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ycow [4]
3 years ago
15

If you are pushing on a crate on a frictionless surface in one direction, and your friend is pushing on the crate in the opposit

e direction with an equal amount of force. Which of the following statements is the most accurate? a. The crate will not move as the forces cancel each other out b. Because the surface is frictionless, the crate will always move regardless of who is pushing c. The crate can continue to move, but it will move at a constant velocity d. The net force is towards the direction that you are pushing, as you started the crate's motion
Physics
1 answer:
liberstina [14]3 years ago
3 0

Answer:

Its not A..

Explanation:

I chose A - was incorrect

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To the nearest square foot, how many square feet are there in an area of 4.4 square meters?
vichka [17]

Answer:

    4.4 square meters = 47 square foot

Explanation:

We have

    1 meter = 3.28084 foot

    1 square meter = 3.28084 x 3.28084 square foot = 10.76 square foot

    4.4 square meters = 4.4 x 10.76 = 47.36 square foot = 47 square foot

    4.4 square meters = 47 square foot

3 0
3 years ago
what does it mean for data to be reproducible? how would you conduct an experiment in such a way as to get data reproducible?
PIT_PIT [208]
Reproducible is the term where your answers are similar to other groups. If every other group has similar results as yours (only if you use the same measurements) your results are reproducible. If your results are different to every other group your results aren't reproducible.  <span />
3 0
3 years ago
Read 2 more answers
Objects with masses of 135 kg and a 435 kg are separated by 0.500 m. Find the net gravitational force exerted by these objects o
IrinaVladis [17]

Answer:

The net gravitational force on the mass is 1.27\times 10^{-5}

Explanation:

We have by Newton's law of gravity the force of attraction between masses m_{1},m_{2}

F_{att}=G\frac{m_{1}m_{2}}{r^{2}}

Applying vales we get

Force of attraction between 135 kg mass and 38 kg mass is

F_{1}=6.67\times 10^{-11}\frac{135\times 38}{(0.25)^{2}}\\\\F_{1}=5.47\times 10^{-6}N

Force of attraction between 435 kg mass and 38 kg mass is

F_{2}=6.67\times 10^{-11}\frac{435\times 38}{(0.25)^{2}}\\\\F_{2}=1.76\times 10^{-5}N

Thus the net force on mass 38.0 kg is F_{2}-F_{1}=1.76\times 10^{-5}-5.47\times 10^{-6}\\\\F_{2}-F_{1}=1.27\times 10^{-5}

8 0
3 years ago
How does pressure affect surface tension
tatyana61 [14]

the effect of pressure on surface tension can be attributed in part to absorption of gas at the surface of the liquid and in part to an intrinsic decrease in density of the liquid in the neighborhood of the surface.

In the case of liquids , Owing to contact forces between the edge of the surface and the vessel, the surface acquires a curvature, and if the liquid rises up at the edges where it meets the vessel, the pressure will be less in the liquid than in the air, for points just below and just above the surface. The vessel exerts an upward force on the liquid. This is simply a matter of looking at the directions of forces acting, knowing that the surface is under tension.

3 0
3 years ago
Interactive Solution 9.37 presents a method for modeling this problem. Multiple-Concept Example 10 offers useful background for
viva [34]

Answer:

21.67 rad/s²

208.36538 N

Explanation:

\omega_f = Final angular velocity = \dfrac{1}{6}78=13\ rad/s

\omega_i = Initial angular velocity = 78 rad/s

\alpha = Angular acceleration

\theta = Angle of rotation

t = Time taken

r = Radius = 0.13

I = Moment of inertia = 1.25 kgm²

From equation of rotational motion

\omega_f=\omega_i+\alpha t\\\Rightarrow \alpha=\dfrac{\omega_f-\omega_i}{t}\\\Rightarrow \alpha=\dfrac{13-78}{3}\\\Rightarrow \alpha=-21.67\ rad/s^2

The magnitude of the angular deceleration of the cylinder is 21.67 rad/s²

Torque is given by

\tau=I\alpha\\\Rightarrow \tau=1.25\times -21.67\\\Rightarrow \tau=-27.0875

Frictional force is given by

F=\dfrac{\tau}{r}\\\Rightarrow F=\dfrac{-27.0875}{0.13}\\\Rightarrow F=-208.36538\ N

The magnitude of the force of friction applied by the brake shoe is 208.36538 N

5 0
3 years ago
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