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Solnce55 [7]
2 years ago
14

High-speed stroboscopic photographs show that the head of a 200 g golf club is traveling at 43.7 m/s just before it strikes a 45

.9 g golf ball at rest on a tee. After the collision, the club head travels (in the same direction) at 30.7 m/s. Find the speed of the golf ball immediately after impact. Answer in units of m/s.
Physics
1 answer:
Helga [31]2 years ago
8 0

Answer:

41.27m/s

Explanation:

According to law of conservation of momentum

m1u1+m2u2 = (m1+m2)v

m1 and m2 are the masses

u1 and u2 are the initial velocities

v is the velocity after impact

Given

m1 = 0.2kg

u1 = 43.7m/s

m2 = 45.9g = 0.0459kg

u2 = 30.7m/s

Required

Velocity after impact v

Substitute the given parameters into the formula

0.2(43.7)+0.0459(30.7) = (0.2+0.0459)v

8.74+1.409 = 0.2459v

10.149 = 0.2459v

v = 10.149/0.2459

v = 41.27m/s

Hence the speed of the golf ball immediately after impact is 41.27m/s

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Nana76 [90]

Answer:

Acceleration=24.9ft^2/s^2

Angular acceleration=1.47rads/s

Explanation:

Note before the ladder is inclined at 30° to the horizontal with a length of 16ft

Hence angular velocity = 6/8=0.75rad/s

acceleration Ab=Aa +(Ab/a)+(Ab/a)t

4+0.75^2*16+a*16

0=0.75^2*16cos30°-a*16sin30°---1

Ab=0+0.75^2sin30°+a*16cos30°----2

Solving equation 1

(0.75^2*16cos30/16sin30)=angular acceleration=a=1.47rad/s

Also from equation 2

Ab=0.75^2*16sin30+1.47*16cos30=24.9ft^2/s^2

6 0
2 years ago
An object executes simple harmonic motion with an amplitude A. (Use any variable or symbol stated above as necessary.) (a) At wh
valentina_108 [34]

Answer:

(a) x=ASin(ωt+Ф₀)=±(√3)A/2

(b) x=±(√2)A/2

Explanation:

For part (a)

V=AωCos(ωt+Ф₀)⇒±0.5Aω=AωCos(ωt+Ф₀)

Cos(ωt+Ф₀)=±0.5⇒ωt+Ф₀=π/3,2π/3,4π/3,5π/3

x=ASin(ωt+Ф₀)=±(√3)A/2

For part(b)

U=0.5E and U+K=E→K=0.5E

E=K(Max)

(1/2)mv²=(0.5)(1/2)m(Vmax)²

V=±(√2)Vmax/2→ωt+Ф₀=π/4,3π/4,7π/4

x=±(√2)A/2

7 0
2 years ago
Katie rolls a toy car off the end of a table.
hoa [83]

Answer: C.

Explanation:

i choose B and got it wrong it told me the answer after on edge.

4 0
2 years ago
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How far could a rabbit run if it ran 36km/h for 5.0min?
Korolek [52]

3 kilometers, it is just 5/60 or 1/12 multiplied by 36.

4 0
2 years ago
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If a sound with frequency fs is produced by a source traveling along a line with speed vs. If an observer is traveling with spee
Alexus [3.1K]

Answer:

457.81 Hz

Explanation:

From the question, it is stated that it is a question under Doppler effect.

As a result, we use this form

fo = (c + vo) / (c - vs) × fs

fo = observed frequency by observer =?

c = speed of sound = 332 m/s

vo = velocity of observer relative to source = 45 m/s

vs = velocity of source relative to observer = - 46 m/s ( it is taking a negative sign because the velocity of the source is in opposite direction to the observer).

fs = frequency of sound wave by source = 459 Hz

By substituting the the values to the equation, we have

fo = (332 + 45) / (332 - (-46)) × 459

fo = (377/ 332 + 46) × 459

fo = (377/ 378) × 459

fo = 0.9974 × 459

fo = 457.81 Hz

7 0
2 years ago
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