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natulia [17]
3 years ago
10

Worth brainliest!! help me on both and explain I will spam ur questions if u answer something dumb

Physics
2 answers:
defon3 years ago
8 0

Answer:

vivi I is here and I will come up was wondering what do you think about this I don't know

romanna [79]3 years ago
3 0

Answer:

Answers are below hopefully they're correct and what u need

Explanation:

1. Step 1: Given

The distance is 600 km, since 100(speed) x 6 (time) = 600

2. Step 2: Formula

s = d/t

d = t x s

t = d/s

3. Step 3: Substitue

s = 600/6

d = 100 x 6

t = 600/100

4. Step 4: Solve

600/6 = 100

100 x 6 = 600

600/100 = 6

Step 5: Answer

Therefore, the distance is 600km, the speed or rate is 100 km/hr, and the time is 6hrs.

NUMBER 2

1. Step 1: Given

Distance = 200 m

Rate or speed = 8m/s

Time = ?

2. Step 2: Formula

s = d/t

d = t x s

t = d/s

3. Step 3: Substitue

t = 200/8

s = 200/25

d = 25 x 8

4. Step 4: Solve

200/8 = 25

25 x 8 = 200

200/25 = 8

5. Step 5: Answer

Therefore, the time is 25sec, distance is 200m, and speed or rate is 8m/s

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An ideal monatomic gas initially has a temperature of T and a pressure of p. It is to expand from volume V1 to volume V2. If the
yawa3891 [41]

Answer:

Isothermal :   P2 = ( P1V1 / V2 ) ,  work-done pdv = nRT * In( \frac{V2}{v1} )

Adiabatic : : P2 = \frac{P1V1^{\frac{5}{3} } }{V2^{\frac{5}{3} } }  , work-done =

W = (3/2)nR(T1V1^(2/3)/(V2^(2/3)) - T1)

Explanation:

initial temperature : T

Pressure : P

initial volume : V1

Final volume : V2

A) If expansion was isothermal calculate final pressure and work-done

we use the gas laws

= PIVI = P2V2

Hence : P2 = ( P1V1 / V2 )

work-done :

pdv = nRT * In( \frac{V2}{v1} )

B) If the expansion was Adiabatic show the Final pressure and work-done

final pressure

P1V1^y = P2V2^y

where y = 5/3

hence : P2 = \frac{P1V1^{\frac{5}{3} } }{V2^{\frac{5}{3} } }

Work-done

W = (3/2)nR(T1V1^(2/3)/(V2^(2/3)) - T1)

Where    T2 = T1V1^(2/3)/V2^(2/3)

3 0
3 years ago
The workdone In pulling a body which weighs 30N along a horizontal plane by a constant force of 20N is 320J. Find the distance m
sp2606 [1]

Answer: The answer is 800 J

Explanation:

4 0
2 years ago
The two forces of a 3rd law pair always act on different bodies. The two forces of a 3rd law pair always act on different bodies
svetoff [14.1K]

1. The two forces of a 3rd law pair always act on different bodies.

TRUE

SO above statement is TRUE because Newton's III law is valid for two different objects where one object will exert force and other body will exert reaction force on it.

2. Part F Given that two bodies interact via some force, the accelerations of these two bodies have the same magnitude but opposite directions. (Assume no other forces act on either body.)

FALSE

This is false because the force of action and Reaction is same on two different objects but for finding acceleration we need to divide the force by mass of two objects and since the mass of two bodies may be different so we can say that acceleration may be different.

3. Part G According to Newton's 3rd law, the force on the (smaller) moon due to the (larger) earth is

(i) greater in magnitude and antiparallel to the force on the earth due to the moon.

(ii) greater in magnitude and parallel to the force on the earth due to the moon.

(iii) equal in magnitude but antiparallel to the force on the earth due to the moon.

(iv) equal in magnitude and parallel to the force on the earth due to the moon.

(v) smaller in magnitude and antiparallel to the force on the earth due to the moon.

(vi) smaller in magnitude and parallel to the force on the earth due to the moon.

Since Earth and moon is an isolated system so here Newton's III law is valid due to which we can say that two forces are equal in magnitude but opposite in sign

so correct answer will be

equal in magnitude but antiparallel to the force on the earth due to the moon.

4 0
3 years ago
What is the name of the process where an object is exposed to nuclear radiation
Murljashka [212]

Answer:

Exposing objects to beams of radiation is called irradiation

6 0
3 years ago
The internal energy of nmoles of an ideal gas depends on a. one state variable T.
VMariaS [17]

Answer:

Correct option a. one state variable T.

Explanation:

In the case of an ideal gas it is shown that internal energy depends exclusively on temperature, since in an ideal gas any interaction between the molecules or atoms that constitute it is neglected, so that internal energy is only kinetic energy, which depends Only of the temperature. This fact is known as Joule's law.

The internal energy variation of an ideal gas (monoatomic or diatomic) between two states A and B is calculated by the expression:

ΔUAB = n × Cv × (TB - TA)

Where n is the number of moles and Cv the molar heat capacity at constant volume. Temperatures must be expressed in Kelvin.

An ideal gas will suffer the same variation in internal energy (ΔUAB) as long as its initial temperature is TA and its final temperature TB, according to Joule's Law, whatever the type of process performed.

3 0
3 years ago
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