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posledela
3 years ago
14

Which of the following situations describes static friction? a. A heavy crate standing on a rough patch of mud and is being push

ed by two workers, but they cannot get it to move b. A large piano resting on a sheet of ice is being pushed by a child, but the child is not strong enough to get it to move
Physics
2 answers:
marin [14]3 years ago
8 0

Answer:

A.

Explanation:

The answer is A because the weight of the crate is sitting on something rough. The roughness is going to resist anything that tries to move against it because the force being exerted by the 2 workers is not strong enough to get the crate to move out of its place on the mud since it's static.

The answer can't be B because ice is smooth and doesn't have any friction, the little girl can't move it because it's to heavy for her to move, not because there's friction acting against it.

Yuliya22 [10]3 years ago
3 0

Answer:

A. A heavy crate standing on a rough patch of mud and is being pushed by two workers, but they cannot get it to move.

Explanation:

I got it correct on the test.

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Which of the following statements are true? (mark all that apply)
kvv77 [185]

Answer:

(a) and (b)

Explanation:

Energy is the capacity to do work, and exists in various forms. These forms can be converted one to another by the use of appropriate means. Some examples are sound, mechanical, solar, light, which causes the sensation of vision, etc. energy is measured in Joules (J).

The rate of transfer of energy is called power.

i.e Power = \frac{energy}{time}

It is measured in Watts (W).

When a white light is disperses into its colors, gray and black are not part of the colors. And a black sometimes could be as a result of the absorption of all other colors of light.

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3 years ago
A spacecraft in the shape of a long cylinder has a length of 100 m, and its mass with occupants is 1 480 kg. It has strayed too
maks197457 [2]

Answer:

2352645198509.9604 m/s²

Explanation:

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

M = Mass of black hole = 90\times 1.989\times 10^{30}\ kg

R_{f} = 10000+100 m

R_{sb} = Distance between the nose and the center of the black hole = 10000 m

The difference in the gravitational field in this system is given by

\Delta F=\dfrac{GMm}{R_{f}^2}-\dfrac{GMm}{R_{sb}^2}\\\Rightarrow \Delta g=GM\left(\frac{1}{R_f^2}-\frac{1}{R_{sb}^2}\right)\\\Rightarrow g=6.67\times 10^{-11}\times 90\times 1.989\times 10^{30}\left(\frac{1}{(10000+100)^2}-\frac{1}{10000^2}\right)\\\Rightarrow \Delta g=-2352645198509.9604\ m/s^2

The acceleration is 2352645198509.9604 m/s²

4 0
2 years ago
Which statement about ammeters and voltmeters is true? A. Current and voltage are the same throughout a circuit so where you pla
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Answer:B

Explanation:

6 0
3 years ago
Read 2 more answers
A 30-g bullet is fired with a horizontal velocity of 460 m/s and becomes embedded in block B which has a mass of 3 kg. After the
ICE Princess25 [194]

Answer:

energy loss due to friction and the impacts = 2.97 J; The impact loss due to AB impacting the carrier is =25.72J; The impact loss at first impact is 6,316.64J

Explanation:

First find the velocity of the bullet after the first impact using

M1V1 + 0 = (M1 + M2)v'

Where M1 is the mass of the bullet

M2 is the mass of the block B

M3 is the mass of the carrier

v' is the velocity

v' = M1V1/(M1 + M2)

v'= (30 × 10^-3 kg)(460 m/s) / (30 × 10^-3 kg + 3 kg)

v' = 13.8/3.03

v'= 4.55m/s

Also calculate final velocity of the carrier v2'

v2' = M1V1/(M1 + M2 + M3)

v2'= (30 × 10 kg)(460 m/s) / (30 × 10 kg + 3 kg + 30kg)

v2' =0.42m/s

Now to calculate energy loss due to friction

Normal force

N= W1 + W2 = (m1 + m2)g

Where W1 and W2 is the weight of the bullet and block respectively

g is gravitational acceleration for taken as 9.81m/s

= (0.030 kg + 3 kg)(9.81 m/s) 29.724 N

Friction force = coefficient of kinetic× normal force

Where coefficient of kinetic = 0.2

Ff = (0.2)(29.724)= 5.945 N

Now

Energy loss due to friction = frictional force × distance

Assume distance is 0.5 m.

Energy loss due to friction = 5.945 N × 0.5 m

= 2.97J

Kinetic energy of block with embedded bullet immediately after first impact:

1/2 × (m1 + m2)(v')^2

1/2 × (30 × 10^-3 kg + 3 kg)(4.55m/s)^2

= 1/2 × (3.03kg) × (4.55m/s)^2

= 31.36 J

Final kinetic energy of bullet, Block, and Carrier together

1/2 × (m1 + m2 + m3)(v2')^2

1/2 × (30 × 10^-3 kg + 3 kg + 30kg) (0.42m/s)^2

1/2 × (33.03kg) × (0.42m/s)^2

= 2.91 J

Therefore

Loss due to friction and stopping impact = Kinetic energy of block with embedded bullet immediately after first impact - Final kinetic energy of bullet, Block, and Carrier together

= 31.36 J - 2.91 J

= 28.69 J

Impact loss due to AB impacting the carrier = loss due to friction- energy due to friction

28.69J - 2.97J

=25.72J

Initial kinetic energy of system ABC = 1/2(m1vo)

=1/2(0.030 kg)(460 m/s)^2 = 6,348J

Therefore

Impact loss at first impact = Initial kinetic energy of system ABC - Kinetic energy of block with embedded bullet immediately after first impact:

= 6,348J - 31.36 J

= 6,316.64J

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3 years ago
What causes the angle of the sun’s rays to change during the year on Earth?
amm1812

The tilt of Earth's axis hope this helps

4 0
3 years ago
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