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devlian [24]
3 years ago
9

The change in momentum for an object is equal to

Physics
1 answer:
koban [17]3 years ago
4 0

Answer:

We conclude that the change in momentum of a body is equal to the impulse experienced by a body.

Explanation:

Considering the equation

F • t = m • Δ v

Here,

m • Δ v is basically a change in momentum of a body which is equal to the mass of the object multiplied by the change in its velocity.

Also,

  • F • t is called the impulse of the object.

In the formula, it is clear that the impulse experienced by a body during the collision is basically a change in the momentum of the body.

In other words, the change in momentum of a body is equal to the impulse experienced by a body.

Therefore, we conclude that the change in momentum of a body is equal to the impulse experienced by a body.

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A 15 kg mass is moving at 7.50 meters per second on a horizontal, frictionless surface. What is the total work that must be done
sashaice [31]
Kinetic energy = (1/2) (mass) x (speed)²

At 7.5 m/s, the object's KE is (1/2) (7.5) (7.5)² = 210.9375 joules

At 11.5 m/s, the object's KE is (1/2) (7.5) (11.5)² = 495.9375 joules

The additional energy needed to speed the object up from 7.5 m/s
to 11.5 m/s is (495.9375 - 210.9375) = <em>285 joules</em>.

That energy has to come from somewhere. Without friction, that's exactly
the amount of work that must be done to the object in order to raise its
speed by that much.
8 0
3 years ago
What can you use to determine whether the SATA port no which you are connecting the drive will also run at 6.0Gbps?
kolbaska11 [484]

Answer:

The answer is "use manual motherboard".

Explanation:

The motherboard is also known as the mainboard, it an electronic circuit board, that can connect with the CPU, RAM, and other networking equipment parts. It is also is known as a chipset, that differ widely in style, context, power source, height and performance (Form Factor).

All the data of the computer is stored memory, which checks into the motherboard, that the SATA port which you are connected to is still going to run at 6.0Gbps or not.

8 0
3 years ago
I need help with this physics question
insens350 [35]
It's impossible to describe WHERE a place is without mentioning ANOTHER place.

... Across the street from -- the bank.
... Next door to -- my house.
... 30 miles west of -- Chicago.
... Up above -- the tree.
... Two days ride out of -- Tulsa.
... Halfway home from -- school.
... Twice as far from Earth as -- the moon is.
... The first seat in -- the second row.
... Behind -- the dog's left ear.
... At the bottom of -- the pool.
... On the tip of -- my tongue.
... In the front seat of -- the car.
... I saw you in -- my dream.
... You're always on -- my mind.

The question is trying to get you to realize that to get from a reference point to a certain position, you have to know

How far
and
In what direction.
4 0
4 years ago
The bird is held in level flight due to the force exerted on it by the air as the bird beats its wings. What is the maximum valu
Cerrena [4.2K]

Answer:

 maximumforce is F = mg

Explanation:

For this case we must use Newton's second law,

     Σ F = m a

bold indicate vectors, so we will write it in its components x and y

 X axis

       Fₓ = maₓ

 Axis y

      Fy - W = m aa_{y}

Now let's examine our case, with indicate that the bird is level, the force of the wings can have a measured angle with respect to the x axis, where the vertical component is responsible for the lift, let's use trigonometry to find the components

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      Fₓ = F cos θ

      sin θ = Fy / F

      Fy = F sin θ

Let's replace and calculate

      F sin θ -w = m a

 

As the bird indicates that leveling at the same height, so the vertical acceleration is zero (ay = 0)

       F sin θ = w = mg

The maximum value of this equation occurs when the sin=1, in this case

      F = mg

3 0
3 years ago
A car wieghing 10,000 newtons is parked in a garage. what is true about the forces acting upon it
choli [55]
Since the car is at rest, the force experienced by car will be normal face(exerted by surface to which car is in contact) and weight of the car.
As the car is at rest, net force on the car should be zero.
4 0
4 years ago
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