1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Helen [10]
3 years ago
7

A ball is attached to a string of length 3 m to make a pendulum. The pendulum is placed at a location that is away from the Eart

h's surface by three times the radius of the Earth. What is the acceleration due to gravity (in m/s2) at that height and what is the period (in s) of the oscillations
Physics
1 answer:
Musya8 [376]3 years ago
4 0

1) 0.61 m/s^2

2) 13.9 s

Explanation:

1)

The acceleration due to gravity is the acceleration that an object in free fall (acted upon the force of gravity only) would have.

It can be calculated using the equation:

g=\frac{GM}{r^2} (1)

where

G is the gravitational constant

M=5.98\cdot 10^{24} kg is the Earth's mass

r is the distance of the object from the Earth's center

The pendulum in the problem is at an altitude of 3 times the radius of the Earth (R), so its distance from the Earth's center is

r=4R

where

R=6.37\cdot 10^6 m is the Earth's radius

Therefore, we can calculate the acceleration due to gravity at that height using eq.(1):

g=\frac{GM}{(4R)^2}=\frac{(6.67\cdot 10^{-11})(5.98\cdot 10^{24})0.}{(4\cdot 6.37\cdot 10^6)^2}=0.61 m/s^2

2)

The period of a simple pendulum is the time the pendulum takes to complete one oscillation. It is given by the formula

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

where

L is the length of the pendulum

g is the acceleration due to gravity at the location of the pendulum

Note that the period of a pendulum does not depend on its mass.

For the pendulum in this problem, we have:

L = 3 m is its length

g=0.61 m/s^2 is the acceleration due to gravity (calculated in part 1)

Therefore, the period of the pendulum is:

T=2\pi \sqrt{\frac{3}{0.61}}=13.9 s

You might be interested in
A 1.1-kg object is suspended from a vertical spring whose spring constant is 120 N/m. (a) Find the amount by which the spring is
andriy [413]

Answer:

e = 0.0898m

v = 2.07m/s

Explanation:

a) According to Hooke's law

F = ke

e is the extension

k is the spring constant

Since F = mg

mg = ke

e = mg/k

Substitute the given value

e = 1.1(9.8)/120

e = 10.78/120

e = 0.0898m

Hence it is stretched by 0.0898m from its unstrained length

2) Total Energy = PE+KE+Elastic potential

Total Energy = mgh +1/2mv²+1/2ke²

Substitute the given value

5.0= 1.1(9.8)(0.2)+1/2(1.1)v²+1/2(120)(0.0898)²

Solve for v

5.0 = 2.156+0.55v²+0.48338

5.0-2.156-0.48338= 0.55v²

2.36 =0.55v²

v² = 2.36/0.55

v² = 4.29

v ,= √4.29

v = 2.07m/s

Hence the required velocity is 9.28m/s

4 0
3 years ago
two circular plates, each with a radius of 8.22 cm, have equal and opposite charges of magnitude 3.052 μc. calculate the electri
PtichkaEL [24]

If the separation distance is doubled, then the electric field decreases by a factor of 4.

<h3>What is the electric field strength?</h3>

We know that the electric field strength is known to depend on the magnitude of the charge and the distance of separation. We know that the electric field refers to the region in which the influence of a charge is felt. Recall that a charge is a specie that is positively or negatively charged. The charge on a specie must always be shown by its sign.

We know that the electric field is the region in space where the influence of a charge can be felt. If a charge is placed in the vicinity of another charge, the second charge would experience a force due to the presence of the first charge. This is because the second charge was brought into the electric field of the first charge.

Thus we know that;

E = Kq/r^2

Where;

E = electric field strength

q = magnitude of charge

r = distance of separation

Now;

E = 9.0* 10^9 * 3.052 * 10^-6/(8.22 * 10^-2)^2

E = 4 N/C

Given that the electric filed strength is inversely proportional to the distance of separation, when the distance between the charges is doubled, the electric field decreases by a factor of 4.

Learn more about electric field strength:brainly.com/question/15170044?

#SPJ1

7 0
1 year ago
A 25.0 g marble sliding to the right at 20.0 cm/s overtakes and collides elastically with a 10.0 g marble moving in the same dir
ikadub [295]
In collision that are categorized as elastic, the total kinetic energy of the system is preserved such that,

   KE1  = KE2

The kinetic energy of the system before the collision is solved below.

  KE1 = (0.5)(25)(20)² + (0.5)(10g)(15)²
  KE1 = 6125 g cm²/s²

This value should also be equal to KE2, which can be calculated using the conditions after the collision.

KE2 = 6125 g cm²/s² = (0.5)(10)(22.1)² + (0.5)(25)(x²)

The value of x from the equation is 17.16 cm/s.

Hence, the answer is 17.16 cm/s. 
6 0
3 years ago
A wooden block with mass 1.60 kg is placed against a compressed spring at the bottom of a slope inclined at an angle of 30.0° (p
andreyandreev [35.5K]

Answer:

The amount of potential energy that was initially stored in the spring is 88.8 J.

Explanation:

Given that,

Mass of block = 1.60 kg

Angle = 30.0°

Distance = 6.55 m

Speed = 7.50 m/s

Coefficient of kinetic friction = 0.50

We need to calculate the amount of potential energy

Using formula of conservation of energy between point A and B

U_{A}+k_{A}+w_{A}=U_{B}+k_{B}

U_{A}+0-fd=mgy+\dfrac{1}{2}mv^2

U_{A}=\mu mg\cos\theta\times d+mg h\sin\theta+\dfrac{1}{2}mv^2

Put the value into the formula

U_{A}=0.50\times1.60\times9.8\cos30\times6.55+1.60\times9.8\times6.55\sin30+\dfrac{1}{2}\times1.60\times(7.50)^2

U_{A}=88.8\ J

Hence, The amount of potential energy that was initially stored in the spring is 88.8 J.

7 0
3 years ago
The distance from Earth to the star Epsilon Eridani is about 10.5 light years. Which of the following statements is true?
Simora [160]
<span>The correct answer is B. - It would take a ray of light 10.5 light years to travel from Earth to Epsilon Eridani, or vice-versa. Using our current technology it would take far longer than 21.0 years for a space ship from Earth to travel that far - I would have to guess many hundreds of years.</span>
4 0
3 years ago
Read 2 more answers
Other questions:
  • Which jet stream flows around the sub-tropical high pressure systems and advects moisture along its poleward periphery?
    7·1 answer
  • Calculate the weight of a 25kg object?
    15·1 answer
  • An 85 kg dog is sitting on a couch what is the weight of the dog
    14·2 answers
  • Self-fulfilling prophecy is a primary example of the capability of language to .
    14·1 answer
  • Please help on this one!!
    5·1 answer
  • Clouds of gas and dust as well as new star formation are typically seen in __________galaxy
    7·1 answer
  • The maximum centripetal acceleration a car can sustain without skidding out of a curved path is 9.40 m/s2 . If the car is travel
    8·1 answer
  • What is the lowest frequency that will resonate in an organ pipe 2.00 m in length, closed at one end? The speed of sound in air
    15·1 answer
  • How do chemical formulas represent compounds? <br> Help PLEASE
    8·2 answers
  • Calculate the electrostatic force between two protons separated by a distance of 0.90 m
    15·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!