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irina [24]
3 years ago
15

In pushing a heavy box across the floor, is the force you need to apply to start the box moving greater than, less than, or the

same as the force needed to keep the box moving? On what are you basing your choice?
How do you think the force of friction is related to the weight of the box? Explain.
Physics
1 answer:
Veseljchak [2.6K]3 years ago
3 0

Answer:

Explanation:

Force needed to apply start the box is greater than the force needed to keep it moving because static friction is greater than the kinetic friction .

A  threshold force is needed to move the box and when box started to move kinetic friction comes into play.      

Friction force is directly related to the weight of the box as the friction force is

coefficient of friction time Normal reaction .

And Normal reaction is equal to the weight of box if no force is applied.

f_r=\mu N

N=mg

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juin [17]
Altitude is the angle measured above the horizon
5 0
3 years ago
Read 2 more answers
A very small object with mass 8.30×10-9 kg and positive charge 6.90×10-9 C is projected directly toward a very large insulating
Rainbow [258]

Answer:

41.4496148484\ m/s

Explanation:

\epsilon_0 = Permittivity of free space = 8.85\times 10^{-12}\ F/m

\sigma = Surface charge density = 5.9\times 10^{-8}\ C/m^2

\Delta x = 0.57-0.26

q = Charge = 6.9\times 10^{-9}\ C

m = Mass of object = 8.3\times 10^{-9}\ kg

Electric field due to a sheet is given by

E=\dfrac{\sigma}{2\epsilon_0}\\\Rightarrow E=\dfrac{5.9\times 10^{-8}}{2\times 8.85\times 10^{-12}}\\\Rightarrow E=3333.33\ V/m

Electric field is given by

E=\dfrac{V}{d}

Voltage is given by

V=E\Delta x

Kinetic energy is given by

K=qV

\dfrac{1}{2}mv^2=qE\Delta x\\\Rightarrow v=\sqrt{\dfrac{2qE\Delta x}{m}}\\\Rightarrow v=\sqrt{\dfrac{2\times 6.9\times 10^{-9}\times 3333.33\times (0.57-0.26)}{8.3\times 10^{-9}}}\\\Rightarrow v=41.4496148484\ m/s

The initial speed of the object is 41.4496148484\ m/s

7 0
3 years ago
In a Rutherford scattering experiment a target nucleus has a diameter of 1.34×10-14 m. The incoming α particle has a mass of 6.6
Rasek [7]

Answer:

E = 2.5 x 10⁻¹⁴ J

Explanation:

given,

diameter = 1.33 x 10⁻¹⁴ m

mass = 6.64 x 10⁻²⁷ kg

wavelength is equal to diameter

de broglie wavelength equal to diameter

         \lambda = \dfrac{h}{mv}

         1.33 \times 10^{-14}= \dfrac{6.626 \times 10^{-34}}{6.64 \times 10^{-27}\times v}

         v= \dfrac{6.626 \times 10^{-34}}{6.64 \times 10^{-27}\times 1.33 \times 10^{-14}}

              v = 7.5 x 10⁶ m/s

Kinetic energy is equal to

     E = \dfrac{1}{2}mv^2

     E = \dfrac{1}{2}\times 6.64 \times 10^{-27}\times (7.5\times 10^6)^2

            E = 2.5 x 10⁻¹⁴ J

8 0
3 years ago
The loaded cab of an elevator has a mass of 3.0 x 10 3 kg and moves 200 m up the shaft in 20 s at constant speed. At what averag
Alexeev081 [22]

The average rate at which the cable does work is 294,000 J/s.

The given parameters:

  • <em>mass, m = 3000 kg</em>
  • <em>height, h = 200 m</em>
  • <em>time of motion, t = 20 s</em>

The average rate at which the cable does work is calculated as follows;

P = \frac{E}{t} \\\\P = \frac{mgh}{t} \\\\P = \frac{3000 \times 9.8 \times 200}{20} \\\\P = 294,000 \ J/s

Thus, the average rate at which the cable does work is 294,000 J/s.

Learn more about energy and power here: brainly.com/question/13387946

4 0
2 years ago
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Wittaler [7]

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3 0
3 years ago
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