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3 years ago

An object increases its velocity from 22 m/s to 36 m/s in 5.0 s. What is the average velocity of the object?

Physics

1 answer:

Luden [163]3 years ago

7 0

Answer:

$a=2.8\ m/s^2$

Explanation:

Given that,

Initial velocity of an object, u = 22 m/s

Final velocity of an object, v = 36 m/s

Time, t = 5 s

It can be assumed to find the average acceleration of the object instead of average velocity.

The change in velocity per unit time is equal to average acceleration of an object. It can be given by :

$a=\dfrac{v-u}{t}\\\\a=\dfrac{36\ m/s-22\ m/s}{5}\\\\\a=\dfrac{14}{5}\ m/s^2\\\\a=2.8\ m/s^2$

So, the acceleration of the object is $2.8\ m/s^2$ .

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Would water molecules in Venus’ atmosphere, whose temperature is 740 K, escape into outer space? A water molecule has a mass tha

Akimi4 [234]

Answer:

The water molecule cannot escape, since the average velocity of the water molecules is less than one sixth of the escape velocity of venus.

Explanation:

The average speed of gas molecules is given by:

$v_{rms}=\sqrt{\frac{3RT}{M}}$

R is the gas constant, T is the temperature and M the molar mass of the gas.

We know that a water molecule has a mass that is 18 times that of a hydrogen atom:

$M_H=1.01*10^{-3}\frac{kg}{mol}\\M_{H2O}=18M_H=0.02\frac{kg}{mol}$

So, we have:

$v_{rms}=\sqrt{\frac{3(8.314\frac{J}{mol \cdot K})740K}{0.02\frac{kg}{mol}}}\\v_{rms}=960.65\frac{m}{s}*\frac{1km}{1000m}=0.96\frac{km}{s}$

The water molecule cannot escape, since the average velocity of the water molecules is less than one sixth of the escape velocity of venus:

$10\frac{km}{s}*\frac{1}{6}=1.6\frac{km}{s}\\0.96\frac{km}{s}$

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3 years ago

two long parrallel wires that are 0.30 m apart carry currents of 5.0 A and 8.0 A in the opposite direction.Find the magnitude of

Vera_Pavlovna [14]

Answer: The magnitude of force per length that each wire exert on the other wire is 2.67×10^-5 N/m.

The force is repulsive.

Explanation: Please see the attachments below

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3 years ago

A 1560-kilogram truck moving with a speed of 28.0 m/s runs into the rear end of a 1070-kilogram stationary car. If the collision

Nimfa-mama [501]

Answer:

Δ KE = 249158.6 kJ

Explanation:

given data

Truck mass M = 1560 Kg

Truck initial speed, u = 28 m/s

mass of car m = 1070 Kg

initial speed of car u1 = 0 m/s

solution

first we get here final speed by using conservation of momentum that is express as

Mu = (M+m) V .......................1

put here value we get

1560 × 28 = (1560 + 1070 ) V

solve it we get

final speed V = 16.60 m/s

and

Change in kinetic energy will be here

Δ KE = $\frac{1}{2} Mu^2 - \frac{1}{2}(M+m)V^2$ .................2

put here value and we get

Δ KE = $\frac{1}{2}\times 1560\times 28^2 - \frac{1}{2}\times (1560 + 1070)\times 16.60^2$

solve it we get

Δ KE = 249158.6 kJ

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3 years ago

A bowling ball is traveling at 3 m/s. It starts to roll up a ramp. How high above the ground will the ball be when it stops roll

Zepler [3.9K]

Answer:

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3 years ago

Read 2 more answers

In a certain chemical process, a lab technician supplies 292 J of heat to a system. At the same time, 68.0 J of work are done on

vekshin1

Answer:

The increase in the internal energy of the system is 360 Joules.

Explanation:

Given that,

Heat supplied to a system, Q = 292 J

Work done on the system by its surroundings, W = 68 J

We need to find the increase in the internal energy of the system. It can be given by first law of thermodynamics. It is given by :

$dE=dQ+dW\\\\dE=292\ J+68\ J\\\\dE=360\ J$

So, the increase in the internal energy of the system is 360 Joules. Hence, this is the required solution.

3 0

3 years ago