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bogdanovich [222]
3 years ago
15

An object’s mass increases its

Physics
1 answer:
STatiana [176]3 years ago
4 0

Answer:  

It is commonly known that, if you accelerate an object, its mass will increase; however, to understand why this phenomenon occurs, we mustn’t think of the object’s mass increasing. Instead, we should think of its energy. In physics, mass is just simply locked up energy. We call this type of mass, ‘inertial mass.’

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Help meeeeeeeeeeeeee! please show work! In a collision, Kara Less who was traveling at 11 m/s while texting in her 1300 kg car,
raketka [301]

The impulse experienced is -18,000 kg m/s

Explanation:

The impulse exerted on an object is equal to the change in momentum of the object. Mathematically:

I=\Delta p = m(v-u)

where

m is the mass of the object

v is the final velocity of the object

u is the initial velocity

\Delta p is the change in momentum

I is the impulse

In the collision in this problem,

m = 1300 kg is the mass of the car

u = 11 m/s is the initial velocity

v = -2.5 m/s is the final velocity (negative, since it is in the opposite direction)

Substituting, we find

I=(1300)(-2.5-11)=-17,550 kg m/s

So the closest choice is

-18,000 kg m/s

Learn more about impulse and change in momentum:

brainly.com/question/9484203

#LearnwithBrainly

7 0
3 years ago
Which change will always result in an increase in the gravitational force between two objects?
Anika [276]
The change that will always result in an increase in the gravitational force between two objects is increasing the masses of the objects and decreasing the distance between the objects.
8 0
3 years ago
Read 2 more answers
What is the uncertainty of the position of the bacterium? express your answer with the appropriate units?
lbvjy [14]
For two un-related quantities, the Heisenberg uncertainty equations holds: the prduct of the two uncertainty quantities is greater than \hbar/2
Example of unrelated quantities are position and momentum, energy and time. 
Thus
\Delta x*\Delta p  \ \textgreater \ \hbar/2
Knowing the speed of the bacteria the uncertainty in its position is
\Delta x \ \textgreater \ \hbar/(2 \Delta p) =\hbar/(2mv)
4 0
3 years ago
Which type of experiment involves changing only one variable at a time
BlackZzzverrR [31]
Controlled Experiment 

6 0
3 years ago
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Help mee pleaseee :)))
Anettt [7]

Answer:

See the explanation below.

Explanation:

Solving the first image question:

C ) The resulting force is defined by Newton's second law which tells us that the sum of the forces on a body is equal to the product of mass by acceleration. That is, there must be a force that acts on a body to produce an acceleration. If there is no acceleration it is because there are no external forces or developed by the body. And if there is no acceleration the body moves at a constant speed, in a straight line, so the response is C.

For the second image, we must remember that weight is defined as the product of mass by gravitational acceleration.

W = m*g

where:

W = weight [N]

m = mass [kg]

g = gravity acceleration [m/s²]

Now we have

m = 50 [kg]

ge = Earth gravity acceleration = 10 [m/s²]

gp = Distant planet gravity acceleration = 4 [m/s²]

We = ge*m

We = 10*50 = 500 [N]

Wp =gp*m

Wp = 4*50 = 200 [N]

Therefore the answer is D

For the third image, The mass is always going to be preserved, regardless of where the body or object is in space, its weight is the only one that changes since the gravitational force is modified. That is, the mass on the moon and on Earth will always be the same.

m = 70 [kg]

First, we must calculate the acceleration, by means of the following equation of kinematics.

v_{f} =v_{o} +a*t

where:

Vf = final velocity = 20 [m/s]

Vo = initial velocity = 0 (because stars from the rest)

a = acceleration [m/s²]

t = time = 4 [s]

20 = 0 + a*4

20 = 4*a

a = 5 [m/s²]

Now using Newton's second law which tells us that the total force acting on a body is equal to the product of mass by acceleration.

F = m*a

where:

F = force [N] (units of Newtons)

m = mass = 2 [kg]

a = acceleration = 5 [m/s²]

F = 2*5

F = 10 [N]

The body of Figure D, since a total force of 25 [N] to the left acts on it, in the rest of cases the force is zero or much less than 25 [N]

50 + 40 - 35 - 30 = F

F = 25 [N]

8 0
3 years ago
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