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kari74 [83]
2 years ago
3

Alex kicks a soccer ball with a force of 250 N. The force is applied to the soccer ball for 0.05 seconds. What is the impulse ap

plied to the ball?
Physics
1 answer:
Bess [88]2 years ago
5 0

Answer:

12.5 Ns.

Explanation:

From the question given above, the following data were obtained:

Force (F) = 250 N.

Time (t) = 0.05 s

Impulse =?

Impulse can be obtained by using the following formula:

Impulse = force × time

Impulse = 250 × 0.05

Impulse = 12.5 Ns

Thus, the Impulse applied to the ball is 12.5 Ns.

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Two 0.40 kg soccer ball collide elastically in a head-on collision. The first ball starts at rest, and the second ball has a spe
yulyashka [42]

Explanation:

Mass of two soccer balls, m_1=m_2=0.4\ kg

Initial speed of first ball, u_1=0

Initial speed of second ball, u_2=3.5\ m/s

After the collision,

Final speed of the second ball, v_2=0

(a) The momentum remains conserved. Using the conservation of momentum to find it as :

m_1u_1+m_2u_2=m_1v_1+m_2v_2

v_1 is the final speed of the first ball

0.4\times 0+0.4\times 3.5=0.4v_1+0.4\times 0

0.4\times 3.5=0.4v_1

v_1=3.5\ m/s

(b) Let E_1 is the kinetic energy of the first ball before the collision. It is given by :

E_1=\dfrac{1}{2}mu_1^2

E_1=\dfrac{1}{2}\times 0.4\times 0

It is at rest, so, the kinetic energy of the first ball before the collision is 0.

(c) After the collision, the second ball comes to rest. So, the kinetic energy of the second ball after the collision is 0.

Hence, this is the required solution.

7 0
2 years ago
Read 2 more answers
To overcome an object's inertia, it must be acted upon by __________. A. gravity B. energy C. force D. acceleration
astra-53 [7]
In order to overcome an object’s inertia (resistance to change), it must be acted upon by an unbalanced force, so the answer to the problem is letter C.
7 0
1 year ago
Read 2 more answers
A 50.0 kg object is moving at 18.2 m/s when a 200 N force
gayaneshka [121]

Answer:

distance = 21.56 m

Explanation:

given data

mass = 50 kg

initial velocity  = 18.2 m/s

force = -200 N ( here force applied to opposite direction )

final velocity = 12.6 m/s

solution

we know here acceleration will be as

acceleration a  = force ÷ mass

a = \frac{-200}{50}   =  -4 m/s²

we get here now required time that is

required time = \frac{V_{(final)} - V_{(initial)}}{a}     ...............1

put here value

required time = \frac{12.6-18.2}{-4}  

so distance will be

distance = \frac{V_{(final)}^2 - V_{(initial)}^2}{2a}    ........2

distance = \frac{12.6}^2 -{18.2}^2}{2\times (-4)}  

distance = 21.56 m

7 0
2 years ago
During an auto accident, the vehicle's air bags deploy and slow down the passengers more gently than if they had hit the windshi
vladimir2022 [97]

Answer:

At a deceleration of 60g, or 60 times the acceleration due to gravity a person will travel a distance of 0.38 m before coing to a complete stop

Explanation:

The maximum acceleration of the airbag = 60 g, and the duration of the acceleration = 36 ms or 36/1000 s or 0.036 s

To find out how far (in meters) does a person travel in coming to a complete stop in 36 ms at a constant acceleration of 60g

we write out the equation of motion thus.

S = ut + 0.5at²

wgere

S = distance to come to complete stop

u = final velocoty = 0 m/s

a = acceleration = 60g = 60 × 9.81

t = time = 36 ms

as can be seen, the above equation calls up the given variable as a function of the required variable thus

S = 0×0.036 + 0.5×60×9.81×0.036² = 0.38 m

At 60g, a person will travel a distance of 0.38 m before coing to a complete stop

7 0
2 years ago
Charge q1 is distance s from the negative plate of a parallel-plate capacitor. Charge q2=q1/3 is distance 2s from the negative p
Svetlanka [38]

Answer:

The ratio (U₁/U₂) = 6

Explanation:

U, the potential energy is given as

U = kqQ/r

k = Coulomb's constant

q = charge we're concerned about

Q = charge of the negative plate of the capacitor

r = distance of q from the negative plate of the capacitor.

For charge q₁

U₁ = kq₁Q/s

U₂ = kq₂Q/2s

But q₂ = q₁/3

U₂ becomes U₂ = kq₁Q/6s

U₁ = kq₁Q/s

U₂ = kq₁Q/6s

(U₁/U₂) = 6

5 0
2 years ago
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