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andrezito [222]
4 years ago
11

A puck moves 2.35 m/s in a -22 degree direction. A hockey stick pushes it for 0.215 s, changing its velocity to 6.42 m/s in a 50

degree direction. What is displacement y?
Physics
1 answer:
Whitepunk [10]4 years ago
5 0

Answer:

Displacement in Y direction is 0.434 m

Explanation:

initial velocity of the puck is 2.35 m/s at -22 degree

so here it is given as

v_i = 2.35 cos22 \hat i - 2.35 sin22 \hat j

v_i = 2.18 \hat i - 0.88 \hat j

final velocity is given as 6.42 m/s at 50 degree

so we have

v_f = 6.42 cos50 \hat i + 6.42 sin50\hat j

v_f = 4.13 \hat i + 4.92\hat j

now displacement in Y direction is given as

\Delta y = \frac{v_f + v_i}{2}t

\Delta y = \frac{-0.88 + 4.92}{2} (0.215)

\Delta y = 0.434 m

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A guitar string is 0.620m long, and oscillates at 234Hz. What is the velocity of the waves in the string? m/s
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Answer:

v = 72.54 m/s

Explanation:

We have,

Length of a guitar string is 0.62 m

Frequency of a guitar string is 234 Hz

For guitar string,

L=2\lambda\\\\\lambda=\dfrac{L}{2}\\\\\lambda=\dfrac{0.62}{2}\\\\\lambda=0.31\ m

The velocity of the wave in the string is given by :

v=f\lambda\\\\v=234\times 0.31\\\\v=72.54\ m/s

So, the velocity of the waves in the string is 72.54 m/s.

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4 years ago
Tectonic plates are large segments of the earth's crust that move slowly. suppose one such plate has an average speed of 4.8 cm
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6 0
4 years ago
The magnetic field perpendicular to a single 16.7-cm-diameter circular loop of copper wire decreases uniformly from 0.750 T to z
sammy [17]

Answer:

1.24 C

Explanation:

We know that the magnitude of the induced emf, ε = -ΔΦ/Δt where Φ = magnetic flux and t = time. Now ΔΦ = Δ(AB) = AΔB where A = area of coil and change in magnetic flux = Now ΔB = 0 - 0.750 T = -0.750 T, since the magnetic field changes from 0.750 T to 0 T.

The are , A of the circular loop is πD²/4 where D = diameter of circular loop = 16.7 cm = 16.7 × 10⁻²m

So, ε = -ΔΦ/Δt = -AΔB/Δt= -πD²/4 × -0.750 T/Δt = 0.750πD²/4Δt.

Also, the induced emf ε = iR where i = current in the coil and R = resistance of wire = ρl/A where ρ = resistivity of copper wire =1.68 × 10⁻⁸ Ωm, l = length of wire = πD and A = cross-sectional area of wire = πd²/4 where d = diameter of wire = 2.25 mm = 2.25 × 10⁻³ m.

So, ε = iR = iρl/A = iρπD/πd²/4 = 4iρD/d²

So,  4iρD/d² = 0.750πD²/4Δt.

iΔt = 0.750πD²/4 ÷ 4iρD/d²

iΔt = 0.750πD²d²/16ρ.

So the charge Q = iΔt

= 0.750π(Dd)²/16ρ

= 0.750π(16.7 × 10⁻²m 2.25 × 10⁻³ m)²/16(1.68 × 10⁻⁸ Ωm)

= 123.76 × 10⁻² C

= 1.2376 C

≅ 1.24 C

4 0
4 years ago
Two charges, qA and qB, are separated by a distance, d, and exert a force, F, on each other. Analyze Coulomb's law and answer th
maria [59]

Answer: a. F doubled

b. F reduced by one-quarter i.e

1/4*(F)

c. 1/9*(F)

d. F increased by a factor of 4 i.e 4*F

e. F reduces 3/4*(F)

Explanation: Coulombs law states the force F of attraction/repulsion experience by two charges qA and qB is directly proportional to thier product and inversely proportional to the square of distance d between them. That is

F = k*(qA*qB)/d²

a. If qA is doubled therefore the force is doubled since they are directly proportional.

b. If qA and qB are half, that means thier new product would be qA/2)*qB/2 =qA*qB/4

Which means the product of charge is divided by 4 so the force would be divided by 4 too since they are directly proportional.

c. If d is tripped that is multiplied by 3. From the formula new d would be (3*d)²=9d² but force is inversely proportional to d² so instead of multiplying by 9 the force will be divided by 9

d. If d is cut into half that is divided by 2. The new d would be (d/2)²=d²/4. So d² is divided by 4 so the force would be multiplied by 4

e. If qA is tripled that is multiplied by 3. F would be multiplied by 3 also, if at the same time d is doubled (2*d)²= 4*d² . Force would be divided by 4 at same time. So we have,

3/4*F

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4 years ago
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