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tamaranim1 [39]
3 years ago
9

Diagram of water waves moving from a shallow water to deep water​

Physics
1 answer:
allsm [11]3 years ago
6 0

Answer:

water cycle

Explanation:

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I need help on 6, 7, 8, and 9
Ivahew [28]
6 is b. part B on 6 is a. 7 is a. partB ON 7 b
5 0
3 years ago
1.      A rocket is forced forward by the ______ force of its engines, expelling gases out the rear of the rocket.​
AURORKA [14]

There are two forces acting on a rocket at the moment of lift off: Thrust pushes the rocket upwards by pushing gases downwards in the opposite direction.Weight is the force due to gravity pulling the rocket downwards towards the centre of the earth.So I'm thinking the answer is THRUST.

6 0
3 years ago
A litre of a gas weigh 2 gram at 300 kelvin temperature and 1 atm pressure,if the pressure is made 75 atm then at which temperat
liberstina [14]

Answer:

45000 K .

Explanation:

Given :

A liter of a gas weigh 2 gram at 300 kelvin temperature and 1 atm pressure

We need to find the temperature in which 1 litre of the same gas weigh 1 gram

in pressure 75 atm.

We know, by ideal gas equation :

PV=nRT

Here , n is no of moles , n=\dfrac{Given \ Weight }{Molecular\ Mass}=\dfrac{w}{M}

Putting initial and final values and dividing them :

\dfrac{P_1V_1}{P_2V_2}=\dfrac{\dfrac{w_1}{M}T_1}{\dfrac{w_2}{M}T_2}

\dfrac{1\times 1}{75\times 1}=\dfrac{\dfrac{2}{M}\times 300}{\dfrac{1}{M}\times T_2}\\ \\T_2=45000\ K.

Hence , this is the required solution.

7 0
3 years ago
Bus starts from rest if the acceleration of the bus is 0.5 MS squared what will be the velocity at the end of two minutes and wh
Nutka1998 [239]

Explanation:

Given that,

Initial speed of the bus, u = 0

Acceleration of the bus, a = 0.5 m/s²

Let v is the velocity at the end of 2 minutes. The change in velocity divided by time equals acceleration.

So,

a=\dfrac{v-u}{t}\\\\v=u+at\\\\v=0+0.5\times 120\\\\v=60\ m/s

Let d is the distance cover during that time. So,

v^2-u^2=2ad\\\\d=\dfrac{v^2-u^2}{2a}\\\\d=\dfrac{(60)^2}{2\times 0.5}\\\\d=3600\ m

So, the final speed is 60 m/s and the distance covered during that time is 3600 m.

4 0
3 years ago
(a) A load of coal is dropped (straight down) from a bunker into a railroad hopper car of inertia 3.0 × 104 kg coasting at 0.50
Firlakuza [10]

Answer:

a) m=20000Kg

b) v=0.214m/s

Explanation:

We will separate the problem in 3 parts, part A when there were no coals on the car, part B when there is 1 coal on the car and part C when there are 2 coals on the car. Inertia is the mass in this case.

For each part, and since the coals are thrown vertically, the horizontal linear momentum p=mv must be conserved, that is, p=m_Av_A=m_Bv_B=m_Cv_C, were each velocity refers to the one of the car (with the eventual coals on it) for each part, and each mass the mass of the car (with the eventual coals on it) also for each part. We will write the mass of the hopper car as m_h, and the mass of the first and second coals as m_1 and m_2 respectively

We start with the transition between parts A and B, so we have:

m_Av_A=m_Bv_B

Which means

m_hv_A=(m_h+m_1)v_B

And since we want the mass of the first coal thrown (m_1) we do:

m_hv_A=m_hv_B+m_1v_B

m_hv_A-m_hv_B=m_1v_B

m_1=\frac{m_hv_A-m_hv_B}{v_B}=\frac{m_h(v_A-v_B)}{v_B}

Substituting values we obtain

m_1=\frac{(3\times10^4Kg)(0.5m/s-0.3m/s)}{0.3m/s}=20000Kg=2\times10^4Kg

For the transition between parts B and C, we can write:

m_Bv_B=m_Cv_C

Which means

(m_h+m_1)v_B=(m_h+m_1+m_2)v_C

Since we want the new final speed of the car (v_C) we do:

v_C=\frac{(m_h+m_1)v_B}{(m_h+m_1+m_2)}

Substituting values we obtain

v_C=\frac{(3\times10^4Kg+2\times10^4Kg)(0.3m/s)}{(3\times10^4Kg+2\times10^4Kg+2\times10^4Kg)}=0.214m/s

5 0
3 years ago
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