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3 years ago

Property of a moving object equal to its mass times it velocity

Physics

1 answer:

Ludmilka [50]3 years ago

7 0

Answer:

Momentum

Explanation:

The property of a moving object having mass m and moving with velocity v is called its momentum. It is given by the mathematical form as follows :

$p=m\times v$

It is a vector quantity. Its SI unit of kg-m/s.

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URGENT Find the total equivalent resistance for the circuit.

Nesterboy [21]

Answer:

$20 + 10 = 30 \\ 30and50 in \: parallel \\ = \frac{50 \times 30}{50 + 30} \\ = \frac{1500}{80} \\ 18.75 \\ 30nd40parallel \\ \frac{30 \times 40}{70} \\ \frac{1200}{70} = 17.143 \\ total = 18.75 + 17.143 = 35.893ohm \\ thank \: u$

3 0

3 years ago

A 0. 060-kg tennis ball, moving with a speed of 5. 82 m/s , has a head-on collision with a 0. 090-kg ball initially moving in th

inn [45]

Final speed of the tennis ball, moving with a speed of 5. 82 m/s , has a head-on collision with a 0. 090-kg ball is 2.964 m/s.

<h3>What is conservation of momentum?</h3>

Momentum of an object is the force of speed of it in motion. Momentum of a moving body is the product of mass times velocity. By the law of conservation of momentum,

$m_1u_1 + m_2u_2 = (m_1+m_2)v$

Here, (m) is the mass, (u) is initial velocity before collision, v is final velocity after collision and (subscript 1, and 2) are used for body 1 and 2 respectively. Rewrite the formula for final velocity as,

$v=\dfrac{m_1u_1 + m_2u_2}{(m_1+m_2)}$

A 0. 060-kg tennis ball, moving with a speed of 5. 82 m/s, has a head-on collision with a 0. 090-kg ball, initially moving in the same direction at a speed of 3.44 m/s. Thus, the initial velocity of the second ball is,

$v_{2f}=5.82+3.44+v_{1f}\\v_{2f}=2.38+v_{1f}$

Let v1f is the final velocity of first ball. Thus, the initial velocity of the first ball is,

$v_{1f}=\dfrac{(0.060)(5.82) + (0.090)(3.44-2.38)}{(0.060)+(0.090)}\\v_{1f}=2.964\rm\; m/s$

Thus, final speed of the tennis ball, moving with a speed of 5. 82 m/s , has a head-on collision with a 0. 090-kg ball is 2.964 m/s.

Learn more about the conservation of momentum here;

brainly.com/question/7538238

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4 0

2 years ago

Students perform a set of experiments by placing a block of mass m against a spring, compressing the spring a distance x along a

Verizon [17]

Increasing the angle of inclination of the plane decreases the velocity of the block as it leaves the spring.

The statement that indicates how the relationship between v and x changes is; As x increases, v increases, but the relationship is no longer linear and the values of v will be less for the same value of x.

Reasons:

The energy given to the block by the spring = $\mathbf{0.5 \cdot k \cdot x^2}$

According to the principle of conservation of energy, we have;

On a flat plane, energy given to the block = $0.5 \cdot k \cdot x^2$ = kinetic energy of

block = $0.5 \cdot m \cdot v^2$

Therefore;

0.5·k·x² = 0.5·m·v²

Which gives;

x² ∝ v²

x ∝ v

On a plane inclined at an angle θ, we have;

The energy of the spring = $\mathbf{0.5 \cdot k \cdot x^2}$

The force of the weight of the block on the string, $F = m \cdot g \cdot sin(\theta)$

The energy given to the block = $0.5 \cdot k \cdot x^2 - m \cdot g \cdot sin(\theta)$ = The kinetic energy of block as it leaves the spring = $\mathbf{0.5 \cdot m \cdot v^2}$

Which gives;

$0.5 \cdot k \cdot x^2 - m \cdot g \cdot sin(\theta) = 0.5 \cdot m \cdot v^2$

Which is of the form;

a·x² - b = c·v²

a·x² + c·v² = b

Where;

a, b, and c are constants

The graph of the equation a·x² + c·v² = b is an ellipse

Therefore;

As x increases, v increases, however, the value of v obtained will be lesser than the same value of x as when the block is on a flat plane.

Please find attached a drawing related to the question obtained from a similar question online

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