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4 years ago

The mechanical engery of an object equals its

1 answer:

4 0

The sum of an object's potential and kinetic energies is called the object's mechanical energy. As an object falls its potential energy decreases, while its kinetic energy increases. The decrease in potential energy is exactly equal to the increase in kinetic energy.

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Me - HS_SCLPhysical Science,

nalin [4]

Answer:

What's the question

Explanation:

3 0

4 years ago

Two horizontal forces are exerted on a large crate. The first force is 317 N to the right. The second force is 173 N to the left

DENIUS [597]

Answer:

A. Refer to the figure.

B. The net force acting on the crate is

$F_{net} = F_1 - F_2 = 317 - 173 = 144N$

C. We should first find the acceleration of the crate using the following kinematics equation:

$v = v_0 + at\\6.5 = 0 + 5a\\a = 1.3 m/s^2$

Mass of the crate can be found by Newton’s Second Law:

$F = Ma\\M = F/a = 144/1.3 = 110.7 kg$

7 0

4 years ago

Xander was traveling at a final velocity of 4.5 m/s for 3.5 seconds. Finley was traveling at a final velocity of 3.6 m/s for 4.2

Dima020 [189]

the answer is Finley, Xander, Max

5 0

3 years ago

Read 2 more answers

Two identical particles, each with a mass of 4.5 mg and a charge of 30 nC, are moving directly toward each other with equal spee

e-lub [12.9K]

Answer:

r₁ = 20.5 cm

Explanation:

In this exercise we can use the conservation of energy

the gravitational power energy is always attractive, the electrical power energy is repulsive if the charges are of the same sign

starting point.

Em₀ = U_g + U_e + K = $-G \frac{m_1m_2}{r} +k \frac{q_1q_2}{r} - 2 ( \frac{1}{2} m v^2)$

the two in the kinetic energy is because they are two particles

final point. When it is detained

Em_f = U_g + U_e = $-G \frac{m_1m_2}{r_1} + k \frac{q_1q_2}{r_1}$

the energy is conserved

Em₀ = em_f

the charges and masses of the two particles are equal

$-G \frac{m^2}{r} + k \frac{q^2}{r} + m v^2 = - G \frac{m^2}{r_1} + k \frac{q^2}{r_1}$

sustitute the values

-6.67-11 (4.5 10-3) ² / 0.25 - 9, 109 (30 10-9) ² / 0.25 + 4.5 10-3 4² = - 6.67 10- 11 (4.5 10-3) ² / r1 -9 109 (30 10-9) ² / r1

-5.4 10⁻¹⁵ + 3.24 10⁻⁵ - 7.2 10⁻⁵ = -1.35 10⁻¹⁵ / r₁ + 8.1 10⁻⁶ / r₁

We can see that the terms that correspond to the gravitational potential energy are much smaller than the terms of the electric power, which is why we depress them.

3.24 10⁻⁵ - 7.2 10⁻⁵ = 8.1 10⁻⁶ / r₁

-3.96 10⁻⁵ = 8.1 10⁻⁶ / r₁

r₁ = 8.1 10⁻⁶ /3.96 10⁻⁵

r₁ = 2.045 10⁻¹ m

r₁ = 20.5 cm

4 0

3 years ago

A ball rolled 12.0 m [E] in 10.0 s, hit an obstacle, and rolled straight back. After the collision, the ball rolled 8.00 m [W] i

Angelina_Jolie [31]

Answer:

the average velocity of the ball is 5 m/s.

Explanation:

Given;

initial position of the ball, x₁ = 12 m East

final position of the ball, x₂ = 8 m West

initial time of motion, t₁ = 10.0 s

final time of motion, t₂ = 6.0 s

The average velocity is calculated as follows;

$v = \frac{\Delta x}{\Delta t} = \frac{x_1 \ - \ x_2}{t_1 \ - \ t_2}$

let the Eastward direction be positive

Let the westward direction be negative

$\frac{x_1 \ - \ x_2}{t_1 \ - \ t_2} = \frac{12 \ - \ (-8)}{10 -6}= \frac{20}{4} = 5 \ m/s$

Therefore, the average velocity of the ball is 5 m/s.

5 0

3 years ago