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A 4.80 Kg watermelon is dropped from rest from the roof of an 18.0 m building. Calculate the work done by gravity on the waterme

Nonamiya [84]

Answer:

Work, W = 846.72 Joules

Explanation:

Given that,

Mass of the watermelon, m = 4.8 kg

It is dropped from rest from the roof of 18 m building. We need to find the work done by the gravity on the watermelon from the roof to the ground. It is same as gravitational potential energy i.e.

W = mgh

$W=4.8\ kg\times 9.8\ m/s^2\times 18\ m$

W = 846.72 Joules

So, the work done by the gravity on the watermelon is 846.72 Joules. Hence, this is the required solution.

7 0

2 years ago

Determine the answer to the equation 30 km/h × 17 h =

Jobisdone [24]

30 km/h * 17 h = 30*17 km/h *h
= 510 km

5 0

2 years ago

Of all the planets in our solar system, Jupiter has the greatest gravitational strength. If a 1.5 kg pair of running shoes would

Andre45 [30]

Answer:

gₓ = 23.1 m/s²

Explanation:

The weight of an object is on the surface of earth is given by the following formula:

$W = mg$

where,

W = Weight of the object on surface of earth

m = mass of object

g = acceleration due to gravity on the surface of earth = strength of gravity on the surface of earth

Similarly, the weight of the object on Jupiter will be given as:

$W_{x} = mg_{x}$

where,

Wₓ = Weight of the object on surface of Jupiter = 34.665 N

m = mass of object = 1.5 kg

gₓ = acceleration due to gravity on the surface of Jupiter = strength of gravity on the surface of Jupiter = ?

Therefore,

$34.65 N = (1.5 kg)g_{x}$

$g_{x} = \frac{34.65 N}{1.5 kg}$

7 0

2 years ago

A golf ball strikes a hard, smooth floor at an angle of 39.8 ° and, as the drawing shows, rebounds at the same angle. The mass o

tangare [24]

Answer:

The magnitude of the impulse is 1.33 kg m/s

Explanation:

please look at the solution in the attached Word file

Download docx

7 0

3 years ago

Read 2 more answers

An insulated thermos contains 106.0 cm3 of hot coffee at a temperature of 80.0 °C. You put in 11.0 g of ice cube at its melting

Andrei [34K]

Answer:

the final temperature is T f = 64.977 ° C≈ 65°C

Explanation:

Since the thermus is insulated, the heat absorbed by the ice is the heat released by the coffee. Thus:

Q coffee + Q ice = Q surroundings =0 (insulated)

We also know that the ice at its melting point , that is 0 °C ( assuming that the thermus is at atmospheric pressure= 1 atm , and has an insignificant amount of impurities ).

The heat released by coffee is sensible heat : Q = m * c * (T final - T initial)

The heat absorbed by ice is latent heat and sensible heat : Q = m * L + m * c * (T final - T initial)

therefore

m co * c co * (T fco - T ico) + m ice * L + m ice * c wat * (T fwa - T iwa) = 0

assuming specific heat capacity of coffee is approximately the one of water c co = c wa = 4.186 J/g°C and the density of coffee is the same as water

d co = dw = 1 gr/cm³

therefore m co = d co * V co = 1 gr / cm³ * 106 cm³ = 106 gr

m co * c wat * (T f - T ico) + m ice * L + m ice * c wat * (T f - T iwa) = 0

m co * c wat * T f+ m ice * c wat * T f = m ice * c wat * T iwa + m co * c wat * Tico -m ice * L

T f = (m ice * c wat * T iwa + m co * c wat * Tico -m ice * L ) /( m co * c wat * + m ice * c wat )

replacing values

T f = (11 g * 4.186 J/g°C * 0°C + 106 g * 4.186 J/g°C*80°C - 11 g * 334 J/gr) / ( 11 g * 4.186 J/g°C + 106 g * 4.186 J/g°C* ) = 64,977 ° C

T f = 64.977 ° C

7 0

3 years ago

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