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3 years ago

Forces that are equal in size but opposite in direction are ____.

Physics

1 answer:

Alchen [17]3 years ago

8 0

Answer: The correct answer is balanced force.

Explanation:

Balanced forces are balanced when the forces are acting equal in magnitude but opposite in direction. The balanced forces will not cause change in the speed of the object.

Unbalanced forces are unbalanced when the forces acting on the object are not equal in magnitude. The combined force is the difference between the forces. It will cause the change in the speed of the body.

Therefore, forces that are equal in size but opposite in direction are balanced.

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A) an electron has an initial speed of 226000 m/s. if it undergoes an acceleration of 4.0 x 1014 m/s2, how long will it take to

KIM [24]

initial speed of 226000 m/s

acceleration of 4.0 x 1014 m/s2,

speed of 781000 m/s

What is Acceleration?

Acceleration is a rate of change of velocity with respect to time with respect to direction and speed.

A point or an object moving in a straight line is accelerated if it speeds up or slows down.

Acceleration formula can be written as,

a = (v - u ) / t m/s²

As we have to find the time taken, the formula can be altered as,

$t = \frac{v-u}{a}$

where, t - time taken to reach a final speed

v - final velocity

u - initial velocity

a - acceleration.

Substituting all the given values,

$t =\frac{781000 - 226000} {4* 1014}$

= 1.3875 × 10⁻⁹ seconds.

So, taken to reach the final speed is found to be 1.3 × 10⁻⁹ 8iH..

7 0

2 years ago

S ship maneuvers to within 2.50x10^3 m of an islands 1.80x10^3 m high mountain peak and fires a projectile at an enemy ship 6.10

lilavasa [31]

velocity of the projectile is given as

$v = 2.50 \times 10^3 m/s$

now the angle is given as 75 degree

so here we will have

$v_x = 2.50 \times 10^3 cos75 = 0.65 \times 10^3 m/s$

$v_y = 2.50 \times 10^3 sin75 = 2.4 \times 10^3 m/s$

now the time taken to reach the peak is given as

$t = \frac{x}{v_x}$

$t = \frac{2.50 \times 10^3}{0.65 \times 10^3} = 3.86 s$

now the height moved by the projectile in same time is given as

$y = v_y t + \frac{1}{2}at^2$

now we have

$y = (2.4 \times 10^3)(3.86) - \frac{1}{2}(9.81)(3.86)^2$

$y = 9.2 \times 10^3 m$

so distance between the peak and projectile is given as

$d = 9.2 \times 10^3 - 1.80 \times 10^3 = 7.4 \times 10^3 m$

5 0

3 years ago

A car traveling at 21 m/s starts to decelerate steadily. It comes to a complete stop in 6 seconds. What is its acceleration? Ans

allsm [11]

-3.5mls squared = a

0mls = 21mls + a(6s)
-21mls/6s

The equation used in this is vf = vi + at

Hope this helps!

4 0

3 years ago

Ling lives 2 miles from school. It took him 15 minutes to bike from school to home. The first half of the distance he biked at a

Fittoniya [83]

Explanation:

As we know that

time = distance/speed

The time used for firs half of the trip was

(1 mi)(12 mi/hour) = 1/12 hours = 5 minutes

The last half of the trip will took 10 minutes, 1/6 hour.

Speed = distance/ time

(1 mil) = (1/6h) = 6 mil/h

so the speed for last half of the trip was = 6mph

the average speed was

(2mil)(1/4 hour) = 8 mil/hour

So the ling's average speed was 8mph.

3 0

4 years ago

A satellite is launched to orbit the Earth at an altitude of 3.25 107 m for use in the Global Positioning System (GPS). Take the

Korolek [52]

Answer:Orbital period =21.22hrs

Explanation:

given that

mass of earth M = 5.97 x 10^24 kg

radius of a satellite's orbit, R= earth's radius + height of the satellite

6.38X 10^6 + 3.25 X10^7 m =3.89 X 10^7m

Speed of satellite, v= $\sqrt GM/R$

where G = 6.673 x 10-11 N m2/kg2

V= \sqrt (6.673x10^-11 x 5.97x10^ 24)/(3.89 X 10^ 7m)

V =10,241082.2

v= 3,200.2m/s

a) Orbital period

$\sqrt GM/R$ = $\frac{2\pi r}{T}$

V= $\frac{2\pi r}{T}$

T= 2 $\pi$ r/ V

= 2 X 3.142 X 3.89 X 10^7m/ 3,200.2m/s

=76,385.1 s

60 sec= 1min

60mins = 1hr

76,385.1s =hr

76,385.1/3600=21.22hrs

3 0

3 years ago