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Sidana [21]
3 years ago
12

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 degrees with the horizontal. No forces other than tension and gravity are acting on the load. Which of the following statements are correct? Choose all that apply. a) The train is an inertial frame of reference b) The train is not an inertial frame of reference c) The train may be instantaneously at rest d) The train may be moving at a constant speed in a straight line e) The train may be moving at a constant speed in a circle f) The train must be speeding up g) The train must be slowing down h) The train must be accelerating
Physics
1 answer:
katrin2010 [14]3 years ago
7 0

Answer:

b

c

e

h

Explanation:

Note that the swing direction was not giving in the question and direction could be sideways (in a turn) or in a track or both

The question show something in common ...acceleration

So let's look at the statements and pick the correct ones

a is false while b is correct as the train is accelerating

c is correct. The train is accelerating even thou the speed could not be ascertained

d is false and not feasible as the train is accelerating

e is true as the train maybe moving at a constant speed in a circle

f is false. This could be constant velocity in a circle. Same as g (false)

h is true. It's accelerating

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If you subtract vector 3.7 cm at 45° North of East from vector 4.5 cm at 57° West of North using a scale drawing, what is the re
Gelneren [198K]

The resultant vector is 5.2 cm at a direction of 12⁰ west of north.

<h3>Resultant of the two vectors</h3>

The resultant of the two vectors is calculated as follows;

R = a² + b² - 2ab cos(θ)

where;

  • θ is the angle between the two vectors =  45° + (90 - 57) = 78⁰
  • a is the first vector
  • b is the second vector

R² = (3.7)² + (4.5)² - (2 x 3.7 x 4.5) cos(78)

R² = 27.02

R = 5.2 cm

<h3>Direction of the vector</h3>

θ = 90 - 78⁰

θ = 12⁰

Thus, the resultant vector is 5.2 cm at a direction of 12⁰ west of north.

Learn more about resultant vector here: brainly.com/question/28047791

#SPJ1

3 0
1 year ago
A uniform cubical crate is 0.740 m on each side and weighs 600 N. It rests on the floor with one edge against a very small, fixe
Sindrei [870]

Answer:

H = 0.673

Explanation:

given,

side of cubical crate = 0.74

weight of the crate = 600 N

magnitude of force = 330 N

the Horizontal distance of its Center of mass

= 0.74/2

= 0.37

Let the required Height be H

By Balancing the Torques, we get

H x 330 N = 0.37 x 600

330 H = 222

H = 0.673

hence, the height above the floor where force is acting is equal to 0.673 m

6 0
3 years ago
Even when shut down after a period of normal use, a large commercial nuclear reactor transfers thermal energy at the rate of 150
Sever21 [200]

Answer:2.89\approx 2.9^{\circ}C/s

Explanation:

Given

Power\left ( P\right )=150 MW

mass of core\left ( m\right )=1.60\times 10^5 kg

Average specific heat \left ( C\right )=0.3349 KJ/kg^{\circ}C

And rate of increase of temperature =\frac{\mathrm{d}T}{\mathrm{d} t}

Now

P=mc\frac{\mathrm{d}T}{\mathrm{d} t}

150\times 10^6=1.60\times 10^5\times 0.3349\times \frac{\mathrm{d}T}{\mathrm{d} t}

Thus \frac{\mathrm{d}T}{\mathrm{d} t}=[tex]\frac{1.60\times 10^5\times 0.3349}{150\times 10^6}

\frac{\mathrm{d}T}{\mathrm{d} t}=2.89\approx 2.9^{\circ}C/s

6 0
3 years ago
If a force of 1.0 N pulled up and parallel to the surface of the incline is required to raise the mass back to the top of the in
wel

The question is incomplete! The complete question along with answer and explanation is provided below.

Question:

A 0.5 kg mass moves 40 centimeters up the incline shown in the figure below. The vertical height of the incline is 7 centimeters.

What is the change in the potential energy (in Joules) of the mass as it goes up the incline?  

If a force of 1.0 N pulled up and parallel to the surface of the incline is required to raise the mass back to the top of the incline, how much work is done by that force?

Given Information:  

Mass = m = 0.5 kg

Horizontal distance = d = 40 cm = 0.4 m

Vertical distance = h = 7 cm = 0.07 m

Normal force = Fn = 1 N

Required Information:  

Potential energy = PE = ?

Work done = W = ?

Answer:

Potential energy = 0.343 Joules

Work done = 0.39 N.m

Explanation:

The potential energy is given by

PE = mgh

where m is the mass of the object, h is the vertical distance and g is the gravitational acceleration.

PE = 0.5*9.8*0.07

PE = 0.343 Joules

As you can see in the attached image

sinθ = opposite/hypotenuse

sinθ = 0.07/0.4

θ = sin⁻¹(0.07/0.4)

θ = 10.078°

The horizontal component of the normal force is given by

Fx = Fncos(θ)

Fx = 1*cos(10.078)

Fx = 0.984 N

Work done is given by

W = Fxd

where d is the horizontal distance

W = 0.984*0.4

W = 0.39 N.m

3 0
3 years ago
A car that is standing still accelerates up a hill with a slope of 6.4% at an average acceleration of 2.93 ft/s^2. The hill is 1
alexgriva [62]

Answer:

213 s

Explanation:

Slope is the ratio of change in vertical distance to change in horizontal distance.

Slope = vertical height / horizontal height

Therefore:

6.4% = vertical height / 12.42

vertical height = 6.4% * 12.42

vertical height = 0.8 miles

The distance travelled by the car (s) is:

s² = 0.8² + 12.42²

s² = 154.9

s = 12.45 miles

Acceleration (a) =  2.93 ft/s^2 = 0.00055 mile/s²

initial velocity (u) = 0, final velocity = 203 mph

Using:

s = ut + 0.5at²

12.45 = 0.5(0.00055)t²

t =213 s

5 0
3 years ago
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