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3 years ago

6. A golf ball is hit a distance of 300 yards in 10 sec. What is the speed of the golf ballo

Physics

1 answer:

katen-ka-za [31]3 years ago

3 0

The speed of the ball is 27.4 m/s

Explanation:

The speed of an object is given by:

$speed=\frac{d}{t}$

where

d is the distance covered

t is the time taken

In this problem, we have:

d = 300 yards is the distance covered by the golf ball

t = 10 s is the time taken

Keeping in mind that

1 yard = 0.914 m

We can convert the distance from yards to meters:

$d = 300 \cdot 0.914 = 274.2 m$

And substituting into the equation, we find the speed of the ball:

$speed=\frac{274.2}{10}=27.4 m/s$

Learn more about speed:

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Answer:

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Explanation:

if you do so, then your mom will have coronavirus.

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3 years ago

What is the GPE of a 15,000 kg airplane sitting on the ground?

Vesna [10]

Answer:

C, it is not moving

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2 years ago

A vacuum cleaner produces sound with a measured sound level of 75.0 dB. (a) What is the intensity of this sound in W/m2? W/m2 (b

jonny [76]

Answer:

Part a)

$I = 3.16 \times 10^{-5} W/m^2$

Part b)

$P_o = 0.162 Pa$

Explanation:

Part a)

Level of sound = 75 dB

now we know that

$L = 10 Log\frac{I}{I_0}$

here we know that

$I_0 = 10^{-12} W/m^2$

now we have

$75 = 10 Log(\frac{I}{10^{-12}})$

$I = 3.16 \times 10^{-5} W/m^2$

Part b)

Intensity of sound wave is given as

$I = \frac{1}{2}\rho A^2\omega^2 c$

here we know that

$A = \frac{P_o}{Bk}$

so we have

$I = \frac{1}{2}\rho(\frac{P_o}{Bk})^2\omega^2 c$

$I = \frac{1}{2}\rho P_o^2 \frac{c^3}{B^2}$

now we know

$\rho = 1.2 kg/m^3$

$c = 340 m/s$

$B = 1.4 \times 10^5 Pa$

now we have

$3.16 \times 10^{-5} = \frac{1}{2}(1.2)P_o^2\frac{340^3}{(1.4\times 10^5)^2}$

$P_o = 0.162 Pa$

5 0

3 years ago

A vertical spring has a length of 0.25 m when a 0.175 kg mass hangs from it, and a length of 0.775 m when a 2.075 kg mass hangs

Contact [7]

Answer:

A) 35.5N/m b) 20.1cm

Explanation:

Using Hooke's law;

F = Ke where F is the weight of the object = mass of the object in kg * acceleration due to gravity in m/s^2 and k if the force constant of the spring in N/m and e is the extension of the spring which original length of the spring - new length after extension in meters

For the first body, m*g = K * (0.25- li)

Where li is the initial length of the spring

0.175*9.81 = k(0.25-li)

1.72 = k(0.25-li) as equation 1

For the second body, m *g = K* ( 0.775-li)

2.075*9.81 = k (0.775-li) equation 2

20.36 = k(0.775-li)

Make li subject of the formula;

li = 0.775 - 20.36/k

Substitute for li in equation 1

1.72 = k(0.25- (0.775 - 20.36/k))

1.72 = k ( 0.25 - 0.775 + 20.36/k)

Open the bracket with k

1.72 = 0.25k - 0.775k + 20.36 (since k cancel k)

Collect the like terms:

1.72 - 20.36 = - 0.525k

- 18.64 = -0.525k

Divide both side by -0.525

-18.64/-0.525 = -0.525/-0.525k

K = 35.5N/m

B) substitute for k in using

li = 0.775 - 20.36/k

li = 0.775 - 20.36/35.5

li = 0.775 - 0.574

li = 0.201 in meters

li = 0.201 * 100 centimeters = 20.1cm

4 0

3 years ago

A rocket is launched straight up from the earth's surface at a speed of 1.60×104 m/s . what is its speed when it is very far awa

AlladinOne [14]

16,000 m/s
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3 years ago