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Oksanka [162]
3 years ago
14

What is the angle of reflection of the incident angle is 35°? 35 25° 55 90

Physics
1 answer:
Anestetic [448]3 years ago
4 0

Answer:

i think its the third one or second one

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2. You will need a magnifying glass and a small piece of scrap paper for this demonstration. a. Go outside on a sunny day and pu
andrew-mc [135]

The focal length of a magnifying glass is the distance between the focal point and optical centre of the magnifying glass.

<h3>Focal length</h3>

The focal length, f is the distance from a lens or mirror to the focal point, F.

This is the distance from a lens or mirror at which parallel light rays will meet for a converging lens or mirror or appear to diverge from for a diverging lens or mirror.

A magnifying glass is a converging lens which produces a enlarged, erect and virtual image when an object is placed between the focal point and optical centre.

A magnifying glass will bring to focus at a point sun rays which can cause the paper to catch fire if it is held in place for long.

This point at which the most concentrated ray of light is shining on the paper, is the focal point for that magnifying glass.

Therefore, the focal length of a magnifying glass is the distance between the focal point and optical centre of the magnifying glass.

Learn more about about focal length at: brainly.com/question/25779311

4 0
2 years ago
If a car has a mass of 1,000 kilograms, and a velocity of 35 m/s. What is its momentum?
fenix001 [56]

\vec{p}=m\vec{v}\\p = 1000 * 35=35000[\frac{kg*m}{s}]

7 0
3 years ago
A 1130-kg car is held in place by a light cable on a smooth (frictionless) ramp. The cable makes an angle of 31.0° above the sur
zubka84 [21]

Answer:

T = 5163.89 N

Explanation:

Newton's first law:

∑F =0 Formula (1)

∑F : algebraic sum of the forces in Newton (N)

We define the x-axis in the direction parallel to the movement of the car on the ramp and the y-axis in the direction perpendicular to it.

Forces acting on the car

W: Weight of the car : In vertical direction

FN : Normal force : perpendicular to the ramp

T :Tension force:  at angle of 31.0° above the surface of the ramp

Calculated of the Weight  of the car (W)

W = m*g   m: mass   g:acceleration due to gravity

W =   1130-kg* 9.8 m/s² = 11074 N

x-y weight components

Wx =  11074 N*sin 25.0° = 4680.07 N

Wy = 11074 N*cos 25.0° = 10036.45 N

x-y Tension components

Tx = T*cos 25.0°

Ty = T*sin 25.0°

Newton's first law:

∑Fx =0 Formula (1)

Tx-Wx = 0

T*cos 25.0° - 4680.07 = 0

T*cos 25.0° = 4680.07

T =  4680.07 / cos 25.0°

T = 5163.89 N

4 0
3 years ago
A 0.40-kg cart with charge 4.0 x 10-5 C starts at rest on a horizontal frictionless surface 0.50 m from a fixed object with char
maw [93]

Answer:

26.82m/s

Explanation:

Given

Mass = m= 0.4kg

Initial Velocity = u = 0

Charge = 4.0E-5C

Distance= d = 0.5m

Object Charge = 2E-4C

First, we'll calculate the initial energy (E)

E = Potential Energy

PE = kQq / d

Where k = coulomb constant = 8.99E9Nm²/C²

Energy is then calculated by;

PE = 8.99E9 * 4E-5 * 2E-4 / 0.5

PE = 143.84J

Energy = Potential Energy = Kinetic Energy

K.E = ½mv² = 143.84J

½mv² = ½ * 0.40 * v² = 143.85

0.2v² = 143.85

v² = 143.85/0.2

v² = 719.25

v = √719.25

v = 26.81883666380777

v = 26.82m/s

Hence, the object is 26.82m/s fast when the cart moving is very far (infinity) from the fixed charge

4 0
4 years ago
PLS HELP!!!! If u help I will mark brainiest thx
tatyana61 [14]

Answer:

Semiconductor. sorry i'm late

Explanation:

8 0
4 years ago
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