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ladessa [460]
3 years ago
15

Part A

Physics
1 answer:
geniusboy [140]3 years ago
4 0

Answer:

a) b = -5

b) slope = 3/2

Explanation:

a) The equation of a line is given as y = mx + b, where m is the slope of the line and b is the intercept on the y axis.

Given that y = 3x + b and it passes through the point (2, 1). Hence when x = 2, y = 1. Therefore, substituting for x and y:

1 = 3(2) + b

1 = 6 + b

b = 1 - 6

b = -5

b) The equation of a line passing through two points (x_1,y_1) and x_2,y_2 is given by:

y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)

The equation of the line passing through the two points (0,3) and (4,9) is:

y-3=\frac{9-3}{4-0}(x-0)\\ \\y-3=\frac{3}{2}x\\ \\y = \frac{3}{2}x+3

Comparing y = (3/2)x + 3 with y = mx + b, the slope (m) is 3/2

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A powerful searchlight shines on a man. The man's cross-sectional area is 0.500m2 perpendicular to the light beam, and the inten
babymother [125]

Answer:

The magnitude of the force the light beam exerts on the man is 5.9 x 10⁻⁵N

(b) the force the light beam exerts is much too small to be felt by the man.

Explanation:

Given;

cross-sectional area of the man, A = 0.500m²

intensity of light, I = 35.5kW/m²

If all the incident light were absorbed, the pressure of the incident light on the man can be calculated as follows;

P = I/c

where;

P is the pressure of the incident light

I is the intensity of the incident light

c is the speed of light

P = \frac{I}{c} =\frac{35500}{3*10^8} = 1.18*10^{-4} \ N/m^2

F = PA

where;

F is the force of the incident light on the man

P is the pressure of the incident light on the man

A is the cross-sectional area of the man

F = 1.18 x 10⁻⁴ x 0.5 = 5.9 x 10⁻⁵ N

The magnitude of the force the light beam exerts on the man is 5.9 x 10⁻⁵ N

Therefore, the force the light beam exerts is much too small to be felt by the man.

8 0
4 years ago
If the kinetic energy of a particle is doubled, by what factor has its speed increased?
Nady [450]
<span> velocity increases by √3</span>
6 0
4 years ago
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What is an atom? Give three 3 examples.​
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Answer:

<u>Atom</u><u> </u><u>is </u><u>particle</u><u> </u><u>of </u><u>matter</u><u> </u><u>that </u><u>uniquely</u><u> </u><u>defines</u><u> </u><u>a</u><u> </u><u>chemical</u><u> </u><u>element</u><u>.</u>

Examples of atom:

  • Hydrogen [ H]
  • Neon [Ne]
  • Argon [A]
  • Iron [Fe]
  • Calcium [Ca]

<h3>About Atom:</h3>

Atom consists of a central nucleus that is usually surrounded by one or more electrons .Each electron is negatively charged. The nucleus is positively charged and contains one or more relatively heavy particles known as protons and neutrons.

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Hope this helps....

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8 0
3 years ago
What is the amount of force required to accelerate a 20 kg object to 5 m/s˛?
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Answer:

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3 0
3 years ago
A 0.500-kg mass suspended from a spring oscillates with a period of 1.36 s. How much mass must be added to the object to change
lord [1]

The mass that must be added is 0.628 kg

Explanation:

The period of a mass-spring system is given by

T=2\pi \sqrt{\frac{m}{k}}

where

m is the mass

k is the spring constant

For the initial mass-spring system in the problem, we have

m = 0.500 kg

T = 1.36 s

Solving for k, we find the spring constant:

k=(\frac{2\pi}{T})^2 m = (\frac{2\pi}{1.36})^2 (0.500)=10.7 N/m

In the second part, we want the period of the same system to be

T = 2.04 s

Therefore, the mass on the spring in this case must be

m=(\frac{T}{2\pi})^2 k =(\frac{2.04}{2\pi})^2 (10.7)=1.128 kg

Therefore, the mass that must be added is

\Delta m = 1.128 - 0.500 = 0.628 kg

Learn more about period:

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#LearnwithBrainly

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