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Novay_Z [31]
3 years ago
15

Two forces that are not equal in size are

Physics
1 answer:
Bumek [7]3 years ago
4 0
They are unbalanced forces ..... Hope this helps :3
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How many pets do you have???!!!??
jasenka [17]

Answer:

Right now I have three.

Explanation: Thanks for the points luv ^-^.

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3 years ago
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A boy kicks a soccer ball, giving it an initial speed of 7.5m/s at an angle of 27° above the horizontal(=ground). How high will
Gelneren [198K]

When the initial speed given is 7.5m/s at an angle of 27° , ball will go

4.637 meters.

Assume no air opposition to the ball ;

Vertical component of ball is sin 27° =  0.453

0.453* 7.5 = 3.404 meters /sec

Time taken to reach ground is :

3.404 = -3.404+9.8*t

t= 6.808/9.8= 0.694 sec

Horizontal component is 7.5*cos27°= 6.682m/s

Distance = speed * time

=6.682 * 0.694

=4.637 meters

Horizontal distance it can cover in 0.694 sec is 4.637 meters

So range of ball is 4.637 meters.

Form of motion experienced by an object or particle that is projected near surface of the earth and moves along a curve is called Projectile motion. Three types of projectile motion are Horizontal projectile motion. Oblique projectile motion and Projectile motion on an inclined plane.

To know more about projectile motion, refer

brainly.com/question/24216590

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3 0
1 year ago
The charge that passes a cross-sectional area A=10-4 m2 varies with time according
Anarel [89]

Answer:

Current = dQ/dt

or I = dQ/dt

Where I represents current.

Which is the rate of flow of charge.

Q=4 + 2t + t²

dQ/dt = 2 + 2t --- This is the relation that gives the instantaneous current.

At time t=2sec

dQ/dt = I = 2 + 2t

= 2 + 2(2)

=2 + 4

= 6A.

6 0
3 years ago
Kinetic or potential energy
chubhunter [2.5K]
Potential, if the bow is released kinetic
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3 years ago
Consider two identical and symmetrical wave pulses on a string. Suppose the first pulse reaches the fixed end of the string and
Nesterboy [21]

Answer:

The amplitude of the resultant wave will be 0.

Explanation:

Suppose the first wave has an amplitude of A. Its angle is given as wt.

The second way will also have the same amplitude as that of first.

After the reflection, a phase shift of π is added So the wave is given as

W_1=W_2=Acos(\omega t)\\W_1^{'}=Acos(\omega t+ \pi)

Adding the two waves give

                               W_1'+W_2=Acos(\omega t+ \pi)+Acos(\omega t)\\W_1'+W_2=-Acos(\omega t)+Acos(\omega t)\\W_1'+W_2=0

So the amplitude of the resultant wave will be 0.

7 0
3 years ago
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