two forces P and Q pass through a point A which is 4 ft to the right of and 3 ft above a moment center O. force P is 200 lb dire
1 answer:
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Answer:
P = 216 Watts
Explanation:
Given that,
Resistance 1,
Resistance 2,
Voltage, V = 36 V
We need to find the power dissipated by the 6 ohms resistor. In case of parallel circuit, voltage throughout the circuit is same while the current is different.
Firstly, lets find current of the 6 ohms resistor such that,
V = IR
The power dissipated by 6 ohms resistor is given by :
So, the power dissipated across 6 ohm resistor is 126 watts.
The gravitational force between two masses m₁ and m₂ is
where
G = 6.67408 x 10⁻¹¹ m³/(kg-s²), the gravitational constant
d = distance between the masses.
Given:
F = 1.5 x 10⁻¹⁰ N
m₁ = 0.50 kg
m₂ = 0.1 kg
Therefore
1.5 x 10⁻¹⁰ N = (6.67408 x 10⁻¹¹ m³/(kg-s²))*[(0.5*0.1)/(d m)²]
d² = [(6.67408x10⁻¹¹)*(0.5*0.1)]/1.5x10⁻¹⁰
= 0.0222
d = 0.1492 m = 149.2 mm
Answer: 149.2 mm
where
G = 6.67408 x 10⁻¹¹ m³/(kg-s²), the gravitational constant
d = distance between the masses.
Given:
F = 1.5 x 10⁻¹⁰ N
m₁ = 0.50 kg
m₂ = 0.1 kg
Therefore
1.5 x 10⁻¹⁰ N = (6.67408 x 10⁻¹¹ m³/(kg-s²))*[(0.5*0.1)/(d m)²]
d² = [(6.67408x10⁻¹¹)*(0.5*0.1)]/1.5x10⁻¹⁰
= 0.0222
d = 0.1492 m = 149.2 mm
Answer: 149.2 mm
The acceleration of gravity on Moon B is D) 0.25 m/s2
Explanation:
The data listed in the problem are not clear: find them in the table attached.
The acceleration of gravity on a planet is given by the following equation:
where:
is the gravitational constant
M is the mass of the planet
R is the radius of the planet
Here we want to find the acceleration of gravity on the surface of Moon B, which has the following data:
(mass)
(radius)
And substituting into the equation, we find:
Learn more about gravity:
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