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Allushta [10]
3 years ago
5

The Kelly family and the Garcia family each used their sprinklers last summer. The water output rate for the Kelly family's spri

nkler was 35L per hour. The water output rate for the Garcia family's sprinkler was 20L per hour. The families used their sprinklers for a combined total of 45 hours, resulting in a total water output of 1350L. How long was each sprinkler used?
Physics
1 answer:
elena55 [62]3 years ago
3 0

Answer:

30 hours

15 hours.

Explanation:

Let the Kelly used sprinkler for t hours and Gracia used the sprinkler for the rest ie (45 - t) hours .

Water output by Kelly = (35 x t) L

Water output by Gracia = 20 ( 45 - t ) L

Total output = 35t + 20(45 - t)

So,  35 t + 20 (45 - t ) = 1350

35 t + 900 - 20 t = 1350

15 t = 450

t = 30 hours

So Kelly used sprinkler for 30 hours and Gracia used it for 45 - 30 = 15 hours.

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the video identifies the force pair produced when an apple falls through the air. which force belongs in a free-body diagram of
trapecia [35]

The free-body diagram of an apple falling through the air has weight of the apple pointing downwards and the air-resistance on the apple acting upwards.

When an object falls from up to the ground, the object falls under in the influence of acceleration due to gravity.

The vertical component of the force on the apple as it falls trough the air is given as;

∑Fy = 0

Fₙ - W = 0

Fₙ = W

where;

  • <em>Fₙ is the frictional force on the apple acting upwards</em>
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The free-body diagram of the apple is represented as follows;

                                         ↑ Fₙ

                                         Ο

                                         ↓ W

Thus, the free-body diagram of an apple falling through the air has weight of the apple pointing downwards and the air-resistance on the apple acting upwards.

Learn more here:brainly.com/question/18770265

6 0
2 years ago
What is the exact meaning of net force?
ZanzabumX [31]
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4 0
3 years ago
A small, 300 g cart is moving at 1.20 m/s on an air track when it collides with a larger, 2.00 kg cart at rest?
stiv31 [10]

Answer:

The speed of the large cart after collision is 0.301 m/s.

Explanation:

Given that,

Mass of the cart, m_1 = 300\ g = 0.3\ kg

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Mass of the larger cart, m_2 = 2\ kg

Initial speed of the larger cart, u_2=0

After the collision,

Final speed of the smaller cart, v_1=-0.81\ m/s (as its recolis)

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The speed of the large cart after collision.

Solution,

Let v_2 is the speed of the large cart after collision. It can be calculated using conservation of momentum as :

m_1u_1+m_2u_2=m_1v_1+m_2v_2

m_1u_1+m_2u_2-m_1v_1=m_2v_2

v_2=\dfrac{m_1u_1+m_2u_2-m_1v_1}{m_2}

v_2=\dfrac{0.3\times 1.2+0-0.3\times (-0.81)}{2}

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4 0
3 years ago
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Answer:

1. a

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3. b

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from the above relation we observe that

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As the collision between the electrons and protons increases the speed of the flow of charges will decrease because the opposite charges attract each other and as we know that electrical current is the rate of flow of charge.

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Heating up of wire due to sunlight will cause lattice vibration in the conductor and will obstruct the movement of the charges which build up electric current, hence increasing the resistance of conductivity.

7 0
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