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3 years ago

When an object is thrown upwards and reaches its maximum height its speed is: a. Greater than the initial

Physics

1 answer:

suter [353]3 years ago

5 0

Answer:

Option d

Explanation:

When we throw an object in the upward direction, we provide it with certain initial velocity due to which it covers a certain distance up to the maximum height.

While the object is moving in the upward direction, its velocity keeps on reducing due to the acceleration due to gravity which acts vertically downwards in the opposite direction thus reducing its velocity.

So, the maximum height attained by the object is the point where this upward velocity of the body becomes zero and after that the object starts to fall down.

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How much heat is needed to melt 13.74 kg of silver that is initially at 20°C?

salantis [7]

Answer:

$Q = 4.63 \times 10^6 J$

Explanation:

As we know that melting point of silver is

T = 961.8 degree C

Latent heat of fusion of silver is given as

L = 111 kJ/kg

specific heat capacity of silver is given as

$s = 240 J/kg C$

now we will have

$Q = ms\Delta T + mL$

$\Delta T = 961.8 - 20$

$\Delta T = 941.8 degree C$

now from above equation

$Q = (13.74)(240)(941.8) + (13.74)(111 \times 10^3)$

$Q = 3.1 \times 10^6 + 1.53 \times 10^6$

$Q = 4.63 \times 10^6 J$

3 0

3 years ago

When light moves from air to glass, part of the light is reflected and part of it is refracted. In the image, which ray shows re

marissa [1.9K]

You have not provided the diagram, therefore, I cannot provide an exact answer.
However, I will try to help by explaining how to solve this problem.

When light moves from air to glass:
1- part of the light is reflected back into the air where the angle of incidence is equal to the angle of reflection
2- part of the light enters the water and refracts. The angle of refraction can be calculated using Snell's law.

In a diagram, the reflected ray would be the one getting back into air while the refracted ray would be the one entering the water.

You can check the attached diagram for further illustrations.

Hope this helps :)

6 0

3 years ago

Read 2 more answers

A 0.095 kg tennis ball is traveling to the right at 40 m/s when it bounces of a wall and travels in the opposite direction it ca

Mashutka [201]

given that

mass of ball = 0.095 kg

initial velocity of ball towards the wall = 40 m/s

final velocity of the ball after it rebound = 30 m/s

now change in momentum is given as

$\Delta P = m(v_f - v_i)$

$\Delta P = 0.095(30 - (-40))$

$\Delta P = 6.65 kg m/s$

So change in momentum will be 6.65 kg m/s

3 0

3 years ago

A hollow cylinder that is rolling without slipping is given a velocity of 5.0 m/s and rolls up an incline to a vertical height o

inysia [295]

Answer:

The hollow cylinder rolled up the inclined plane by 1.91 m

Explanation:

From the principle of conservation of mechanical energy, total kinetic energy = total potential energy

$M.E_T = \frac{1}{2}mv^2 + \frac{1}{2} I \omega^2 + mgh$

The total energy at the bottom of the inclined plane = total energy at the top of the inclined plane.

$\frac{1}{2}mv_i^2 + \frac{1}{2} I \omega_i^2 + mg(0) = \frac{1}{2}mv_f^2 + \frac{1}{2} I \omega_f^2 + mgh$

moment of inertia, I, of a hollow cylinder = ¹/₂mr²

substitute for I in the equation above;

$\frac{1}{2}mv_i^2 + \frac{1}{2} (\frac{1}{2}mr^2 \omega_i^2) = \frac{1}{2}mv_f^2 + \frac{1}{2} (\frac{1}{2}mr^2 \omega_f^2) + mgh\\\\ but \ v = r \omega\\\\\frac{1}{2}mv_i^2 + \frac{1}{2} (\frac{1}{2}m v_i^2 ) = \frac{1}{2}mv_f^2 + \frac{1}{2} (\frac{1}{2}m v_f^2) + mgh\\\\\frac{1}{2}mv_i^2 +\frac{1}{4}mv_i^2 = \frac{1}{2}mv_f^2 +\frac{1}{4}mv_f^2 +mgh\\\\\frac{3}{4}mv_i^2 = \frac{3}{4}mv_f^2 +mgh\\\\mgh = \frac{3}{4}mv_i^2 - \frac{3}{4}mv_f^2\\\\gh = \frac{3}{4}v_i^2 - \frac{3}{4}v_f^2\\\\$