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olga2289 [7]
3 years ago
10

A person lifts a 25 kg box of old sports equipment at an angle of 75 degrees to the vertical to a shelf 2.6 meters above. How mu

ch work was done by the person?
Physics
1 answer:
kogti [31]3 years ago
6 0

Answer:

164.87 J

Explanation:

From the question,

Work done (W) = mghcosθ........................ Equation 1

Where m = mass of the box, h = height, g = acceleration due to gravity, θ = angle to the vertical

Given: m = 25 kg, h = 2.6 meters, θ = 75°.

Constant: g = 9.8 m/s²

Substitute these value into equation 1

W = 25×9.8×2.6×cos75°

W = 164.87 J.

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A catapult can project a projectile at a speed as high as 37.0 m/s. (a) if air resistance can be ignored, how high (in m) would
motikmotik

Then the maximum height of the catapult will be 69.78 meters.

<h3>What is kinematics?</h3>

The study of motion without considering the mass and the cause of the motion. The equation of motion is given below.

v = u + at

s = ut + (1/2)at²

v² = u² + 2as

A catapult can project a projectile at a speed as high as 37.0 m/s. (a) if air resistance can be ignored.

a = - 9.81 m/s²

u = 37 m/s

v = 0

s = h

Then the maximum height of the catapult will be given by the third equation.

v² = u² + 2as

0² = 37² - 2 × 9.81 × h

h = 37² / (2 × 9.81)

h = 69.78 meters

Then the maximum height of the catapult will be 69.78 meters.

More about the kinematics link is given below.

brainly.com/question/7590442

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4 0
1 year ago
How much heat energy must be added to 52kg Of water at 68°F to raise the temperature to 212°F? The specific heat capacity for wa
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Answer:

The amount of energy added to rise the temperature Q = 17413.76 KJ

Explanation:

Mass of water = 52 kg

Initial temperature T_{1} = 68 °F = 20° c

Final temperature T_{2} = 212 °F = 100° c

Specific heat of water  C = 4.186 \frac{KJ}{kg c}

Now heat transfer Q = m × C × ( T_{2}  - T_{1} )

⇒ Q = 52 × 4.186 × ( 100 - 20 )

⇒ Q = 17413.76 KJ

This is the amount of energy added to rise the temperature.

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