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Talja [164]
3 years ago
8

Tính số phần trăm phân tử khí trong trọng trường của Trái Đất, có thế năng ep, lớn hơn động năng trung bình chuyển động tịnh tiế

n của chúng. Giả sử rằng nhiệt độ của khí và gia tốc của trọng lực không phụ thuộc vào độ cao
Physics
1 answer:
vekshin13 years ago
4 0

Answer:

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Awnser these pls they r for middle schoolers lol
soldi70 [24.7K]

Answer:

Find the set of value of x Find the set of value of x which satisfy the inequality 2r2- 5x 2 18which satisfy the inequality 2r2- Find the set of value of x which satisfy the inequality 2r2- 5x 2 185x 2 18

8 0
3 years ago
An automobile engine delivers 47.4 hp. how much time will it take for the engine to do 6.82 × 105 j of work? one horsepower is e
kogti [31]
1 horsepower is equal to 746 W, so the power of the engine is
P=47.4 hp \cdot 746  \frac{W}{hp}=35360 W
The power is also defined as the energy E per unit of time t:
P= \frac{E}{t}
Where the energy corresponds to the work done by the engine, which is E=6.82 \cdot 10^5 J. Re-arranging the formula, we can calculate the time t needed to do this amount of work:
t =  \frac{E}{P}= \frac{6.82 \cdot 10^5 J}{35360 W}=19.3 s
8 0
3 years ago
What is the main reason why many nuclear power plants are located near bodies of water? to wash wastes out of the power plant to
Natalija [7]

Answer: to avoid problems with water supply

Explanation: power plant needs water to run

7 0
3 years ago
Read 2 more answers
I have no idea of how to approach this problem
nataly862011 [7]

Answer:

p=1

Explanation:

Well me know that v=m/s

and that a=m/s^2

so

(m/s)^{2} = (m/s^2)(x^p)\\\\(m^2/s^2)/(m/s^2)=x^p\\\\m^2s^2/ms^2=x^p\\\\(m^2/m)(s^2/s^2)=x^p\\\\m=x^p\\\\p=1

Note: We don't take into account 2 because it's a scalar, it doesn't have units so it doesn't add anything to the equation.

4 0
3 years ago
A boat of mass 250 kg is coasting, with its engine in neutral, through the water at speed 1.00 m/s when it starts to rain with i
Vera_Pavlovna [14]

Answer:

Explanation:

1. We use the conservation of momentum for before the raining and after. And also we take into account that in 0.5h the accumulated water is

100kg/h*0.5h = 50kg

p_{b}=p_{a}\\M_{b}v_{b}=(M_{b}+m_{r})v_{a}\\v_{a}=\frac{M_{b}v_{b}}{M_{b}+m_{r}}=\frac{250kg*1\frac{m}{s}}{250kg+50kg}=0.83\frac{m}{s}

2. the momentum does not conserve because the drag force of water makes that the boat loses velocity

3. If we assume that the force of the boat before the raining is

F=ma=m\frac{v-v_{0}}{t-t_{0}}=250kg\frac{1m}{s^{2}}=250N

where we have assumed that the acceleration of the boat is 1m/s{2} just before the rain starts

And if we take the net force as

F_{net}=M_{b}a_{net}=F-F_{d}=250N-bv^{2}\\F_{net}=250N-0.5N\frac{s^{2}}{m^{2}}(1\frac{m}{s})^{2}=249.5N\\a_{net}=\frac{249.5N}{M_{b}}=\frac{249.5N}{250kg}=0.99\frac{m}{s^{2}}

where we take v=1m/s because we are taking into account tha velocity just after the rain stars.

I hope this is useful for you

regards

5 0
3 years ago
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