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velikii [3]
2 years ago
13

A 62.3 N force pulls an object at an angle of 9 = 22.0° to its direction of horizontal motion. It moves a

Physics
1 answer:
Debora [2.8K]2 years ago
8 0

W = F cos α . d

W = 62.3 cos 22 . 12.1

W = 698.94 J

You might be interested in
What happen when a star dies?
Vera_Pavlovna [14]
It depends on the size of the star. If it's size was normal then it cools down into White dwarf, then a black dwarf. If a really huge star dies, then we can see a "Supernova" from that.

Hope this helps!!
7 0
3 years ago
As you move away from a positive charge distribution, the electric field:
GalinKa [24]

Answer:

The electric field always decreases.

Explanation:

The electric field due to a point charge is given by :

E=\dfrac{kq}{r^2}

Where

k = electric constant

q = charge

r = distance from the charge

It is clear from the above equation that as the distance from the charge particle increases the electric field decreases. As you move away from a positive charge distribution, the electric field always decreases. Hence, the correct option is (c) "Always decreases".

3 0
3 years ago
A soccer ball with mass 0.450 kg is initially moving with speed 2.20 m/s. A soccer player kicks the ball, exerting a constant fo
Alinara [238K]

Answer:

0.187 m

Explanation:

We'll begin by calculating the acceleration of the ball. This can be obtained as follow:

Mass (m) = 0.450 Kg

Force (F) = 38 N

Acceleration (a) =?

F = m × a

38 = 0.450 × a

Divide both side by 0.450

a = 38 / 0.450

a = 84.44 m/s²

Finally, we shall determine the distance. This can be obtained as follow:

Initial velocity (u) = 2.20 m/s.

Final velocity (v) = 6 m/s

Acceleration (a) = 84.44 m/s²

Distance (s) =?

v² = u² + 2as

6² = 2.2² + (2 × 84.44 × s)

36 = 4.4 + 168.88s

Collect like terms

36 – 4.84 = 168.88s

31.52 = 168.88s

Divide both side by 168.88

s = 31.52 / 168.88

s = 0.187 m

Thus, the distance is 0.187 m

6 0
3 years ago
A jet accelerates from rest down a runway at 1.75m/s² for a distance of 1500 m before takeoff.
babunello [35]
A. Using the third equation of motion:
v2 = u2 + 2as
from the question;
the jet was initially at rest
hence u = 0
a = 1.75m/s2
s = 1500m
v2 = 02 + 2(1.75)(1500)
v2 = 5250
v = √5250
v = 72.46m/s
hence it moves with a velocity of 72.46m/s.
b. s = ut + 1/2at2
1500 = 0(t) + 1/2(1.75)t2
1500 × 2 = 2× 1/2(1.75)t2
3000 = 1.75t2
1714.29 = t2
41.4 = t
hence the time taken for the plane to down the runway is 41.4s.


Read more on Brainly.com - brainly.com/question/18743384#readmore
8 0
3 years ago
A racing car accelerates at the end of the race from a speed of 100
nevsk [136]

4/3 m/s ( approximately 1.3333... m/s)

5 0
2 years ago
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