A 3500kg suv is driving north at 36m/s what the momentum

Three light bulbs are connected to a battery in a series circuit. How will the bulbs behave if the circuit is closed?

stiv31 [10]

all three will glow A

6 0

2 years ago

A 1500 kg car traveling at 80.0 km/h comes to a screeching halt in a time of 4.00 seconds. Calculate the force of friction exper

amm1812

Answer:

F=m x a

(F is force ,M is mass and A is acceleration)

in thisncase the Mass is given but we need to find ou the acceleration

Formula for acceleration-

a=(v - u)/t

(v is final velocity , u is initiatal velocity and t is time)

a = (0 - 80)/4

a= -80/4

a= -20

By substituting the values-

F= m x a

F= 1500 x -20

F=-30000N

Thus the force acted is -30000N

hope this helps

7 0

2 years ago

Use the fact that 1inch=2.54cm and convert 52 inches into the equivalent length in centimeters.

Lelu [443]

Answer:

To convert inches to centimeters, use an easy formula and multiply the length by the conversion ratio.

Since one inch is equal to 2.54 centimeters, this is the inches to cm formula to conver

Explanation:

4 0

1 year ago

The law of conservation of momentum states that the total momentum of interacting objects does not change . This means the total

pickupchik [31]

Answer:

The momentum of an object is equal to the product of its mass and its velocity.

Explanation:

Consider an object of mass $m$ travelling at a velocity $\vec{v}$ . The momentum $\vec{p}$ of this object would be:

$\vec{p} = m \cdot \vec{v}$ .

For the law of conservation of momentum, consider two objects: object $\rm a$ and object $\rm b$ . Assume that these two objects collided with each other.

Let $m_{\rm a}$ and $m_{\rm b}$ denote the mass of the two objects.
Let $\vec{v}_{\rm a}(\text{initial})$ and $\vec{v}_{\rm b}(\text{initial})$ denote the velocity of the two object right before the interaction.
Let $\vec{v}_{\rm a}(\text{final})$ and $\vec{v}_{\rm b}(\text{final})$ denote the velocity of the two objects right after the interaction.

The momentum of the two objects right before the collision would be $m_{\rm a}\cdot \vec{v}_{\rm a}(\text{initial})$ and $m_{\rm b}\cdot \vec{v}_{\rm b}(\text{initial})$ , respectively.
The momentum of the two objects right after the collision would be $m_{\rm a}\cdot \vec{v}_{\rm a}(\text{final})$ and $m_{\rm b}\cdot \vec{v}_{\rm b}(\text{final})$ , respectively.

The sum of the momentum of the two objects would be:

$m_{\rm a}\cdot \vec{v}_{\rm a}(\text{initial}) + m_{\rm b}\cdot \vec{v}_{\rm b}(\text{initial})$ right before the collision, and
$m_{\rm a}\cdot \vec{v}_{\rm a}(\text{final}) + m_{\rm b}\cdot \vec{v}_{\rm b}(\text{final})$ right after the collision.

Assume that the system of these two objects is isolated. By the law of conservation of momentum, the sum of the momentum of these two objects should be the same before and after the collision. That is:

$m_{\rm a}\cdot \vec{v}_{\rm a}(\text{initial}) + m_{\rm b}\cdot \vec{v}_{\rm b}(\text{initial}) = m_{\rm a}\cdot \vec{v}_{\rm a}(\text{final}) + m_{\rm b}\cdot \vec{v}_{\rm b}(\text{final})$ .

4 0

3 years ago

When mantle rocks near the radioactive core are heated, they become less dense than the cooler, upper mantle rocks. These warmer

avanturin [10]

C.convection cells is the right answer

4 0

2 years ago

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