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Tasya [4]
3 years ago
7

Consider what happens at the moment when the block leaves the surface of the globe. Which of the following statements are correc

t?
- The net acceleration of the block is directed straight down.
- The component of the force of gravity toward the center of the globe is equal to the magnitude of the normal force.
- The force of gravity is the only force acting on the block.
Physics
2 answers:
Musya8 [376]3 years ago
8 0

Answer:

Option (3) is correct.

Explanation:

when the block leaves the surface, the normal reaction acting on the block is zero and the acceleration is acceleration due to gravity which acts towards the centre of globe.

So, only the force of gravity acts on the block which acts towrads the centre of the globe.

Option (3) is correct.

erma4kov [3.2K]3 years ago
6 0

Answer:

Statement 1 and 3 are correct.

Explanation:

1. The mass moves  downward, so the net acceleration of the block is straight downward.

2.The mass is sliding through the globe, so only the force of gravity is acting on the mass which pulls it in downward direction. The force of gravity has two components [mg sin∅] and [mg cos∅].

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4vir4ik [10]
  • P_1=+3D
  • P_2=-2D

\\ \bull\tt\dashrightarrow P=P_1+P_2

\\ \bull\tt\dashrightarrow P=+3D-2D=+1D

Now

\\ \bull\tt\dashrightarrow f=\dfrac{1}{P}

\\ \bull\tt\dashrightarrow f=\dfrac{1}{1}

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3 0
2 years ago
In a 100 mm diameter horizontal pipe, a venturimeter of 0.5 contraction ratio has been fitted. The head of water on the meter wh
horrorfan [7]

Answer:

the rate of flow = 29.28 ×10⁻³ m³/s or 0.029 m³/s

Explanation:

Given:

Diameter of the pipe = 100mm = 0.1m

Contraction ratio = 0.5

thus, diameter at the throat of venturimeter = 0.5×0.1m = 0.05m

The formula for discharge through a venturimeter is given as:

Q=C_d\frac{A_1A_2}{\sqrt{A_1^2-A_2^2}}\sqrt{2gh}

Where,

C_d is the coefficient of discharge = 0.97 (given)

A₁ = Area of the pipe

A₁ = \frac{\pi}{4}0.1^2 = 7.85\times 10^{-3}m^2

A₂ = Area at the throat

A₂ = \frac{\pi}{4}0.05^2 = 1.96\times 10^{-3}m^2

g = acceleration due to gravity = 9.8m/s²

Now,

The gauge pressure at throat = Absolute pressure - The atmospheric pressure

⇒The gauge pressure at throat = 2 - 10.3 = -8.3 m (Atmosphric pressure = 10.3 m of water)

Thus, the pressure difference at the throat and the pipe = 3- (-8.3) = 11.3m

Substituting the values in the discharge formula we get

Q=0.97\frac{7.85\times 10^{-3}\times 1.96\times 10^{-3}}{\sqrt{7.85\times 10^{-3}^2-1.96\times 10^{-3}^2}}\sqrt{2\times 9.8\times 11.3}

or

Q=\frac{0.97\times15.42\times 10^{-6}\times 14.88}{7.605\times 10^{-3}}

or

Q = 29.28 ×10⁻³ m³/s

Hence, the rate of flow = 29.28 ×10⁻³ m³/s or 0.029 m³/s

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3 years ago
A car goes from 0 to 26.8 m/s in 6.2 s. What is the average acceleration of the car?
Nutka1998 [239]

Answer:

0 - 60 mph = 0 - 26.8 m/s = 0 - 96.6 km/h; 0 - 100 km/h = 0 - 27.8 m/s = 0 - 62.1 mph.

Explanation:

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omeli [17]

Answer:

In physical changes no new materials are formed and the particles do not change apart from gaining or losing energy. ... Particles stay the same unless there is a chemical change whether the matter is solid, liquid or gas. Only their arrangement, energy and movement changes.

Explanation:

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