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2 years ago

Dos pelotas de tenis se lanzan verticalmente, hacia arriba y desde el mismo punto, con 2 s de

Physics

1 answer:

Angelina_Jolie [31]2 years ago

4 0

Answer:

Calcula la altura a la que se cruzan, si se han lanzado desde un punto situado a 1 m del

Explanation:

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A graduated cylinder contains 17.5 ml of water. When a metal cube is placed onto the cylinder, its water level rises to 20.3 ml

pogonyaev

If you are asking what the volume of the cube is it would be 20.3 - 17.5 ml so 2.8 ml.

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2 years ago

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Which of the following represents a convex lens?<br> A. +f<br> B. -di<br> C. -f<br> D. +di

Aleonysh [2.5K]

F is the focal length of the lens. A positive f indicates a converging optical device, like a convex lens or a concave mirror.

Choice A

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2 years ago

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A father demonstrates projectile motion to his children by placing a pea on his fork's handle and rapidly depressing the curved

MariettaO [177]

Answer:

4.17 m/s

Explanation:

To solve this problem, let's start by analyzing the vertical motion of the pea.

The initial vertical velocity of the pea is

$u_y = u sin \theta = (7.39)(sin 69.0^{\circ})=6.90 m/s$

Now we can solve the problem by applying the suvat equation:

$v_y^2-u_y^2=2as$

where

$v_y$ is the vertical velocity when the pea hits the ceiling

$a=g=-9.8 m/s^2$ is the acceleration of gravity

s = 1.90 is the distance from the ceiling

Solving for $v_y$ ,

$v_y = \sqrt{u_y^2+2as}=\sqrt{(6.90)^2+2(-9.8)(1.90)}=3.22 m/s$

Instead, the horizontal velocity remains constant during the whole motion, and it is given by

$v_x = u cos \theta = (7.39)(cos 69.0^{\circ})=2.65 m/s$

Therefore, the speed of the pea when it hits the ceiling is

$v=\sqrt{v_x^2+v_y^2}=\sqrt{2.65^2+3.22^2}=4.17 m/s$

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2 years ago

The famous cliff divers of Acapulco leap from a perch 35 m above the ocean. How fast are they moving when they reach the surface

Rus_ich [418]

1) 26.2 m/s

The mechanical energy of the divers at any point of their vertical motion is sum of the kinetic energy and the gravitational potential energy:

$E=K+U = \frac{1}{2}mv^2 + mgh$

where

m is the mass of the diver

v is the speed

g = 9.8 m/s^2 is the acceleration due to gravity

h is the height above the water

When the diver is on the cliff, v = 0 (he is at rest), so K=0 and the initial mechanical energy is just potential energy:

$E_i = mgh$

where h=35 m is the height of the cliff.

When the diver hits the water above, h = 0, so U=0 and the final mechanical energy is just kinetic energy:

$E_f = \frac{1}{2}mv^2$

since the total mechanical energy is conserved, we have

$E_i = E_f\\mgh = \frac{1}{2}mv^2$

And solving the equation for v, we find the speed when they reach the surface of the water:

$v=\sqrt{2gh}=\sqrt{2(9.8 m/s^2)(35 m)}=26.2 m/s$

2) It is converted into thermal energy of the water

When the diver enters the water, he suddenly feels another force acting against the motion of the diver: the resistance of the water. The resistance of the water acts upward, slowing down the diver until he stops.

In this process, the speed of the diver (v) decreases, and therefore the kinetic energy of the diver decreases as well, until it becomes zero.

However, this does not mean that the conservation of energy has been violated. In fact, the kinetic energy of the diver has been converted into thermal energy of the molecules of water surrounding the diver.

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2 years ago

A 5.00-pF, parallel-plate, air-filled capacitor with circular plates is to be used in a circuit in which it will be subjected to

Ne4ueva [31]

Answer:

a) r=4.24cm d=1 cm

b) $Q=5x10^{-10} C$