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4 years ago

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Find the magnitude of the impulse delivered to a soccer ball when a player kicks it with a force of 1450 N . Assume that the pla

yer's foot is in contact with the ball for 5.50×10−3 s .

1 answer:

defon4 years ago

5 0

Answer:

7.98 Newton Seconds

Explanation:

J = F * (Δt)

J = 1450 (5.5 * 10^-3)

J = 7.975

We only use three significant digits, so our answer becomes 9.98

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What two factors keep Earth in orbit and around the sun and the moon in orbit around Earth?

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D because bits makes the most sense logically

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1. Which of the following parts of a sound wave corresponds to a crest in a transverse wave?. A. peak. B. compression C. rarefac

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You are standing on a train station platform as a train goes by close to you. As the train approaches, you hear the whistle soun

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Answer: v = (33320 - 340f2)/( f2 + 98)

Explanation: this question refers to Doppler effect, hence,

f1 = { V / (V - v)} x f2

f1 = 98Hz, f2 = 76 Hz, V = 340m/s, v = ?, fs = fsourcs

For f1,

98 = { 340 / (340 + v)} x fs ...(i)

for f2,

f2 = { 340 / (340 - v)} x fs ... (ii)

Divide (ii) by (i)

98/f2 = [{ 340 / (340 + v)} x fs]÷ [ 340 / (340 - v)} x fs]

98/f2 = {340 / (340 + v)} x fs x (340 - v)} / 340 x fs

98/f2 = (340 + v)/ (340 - v)

Cross multiplying

98(340 - v) = f2 (340 + v)

33320 - 98v = 340f2 + vf2

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vf2 +98v = 33320 - 340f2

v(f2 +98) = 33320 - 340f2

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4 years ago

A spaceship is traveling through deep space towards a space station and needs to make a course correction to go around a nebula.

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Answer:

Magnitude = 3.64 × $10^6$

စ = 43.9°

Explanation:

given data

ship to travel = 1.7 × $10^6$ kilometers

turn = 70°

travel an additional = 2.7 × $10^6$ kilometers

solution

we will consider here

Px = 1.7 × $10^6$

Py = 0

Qx =2.7 × $10^6$ cos(70)

Qy= 2.7 × $10^6$ sin(70)

so that

Hx = Px + Qx ............1

Hx = 2.62 × $10^6$

and

Hy = Py + Qy ..........2

Hy = 2.53 × $10^6$

so Magnitude = $\sqrt{((2.62 \times 10^6 )^2+(2.53 \times 10^6)^2)}$

Magnitude = 3.64 ×

so direction will be

tan စ = Hy ÷ Hx ......................3

tan စ = $\frac{2.53}{2.62}$

tan စ = 0.9656

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3 0

3 years ago

Two protons move with uniform circular motion in the presence of uniform magnetic fields. Proton one moves twice as fast as prot

lubasha [3.4K]

Answer:

r₂ = 4 r

Explanation:

For this exercise let's use Newton's second law with the magnetic force

F = q v x B

bold letters indicate vectors, the magnitude of this expression is

F = q v B sin θ

in this case we assume that the angle is 90º between the speed and the magnetic field.

If we use the rule of the right hand with the positive charge, the thumb in the direction of the speed, the fingers extended in the direction of the magnetic field, the palm points in the direction of the force, which is towards the center of the circle, therefore the force is radial and the acceleration is centripetal

a = v² / r

let's use Newton's second law

F = ma

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r = $\frac{qB}{mv}$

Let's apply this expression to our case.

Proton 1

r = \frac{qB_1}{mv_1}

Proton 2

r₂ = $\frac{q \ B_2}{m \ v_2}$

in the exercise indicate some relationships between the two protons

* v₁ = 2 v₂

v₂ = v₁ / 2

* B₂ = 2B₁

we substitute

r₂ = $\frac{q \ 2B_1}{m \ \frac{v_1}{2} }$

r₂ = 4 $\frac{qB_1}{mv_1}$

r₂ = 4 r

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3 years ago