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Neporo4naja [7]
3 years ago
14

A ball is thrown at 20 m/s from the ground upwards at an angle of elevation of 30°. How far away does it land? 35.35 m

Physics
1 answer:
mestny [16]3 years ago
8 0

Answer:

35.35 m

Explanation:

The following data were obtained from the question:

Initial velocity (u) = 20 m/s

Angle of projection (θ) = 30°

Acceleration due to gravity (g) = 9.8 m/s²

Range (R) =.?

The range (i.e how far away) of the ball can be obtained as follow:

R = u² Sine 2θ /g

R = 20² Sine (2×30) / 9.8

R = 400 Sine 60 / 9.8

R = (400 × 0866) / 9.8

R = 346.4 / 9.8

R = 35.35 m

Therefore, the range (i.e how far away) of the ball is 35.35 m

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An 80kg parachutist jumps out of an airplane. Neglecting air friction, how fast will he be going after a 10 second free fall?
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A 0.500 kg bullet is fired from a gun at 25.0 m/s, how much kinetic energy does it have?
slamgirl [31]

Considering the definition of kinetic energy, the bullet has a kinetic energy of 156.25 J.

<h3>Kinetic energy</h3>

Kinetic energy is a form of energy. It is defined as the energy associated with bodies that are in motion and this energy depends on the mass and speed of the body.

Kinetic energy is defined as the amount of work necessary to accelerate a body of a given mass and in a rest position, until it reaches a given speed. Once this point is reached, the amount of accumulated kinetic energy will remain the same unless there is a change in speed or the body returns to its rest state by applying a force to it.

The kinetic energy is represented by the following expression:

Ec= ½ mv²

Where:

  • Ec is the kinetic energy, which is measured in Joules (J).
  • m is the mass measured in kilograms (kg).
  • v is the speed measured in meters over seconds (m/s).

<h3>Kinetic energy of a bullet</h3>

In this case, you know:

  • m= 0.500 kg
  • v= 25 m/s

Replacing in the definition of kinetic energy:

Ec= ½ ×0.500 kg× (25 m/s)²

Solving:

<u><em>Ec= 156.25 J</em></u>

Finally, the bullet has a kinetic energy of 156.25 J.

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An electric furnace is to melt 40 kg of aluminium/hour. The initial temperature of aluminium is 32°C. Given that aluminium has s
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Answer:

Part a)

P = 13.93 kW

Part b)

R = 8357.6 Cents

Explanation:

Part A)

heat required to melt the aluminium is given by

Q = ms\Delta T + mL

here we have

Q = 40(950)(680 - 32) + 40(450 \times 10^3)

Q = 24624 kJ + 18000 kJ

Q = 42624 kJ

Since this is the amount of aluminium per hour

so power required to melt is given by

P = \frac{Q}{t}

P = \frac{42624}{3600} kW

P = 11.84 kW

Since the efficiency is 85% so actual power required will be

P = \frac{11.84}{0.85} = 13.93 kW

Part B)

Total energy consumed by the furnace for 30 hours

Energy = power \times time

Energy = 13.93 kW\times 30 h

Energy = 417.9 kWh

now the total cost of energy consumption is given as

R = P \times 20 \frac{Cents}{kWh}

R = 417.9 kWh\times  20 \frac{cents}{kWh}

R = 8357.6 Cents

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