Find the energy in Joules required to lift a 55.0 Megagram object a distance of 500 cm.

1 answer:

7 0

Energy to lift something =

(mass of the object) x (gravity) x (height of the lift).

BUT ...

This simple formula only works if you use the right units.

Mass . . . kilograms
Gravity . . . meters/second²
Height . . . meters

For this question . . .

Mass = 55 megagram = 5.5 x 10⁷ grams = 5.5 x 10⁴ kilograms

Gravity (on Earth) = 9.8 m/second²

Height = 500 cm = 5.0 meters

So we have ...

Energy = (5.5 x 10⁴ kilogram) x (9.8 m/s²) x (5 m)

= 2,696,925 joules .

That's quite a large amount of energy ... equivalent to
straining at the rate of 1 horsepower for almost exactly an
hour, or burning a 100 watt light bulb for about 7-1/2 hours.

The reason is the large mass that's being lifted.
On Earth, that much mass weighs about 61 tons.

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4. Brandon throws a tennis ball vertically upward. The ball returns to the point of release after 4.0 s. What is the

vekshin1

Answer:

-39.2m/s

Explanation:

Using the equation of motion;

v = u + at

Since the ball is thrown upward, the acceleration due to gravity acting on it will be negative, hence a = -g

v = u - gt

Since g = 9.8m/s²

t = 4.0s

u = 0m/s

v = 0 + (-9.8)(4)

v = -39.2m/s

Hence the speed of the ball before release is -39.2m/s

6 0

3 years ago

Salmon often jump waterfalls to reach their

PilotLPTM [1.2K]

The minimum velocity of the Salmon jumping at the given angle is 12.3 m/s.

The given parameters;

height of the waterfall, h = 0.432 m
distance of the Salmon from the waterfall, s = 3.17 m
angle of projection of the Salmon, = 30.8º

The time of motion to fall from 0.432 m is calculated as;

$h = v_0_y + \frac{1}{2} gt^2\\\\0.432 = 0 + (0.5\times 9.8)t^2\\\\0.432 = 4.9t^2\\\\t^2 = \frac{0.432}{4.9} \\\\t^2 = 0.088\\\\t = \sqrt{0.088} \\\\t = 0.3 \ s$