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Nesterboy [21]
3 years ago
7

The period of a simple pendulum in a grandfather clock on another planet is 1.80 s. What is the acceleration due to gravity (in

m/s2) on this planet if the length of the pendulum is 1.00?
Physics
1 answer:
weeeeeb [17]3 years ago
4 0

Answer:

12.17 m/s²

Explanation:

The formula of period of a simple pendulum is given as,

T = 2π√(L/g)........................ Equation 1

Where T = period of the simple pendulum, L = length of the simple pendulum, g = acceleration due to gravity of the planet. π = pie

making g the subject of the equation,

g = 4π²L/T²................... Equation 2

Given: T = 1.8 s, l = 1.00 m

Constant: π = 3.14

Substitute into equation 2

g = (4×3.14²×1)/1.8²

g = 12.17 m/s²

Hence the acceleration due to gravity of the planet = 12.17 m/s²

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proton is accelerated to a speed of 0.93c, what is the 1. A proton has a mass of 1.673 x 10-27 kg. If the proton is accelerated
coldgirl [10]

Answer : The correct option is, (B) 1.3\times 10^{-18}kg.m/s

Explanation : Given,

Mass of proton = 1.673\times 10^{-27}kg

Speed of proton = 0.93 c

Formula used for relativistic momentum of the proton is:

p=\frac{m_ov}{\sqrt{1-\frac{v^2}{c^2}}}

where,

p = relativistic momentum of the proton

m_o = mass of proton

v = speed of proton

c = speed of light = 3\times 10^8m/s

Now put all the given values in the above formula, we get:

p=\frac{(1.673\times 10^{-27}kg)\times (0.93c)}{\sqrt{1-\frac{(0.93c)^2}{c^2}}}

p=\frac{(1.55589\times 10^{-27}c)}{0.367}

p=\frac{(1.55589\times 10^{-27})\times (3\times 10^8)}{0.367}

p=1.272\times 10^{-18}kg.m/s\approx 1.3\times 10^{-18}kg.m/s

Therefore, the relativistic momentum of the proton is, 1.3\times 10^{-18}kg.m/s

5 0
3 years ago
Based on Kepler's work, which best describes the orbit If a planet around the Sun?
krek1111 [17]

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an ellipse with the Sun at one focus  or D

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Answer for edgenuity

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3 years ago
A student slides her 80.0-kg desk across the level floor of her dormitory room a distance 4.00 m at constant speed. If the coeff
zysi [14]

Answer:

Work done is equal to 125.44 J

Explanation:

We have given mass of the student m = 80 kg

Distance moved d = 4 m

Acceleration due to gravity g=9.8m/sec^2

Coefficient of kinetic friction \mu =0.4

So normal force exerted by student F=\mu mg=0.4\times 80\times 9.8=313.6N

We know that work done is equal to multiplication of force and distance

So work done W=Fd=313.6\times 0.4=125.44J

7 0
3 years ago
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