to 3.6 MJ. If energy is being transmitted or used at a constant rate (power) over a period of time, the total energy in kilowatt-hours is the product of the power in kilowatts and the time.
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In beta decay, what is emitted?
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An electron is emitted in both positive and negative beta decay, although the positive one is called positron emission.
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Calculate frequency if wavelength is 35 m and wave speed is 546 m/s.
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2 years ago
An engine causes a car to move 10 meters with a force of 100 N. The engine produces 10,000 J of energy. What is the efficiency o
kodGreya [7K]

Answer:

Part A

The efficiency of the engine is 10%

Part B

The change in internal energy is 300 J

Part C

The change in volume is 1 m³ which is one cubic meter

Explanation:

Part A

Efficiency is defined as the ratio of energy output to energy input;

The given parameters are;

The distance the car is moved, d = 10 meters

The force which moves the car, F = 100 N

The amount of energy produced by the engine = 1,000 J

Therefore, we have;

The energy output to the car = The work done on the car = Force applied to the car, F × The distance the car moves, d;

∴ The energy output to the car by the engine = F × d = 100 N × 10 m = 1,000 J

The energy input from the engine = The energy produced by the engine = 10,000 J

The efficiency of the engine = (The energy output)/(The energy input)= 1,000J/10,00J = 0.1

The efficiency in percentage = 0.1 × 100 = 10 %

The efficiency of the engine = 10%

Part B

The amount of heat added to the substance, ΔQ = 1,000J

The amount of work the substance does on the atmosphere, W = 700 J

The change in internal energy, ΔU is given as follows;

ΔQ = W + ΔU

∴ ΔU = ΔQ - W

For the substance, we have;

The change in internal energy, ΔU = 1,000 J - 700 J = 300 J

Part C

The work done by the piston, W = 1,000 J

The pressure, in the piston, P = 1,000 Pa = constant

The work done by the piston in a constant pressure process, W = P × ΔV

Where;

W = The work done

P = The constant pressure applied

ΔV = The change in volume = V₂ - V₁

V₂ = The final volume

V₁ = The initial volume

∴The change in volume ΔV = W/P = 1,000 J/(1,000Pa) = 1 m³

The change in volume ΔV = 1 m³

3 0
2 years ago
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