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3 years ago

What use do we have for motion diagrams?

Physics

1 answer:

mamaluj [8]3 years ago

7 0

Answer:

A motion diagram represents the motion of an object by displaying its location at various equally spaced times on the same diagram. Motion diagrams are a pictorial description of an object's motion. They show an object's position and velocity initially and present several spots in the center of the diagram.

Explanation:

You might be interested in

If the frequency of an electromagnetic wave increases, does the number of waves passing by you increase, decrease, or stay the s

jarptica [38.1K]

Answer:

option(d)

Explanation:

The frequency of a wave is the property of the source of wave.

The velocity of all the electromagnetic waves is same as the speed of light. It only changes as the light passes through one medium to another.

The frequency is defined as the number of waves coming out from the source in 1 second.

As the frequency of wave increases, the number of wave coming per second increases.

So, the number of waves passing by increases but the speed remains same.

Option (d)

3 0

3 years ago

The light bulb in your room is not very bright. You want to use a concave mirror to direct its illumination into a tight, parall

ehidna [41]

A point source is to be used with a concave mirror to produce a beam of parallel light. The source should be placed midway between the center of curvature and the mirror. To solve this question you have to draw a ray diagram. Apply the mirror equation to determine the image distance and finally make sure that the first two steps are consistent with each other.

<span>I hope this helps, Regards.</span>

5 0

3 years ago

1. Faça as transformações:

zhannawk [14.2K]

Answer:

$seconds (s) = hours (h) *3,600 ; h = \frac{s}{3,600} \\g = kg * 1,000; kg = \frac{g}{1,000} \\cm = \frac{m}{100};m = cm * 100$

1. a) 0.5 h = 1,800 s

h) 20 cm = 0.2 m

b) 2.0 h = 7,200 s

i) 5.0 kg = 5,000 g

c) 3.5 h = 12,600 s

j) 1.5 kg = 1,500 g

d) 1/4 h = 900 s

k) 450.0 g = 0.45 kg

e) 3.0 m = 300 cm

l) 20.0 g = 0.02 kg

f) 2.5 m = 250 cm

m) 500.0 g = 0.5 kg

g) 0.5 m = 500 mm

n) 1000.0 g = 1 kg

3 0

3 years ago

Consider four point charges arranged in a square with sides of length L. Three of the point charges have charge q and one of the

nydimaria [60]

Answer: $F_{net}=\frac{kq^2}{(L)^2}\left [ \frac{1}{2}+\sqrt{2}\right ]$

Explanation:

Given

Three charges of magnitude q is placed at three corners and fourth charge is placed at last corner with -q charge

Force due to the charge placed at diagonally opposite end on -q charge

$F_1=\frac{kq(-q)}{(L\sqrt{2})^2}$

where $L\sqrt{2}=$ Distance between the two charges

$F_1=-\frac{kq^2}{2L^2}$

negative sign indicates that it is an attraction force

Now remaining two charges will apply the same amount of force as they are equally spaced from -q charge

$F_2=\frac{kq(-q)}{(L)^2}$

The magnitude of force by both the charge is same but at an angle of $90^{\circ}$

thus combination of two forces at 2 and 3 will be

$F'=\sqrt{2}\frac{kq^2}{2L^2}$

Now it will add with force due to 1 charge

Thus net force will be

$F_{net}=\frac{kq^2}{(L)^2}\left [ \frac{1}{2}+\sqrt{2}\right ]$

6 0

3 years ago

As a laudably skeptical physics student, you want to test Coulomb's law. For this purpose, you set up a measurement in which a p

kherson [118]

Explanation:

The Coulomb's law states that the magnitude of each of the electric forces between two point-at-rest charges is directly proportional to the product of the magnitude of both charges and inversely proportional to the square of the distance that separates them:

$F=\frac{kq_1q_2}{d^2}$

In this case we have an electron (-e) and a proton (e), so:

$F=-\frac{ke^2}{d^2}\\F=-\frac{8.99*10^9\frac{N\cdot m^2}{s^2}(1.6*10^{-19}C)^2}{(933*10^{-9}m)^2}\\F=-2.64*10^{-16}N$

In this case, the electric force is negative, therefore, the force is repulsive and its magnitude is:

$F=2.64*10^{-16}N$

3 0

3 years ago