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andrew11 [14]
2 years ago
5

A person located on the equator is orbiting the center of the Earth to the East at 1670 km/s.

Physics
1 answer:
slega [8]2 years ago
3 0

Moving at the same velocity to the East which means that they are both on the same planet in space.

<h3>Movement and velocity of planet </h3>

Relative to the center of the Earth, a person on the North Pole is moving at the same velocity to the East because both are present on the same planet in space.

The whole planet of earth spins in the same direction as well as with the same velocity so we can conclude that option A is the correct statement about direction and velocity of the earth.

Learn more about velocity here: brainly.com/question/25749514

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11. You want to calculate the displacement of an object thrown over a bridge. Using -10 m/s2 for acceleration due to gravity, wh
IrinaVladis [17]

Answer:

The displacement was 320 meters.

Explanation:

Assuming projectile motion and zero initial speed (i.e., the object was dropped, not thrown down), you can calculate the displacement using the kinematic equation:

d = \frac{1}{2}gt^2=\frac{1}{2}10\frac{m}{s^2}\cdot 8^2 s^2=320 m

The displacement was 320 meters.

7 0
3 years ago
Two uncharged metal spheres, spaced 25.0 cm apart, have a capacitance of 26.0 pF. How much work would it take to move 12.0 nC of
viktelen [127]

Answer:

W=2.76\times 10^{-6}\ J

Explanation:

Given that,

The distance between two spheres, r = 25 cm = 0.25 m

The capacitance, C = 26 pF = 26×10⁻¹² F

Charge, Q = 12 nC = 12 × 10⁻⁹ C

We need to find the work done in moving the charge. We know that, work done is given by :

U=\dfrac{Q^2}{2C}

Put all the values,

U=\dfrac{(12\times 10^{-9})^2}{2\times 26\times 10^{-12}}\\\\U=2.76\times 10^{-6}\ J

So, the work done is 2.76\times 10^{-6}\ J.

8 0
3 years ago
How high was a brick dropped from if if falls in 2.5 seconds?
jenyasd209 [6]
Using the kinematic equation d =  V_0 * t  + 1/2 * a * t^2, where d is height you can rewrite this to be d = 1/2*g*t^2 or 4.9t^2
g = a because this is a free fall 
d = 1/2 * 9.81m/s^2 * 2.5^2
d = 30.65625m
d = 30.7m
7 0
3 years ago
Which process requires more energy: completely vaporizing 1 kg of saturated liquid water at 1 atm pressure or completely vaporiz
Lilit [14]
The correct answer would be the first option. The process that would need more energy would be vaporizing 1 kg of saturated liquid water at a pressure of 1 atmosphere. This can be seen from the latent heat of vaporization of each system. For the saturated water at 1 atm, the latent heat is equal to 40.7 kJ per mole while, at 8 atm, the latent heat is equal to 36.4 kJ per mole. The latent heat of vaporization is the amount of heat needed in order to vaporize a specific amount of substance without any change in the temperature. As we can observe, more energy is needed by the liquid water at 1 atm. 
7 0
3 years ago
A mass-spring system has k = 56.8 N/m and m = 0.46 kg.
andriy [413]

Answer:

A. \omega=11.1121\ rad.s^{-1}

B. f=1.7685\ Hz

C. T=0.5654\ s

Explanation:

Given:

  • spring constant, k=56.8\ N.m^{-1}
  • mass attached, m=0.46\ kg

A)

for a spring-mass system the frequency is given as:

\omega=\sqrt{\frac{k}{m} }

\omega=\sqrt{\frac{56.8}{0.46}}

\omega=11.1121\ rad.s^{-1}

B)

frequency is given as:

f=\frac{\omega}{2\pi}

f=\frac{11.1121}{2\pi}

f=1.7685\ Hz

C)

Time period of a simple harmonic motion is given as:

T=\frac{1}{f}

T=0.5654\ s

7 0
3 years ago
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