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dexar [7]
2 years ago
7

A student pushes a block from rest across a frictionless surface while the block is in front of a motion detector for 3 seconds.

The block first speeds up for the 1st second, then moves at constant velocity for the 2nd second, and finally slows down during the final second.
Which statement best explains the motion of the block using Newton's 1st Law?
There is a net force on the block during the 2nd second as a net force keeps the motion constant.
There is a net force on the block during the 3rd second as a net force keeps the motion constant.
There is no net force on the block during the 2nd second as zero net force keeps the motion constant.
There is no net force on the block during the 3rd second as zero net force causes the motion to accelerate.​
Physics
1 answer:
Furkat [3]2 years ago
5 0

The true statement according to Newton's laws of motion is; "there is no net force on the block during the 2nd second as zero net force keeps the motion constant."

According to the Newton first law of motion, an object will continue in its state of rest or uniform motion unless it is acted upon by a net force. The action of a net force leads to an acceleration.

In the 2nd second of the journey, there is no net force on the block hence the block continues to move at constant velocity.

Learn more: brainly.com/question/13678295

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What do each of the variables mean ? <br> F=__________ m=_________<br>a=___________<br>​
wariber [46]

Answer:

F = Force (Measured in Newtons, N), m = Mass (Measured in kilograms, kg), and a = acceleration (Measured in metres per second squared, m/s^{2}

Explanation:

This is Newton's Second Law!

Hope this helps!

PLS mark as brainliest, hope this helps!

8 0
3 years ago
What line of code will call force with a value of 10 for mass and a value of 9.81 for acceleration?
Kaylis [27]

Line of code will call force with a value of 10 for mass and a value of 9.81 for acceleration is force(10, 9.81).

<h3 /><h3>Line of code for force and acceleration</h3>
  • In mechanics, acceleration refers to the rate at which an object's velocity with respect to time varies.
  • Acceleration is a vector quantity (in that they have magnitude and direction).
  • The direction of an object's acceleration is determined by the direction of the net force acting on it.
  • Newton's Second Law states that the combined effect of two factors determines how much an item accelerates.
  • The size of the net balance of all external forces acting on the object is, in accordance with the materials used to create it.
  • It inversely proportional to its mass, whereas the magnitude of the net resultant force is directly proportional to the net force.

def force(mass, acceleration):

force_val = mass*acceleration

return force_val

10 is assigned to mass and 9.81 is assigned to acceleration

def force(10, 9.81)

So, Line of code will call force with a value of 10 for mass and a value of 9.81 for acceleration is force(10, 9.81).

Learn more about acceleration here:

brainly.com/question/460763

#SPJ4

7 0
1 year ago
a tire is rolling along a road, without slipping with a velocity v. a piece of tape is attached to the tire. When the tape is op
Kryger [21]

Answer:

The right solution will be the "2v".

Explanation:

For something like an object underneath pure rolling the speed at any point is calculated by:

⇒  v_{rolling}=v_{translational}+v_{rotational}

Although the angular velocity was indeed closely linked to either the transnational velocity throughout particular instance of pure rolling as:

⇒  \omega=\frac{v_{translational}}{r}

Significant meaning is obtained, as speeds are in the same direction. Therefore the speed of rotation becomes supplied by:

⇒  v_{rotational}=\omega \times r

On substituting the estimated values, we get

⇒                   =\frac{v_{translational}}{r} \times r

⇒                   =v_{translational}

So that the velocity will be:

⇒  v_{rolling}=v+v

⇒              =2v

4 0
3 years ago
(hrw8c10p78) Two uniform solid spheres have the same mass, 1.65 kg, but one has a radius of 0.206 m while the other has a radius
Burka [1]

Complete question is:

Two uniform solid spheres have the same mass of 1.65 kg, but one has a radius of 0.206 m and the other has a radius of 0.804 m. Each can rotate about an axis through its center. (a) What is the magnitude τ of the torque required to bring the smaller sphere from rest to an angular speed of 367 rad/s in 14.5 s? (b) What is the magnitude F of the force that must be applied tangentially at the sphere’s equator to give that torque? What are the corresponding values of (c) τ and (d) F for the larger sphere?

Answer:

A) τ = 0.709 N.m

B) F = 3.44 N

C) τ = 10.8 N.m

D) F = 13.43N

Explanation:

We are given;

Mass if each sphere = 1.65kg

Radius of the first sphere; r1 = 0.206m

Radius of second sphere; r2 = 0.804m

A) initial angular speed of smaller sphere; ω_i = 0 rad/s

Final angular speed of smaller sphere; ω_f = 367 rad/s

Time;t = 14.5 s

The constant angular acceleration is calculated from;

ω_f = ω_i + αt

367 = 0 + α(14.5)

Thus,

α = 367/14.5 = 25.31 rad/s²

The torque is given by the formula;

τ = Iα

Where τ is torque ; I is moment of inertia given as (2/5)Mr²

α is angular acceleration

Thus;

τ = (2/5)(1.65)(0.206)² x 25.31

τ = 0.709 N.m

B) The magnitude of the force that must be applied to give the torque τ is gotten from the formula;

τ = F•r•sin90°

0.709 = F x 0.206 x 1

F = 0.709/0.206

F = 3.44 N

C) Now for the larger sphere, we'll repeat the same procedure in a above. Thus;

The torque is given by the formula;

τ = Iα

Where τ is torque ; I is moment of inertia given as (2/5)Mr²

α is angular acceleration

Thus;

τ = (2/5)(1.65)(0.804)² x 25.31

τ = 10.8 N.m

D) Now for the larger sphere, we'll repeat the same procedure in b above. Thus, τ is gotten from the formula;

τ = F•r•sin90°

10.8 = F x 0.804 x 1

F = 10.8/0.804

F = 13.43N

6 0
3 years ago
How can you make work output of a machine greater than the work input?
I am Lyosha [343]
Input current passing easy to can you make work output of a machine greater than the work
6 0
3 years ago
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