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miskamm [114]
3 years ago
7

High speed stroboscopic photographs show that the head of a 244 g golf club is traveling at 57.6 m/s just before it strikes a 45

.2 g golf ball at rest on a tee. After the collision, the club head travels (in the same direction) at 39.9 m/s. Find the speed of the golf ball just after impact.
Physics
1 answer:
almond37 [142]3 years ago
6 0

Answer:

The speed will be "1.06 m/s".

Explanation:

The given values are:

Momentum,

m1 = 244 g

m2 = 45.2 g

On applying momentum conservation ,

Let v2 become the final golf's speed.  

From Momentum Conservation

⇒  Total \ initial \ momentum = Total \ final \ momentum

⇒  m1\times u1 + m2\times u2 = m1\times v1 + m2\times v2

On putting the estimated values, we get

⇒  0.244\times 57.6+0=0.244\times 39.9+45.2\times v2

⇒  57.844+0=9.7356+45.2\times v2

⇒  48.1084=45.2\times v2

⇒  v2=\frac{48.1084}{45.2}

⇒  v2=1.06 \ m/s

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An archer puts a 0.30-kg arrow to the bowstring. An average force of 201 N is exerted to draw the string back 1.3 m. Assuming th
vovikov84 [41]

Answer:

41.74 m/s

Explanation:

The energy used to draw the bowstring = the kinetic energy of the arrow.

Fd = 1/2mv²................................ Equation 1

Where F = force, d = distance move string, m = mass of the arrow, v = speed of the arrow.

make v the subject of the equation

v = √(2Fd/m)...................... Equation 2

Given: F = 201 N, m = 0.3 kg, d = 1.3 m.

Substitute into equation 2

v = √(2×201×1.3/0.3)

v = √(1742)

v = 41.74 m/s.

Hence the arrow leave the bow with a speed of 41.74 m/s

3 0
3 years ago
Read 2 more answers
7. En la sala de una casa hay una gran ventana de vidrio, por la que se presenta una pérdida significativa de calor; las medidas
jonny [76]

Answer:

I DON'T SPEAK TACO BELL!

Explanation:

7 0
3 years ago
Which of the following is best described as an inclined plane twisted into a spiral?
Margarita [4]

the answer is A or B

5 0
3 years ago
A stretched string has a mass per unit length of 5.40 g/cm and a tension of 17.5 N. A sinusoidal wave on this string has an ampl
kondaur [170]

Answer:

Part a)

y_m = 0.157 mm

part b)

k = 101.8 rad/m

Part c)

\omega = 579.3 rad/s

Part d)

here since wave is moving in negative direction so the sign of \omega must be positive

Explanation:

As we know that the speed of wave in string is given by

v = \sqrt{\frac{T}{m/L}}

so we have

T = 17.5 N

m/L = 5.4 g/cm = 0.54 kg/m

now we have

v = \sqrt{\frac{17.5}{0.54}}

v = 5.69 m/s

now we have

Part a)

y_m = amplitude of wave

y_m = 0.157 mm

part b)

k = \frac{\omega}{v}

here we know that

\omega = 2\pi f

\omega = 2\pi(92.2) = 579.3 rad/s

so we  have

k = \frac{579.3}{5.69}

k = 101.8 rad/m

Part c)

\omega = 579.3 rad/s

Part d)

here since wave is moving in negative direction so the sign of \omega must be positive

4 0
3 years ago
Question 2 (1 point)
Tju [1.3M]

Answer:

I know someone anwsered but it would be 400M

Explanation:

i initial velocity (u)=10m/s

acceleration (a)=0

time taken (t) =40s

then distance (s)=u t +1/2 a t^2

s= u t +0 (as a is 0)

s= 10 x 40

s= 400M

7 0
3 years ago
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