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3 years ago

What wa can friction do like brake pads

1 answer:

4 0

Friction is the force in which two objects collide with each other to Dollie down the object, surfaces such as oil and wood floors have less friction than surfaces such as carpet and grass. Brake pads are used to slow objects such as cars quickly so that it doesn't take 2 minutes for a 20 mph car to slow down naturally

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I WILL GIVE BRAINLIEST IF SOMEONE GETS THIS......

pav-90 [236]

Answer:

Explanation:

Firstly to calculate the total mass of the can before the metal was lowered we need to add the mass of the eureka can and the mass of the water in the can. We don't know the mass of the water but we can easily find if we know the volume of the can. In order to calculate the volume we would have to multiply the area of the cross section by the height. So we do the following.

100 $cm^{2}$ x 10cm = 1000 $cm^{3}$

Now in order to find the mass that water has in this case we have to multiply the water's density by the volume, and so we get....

$\frac{1g}{cm^{3} }$ x 1000 $cm^{3}$ = 1000g or 1kg

Knowing this, we now can calculate the total mass of the can before the metal was lowered, by adding the mass of the water to the mass of the can. So we get....

1000g + 100g = 1100g or 1.1kg

The volume of the water that over flowed will be equal to the volume of the metal piece (since when we add the metal piece, the metal piece will force out the same volume of water as itself, to understand this more deeply you can read the about "Archimedes principle"). Knowing this we just have to calculate the volume of the metal piece an that will be the answer. So this time in order to find volume we will have to divide the total mass of the metal piece by its density. So we get....

20g ÷ $\frac{8g}{cm^{3} }$ = 2.5 $cm^{3}$

Now to find out the total mass of the can after the metal piece was lowered we would have to add the mass of the can itself, mass of the water inside the can, and the mass of the metal piece. We know the mass of the can, and the metal piece but we don't know the mass of the water because when we lowered the metal piece some of the water overflowed, and as a result the mass of the water changed. So now we just have to find the mass of the water in the can keeping in mind the fact that 2.5 $cm^{3}$ overflowed. So now we the same process as in number a) just with a few adjustments.

$\frac{1g}{cm^{3} }$ x (1000 $cm^{3}$ - 2.5 $cm^{3}$ ) = 997.5g

So now that we know the mass of the water in the can after we added the metal piece we can add all the three masses together (the mass of the can. the mass of the water, and the mass of the metal piece) and get the answer.

100g + 997.5g + 20g = 1117.5g or 1.1175kg

5 0

3 years ago

A satellite in Earth orbit has a mass of 100 kg and is at an altitude of 2.00 × 10⁶m. (c) What If? What force, if any, does the

Montano1993 [528]

The satellite exerts the same force as the Earth of magnitude 569.3N upward direction.

A physical body's overall composition is measured by its mass. As a measure of the body's resistance to acceleration (change in velocity) in the presence of a net force, it also measures the body's inertia. The mass of a thing also affects how strongly it attracts other bodies through gravity.

The SI unit of mass exists the kilogramme (kg). In the realm of science and technology, the weight of a body in a certain reference frame is the force that pushes it toward an acceleration that is equal to the local acceleration of free fall in that frame.

The satellite exerts the same force as the Earth of magnitude 569.3N upward direction.

To learn more about force, refer to:

brainly.com/question/25573309

#SPJ4

3 0

2 years ago

A 143-g baseball is flying through the air with a speed of 180 km/hr just after it is hit by a bat. If its velocity is at an ang

Svetllana [295]

To solve the problem it is necessary to apply the concepts related to the conservation of linear Moment, that is to say

$P=mv$

Where,

m = Mass

v = Velocity

P = Linear momentum

For the given data we have to:

$v_1 = 180\frac{km}{h}(\frac{1h}{3600s})(\frac{1000m}{1km})$

$v_1 = 50m/s$

The components of this force would be given by,

$v_z = 50*sin37 = 30.09m/s$

$v_{x,y} = 50*cos37=39.93m/s \rightarrw$ According to the definition given at the end of the problem, this component corresponds to that expressed for x and y.