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Kazeer [188]
3 years ago
14

What exactly is meant by a "freely falling" object? (i.e. what is the definition of free fall?)?

Physics
1 answer:
Jobisdone [24]3 years ago
5 0
Free fall is when an object is moving down with only the force of gravity.
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The -is the process by which gases in the atmosphere absorb and reradiate heat?
Nikolay [14]
Water vapor and carbon dioxide

3 0
2 years ago
1. What is the kinetic energy of a 1.75 kg ball travelling at a speed of 54 m/s?
Over [174]

Answer:

We conclude that the kinetic energy of a 1.75 kg ball traveling at a speed of 54 m/s is 2551.5 J.

Explanation:

Given

  • Mass m = 1.75 kg
  • Velocity v = 54 m/s

To determine

Kinetic Energy (K.E) = ?

We know that a body can possess energy due to its movement — Kinetic Energy.

Kinetic Energy (K.E) can be determined using the formula

K.E=\frac{1}{2}mv^2

where

  • m is the mass (kg)
  • v is the velocity (m/s)
  • K.E is the Kinetic Energy (J)

now substituting m = 1.75, and v = 54 in the formula

K.E=\frac{1}{2}mv^2

K.E=\frac{1}{2}\left(1.75\right)\left(54\right)^2

K.E=1458\times 1.75

K.E=2551.5 J

Therefore, the kinetic energy of a 1.75 kg ball traveling at a speed of 54 m/s is 2551.5 J.

7 0
2 years ago
Sketch the resultant field pattern around the following current carrying conductors and
Novay_Z [31]
Cccccccvvvvvcjjjjjjjjjjkllk ki e
8 0
2 years ago
A sealed box contains a monatomic ideal gas. The number of gas atoms per unit volume is 5.00 * 1020 atoms>cm3, and the averag
Veronika [31]

Answer:

Pressure,P=6×10^3Pa

Explanation:

The gas has an ideal gas behaviour and ideal gas equation

PV=NKT

T= V/N p/K ...eq1

Average transitional kinetic energy Ktr=1.8×10-23J

Ktr=3/2KT

T=2/3Ktr/K....eq2

Equating eq1 and 2

V/N p/K = 2/3Ktr/K

Cancelling K on both sides

P= 2/3N/V( Ktr)

Substituting the value of N/V and dividing by 10^-6 to convert cm^3 to m^3

P = 2/3 (5.0×10^20)/10^-6 × 1.8×10^-23

P= 6 ×10^3Pa

6 0
3 years ago
Select the correct answer.
andrew-mc [135]
I think the answer would be C
7 0
3 years ago
Read 2 more answers
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