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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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What do you mean by levels of professions​
emmasim [6.3K]

Answer:

your answer is.

I hope it will helpful for you

thank you

mark as brainest answer

8 0
3 years ago
Jake is helping Fin push a box at a constant velocity up an incline that makes an angle of 30.0° above the horizontal by applyin
andre [41]

Given data

The angle of inclination of the plane is theta = 30 degree

The applied force in the inclined plane is F = 94 N

The distance moved in the inclined plane is d = 2.30 m

The coefficient of kinetic friction is u_k = 0.280

The free-body diagram of the above configuration is shown below:

Here, the normal reaction force on the box is N, the acceleration due to gravity is denoted as g, the friction force on the box is F_f, and the mass of the box is denoted as m.

(a)

The expression for the work done by the pushing force is given as:

W=Fd

Substitute the value in the above equation.

\begin{gathered} W=94\text{ N}\times2.30\text{ m} \\ W=216.2\text{ J} \end{gathered}

Thus, the work done by the pushing force is 216.2 J.

(b)

The box is moving at the constant velocity, therefore, the pushing force will be equal to the frictional force and the component of the gravitational force in the inclined plane.

\begin{gathered} F=F_f+mg\sin \theta \\ F=\mu_kN+mg\sin \theta \end{gathered}

The expression for the normal reaction force is given as:

N=mg\cos \theta

The expression for the mass of the box is given as:

\begin{gathered} F=\mu_k\times mg\cos \theta+mg\sin \theta \\ m=\frac{F}{\mu_kg\cos \theta+g\sin \theta} \end{gathered}

Substitute the value in the above equation.

\begin{gathered} m=\frac{94\text{ N}}{0.28\times9.8m/s^2\times\cos 30^o+9.8m/s^2\times\sin 30^0} \\ m=12.9\text{ kg} \end{gathered}

Thus, the mass of the box is 12.9 kg.

7 0
1 year ago
Light from a laser forms a 1.31-mm diameter spot on a wall. If the light intensity in the spot is what is the power output of th
Wewaii [24]

Answer:

0.04973 W

Explanation:

I = Intensity of laser = 3.69\times 10^4\ W/m^2 (assumed, as it is not given)

d = Diameter of spot = 1.31 mm

r = Radius = \dfrac{d}{2}=\dfrac{1.31}{2}\ mm

A = Area = \pi r^2

Power is given by

P=IA\\\Rightarrow P=3.69\times 10^4\times \pi \dfrac{(1.31\times 10^{-3})^2}{4}\\\Rightarrow P=0.04973\ W

The power output of the laser is 0.04973 W

5 0
3 years ago
Anyone know the answer to this question?
LiRa [457]

Answer: neither accurate nor precise

Explanation:

Showing that it should be near 9.8 m/s and the trials aren't near that accuracy,  I would say neither.

4 0
3 years ago
A motorcycle accelerated from 10 metre per second to 30 metre per second in 6 seconds. How far did it travel in this time ?
Ber [7]
U=10 m/s
v=30 m/s
t=6 sec

therefore, a=(v-u)/t
                   =(30-10)/6
                   =(10/3) ms^-2

now, displacement=ut+0.5*a*t^2
                              =60+ 0.5*(10/3)*36
                              =120 m
And you can solve it in another way:

v^2=u^2+2as
or, s=(v^2-u^2)/2a
       =(900-100)/6.6666666.......
       =120 m

7 0
4 years ago
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