If the mass of a moving object decreases from 100 grams to 25 grams, what happens to its momentum

Sunitha can type 1800 words in half an hour. What is her typing speed in words per minute?

Andre45 [30]

Answer:

60words/minute

Explanation:

If Sunitha can type 1800 words in half an hour, this can be expressed as;

1800 words = 30 minutes

To get her typing speed per minute, we will use the formula

Speed = Number of words/Time used

Typing speed = 1800/30

Typing speed = 60words/minute

Hence her typing speed in words per minute is 60words/minute

6 0

2 years ago

What are the two factors that affect the friction force between two surfaces

Masja [62]

Answer:

coefficient of static friction of the surface and the normal force

Explanation:

The coefficient of static friction of the surface and the normal force exerted on the surface given by equation F = μR

5 0

3 years ago

An industrial laser is used to burn a hole through a piece of metal. The average intensity of the light is W/m². What is the rms

inn [45]

I think your question should be:

An industrial laser is used to burn a hole through a piece of metal. The average intensity of the light is

$S = 1.23*10^9 W/m^2$

What is the rms value of (a) the electric field and

(b) the magnetic field in the electromagnetic wave emitted by the laser

Answer:

a) $6.81*10^5 N/c$

b) $2.27*10^3 T$

Explanation:

To find the RMS value of the electric field, let's use the formula:

$E_r_m_s = sqrt*(S / CE_o)$

Where

$C = 3.00 * 10^-^8 m/s$ ;

$E_o = 8.85*10^-^1^2 C^2/N.m^2$ ;

$S = 1.23*10^9 W/m^2$

Therefore

$E_r_m_s = sqrt*{(1.239*10^9W/m^2) / [(3.00*10^8m/s)*(8.85*10^-^1^2C^2/N.m^2)]}$

$E_r_m_s= 6.81 *10^5N/c$

b) to find the magnetic field in the electromagnetic wave emitted by the laser we use:

$B_r_m_s = E_r_m_s / C$ ;

$= 6.81*10^5 N/c / 3*10^8m/s$ ;

$B_r_m_s = 2.27*10^3 T$

8 0

3 years ago

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If a metal ball suspended by a rod is at rest, which force is responsible for balancing the force due to gravity?

Viktor [21]

There are different kinds of forces; applied force, force of gravity, friction force, normal force, tension force. We will focus on the common forces, applied force and force of gravity. An applied force is a force that is applied to an object by another object. The force of gravity is the force with which massively large objects such as the earth attracts another object towards itself. All objects of the earth exert a gravity that is directed towards the center of the earth. Therefore, the force of gravity of the earth is equal to the weight of the object.

5 0

3 years ago

A group of students performed a compression experiment where they placed weights on top of a cylinder of material and measured t

kolbaska11 [484]

The material that the cylinder is made from is Butyl Rubber.

<h3>What is Young's modulus?</h3>

Young's modulus, or the modulus of elasticity in tension or compression, is a mechanical property that measures the tensile or compressive strength of a solid material when a force is applied to it.

<h3>Area of the cylinder</h3>

A = πr²

$A = \pi \times (0.02)^2 = 0.00126 \ m^2$

<h3>Young's modulus of the cylinder</h3>

$E = \frac{stress}{strain} \\\\E = \frac{F/A}{e/l} \\\\E = \frac{Fl}{Ae} \\\\$

Where;

e is extension

When 5 kg mass is applied, the extension = 10 cm - 9.61 cm = 0.39 cm = 0.0039 m.

$E = \frac{(5\times 9.8) \times 0.1}{0.00126 \times 0.0039} \\\\E = 9.97 \times 10^5 \ N/m^2\\\\E = 0.000997 \times 10^9 \ N/m^2\\\\E = 0.000997 \ GPa\\\\E \approx 0.001 \ GPa$

When the mass is 50 kg,

extension = 10 cm - 7.73 cm = 2.27 cm = 0.0227 m

$E = \frac{(50\times 9.8) \times 0.1}{0.00126 \times 0.0227} \\\\E = 1.7 \times 10^6 \ N/m^2\\\\E = 0.0017 \times 10^9 \ N/m^2\\\\E = 0.0017 \ GPa\\\\E \approx 0.002 \ GPa$

The Young's modulus is between 0.001 GPa to 0.002 GPa

Thus, the material that the cylinder is made from is Butyl Rubber.

Learn more about Young's modulus here: brainly.com/question/6864866

3 0

2 years ago

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