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Andreas93 [3]
3 years ago
10

What effect does an unbalanced force have on an object?

Physics
2 answers:
Vladimir79 [104]3 years ago
7 0

<u>Answer:</u> This force will cause the motion in an object.

<u>Explanation:</u>

Unbalanced force is defined as the forces which are not equal in magnitude. When two forces with different magnitudes are applied on an object from opposite directions, the object will start moving in the direction of the net force.

These forces result in the motion of an object.

Balanced forces are the forces which are equal in magnitude. When two forces with equal magnitude are applied on an object from opposite directions, the object will not start moving.

These forces does not result in the motion of an object.

Hence, this force will cause the motion in an object.

Mars2501 [29]3 years ago
5 0
<span>it either changes the direction of the object or it stops the object in motion.</span>
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Convert the mass to kgs thanks
Natali [406]
Memorize this and you'll be able to do ALL of these:  <em>1  kg = 1,000 g</em>

So if you have some grams, divide the number by 1,000 to get kilograms.

1,000 g = 1.000 kg

500 g = 0.500 kg

100 g = 0.100 kg

50 g = 0.050 kg

20 g = 0.020 kg

10 g = 0.010 kg
4 0
3 years ago
3. Compare the slope of the velocity-time graph to the average of all your acceleration values. Are they close? What does the sl
Marina86 [1]

The slope of a speed-time graph is the acceleration represented by the graph.

All other parts of this question refer to a lab experiment or exercise
where I was not present, but Zeesam16 was.  Therefore I have no data
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6 0
3 years ago
A camera with a 50.0-mm focal length lens is being used to photograph a person standing 3.00 m away. (a) How far from the lens m
kirill [66]

a) 50.8 mm

b) The whole image (1:1)

c) It seems reasonable

Explanation:

a)

To project the image on the film, the distance of the film from the lens must be equal to the distance of the image from the lens. This can be found by using the lens equation:

\frac{1}{f}=\frac{1}{p}+\frac{1}{q}

where

f is the focal length of the lens

p is the distance of the object from the lens

q is the distance of the image from the lens

In this problem:

f = 50.0 mm = 0.050 m is the focal length (positive for a convex lens)

p = 3.00 m is the distance of the person from the lens

Therefore, we can find q:

\frac{1}{q}=\frac{1}{f}-\frac{1}{p}=\frac{1}{0.050}-\frac{1}{3.00}=19.667m^{-1}\\q=\frac{1}{19.667}=0.051 m=50.8 mm

b)

Here we need to find the height of the image first.

This can be done by using the magnification equation:

\frac{y'}{y}=-\frac{q}{p}

where:

y' is the height of the image

y = 1.75 m is the height of the real person

q = 50.8 mm = 0.0508 m is the distance of the image from the lens

p = 3.00 m is the distance of the person from the lens

Solving for y', we find:

y'=-\frac{qy}{p}=-\frac{(0.0508)(1.75)}{3.00}=-0.0296 m=-29.6mm

(the negative sign means the image is inverted)

Therefore, the size of the image (29.6 mm) is smaller than the size of the film (36.0 mm), so the whole image can fit into the film.

c)

This seems reasonable: in fact, with a 50.0 mm focal length, if we try to take the picture of a person at a distance of 3.00 m, we are able to capture the whole image of the person in the photo.

3 0
3 years ago
The measure of a spring’s resistance to being compressed or stretched is the
larisa86 [58]

<u>Answer;</u>

<em>Spring constant </em>

<u>Explanation;</u>

The measure of a spring’s resistance to being compressed or stretched is the <u>spring constant</u>.

  • The symbol of spring constant is K, since it is a constant. From the Hooke's law,for a helical spring or any elastic material, the extension force is directly proportional to the extension provided the elastic limit is not exceeded.
  • Therefore; the spring constant = Force/extension. That is; K = F/e; where k is the spring constant, F is the extension force and e is the extension.
  • Spring constant depicts the resistance of the spring to compressional and stretching forces.
7 0
3 years ago
Read 2 more answers
A wire with a length of 150 m and a radius of 0.15 mm carries a current with a uniform current density of 2.8 x 10^7A/m^2. The c
Mrac [35]

Answer:

The current is 2.0 A.

(A) is correct option.

Explanation:

Given that,

Length = 150 m

Radius = 0.15 mm

Current densityJ=2.8\times10^{7}\ A/m^2

We need to calculate the current

Using formula of current density

J = \dfrac{I}{A}

I=J\timesA

Where, J = current density

A = area

I = current

Put the value into the formula

I=2.8\times10^{7}\times\pi\times(0.15\times10^{-3})^2

I=1.97=2.0\ A

Hence, The current is 2.0 A.

7 0
3 years ago
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