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Naya [18.7K]
1 year ago
15

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 10∘ with the vertical. No forces are acting on the load. Which of the following statements are correct?
Select all 3 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
Physics
1 answer:
zhannawk [14.2K]1 year ago
4 0

The train is not an inertial frame of reference, and the train may be moving at a constant speed in a straight line or the train may be instantaneously at rest.

<h3>What is uniform speed and Acceleration?</h3>

The pace at which an object's position changes about relation to a frame of reference and time is known as its velocity.

Uniform velocity is the state in which a body travels the same distance in the same amount of time. When the amplitude and direction of a body's average velocity (v) over time (t) do not change over time, the body is said to be moving at a uniform velocity. An illustration of uniform motion is the moon's orbit around the earth.

As the rate of change of velocity, acceleration is an illustration of a vector quantity.

When an item is moving in a straight path with an increase in velocity occurring at regular intervals of time, it is said to be experiencing uniform acceleration. The uniform acceleration of an object during free fall is one example.

To get more information about speed and Acceleration :

brainly.com/question/9282582

#SPJ1

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You are driving your car, and the traffic light ahead turns red. you apply the brakes for 2.26 s, and the velocity of the car de
Lorico [155]

Let the initial velocity of the car be u.

Final velocity of the car (v) = 5.43 m/s

deceleration (a) = - 2.78 m/s^2

Time taken (t) = 2.26 s

Using the first equation of motion:

v = u + at

u = v - at

u = 5.43 - (-2.78 \times 2.26)

u = 5.43 + 6.28

u = 11.71 m/s

Let the car's displacement be x.

Using second equation of motion:

x = ut + \frac{1}{2}at^2

x = 11.71 \times 2.26 - \frac{1}{2} \times 2.78 \times 2.26^2

x = 26.4646 - 7.0995

x = 19.3651 meters

Hence, the displacement of the car is 19.36 meters

6 0
3 years ago
Which of the following statements is true? *
lozanna [386]

Answer: B Genes make up chromosomes

Explanation: hope it helps

7 0
3 years ago
The coefficient of static friction between car's tires and a level road is 0.80 If car to be stopped in time of 3.0 sec its spee
Vlad1618 [11]

Answer:

23.5 m/s

Explanation:

The velocity of the car in decelerated motion is given by

v = u + at

where

v = 0 is the final velocity

u is the initial velocity

a is the acceleration of the car

t = 3.0 s is the time it takes for the car to stop

The acceleration of the car is given by the frictional force, which is the only force acting on the car along the direction of motion, so:

ma = -\mu mg\\a = -\mu g = -(0.80)(9.8 m/s^2)=-7.84 m/s^2

where

\mu=0.80 is the coefficient of friction

Solving the previous equation for u, we find the initial velocity:

u=v-at=0-(-7.84 m/s^2)(3.0 s)=23.5 m/s

4 0
4 years ago
A load of 500N is carried by 200N effort in a simple machine having load distance 3m Calculate effort distance.​
vlada-n [284]

Answer:

<h2>2.5 m</h2>

Explanation:

Load ( L ) = 500 N

Effort ( E ) = 200 N

Load distance ( LD ) = 3 m

Effort distance ( ED ) = ?

Now, Let's find the Effort distance ( ED )

We know that,

Output work = Input work

i.e L × LD = E × ED

plug the values

500 \times 3 = 200 \times ED

multiply the numbers

1500 = 200 \times ED

Swipe the sides of the equation

200 \: ED \:  = 500

Divide both sides of the equation by 200

\frac{200 \: ED}{200}  =  \frac{500}{200}

Calculate

ED\:  = 2.5 \: m

Hope this helps..

best regards!!

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AS Hitler once said, Static Electricity my dear watson
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