Light passing through the center of a lens will carry on undeviated.

an object with a mass of 2.4 kg has a force of 12.6 N applied to it. What is the resulting acceleration of the object? WITH PROO

Scorpion4ik [409]

Newton’s Law: F = MA
A = F/M (change equation)
12.6 N/ 2.4 kg = 5.25
Answer: acceleration is 5.25 m/s^2

3 0

3 years ago

An acorn with a mass of 0.300 kilograms falls from a tree. What is its kinetic energy when it reaches a velocity of 5.oo m/s?

vladimir1956 [14]

Answer: 3.75 joules

Explanation:

Given that:

Mass of acorn = 0.300 kilograms

velocity = 5.oo m/s

Kinetic energy = ?

Since, kinetic energy is the energy possessed by a moving object, its value depends on the mass M and velocity V of the acorn.

Thus, Kinetic energy = 1/2 x mv^2

= 1/2 x 0.300kg x (5.00m/s)^2

= 0.5 x 0.3kg x (5.00m/s)^2

= 0.15 x (5.00m/s)^2

= 3.75 joules

Thus, the kinetic energy of the falling acorn is 3.75 joules

5 0

3 years ago

A wave x meters long has a speed of y meters per second. The frequency of the wave is

Sergio [31]

The correct answer is (b.) y/x hertz. That is because the formula to get the frequency is f = v / w. The following values (v=y meters / second; wavelength = x meters) must be substituted to the equation, which leaves you y/x hertz.

6 0

3 years ago

Density of water is 1000 kg/m3 and that of glass is 2500 kg/m3 . The speed of light will be maximum in

dalvyx [7]

Water I believe sorry if wrong

6 0

3 years ago

A 50.-kilogram rock rolls off the edge of a cliff. if it is traveling at a speed of 24.2 m/s when it hits the ground, what is th

ElenaW [278]

The correct answer to the question is : 29.88 m.

EXPLANATION :

As per the question, the mass of the rock m = 50 Kg.

The rock is rolling off the edges of the cliff.

The final velocity of the rock when it hits the ground v = 24 .2 m/s.

Let the height of the cliff is h.

The potential energy gained by the rock at the top of the cliff = mgh.

Here, g is known as acceleration due to gravity, and g = $9.8\ m/s^2$

When the rock rolls off the edge of the cliff, the potential energy is converted into kinetic energy.

When the rock hits the ground, whole of its potential energy is converted into its kinetic energy.

The kinetic energy of the rock when it touches the ground is given as -

Kinetic energy K.E = $\frac{1}{2}mv^2$ .

From above we know that -

Kinetic energy at the bottom of the cliff = potential energy at a height h

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

⇒ $v^2=\ 2gh$

⇒ $h=\ \frac{v^2}{2g}$

⇒ $h=\ \frac{(24.2)^2}{2\times 9.8}$

⇒ $h=\ 29.88\ m$

Hence, the height of the cliff is 29.88 m

5 0

3 years ago