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Nataly_w [17]
3 years ago
10

A mass M suspended by a spring with force constant k has aperiod T when set into oscillations on Earth. Its period on Mars,whose

mass is about 1/9 and radius 1/2 that of Earth, is mostnearly
A) 1/3 T
B) 2/3T
C)T
D) 3/2T
E) 3 T
Physics
1 answer:
olchik [2.2K]3 years ago
6 0

Answer:

C)T

Explanation:

The period of a mass-spring system is:

T=2\pi\sqrt\frac{m}{k}

As can be seen, the period of this simple harmonic motion, does not depend at all on the gravitational acceleration (g), neither the mass nor the spring constant depends on this value.

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The bar graph shows energy data taken from a roller coaster at a theme park. Analyze the data and assess its validity. In 3–5 se
alexdok [17]

The data given in the bar graph is valid because it follows the law of conservation of energy, since the GPE at top of 2nd hill plus KE at top of 2nd hill equals KE at bottom of 1st hill.

<h3>What is law of conservation of energy?</h3>

The law of conservation of energy states that energy can neither be created nor destroyed but can be transformed from one form to another.

Based on the law of conservation of energy, kinetic energy of a roller coaster can be converted into potential energy of the roller coaster and vice versa.

ΔK.E = ΔP.E

where;

  • ΔK.E is change in kinetic energy
  • ΔP.E is change in potential energy

The kinetic energy of the coaster is greatest at the bottom of the hill, as the coaster moves upward, the kinetic energy decreases and will be converted into potential energy. The potential energy of the coaster increases as the coaster moves up the hill and will become maximum at the highest point of the hill.

From the given data;

GPE at top of 2nd hill + KE at top of 2nd hill = KE at bottom of 1st hill

Learn more about conservation of energy here: brainly.com/question/166559

#SPJ1

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1 year ago
Fred has a very detailed model of the solar system where each planet is made out of granite rock. Since his little sister really
eimsori [14]
C or "Density" The granite rock has a much higher density than the Styrofoam. 
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3 years ago
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Two point charges, a +45nC charge X and a +12nC charge Y are separated by a distance of 0.5m.
Gnoma [55]

A) Calculate the resultant electric field strength at the midpoint between the charges.

Qx is the charge at X and Qy is the charge at Y.

E at midpoint = k×Qx/0.25² - k×Qy/0.25²

k = 9×10⁹Nm²C⁻², Qx = 45nC, Qy = 12nC

E = 4752N/C

Well done.

B) Calculate the distance from X at which the electric field strength is zero.

Let D be some point between X and Y for which the net E field is 0.

Let d be the distance from X to D.

Set up the following equation:

E at D = k×Qx/d² - k×Qy/(0.5-d)² = 0

Do some algebra to solve for d:

k×Qx/d² = k×Qy/(0.5-d)²

Qx/d² = Qy/(0.5-d)²

Qx(0.5-d)² = Qyd²

(0.5-d)√Qx = d√Qy

0.5√Qx-d√Qx = d√Qy

d(√Qx+√Qy) = 0.5√Qx

d = (0.5√Qx)/(√Qx+√Qy)

Plug in Qx = 45nC, Qy = 12nC

d ≈ 330mm

C) Calculate the magnitude of the electric field strength at the point P on the diagram below.

First determine the angles of the triangle. The sides of the triangle are 0.3m, 0.4m, and 0.5m, so this is a right triangle where the angle between the 0.3m and 0.4m sides is 90°

∠Y = tan⁻¹(0.4/0.3) = 53.13°

∠X = 90-∠Y = 36.87°

Determine the horizontal component of E at P:

Ex = E from Qx × cos(∠X) - E from Qy × cos(∠Y)

Ex = k×Qx/0.4²×cos(36.87°) - k×Qy/0.3²×cos(53.13°)

Ex = 1305N/C

Determine the vertical component of E at P:

Ey = E from Qx × sin(∠X) - E from Qy × sin(∠Y)

Ey = k×Qx/0.4²×sin(36.87°) - k×Qy/0.3²×sin(53.13°)

Ey = 2479N/C

Use the Pythagorean theorem to determine the magnitude of E at P:

E = √(Ex²+Ey²)

E ≈ 2802N/C

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From the geometry of the problem, the 20 m-long cable creates the hypotenuse of a right triangle, with the extended of the other two sides of size 20 m * cos(30 deg), which is around 17.3 m. Therefore, the ball has increased by 20 m - 17.3 m = 2.7 m. 

The potential energy will have altered by m*g*h, which is 1400 kg * 9.8 m/s^2 * 1.6 m , or about 37044 joules.

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pickupchik [31]
The time required for a moon to orbit around the earth is about 27-28 days

In order for lunar eclipse to occur the line that should be formed is:
Sun-Earth-Moon
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in order for solar eclipse to occur the line is now:
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Therefore moon needs to make half of its orbit to go from behind the earth to in front of the earth.

28/2 = 14

Answer is 14
5 0
3 years ago
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