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Ludmilka [50]
2 years ago
12

A 45000 watt crane operating at full power lifts a 2100 kg object vertically for 17.4 seconds. How high has the crane lifted the

object?
Physics
1 answer:
LenKa [72]2 years ago
8 0

Answer:

Explanation:

We need the power equation here, which is:

Power = (F * Δx)/time

where F * Δx is the amount of work done.

F is a force which is measured in Newtons. We are given the mass of the object, but since we need a Force measure, we need the weight of the object:

F = 2100(9.0)

F = 21000 to the correct number of sig dig.

Now we can plug in the values we have and solve for the displacement, Δx:

45000=\frac{21000x}{17.4} and isolating x:

\frac{17.4(45000)}{21000}=x so

x = 37 m

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A person walks 9.0 km directly east and then turns left and heads directly north for 12.0 km. What is his displacement from the
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A person walks 9.0 km directly east and then turns left and heads directly north for 12.0 km.

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3 years ago
The volume of water in the Pacific Ocean is about 7.0 × 10 8 km 3 . The density of seawater is about 1030 kg/m3. (a) Determine t
Novay_Z [31]

To solve the problem it is necessary to consider the concepts related to Potential Energy and Kinetic Energy.

Potential Energy because of a planet would be given by the equation,

PE=\frac{GMm}{r}

Where,

G = Gravitational Universal Constant

M = Mass of Ocean

M = Mass of Moon

r = Radius

From the data given we can calculate the mass of the ocean water through the relationship of density and volume, then,

m = \rho V

m = (1030Kg/m^3)(7*10^8m^3)

m = 7.210*10^{11}Kg

It is necessary to define the two radii, when the ocean is far from the moon and when it is facing.

When it is far away, it will be the total diameter from the center of the earth to the center of the moon.

r_1 = 3.84*10^8 + 6.4*10^6 = 3.904*10^8m

When it's near, it will be the distance from the center of the earth to the center of the moon minus the radius,

r_2 = 3.84*10^8-6.4*10^6 - 3.776*10^8m

PART A) Potential energy when the ocean is at its furthest point to the moon,

PE_1 = \frac{GMm}{r_1}

PE_1 = \frac{(6.61*10^{-11})*(7.21*10^{11})*(7.35*10^{22})}{3.904*10^8}

PE_1 = 9.05*10^{15}J

PART B) Potential energy when the ocean is at its closest point to the moon

PE_2 = \frac{GMm}{r_2}

PE_2 = \frac{(6.61*10^{-11})*(7.21*10^{11})*(7.35*10^{22})}{3.776*10^8}

PE_2 = 9.361*10^{15}J

PART C) The maximum speed. This can be calculated through the conservation of energy, where,

\Delta KE = \Delta PE

\frac{1}{2}mv^2 = PE_2-PE_1

v=\sqrt{2(PE_2-PE_1)/m}

v = \sqrt{\frac{2*(9.361*10^{15}-9.05*10^{15})}{7.210*10^{11}}}

v = 29.4m/s

8 0
3 years ago
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