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Karo-lina-s [1.5K]
3 years ago
15

A boy and his skateboard have a combined mass of 65 kg what is the speed of the boy and skateboard if they have a momentum of 27

5 kg•m/s
Physics
1 answer:
kvasek [131]3 years ago
3 0
Let m = 65 kg, the mass of the boy and skateboard.
Let v = speed of the boy and skateboard.

The momentum is
P = mv = (65 kg)*(v m/s) = 275 (kg-m)/s²
v = 275/65 = 4.23 m/s

Answer:  4.23 m/s
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The speed of sound in room temperature (20°C) air is 343 m/s; in room temperature helium, it is 1010 m/s. The fundamental freque
Lera25 [3.4K]

Answer: f = 927.55Hz

Explanation: Since the the tube is open-closed, the length of air and the wavelength of sound passing through the tube is given below

L = λ/4 where λ = wavelength.

speed of sound in air = v = 343m/s.

fundamental frequency of open closed tube = 315Hz

λ = 4L.

v = fλ

343 = 315 * 4L

343 = 1260 * L

L = 343/ 1260

L = 0.27m

In the same tube of length L = 0.27m but different medium ( helium), the speed of sound is 1010m/s.

The length of tube and wavelength are related by the formulae below

L = λ/4, λ=4L

λ = 4 * 0.27

λ = 1.087m.

v = fλ

1010 = f * 1.087

f = 1010/1.807

f = 927.55Hz

4 0
3 years ago
A student tosses a ball horizontally from a balcony to a friend 3.8 meters down below them. How long does the ball take to reach
Vsevolod [243]

Answer:

The time it takes the ball to fall 3.8 meters to friend below is approximately 0.88 seconds

Explanation:

The height from which the student tosses the ball to a friend, h = 3.8 meters above the friend

The direction in which the student tosses the ball = The horizontal direction

Given that the ball is tossed in the horizontal direction, and not the vertical direction, the initial vertical component of the velocity of the ball = 0

The equation of the vertical motion of the ball can therefore, be represented by the free fall equation as follows;

h = 1/2 × g × t²

Where;

g = The acceleration due gravity of the ball = 9.81 m/s²

t = The time of motion to cover height, h

Then height is already given as h = 3.8 m

Substituting gives;

3.8 = 1/2 × 9.81 × t²

t² = 3.8/(1/2 × 9.81) ≈ 0.775 s²

∴ t = √0.775 ≈ 0.88 seconds

The time it takes the ball to fall 3.8 meters to friend below is t ≈ 0.88 seconds.

8 0
3 years ago
Tell the value of the underlined digit 843,208,732,833 eight is underlined
devlian [24]
The value of the underlined 8 is, hundred billion's. Hope this helped!
4 0
4 years ago
Read 2 more answers
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s2008m [1.1K]

Answer:

fibrous =potato

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5 0
3 years ago
A 46.8-g golf ball is driven from the tee with an initial speed of 58.8 m/s and rises to a height of 24.7 m. (a) Neglect air res
Andre45 [30]

Answer:

a) the kinetic energy of the ball at its highest point is 69.58 J

b) its speed when it is 8.11 m below its highest point is 55.97 m/s

Explanation:

Given that;

mass of golf ball m = 46.8 g = 0.0468 kg

initial speed of the ball v₁ = 58.8 m/s

height h = 24.7 m

acceleration due to gravity = 9.8 m/s²

the kinetic energy of the ball at its highest point = ?

from the conservation of energy;

Kinetic energy at the highest point will be;

K.Ei + P.Ei = KEf + PEf

now the Initial potential energy of the ball P.Ei = 0 J

so

1/2mv² + 0 J = KEf + mgh

K.Ef = 1/2mv² - mgh

we substitute

K.Ef = [1/2 × 0.0468 × (58.8 )²] - [0.0468 × 9.8 × 24.7]

K.Ef  = 80.904 - 11.3284

K.Ef = 69.58 J

Therefore, the kinetic energy of the ball at its highest point is 69.58 J

b) when the ball is 8.11 m below the highest point, speed = ?

so our raw height h' will be ( 24.7 m - 8.11 m) = 16.59 m

so our velocity will be v₂

also using the principle of energy conservation;

K.Ei + P.Ei = KEh + PEh

1/2mv² + 0 J = 1/2mv₂² + mgh'

1/2mv₂² = 1/2mv² - mgh'

multiply through by 2/m

v₂² = v² - 2gh'

v₂ = √( v² - 2gh' )

we substitute

v₂ = √( (58.8)² - 2×9.8×16.59 )

v₂ = √( 3457.44 - 325.164 )  

v₂ = √( 3132.276 )

v₂ = 55.97 m/s

Therefore, its speed when it is 8.11 m below its highest point is 55.97 m/s

5 0
3 years ago
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