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Sholpan [36]
2 years ago
4

Students connect a spring scale to a block on a rough horizontal surface. The students use the spring scale to measure the magni

tude of the horizontal force needed to pull the block at a constant speed. Which of the following statements explains why two forces exerted between objects ar equal in magnitude?
a. The gravitational and normal forces exerted on the block, because they are a Newton's third-law pair.
b. The frictional force and force exerted by the spring scale on the block, because they are a Newton's third-law pair.
c. The normal force and the frictional force exerted on the block, because objects always exert forces of equal magnitude on each other.
d. The frictional forces that the block and the surface exert on each other, because objects always exert forces of equal magnitude on each other.
Physics
1 answer:
timofeeve [1]2 years ago
6 0

For there to be a reaction there must also be action. In the horizontal movement there is a balance in which the magnitude of the Forces in opposite directions must be in total 0. The only horizontal forces are friction and the force exerted by the spring scale. For this reason the correct answer is B:

<em>The frictional force and force exerted by the spring scale on the block, because they are a Newton's third-law pair. </em>

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Answer:

Our solar system has total eight planets out of which four are inner planets and four are outer planets. The four outer planets are Jupiter, Saturn, Uranus and Neptune. The common characteristics of outer planets is that they are gaseous planets. They are larger on size than the inner rocky planets and are faraway from Sun. They have larger period of revolution around the Sun.

Uranus is a gaseous planet and lies far from Sun and hence has large period of revolution. It takes 84 Earth years to revolve around Sun. This data indicates that Uranus resides in the outer region of the Solar System.

4 0
2 years ago
As a person pushes a box across a floor, the energy from the person’s moving arm is transferred to the box, and the box and the
san4es73 [151]

Answer:

conserved

Explanation:

During this process the energy is conserved

3 0
2 years ago
Read 2 more answers
A 60.0-kg mass person wishes to push a 120-kg mass box across a level floor. The coefficient of static friction between the pers
Aneli [31]

Answer:

μ = 0.350

Explanation:

For the person to able to move the box, the force exerted by the person on the box must equal the force exerted by the box:

F_{p} = F_{b}

In this case, force can be calculated as a product of mass (m) by the acceleration of gravity (g) and the coefficient of static friction (μ):

m_{p}*g*\mu_{p}=m_{b}*g*\mu_{b}\\m_{p}*\mu_{p}=m_{b}*\mu_{b}\\60*0.7=120*\mu_{b}\\\mu_{b}= 0.35

Therefore, for the person to be able to push the box horizontally, the coefficient of static friction between the box and the floor should not be higher than 0.350.

8 0
2 years ago
A 250 GeV beam of protons is fired over a distance of 1 km. If the initial size of the wave packet is 1 mm, find its final size
Margarita [4]

Answer:

The final size is approximately equal to the initial size due to a very small relative increase of 1.055\times 10^{- 7} in its size

Solution:

As per the question:

The energy of the proton beam, E = 250 GeV =250\times 10^{9}\times 1.6\times 10^{- 19} = 4\times 10^{- 8} J

Distance covered by photon, d = 1 km = 1000 m

Mass of proton, m_{p} = 1.67\times 10^{- 27} kg

The initial size of the wave packet, \Delta t_{o} = 1 mm = 1\times 10^{- 3} m

Now,

This is relativistic in nature

The rest mass energy associated with the proton is given by:

E = m_{p}c^{2}

E = 1.67\times 10^{- 27}\times (3\times 10^{8})^{2} = 1.503\times 10^{- 10} J

This energy of proton is \simeq 250 GeV

Thus the speed of the proton, v\simeq c

Now, the time taken to cover 1 km = 1000 m of the distance:

T = \frac{1000}{v}

T = \frac{1000}{c} = \frac{1000}{3\times 10^{8}} = 3.34\times 10^{- 6} s

Now, in accordance to the dispersion factor;

\frac{\delta t_{o}}{\Delta t_{o}} = \frac{ht_{o}}{2\pi m_{p}\Delta t_{o}^{2}}

\frac{\delta t_{o}}{\Delta t_{o}} = \frac{6.626\times 10^{- 34}\times 3.34\times 10^{- 6}}{2\pi 1.67\times 10^{- 27}\times (10^{- 3})^{2} = 1.055\times 10^{- 7}

Thus the increase in wave packet's width is relatively quite small.

Hence, we can say that:

\Delta t_{o} = \Delta t

where

\Delta t = final width

3 0
2 years ago
23. While sliding a couch across a floor, Andrea and Jennifer exert forces F → A and F → J on the couch. Andrea’s force is due n
PSYCHO15rus [73]

Answer:

a)  (95.4 i^ + 282.6 j^) N , b) 298.27 N  71.3º and c)   F' = 298.27 N   θ = 251.4º

Explanation:

a) Let's use trigonometry to break down Jennifer's strength

      sin θ = Fjy / Fj

      cos θ = Fjx / Fj

Analyze the angle is 32º east of the north measuring from the positive side of the x-axis would be

          T = 90 -32 = 58º

         Fjy = Fj sin 58

         Fjx = FJ cos 58

         Fjx = 180 cos 58 = 95.4 N

         Fjy = 180 sin 58 = 152.6 N

Andrea's force is

         Fa = 130.0 j ^

We perform the summary of force on each axis

X axis

       Fx = Fjx

       Fx = 95.4 N

Axis y

       Fy = Fjy + Fa

       Fy = 152.6 + 130

       Fy = 282.6 N

       F = (95.4 i ^ + 282.6 j ^) N

b) Let's use the Pythagorean theorem and trigonometry

       F² = Fx² + Fy²

       F = √ (95.4² + 282.6²)

       F = √ (88963)

       F = 298.27 N

       tan θ = Fy / Fx

       θ = tan-1 (282.6 / 95.4)

       θ = tan-1 (2,962)

       θ = 71.3º

c) To avoid the movement they must apply a force of equal magnitude, but opposite direction

       F' = 298.27 N

       θ' = 180 + 71.3

       θ = 251.4º

4 0
2 years ago
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