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3 years ago

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If you know that the period of a pendulum is 1.87 seconds, what is the length of that pendulum? (Assume that we are on Earth and

that gravity is 9.81 meters/second².) Select one of the options below as your answer: A. 0.87 centimeters B. 2.1 meters C. 1.6 meters D. 0.87 meters E. 8.3 meters

2 answers:

shepuryov [24]3 years ago

8 0

Period of an ideal simple pendulum = 2π √(L / G)

1.87 = 2π √ (L / 9.81)

Divide each side by 2π : (1.87 / 2π) = √ (L / 9.81)

Square each side: (1.87 / 2π)² = L / 9.81

Multiply each side by 9.81 : L = (9.81) (1.87 / 2π)² = <em> 0.869 meter</em>

Choice 'D' is the closest one.

mote1985 [20]3 years ago

8 0

Answer : The correct option is, (D) 0.87 meters

Solution :

Formula used :

$T=2\pi \times \sqrt{\frac{L}{g}}$

where,

T = time period of a pendulum = 1.87 seconds

L = length of the pendulum = ?

g = gravity on earth = $9.8m/s^2$

Now put all the given values in the above formula, we get the length of the pendulum.

$1.87s=2\times \frac{22}{7}\times \sqrt{\frac{L}{9.8m/s^2}}$

$0.2975=\sqrt{\frac{L}{9.8m/s^2}}$

Now squaring on both the sides, we get

$L=0.868m=0.87m$

Therefore, the length of the pendulum is, 0.87 meters.

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2 years ago

A 4000 kg satellite is placed 2.60 x 10^6 m above the surface of the Earth.

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a) The acceleration of gravity is $4.96 m/s^2$

b) The critical velocity is 6668 m/s (24,006 km/h)

c) The period of the orbit is 8452 s

d) The satellite completes 10.2 orbits per day

e) The escape velocity of the satellite is 9430 m/s

f) The escape velocity of the rocket is 11,191 m/s

Explanation:

a)

The acceleration of gravity for an object near a planet is given by

$g=\frac{GM}{(R+h)^2}$

where

G is the gravitational constant

M is the mass of the planet

R is the radius of the planet

h is the height above the surface

In this problem,

$M=5.98\cdot 10^{24} kg$ (mass of the Earth)

$R=6.37\cdot 10^6 m$ (Earth's radius)

$h=2.60\cdot 10^6 m$ (altitude of the satellite)

Substituting,

$g=\frac{(6.67\cdot 10^{-11})(5.98\cdot 10^{24}}{(6.37\cdot 10^6 + 2.60\cdot 10^6)^2}=4.96 m/s^2$

b)

The critical velocity for a satellite orbiting around a planet is given by

$v=\sqrt{\frac{GM}{R+h}}$

where we have again:

$M=5.98\cdot 10^{24} kg$ (mass of the Earth)

$R=6.37\cdot 10^6 m$ (Earth's radius)

$h=2.60\cdot 10^6 m$ (altitude of the satellite)

Substituting,

$v=\sqrt{\frac{(6.67\cdot 10^{-11})(5.98\cdot 10^{24}}{(6.37\cdot 10^6 + 2.60\cdot 10^6)}}=6668 m/s$

Converting into km/h,

$v=6668 m/s \cdot \frac{3600 s/h}{1000 m/km}=24,006 km/h$

c)

The period of the orbit is given by the circumference of the orbit divided by the velocity:

$T=\frac{2\pi (R+h)}{v}$

where

$R=6.37\cdot 10^6 m$

$h=2.60\cdot 10^6 m$

v = 6668 m/s

Substituting,

$T=\frac{2\pi (6.37\cdot 10^6 + 2.60\cdot 10^6)}{6668}=8452 s$

d)

One day consists of:

$t = 24 \frac{hours}{day} \cdot 60 \frac{min}{hours} \cdot 60 \frac{s}{min}=86400 s$

While the period of the orbit is

T = 8452 s

So, the number of orbits completed by the satellite in one day is

$n=\frac{t}{T}=\frac{86400}{8452}=10.2$

e)

The escape velocity for an object in the gravitational field of a planet is given by

$v=\sqrt{\frac{2GM}{R+h}}$

where here we have:

$M=5.98\cdot 10^{24} kg$

$R=6.37\cdot 10^6 m$

$h=2.60\cdot 10^6 m$

Substituting, we find

$v=\sqrt{\frac{2(6.67\cdot 10^{-11})(5.98\cdot 10^{24}}{(6.37\cdot 10^6 + 2.60\cdot 10^6)}}=9430 m/s$

f)

We can apply again the formula to find the escape velocity for the rocket:

$v=\sqrt{\frac{2GM}{R+h}}$

Where this time we have:

$M=5.98\cdot 10^{24} kg$

$R=6.37\cdot 10^6 m$

$h=0$ , because the rocket is located at the Earth's surface, so its altitude is zero.

And substituting,

$v=\sqrt{\frac{2(6.67\cdot 10^{-11})(5.98\cdot 10^{24}}{(6.37\cdot 10^6)}}=11,191 m/s$

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The vertical component of the Earth’s magnetic field is, $B=6.5\times 10^5\ T$

We need to find the induced emf between the tip of a blade and the hub. The induced emf in terms of angular velocity of an rotating object is given by :

$\epsilon=\dfrac{1}{2}B\omega l^2$

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