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2 years ago

2 ways to change frictional force between 2 objects

2 answers:

8 0

It can be changed by shorter distances and by the amount of weight it has or the amount of force that is pushing that object to go however distance it can.

Zanzabum2 years ago

5 0

In order to change the frictional force between two solid surfaces, it can be changed by shorter distances and by the amount of weight it has or the amount of force that is pushing that object to go however distance it can.

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Which has more kinetic energy a bowling ball or a soccer ball

lawyer [7]

A bowling ball because it is heavier and it has more air force going against it

6 0

2 years ago

Read 2 more answers

Gases, just like liquids are made of particles which can be further classified as atoms or molecules,

olga_2 [115]

Answer:

True

Explanation:

Solids, Liquids, and Gases are all make up of atoms and molecules.

8 0

3 years ago

A wave in which particles of the medium move at right angles to the direction of the wave is called a

Licemer1 [7]

Answer:

transverse wave: A wave in which particles of the medium move at right angles to the direction of the wave is called a transverse wave.

5 0

3 years ago

A block rests on a frictionless table on Earth. After a 40-N horizontal force is applied to the block, it accelerates at 9.2 m/s

Paul [167]

Answer:

4.6 $\frac{m}{s^2}$

Explanation:

Since the table is frictionless, there is no force of dynamic friction between table an block when the horizontal force is applied to it on Earth. Exactly the same is true when the table is taken to the Moon. Therefore, the Net Force acting on the object in both cases when the object accelerates, is the external horizontal force.

Notice that on Earth and on the Moon, the weight of the object (vertical and pointing up) is compensated by the normal force of the table on the object (pointing up and of the same magnitude as the weight) that precludes movement in the vertical direction. So in both cases, its acceleration will only be due to the horizontal force.

We use the equation for Net Force to find the mass of the object:

$F=m*a\\40 N =m * 9.2 \frac{m}{s^2}\\\frac{40}{9.2} kg=m\\m=\frac{40}{9.2} kg$

We use this mass (since the mass of the object is a constant independent of where the object is) to find the acceleration the object will experience when the 20 N horizontal force is applied on it on the Moon:

$f=m*a\\20N=\frac{40}{9.2} kg*a\\20*9.2=40*a\\\frac{20*9.2}{40} =a\\a=\frac{9.2}{2} \frac{m}{s^2} \\ a=4.6 \frac{m}{s^2}$

3 0

3 years ago

The graph shows position versus time for an object moving with the constant acceleration. what is the final velocity of the obje

natali 33 [55]

Answer:

3.0 m/s

Explanation:

The equation of motion for constant acceleration (a) is ...

$x(t)=\dfrac{1}{2}at^2 +v_0t+x_0$

The problem statement tells us v₀ = 1, and we read from the graph that x₀=1. We also read from the graph that x(10) = 21. Filling these values into the equation, we can find a and x'(10).

$21 = \dfrac{1}{2}a(10^2)+1(10)+1\\\\10=50a\qquad\text{subtract 11}\\\\a=\dfrac{1}{5}\\\\x'(t)=at+v_0\\\\x'(10)=\dfrac{1}{5}\cdot 10+1=3.0$

The final velocity of the object at t=10 s is about 3.0 m/s.

_____

Comment on the graph

We note this graph better represents increasing acceleration than it does constant acceleration. x(2) = 3.4 per the equation. It is graphed as about 4.

8 0

3 years ago