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3 years ago

8. A person with a mass of 15 kg is walking on a flat surface at a velocity of 5 m/s. What is the walker's momentum?

Physics

1 answer:

Alex3 years ago

3 0

Answer: The correct answer is 75 kg m/s.

Explanation:

The expression of the momentum in terms of velocity and mass is as follows;

p=mv

Here, m is the mass of the object and v is the velocity.

It is given in the problem that a person with a mass of 15 kg is walking on a flat surface at a velocity of 5 m/s.

Calculate the walker's momentum.

Put m= 15 kg and v= 5 m/s in the above expression.

p=(15)(5)

p= 75 kg m/s

Therefore, the momentum of the speedboat is 75 kg m/s.

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7nadin3 [17]

Answer:

<h2>C. 0.55 m/s towards the right</h2>

Explanation:

Using the conservation of law of momentum which states that the sum of momentum of bodies before collision is equal to the sum of the bodies after collision.

Momentum = Mass (M) * Velocity(V)

BEFORE COLLISION

Momentum of 0.25kg body moving at 1.0m/s = 0.25*1 = 0.25kgm/s

Momentum of 0.15kg body moving at 0.0m/s(body at rest) = 0kgm/s

AFTER COLLISION

Momentum of 0.25kg body moving at x m/s = 0.25* x= 0.25x kgm/s

x is the final velocity of the 0.25kg ball

Momentum of 0.15kg body moving at 0.75m/s(body at rest) =

0.15 * 0.75kgm/s = 0.1125 kgm/s

Using the law of conservation of momentum;

0.25+0 = 0.25x + 0.1125

0.25x = 0.25-0.1125

0.25x = 0.1375

x = 0.1375/0.25

x = 0.55m/s

Since the 0.15 kg ball moves off to the right after collision, the 0.25 kg ball will move at 0.55 m/s towards the right

4 0

2 years ago

Could anyone help with this question I do

ra1l [238]

which of those can we not get back once it has been used, the answer is oil or petroleum

8 0

3 years ago

Read 2 more answers

Based on the following equation, answer the questions below. ρ = (2γϕ + ψ)/rg where ρ [=] moles per cubic foot [mol/ft3] γ [=] j

AlekseyPX

1) Fundamental units of $\Psi$ are $[\frac{mol}{m\cdot s^2}]$

2) Fundamental units of $\Phi$ are $[\frac{mol}{m^3}]$

Explanation:

The equation for the variable $\rho$ is

$\rho =\frac{2\gamma \Phi+\Psi}{rg}$

where we have:

$\rho$ measured in $[\frac{mol}{ft^3}]$

$\gamma$ measured in $[\frac{J}{kg}]$

$r$ measured in $[in]$

$g$ measured in $[\frac{m}{s^2}]$

We can re-write the equation as

$\rho rg = 2\gamma \Phi + \Psi$

And we notice that the units of the term on the left must be equal to the units of the term on the right.

This means that:

1) First of all, $\Psi$ must have the same units of $\rho r g$ . So,

$[\rho r g]=[\frac{mol}{ft^3}][in][\frac{m}{s^2}]$

However, both ft (feet) and in (inches) are not fundamental dimensions: this means that they can be expressed as meters. Therefore, the fundamental units of $\Psi$ are

$[\Psi]=[\frac{mol}{m^3}][m][\frac{m}{s^2}]=[\frac{mol}{m\cdot s^2}]$