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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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Veronika [31]

Answer:

A mass of 4 Kg rest on the horizontal plane. The plane is gradually inclined until at an angle of θ=15

∘

 with the horizontal,the mass just being to slide what is the coffeficient  of static friction between the block & the surface.

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7 0
2 years ago
7. If 8 million kg of water flows over Niagara Falls each second, calculate the power available at the bottom of the falls.
Alexxx [7]

Answer:

The power will be "3.92×10⁹ Watts". A further explanation is given below.

Explanation:

The given values as per the question,

Rate,

= 8 million kg

Distance,

= 50 m

Gravity,

= 9.8 m/s²

As we know,

The power will be:

⇒ Power = Rate\times Distance\times  Gravity

On putting the values, we get

⇒             =  8\times 10^6\times 50\times 9.8

⇒             =3.92\times 10^9 \  Watts

7 0
2 years ago
An object is dropped from a height of 75.0 m above ground level. (a) Determine the distance traveled during the first second. (b
lys-0071 [83]

Answer:

a)Distance traveled during the first second = 4.905 m.

b)Final velocity at which the object hits the ground = 38.36 m/s

c)Distance traveled during the last second of motion before hitting the ground = 33.45 m

Explanation:

a) We have equation of motion

             S = ut + 0.5at²

     Here u = 0, and a = g

              S = 0.5gt²

    Distance traveled during the first second ( t =1 )

              S = 0.5 x 9.81 x 1² = 4.905 m

   Distance traveled during the first second = 4.905 m.

b)  We have equation of motion

            v² = u² + 2as

      Here u = 0, s= 75 m and a = g

           v² = 0² + 2 x g x 75 = 150 x 9.81

           v = 38.36 m/s

      Final velocity at which the object hits the ground = 38.36 m/s

c) We have S = 0.5gt²

                   75 = 0.5 x 9.81 x t²

                    t = 3.91 s

   We need to find distance traveled last second

   That is

          S = 0.5 x 9.81 x 3.91² - 0.5 x 9.81 x 2.91² = 33.45 m

   Distance traveled during the last second of motion before hitting the ground = 33.45 m

       

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3 years ago
Un cable está tendido sobre dos postes colocados con una separación de 10 m. A la mitad del cable se cuelga un letrero que provo
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Answer:

El peso del cartel es 397,97 N

Explanation:

La tensión dada en cada segmento del cable = 2000 N

El desplazamiento vertical del cable = 50 cm = 0,5 m

La distancia entre los polos = 10 m

La posición del letrero en el cable = En el medio = 5

El ángulo de inclinación del cable a la vertical = tan⁻¹ (0.5 / 5) = 5.71 °

El peso del letrero = La suma del componente vertical de la tensión en cada lado del letrero

El peso del signo = 2000 × sin (5.71 grados) + 2000 × sin (5.71 grados) = 397.97 N

El peso del signo = 397,97 N.

8 0
3 years ago
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