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frutty [35]
3 years ago
5

A pendulum is taken to another planet where it swings back and forth exactly 17 times every 35.66 seconds. The pendulum's arm is

0.74 m long. What is the acceleration due to gravity on this planet in units of m/s?
Physics
1 answer:
Troyanec [42]3 years ago
8 0

Answer: g = 6.65 m/s²

Explanation: The frequency of a simple pendulum is given by the formulae below

f = 1/2π *(√g/l)

Where f = frequency = number of oscillation /time =17/35.66 = 0.477Hz

g = acceleration due gravity in another planet

l = length of simple pendulum = 0.74m

0.477 = 1/2π * (√g/0.74)

0.477 = 1/2 * 3.142 * (√g/0.74)

0.477 = 1/ 6.284 * (√g/0.74)

0.477 = 0.1591 * (√g/0.74)

0.477/0.1591 = (√g/0.74)

2.998 = (√g/0.74)

By taking the square of both sides

8.988 = g/ 0.74

g = 6.65 m/s²

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A basketball player throws a ball horizontally toward another player 10 meters away. If the ball is
Nina [5.8K]

Answer:

It will take 0.46 seconds to reach home plate by ball.

Explanation:

7 0
3 years ago
Particle A of charge 3.06 10-4 C is at the origin, particle B of charge -5.70 10-4 C is at (4.00 m, 0), and particle C of charge
Alex

Answer:

F_net = 26.512 N

Explanation:

Given:

Q_a = 3.06 * 10^(-4 ) C

Q_b = -5.7 * 10^(-4 ) C

Q_c = 1.08 * 10^(-4 ) C

R_ac = 3 m

R_bc = sqrt (3^2 + 4^2) = 5m

k = 8.99 * 10^9

Coulomb's Law:

F_i = k * Q_i * Q_j / R_ij^2

Compute F_ac and F_bc :

F_ac = k * Q_a * Q_c / R^2_ac

F_ac =  8.99 * 10^9* ( 3.06 * 10^(-4 ))* (1.08 * 10^(-4 )) / 3^2

F_ac = 33.01128 N

F_bc = k * Q_b * Q_c / R^2_bc

F_bc =  8.99 * 10^9* ( 5.7 * 10^(-4 ))* (1.08 * 10^(-4 )) / 5^2  

F_bc = - 22.137 N

Angle a is subtended between F_bc and y axis @ C

cos(a) = 3 / 5

sin (a) = 4 / 5

Compute F_net:

F_net = sqrt (F_x ^2 + F_y ^2)

F_x = sum of forces in x direction:

F_x = F_bc*sin(a) = 22.137*(4/5) = 17.71 N

F_y = sum of forces in y direction:

F_y = - F_bc*cos(a) + F_ac = - 22.137*(3/5) + 33.01128 = 19.72908 N

F_net = sqrt (17.71 ^2 + 19.72908 ^2) = 26.5119 N

Answer: F_net = 26.512 N

5 0
4 years ago
g a stone with mass m=1.60 kg IS thrown vertically upward into the air with an initial kinetic energy of 470 J. the drag force a
otez555 [7]

Answer:

Height reached will be 28.35 m

Explanation:

Here we can use the work energy theorem to find the maximum height

As we know by work energy theorem

Work done by gravity + work done by friction = change in kinetic energy

-mgh - F_f h = 0 - \frac{1}{2}mv_i^2

now we will have

-1.60(9.8)(h) - 0.900(h) = - 470

-16.58 h = -470

h = 28.35 m

so here the height raised by the stone will be 28.35 m from the ground after projection in upward direction

5 0
3 years ago
Which energy transfer occurs when a cube of ice is placed in a glass of<br> water?
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Acetic acid is a weak acid and sodium hydroxide is strong base. Salts of the two will hydrolyse to give basic solution. So, at neutral point, pH of the solution will be greater than 8.
5 0
3 years ago
An electromagnet cannot be shut off once it is started. <br> True <br> or<br> False
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False the electro magnet is controlled electrically so its like turning on/off a switch thing of the magnets they have at junkyards.
6 0
4 years ago
Read 2 more answers
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