3 years ago

Sohan said to Geeta I have done my work change the following sentence into indirect

Physics

1 answer:

12345 [234]3 years ago

8 0

Answer:

Sohan told Geeta that i had done my work

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If you were to mix 3 liters of 20 degrees celsius water with 4 liters of 30 degrees water at 7 liters of water what temperature

Debora [2.8K]

Lets say final temperature is @ celsius
3c(@-20)= 4c([email protected])
[email protected]=180
@=180/7

3 0

3 years ago

1. A student notices that wearing darker colors in sunlight makes him feel warmer, so he decides to conduct an experiment. He ta

liraira [26]

Answer: the color of the cloth

Explanation:

8 0

2 years ago

A cyclist starts from rest and coasts down a 4.0∘ hill. The mass of the cyclist plus bicycle is 85 kg. Ignore air resistance and

Angelina_Jolie [31]

A) change in ht after 180m = 180 * sin(4-deg.) = 12.56m

net work done by gravity on the cyclist = mass * gravity * height diff.

= 85 * 9.8 * 12.56

= 10470J

= 10.5kJ

B) Kinetic energy = 1/2 * mass * vel.^2 = work done by gravity = 10470J

vel.^2 = 10470 * 2 / 85 = 246.4

vel. = 15.7m/s

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3 years ago

There are 8.8 liters of gas in a piston at a pressure of 1.75 atmospheres. The temperature remains constant, and the gas is comp

saveliy_v [14]

The new pressure P2 is 2.48 atmosphere.

<u>Explanation:</u>

Here, the one of the product of pressure and volume is equal to the products of pressure and volume of other.

By using Boyles's law,

pressure is inversely proportional to volume,

P1 V1 = P2 V2

where P1, V1 represents the first pressure and volume,

P2, V2 represents the second pressure and volume

P2 = (P1 V1) / V2

= (1.75 $\times$ 8.8) / 6.2

P2 = 2.48 atmosphere.

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3 years ago

Object A of mass M is moving east at speed v. It collides with object B of mass 2M that was initially at rest. The motion of the

Firlakuza [10]

Answer:

$v_B=\frac{v}{3}$

Explanation:

Given that:

mass of object A, $m_A=M$

mass of object B, $m_B=2M$

speed of object A, $v_A=v$

So, according to the conservation of momentum, the momentum before collision is equal to the momentum after conservation.

$m_A.v+m_B\times 0=m_A\times v_A +m_B\times v_B$

$M\times v+0 = M\times \frac{v}{3}+2M\times v_B$

$2M\times v_B= \frac{2M\times v}{3}$

$v_B=\frac{v}{3}$

3 0

3 years ago

Sohan said to Geeta I have done my work change the following sentence into indirect​

Sohan said to Geeta I have done my work change the following sentence into indirect