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nalin [4]
3 years ago
5

You are conducting an experiment inside a train car that may move along level rail tracks. A load is hung from the ceiling on a

string. The load is not swinging, and the string is observed to make a constant angle of 45∘ with the horizontal. No forces other than tension and gravity are acting on the load. Which of the following statements are correct?Statements include: 1. The train is an inertial frame of reference. 2. The train is not an inertial frame of reference. 3. The train may be at rest. 4. The train may be moving at a constant speed in a straight line. 5.The train may be moving at a constant speed in a circle. 6. The train must be speeding up. 7. The train must be slowing down. 8. The train must be accelerating.
Physics
1 answer:
sashaice [31]3 years ago
8 0

Answer:

correct answers will be

2. The train is not an inertial frame of reference.

5.The train may be moving at a constant speed in a circle.

8. The train must be accelerating.

Explanation:

As we know that string makes an angle with the horizontal

so we can write the force equation for the given ball

F_x = Tcos\theta

ma = Tcos45

also in Y direction

mg = Tsin45

now we have

\frac{ma}{mg} = cot 45

a = g cot45

now we can conclude that train is accelerating with acceleration same as gravity

now correct answers will be

2. The train is not an inertial frame of reference.

5.The train may be moving at a constant speed in a circle.

8. The train must be accelerating.

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4. A meter has a resistance of 100 Ω and gives a full scale deflection when it carries a current of 25 μA. (a) What resistor, Rx
frez [133]

Answer:

A=50mΩ

B≅50mΩ

Explanation:

A) To answer this question we have to use the Current Divider Rule. that rule says:

Ix=.\frac{Req}{Rx} *Itotal (1)

Itotal represents the new maximun current, 50mA, Ix is the current going through the 100 ohms resistor, and Req. is the equivalent resitor.

We now have a set of two resistor in parallel, so:

Req.=\frac{1}{\frac{1}{R1}+\frac{1}{R2}  } (2)

where R1 is the resitor we have to calculate, and R2 is the 100 ohms resistor (25 uA).

substituting and rearranging (2)

Req.=\frac{ 100*R1}{R1+100} (3)

Now substituting (3) in (1).

25*10^{-6} =\frac{\frac{ 100*R1}{R1+100}}{100} *50*10^{-3}

solving this, The value of R1 is: 50mΩ

This value of R1 will guaranty that the ammeter full reflection willl be at 50mA.

Given that R2 (100ohm) it too much bigger than 50mΩ, the equivalent resistor will tend to 50mΩ

If you substitude this values on (2) Req. will be 49.97 mΩ.

7 0
3 years ago
what is the average acceleration of a four wheeler that speeds up from 3 miles per hour to 15 miles per hour in 0.6 hours
den301095 [7]
A= change in velocity / change in time

a  =  \frac{change \: in \: velocity}{change \: in \: time}
a =  \frac{15 - 3}{0.6}
a = 20 \:  \frac{m}{ {s}^{2} }
6 0
4 years ago
Male Rana catesbeiana bullfrogs are known for their loud mating call. The call is emitted not by the frog's mouth but by its ear
Sidana [21]

Answer:

The amplitude of the eardrum's oscillation is 6.65×10^-13 m.

Explanation:

Given data:

The sound has a frequency of 262 Hz

The sound level is 84 dB

The air density is 1.21 kg/m^3

The speed of sound is 346 m/s

Solution:

As, Intensity of sound is given by,

I = Io×10^(s/10 db)

I = 2×π^2×ρ×v×f^2×Sm^2

Thus,

Sm = √(Io×10^(s/10 db)) / √( 2×π^2×ρ×v×f^2)

Now, put the values,

Sm = √( 10^-12 × 10^(84/10) ) / √( 2×(3.14)^2×1.21×346×(262)^2 )

Sm = √(2.51×10^-4 / 5.66×10^8)

Sm = √0.443×10^-12

Sm = 6.65×10^-13 m.

8 0
3 years ago
A student is pushing a box across the room. To push the box three times farther, the student needs to do how much work?
Travka [436]

Answer:

Removing some of the books reduced the mass of the box, and less force was needed to push it across the floor.

8 0
3 years ago
PLEASE HELP I GIVE BRAINLIEST AND 15 POINTS
deff fn [24]
The mass of the man would remain the same…
His weight would change. Assuming the man’s weight on earth is 60N.
Since weight is a force, according to Newton’s second law, F = ma ( m = mass, a = acceleration due to gravity) First lets find the mass of the man, as it is required to find his weight on the moon.
F=ma[taking a of earth as10m/s
2
]
60=m.10[divide10onbothsides]
m=
10
60
​
= 6 Kg
Acceleration due to gravity on the moon is83%less than the acceleration due to gravity on earth(1.622m/s
2
).
F=ma
F=6.1.622=9.732 N
So a person weighting 60N on earth would approximately weight around 10Non the moon.
8 0
3 years ago
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