Can someone please answer this question I'm doing the test now?
vf ^2 = kx^2/m = 56(0.75)^2 / 2.5 = 12.6
Therefore, v= 3.5 m/s.
Answer:
4.45×10¯¹¹ N
Explanation:
From the question given above, the following data were obtained:
Mass of ball (M₁) = 4 Kg
Mass of bowling pin (M₂) = 1.5 Kg
Gravitational constant (G) = 6.67×10¯¹¹ Nm²/Kg²
Distance apart (r) = 3 m
Force of attraction (F) =?
The force of attraction between the ball and the bowling pin can be obtained as follow:
F = GM₁M₂ / r²
F = 6.67×10¯¹¹ × 4 × 1.5 / 3²
F = 4.002×10¯¹⁰ / 9
F = 4.45×10¯¹¹ N
Therefore, the force of attraction between the ball and the bowling pin is 4.45×10¯¹¹ N
Answer:
(a) V = 0.75 m/s
(b) V = 0.125 m/s
Explanation:
The speed of the flow of the river can be given by following formula:
V = Q/A
V = Q/w d
where,
V = Speed of Flow of River
Q = Volume Flow Rate of River
w = width of river
d = depth of river
A = Area of Cross-Section of River = w d
(a)
Here,
Q = (300,000 L/s)(0.001 m³/1 L) = 300 m³/s
w = 20 m
d = 20 m
Therefore,
V = (300 m³/s)/(20 m)(20 m)
<u>V = 0.75 m/s</u>
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(b)
Here,
Q = (300,000 L/s)(0.001 m³/1 L) = 300 m³/s
w = 60 m
d = 40 m
Therefore,
V = (300 m³/s)/(60 m)(40 m)
<u>V = 0.125 m/s</u>