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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

1 answer:

vekshin13 years ago

4 0

Answer:

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A thin ring of radius 73 cm carries a positive charge of 610 nC uniformly distributed over it. A point charge q is placed at the

kow [346]

Answer:

q = - 93.334 nC

Explanation:

GIVEN DATA:

Radius of ring 73 cm

charge on ring 610 nC

ELECTRIC FIELD p FROM CENTRE IS AT 70 CM

E = 2000 N/C

Electric field due tor ring is guiven as

$E = \frac{KQx}{[x^2+ R^2]^{3/2}}$

$E = \frac{9\time 10^9 \times 610\times 10^[-9} 0.70}{(0.70^2 + 0.73^2)^{3/2}}$

E1 = 3714.672 N/C

electric field due to point charge q

$E =\frac[kq}{x^2}$

$E = \frac{9\times 10^9 \times q}{0.70^2}$

$E2 = 1.837\times 10^{10}\times q$

now the eelctric charge at point P is

E = E1 + E2 $2000 = 3714.672 + 1.837\times 10[10} \times q$

solving for q

q = - 93.334 nC

7 0

3 years ago

What is the kinetic energy of a vehicle that has a mass of 3,500 kg and is moving at 40 m/s

stiv31 [10]

Answer:2800000j

Explanation:

For us to know the kinetic energy of the vehicle,

Where m is the mass

And v is the velocity

Then, K.E=1/2mv^2

While, K.E=1/2×3500×40^2

Therefore, our answer will now be

K.E=2800000j

5 0

3 years ago

When an automobile moves with constant speed down a highway, most of the power developed by the engine is used to compensate for

uysha [10]

Answer:

F=5449 N

Explanation:

Work done is a product of force and displacement ie

Work done, W, = Force*Displacement

Power, P, is Work done/Time

$P=\frac {W}{t}=\frac {FS}{t}$ where P is power, W is work done, F is force, S is displacement and t is time

In this case, F is the frictional force. Converting the power from hp to W, we multiply by 746 hence P=746*168=125328 W

Since displacement/time is velocity, then

P=FV where V is velocity in m/s

Making F the subject

$F=\frac {P}{V}$

$F=\frac {125328}{23}=5449.043478 N$

F=5449 N

7 0

3 years ago

A shell of mass m and speed v explodes into two identical fragments. If the shell was moving horizontally (the positive x direct

-Dominant- [34]

Answer:

The velocity of the other fragment immediately following the explosion is v .

Explanation:

Given :

Mass of original shell , m .

Velocity of shell , + v .

Now , the particle explodes into two half parts , i.e $\dfrac{m}{2}$ .

Since , no eternal force is applied in the particle .

Therefore , its momentum will be conserved .

So , Final momentum = Initial momentum

$mv=\dfrac{mv}{2}+\dfrac{mu}{2}\\\\u=v$

The velocity of the other fragment immediately following the explosion is v .

4 0

3 years ago

A constant force of 8N acting on an object displaces it through a distance of 3.0 m in the direction of force. Calculate work-do

sweet-ann [11.9K]

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{Work\:done=24\:J}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline{\bold{Given :}}} \\ \tt: \implies Constant \: force(F) = 8 \: N \\ \\ \tt: \implies Displacement(s) = 3 \: m \\ \\ \red{\underline{\bold{To \: Find : }}} \\ \tt: \implies Work \: Done(W.D) = ?$

• <u>According to given question</u> :

$\green{ \star} \tt \: \theta \: = 0 \degree \: \: \: \: (Angle \: between \: force \: and \: displacement) \\ \\ \bold{As \: we \: know \: that} \\ \tt: \implies Work \: Done = FS \: cos \: \theta \\ \\ \tt: \implies Work \: Done = 8 \times 3 \times cos \:0 \degree \\ \\ \green{ \circ} \tt \: cos \: 0 \degree = 1 \\ \\ \tt: \implies Work \: Done =24 \times 1 \\ \\ \green{\tt: \implies Work \: Done =24 \: J}$

5 0

2 years ago