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Citrus2011 [14]
3 years ago
12

A rock with a mass of 16 kilograms is put aboard an airplane in New York City and flown to Boston. How much work does the gravit

ational field of the earth do on the rock
Physics
1 answer:
Law Incorporation [45]3 years ago
7 0

The work done by the gravitational field of the earth on the rock is 9.998 x 10⁸ J.

<u>The given parameter include:</u>

the mass of the object, m₁ = 16 kg

Note: the mass of the earth, m₂ = 5.972 x 10²⁴ kg

  • The work done by the gravitational field of the earth is given as;

Work done = gravitational force (F) x radius of the earth (R)

Work \ done = \frac{Gm_1m_2}{R^2} \times R\\\\Work \ done = \frac{Gm_1m_2}{R} \\\\where;\\\\R \ is \ the \ radius \ of \ the \ earth = 6,378 \ km = 6,378,000 \ m\\\\G \ is \ the \ universal \ gravitation \ constant = 6.674 \times 10^{-11} Nm^2/kg^2\ \\\\Work \ done = \frac{(6.674 \times 10^{-11} ) \times (5.972\times 10^{24}) \times (16)}{6,378,000 } \\\\Work \ done = 9.998 \times 10^{8} \ J

Therefore, the work done by the gravitational field of the earth on the rock is 9.998 x 10⁸ J.

<u>To learn more about work done by gravitational field of the earth visit</u>: brainly.com/question/13934028

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A car battery has a rating of 170 ampere-hours. This rating is one indication of the total charge that the battery can provide t
Pavlova-9 [17]

Answer:

612000 C

Explanation:

Current, I, is given as the rate of flow of charge, that is:

I = Δq / Δt

where q = electric charge

t = time taken

This implies that:

Δq = I * Δt

The battery rating is 170 Ampere-hours, therefore:

Δq = 170 * 1 hour

But 1 hour = 3600 seconds;

=> Δq = 170 * 3600 = 612000 C

The total charge that the battery can provide is 612000 C.

8 0
3 years ago
About how many weeks does it take the Moon to go from a full Moon phase to a third quarter phase? A. 1 B. 2 C. 3 D. 4
Lostsunrise [7]
The answer is D) About 4.
~Deceptiøn
3 0
3 years ago
The total energy of a block—spring system is 0.18 J. The amplitude is 14.0 cm and the maximum speed is 1.25 m/s. Find: (a) the m
algol13

a) The mass is 0.23 kg

b) The spring constant is 1.25 N/m

c) The frequency is 1.42 Hz

d) The speed of the block is 1.08 m/s

Explanation:

a)

We can find the mass of the block by applying the law of conservation of energy: in fact, the total mechanical energy of the system (which is sum of elastic potential energy, PE, and kinetic energy, KE) is constant:

E=PE+KE=const.

The potential energy is given by

PE=\frac{1}{2}kx^2

where k is the spring constant and x is the displacement. When the block is crossing the position of equilibrium, x = 0, so all the energy is kinetic energy:

E=KE_{max}=\frac{1}{2}mv_{max}^2 (1)

where

m is the mass of the block

v_{max}=1.25 m/s is the maximum speed

We also know that the total energy is

E=0.18 J

Re-arranging eq.(1), we can find the mass:

m=\frac{2E}{v_{max}^2}=\frac{2(0.18)}{(1.25)^2}=0.23 kg

b)

The maximum speed in a spring-mass system is also given by

v_{max} =\sqrt{\frac{k}{m}} A

where

k is the spring constant

m is the mass

A is the amplitude

Here we have:

v_{max}=1.25 m/s is the maximum speed

m = 0.23 kg is the mass

A = 14.0 cm = 0.14 m is the amplitude

Solving for k, we find the spring constant

k=\frac{v_{max}^2}{A^2}m=\frac{1.25^2}{0.14^2}(0.23)=18.3 N/m

c)

The frequency in a spring-mass system is given by

f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}

where

k is the spring constant

m is the mass

In this problem, we have:

k = 18.3 N/m is the spring constant (found in part b)

m = 0.23 kg is the mass (found in part a)

Substituting and solving for f, we find the frequency of the system:

f=\frac{1}{2\pi}\sqrt{\frac{18.3}{0.23}}=1.42 Hz

d)

We can solve this part by using the law of conservation of energy; in fact, we have

E=PE+KE=\frac{1}{2}kx^2 + \frac{1}{2}mv^2

Where v is the speed of the system when the displacement is equal to x.

We know that the total energy of the system is

E = 0.18 J

Also we know that

k = 18.3 N/m is the spring constant

m = 0.23 kg is the mass

Substituting

x = 7.00 cm = 0.07 m

We can solve the equation to find the corresponding speed v:

v=\sqrt{\frac{2E-kx^2}{m}}=\sqrt{\frac{2(0.18)-(18.3)(0.07)^2}{0.23}}=1.08 m/s

#LearnwithBrainly

3 0
3 years ago
An electric field of 280000 N/C points due west at a certain spot. What is the magnitude of the force that acts on a charge of -
nordsb [41]
<h2>The magnitude of the force that acts on a charge of -7.9C at this spot is 2.21 x 10⁶ N.</h2>

Explanation:

Electric field is the ratio of force and charge.

Electric field, E = 280000 N/C

Charge, q = -7.9 C

We have

                 E=\frac{F}{q}\\\\280000=\frac{F}{7.9}\\\\F=280000\times 7.9\\\\F=2.21\times 10^6N

The magnitude of the force that acts on a charge of -7.9C at this spot is 2.21 x 10⁶ N.

4 0
4 years ago
4) Which of the following gases are typically used for colorful lighting when an electric current is applied ?
PIT_PIT [208]
The answer is B. Hope this Helps!
7 0
3 years ago
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