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Allushta [10]
4 years ago
11

True or false a force cannot set a motionless object in motion

Physics
2 answers:
Xelga [282]4 years ago
7 0

Answer:

False

Explanation:

Anything that sits and does nothing will keep doing nothing until an outside force moves it.

Lyrx [107]4 years ago
4 0

If that statement is true, then NO motionless object could EVER start moving.

Yet I've seen several that did.  So it must be <em>false</em> .

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Caleb is investigating the effect of friction on the motion of an object. He uses the following supplies for the investigation:
Annette [7]
Part a
Place the ramps on the floor and time each ramp for time it takes for wooden block too reach thee floor. The times will be different and thus you can conclude that friction will increase the time
Part b
Caleb uses the same block as this is his control variable that he wants to keep same so that the results are correct and accurate
Part c
The independent variable is the thing you change before you test so it will be the surface on which the wooden block is sliding
The dependent variable is the variable you measure so that will be the time it takes for the block to slide to the bottom.
7 0
3 years ago
What is kinetic energy and what is net force formula
Over [174]

Answer:

Here's the equation for net force: F = ma. The work done on the plane, which becomes its kinetic energy, equals the following: Net force F equals mass times acceleration. Assume that you're pushing in the same direction that the plane is going; in this case, cos 0 degrees = 1, so.

Explanation:

In physics, the kinetic energy of an object is the energy that it possesses due to its motion. It is defined as the work needed to accelerate a body of a given mass from rest to its stated velocity. Having gained this energy during its acceleration, the body maintains this kinetic energy unless its speed changes

Hope this help also looking it up helps ;)

6 0
3 years ago
Which of the following statements about sidewall markings is correct? a. The load index is given as a letter. b. The traction an
vagabundo [1.1K]

Answer: all the above options are correct.

Explanation:

In sidewall markings,the load index is given as a letter,traction and temperature ratings are based on the speed rating of the tire,the tire's recommended inflation pressure and load are indicated and the DOT code indicates when and where the tire was made.

8 0
3 years ago
Suppose that the dipole moment associated with an iron atom of an iron bar is 2.8 × 10-23 J/T. Assume that all the atoms in the
masya89 [10]

To solve this exercise it is necessary to apply the equations related to the magnetic moment, that is, the amount of force that an image can exert on the electric currents and the torque that a magnetic field exerts on them.

The diple moment associated with an iron bar is given by,

\mu = \alpha *N

Where,

\alpha = Dipole momento associated with an Atom

N = Number of atoms

\alpha y previously given in the problem and its value is 2.8*10^{-23}J/T

L = 5.8cm = 5.8*10^{-2}m

A = 1.5cm^2 = 1.5*10^{-4}m^2

The number of the atoms N, can be calculated as,

N = \frac{\rho AL}{M_{mass}}*A_n

Where

\rho = Density

M_{mass} = Molar Mass

A = Area

L = Length

A_n =Avogadro number

N = \frac{(7.9g/cm^3)(1.5cm)(5.8cm^2)}{55.9g/mol}(6.022*10^{23}atoms/mol)

N = 7.4041*10^{23}atoms

Then applying the equation about the dipole moment associated with an iron bar we have,

\mu = \alpha *N

\mu = (2.8*10^{-23})*(7.4041*10^{23})

\mu = 20.72Am^2

PART B) With the dipole moment we can now calculate the Torque in the system, which is

\tau = \mu B sin(90)

\tau = (20.72)(2.2)

\tau = 45.584N.m

<em>Note: The angle generated is perpendicular, so it takes 90 ° for the calculation made.</em>

3 0
3 years ago
A tennis ball (m=0.060 kg) is moving horizontally at 20 m/s toward a tennis player who hits it straight back at 26 m/s. What is
Rus_ich [418]

Answer:

0.36 kg-m/s

Explanation:

Given that,

Mass of a ball, m = 0.06 kg

Initial velocity of the ball, u = 20 m/s

Final velocity of the ball, v = 26 m/s

We need to find the change in momentum of the tennis ball. It is equal to the final momentum minus initial momentum

\Delta p=m(v-u)\\\\=0.06\times (26-20)\\\\=0.36\ kg-m/s

So, the change in momentum of the ball is 0.36 kg-m/s.

4 0
3 years ago
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