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Svetach [21]
1 year ago
10

What type of acceleration does an object moving with constant speed in a circular path experience?Select one:A) constant acceler

ation.B) terminal acceleration.C) linear acceleration.D) centripetal acceleration.E) free fall.
Physics
1 answer:
olga_2 [115]1 year ago
3 0

ANSWER:

D) centripetal acceleration.

STEP-BY-STEP EXPLANATION:

When a body performs a uniform circular motion, the direction of the velocity vector changes at every instant. This variation is experienced by the linear vector, due to a force called centripetal, directed towards the center of the circumference that gives rise to the centripetal acceleration.

Therefore, the answer is centripetal acceleration.

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A student fills a tank of radius r with water to a height of h1 and pokes a small, 1.0 cm diameter hole at a distance h2 from th
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when a hole is made at the bottom of the container then water will flow out of it

The speed of ejected water can be calculated by help of Bernuolli's equation and Equation of continuity.

By Bernoulli's equation we can write

Po + \frac{1}{2}\rho v_1^2 + \rho g h = Po + \frac{1}{2}\rho v_2^2 + \rho g *0

Now by equation of continuity

A_1v_1 = A_2v_2

\pi (0.2)^2 v_1 = \pi (0.01)^2 v_2

from above equation we can say that speed at the top layer is almost negligible.

v_1 = 0

now again by equation of continuity

\rho g h = \frac{1}{2} \rho v^2

v = \sqrt{2 g h}

here we have

h = h_1 - h_2

h = 0.50 - 0.03 = 0.47m

now speed is given by

v = \sqrt{2* 9.8 * 0.47}

v = 3.03 m/s

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3 years ago
2.15 Un móvil en trayectoria rectilinea ocupa las posiciones indicadas en la tabla para cada minuto:
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To make sure I don’t has 35 012345 bishops
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The energy that a substance contains is
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The energy that substance contain is. B.heat
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At a particular instant the magnitude of the momentum of a planet is 2.60 × 10^29 kg·m/s, and the force exerted on it by the sta
aleksley [76]

Answer:

F=(-4.8*10^22,0,0) N

Explanation:

<u>Given  :</u>

We are given the magnitude of the momentum of the planet and let us call this momentum (p_now) and it is given by p_now = 2.60 × 10^29 kg·m/s. Also, we are given the force exerted on the planet F = 8.5 × 10^22 N. and the angle between the planet and the star is Ф = 138°

Solution :

We are asked to find the parallel component of the force F The momentum here is not constant, where the planet moving along a curving path with varying speed where the rate change in momentum and the force may be varying in magnitude and direction. We divide the force here into two parts: a parallel force F to the momentum and a perpendicular force F' to the momentum.  

The parallel force exerted to the momentum will speed or reduce the velocity of the planet and does not change its moving line. Let us apply the direction cosines, we could obtain the parallel force as next  

F=|F|cosФp            (1)

Where the parallel force F is in the opposite direction of p as the angle between them is larger than 90°. Now we can plug our values for 0 and I F I into equation (1) to get the parallel force to the planet  

F=|F|cosФp

 =-4.8*10^22 N*p

<em>As this force is in one direction, we could get its vector as next  </em>

F=(-4.8*10^22,0,0) N

F=(0,-4.8*10^22,0) N

F=(0,0-4.8*10^22) N

The cosine of 138°, the angle between F and p is, is a negative number, so F is opposite to p. The magnitude of the planet's momentum will decrease.

8 0
3 years ago
After pouring water into a cup with holes in the bottom what happened to the water
Nesterboy [21]
It goes through the holes in the cup . 
3 0
3 years ago
Read 2 more answers
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