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3 years ago

Why does gravity on Earth have a stronger attraction with you than the sun has with you? Plsssssss help I really need this rn

Physics

1 answer:

n200080 [17]3 years ago

5 0

Answer:

Although the Earth has a lesser mass than the Sun, it is far closer to you, allowing for a stronger pull.

Explanation:

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Anna is conducting an experiment to determine how weather affects cell phone reception. She is trying to decide the best way to

Anestetic [448]

Your answer should be D.
Hope this helps :)

5 0

3 years ago

The electric field in a region of space has the components Ey = Ez = 0 and Ex = (4.00 N/C · m) x. Point A is on the y axis at y

igor_vitrenko [27]

Answer:

$V_{b}-V_{a} = -38.72$

Explanation:

Consider the axis diagram attached.

Given:

Ey = Ez = 0

Eₓ = - 4x N/C · m

Since electric field is in x direction, potential difference would be:

$V_{b} - V_{a} =-\int\limits^{4.40}_{0} {E_{x} } \, dx$

Here we integrate between limits 0 and 4.40 which is distance between A and B along x-axis.

$V_{b} - V_{a} = -4 \left[\begin{array}{ccc}\frac{x^{2} }{2} \end{array}\right]^{4.40}_{0}$

$V_{b}-V_{a} = -4 (9.68)\\V_{b}-V_{a} = -38.72$

5 0

3 years ago

What effect does urbanization have on land

GrogVix [38]

Urbanization can cause a lot of problems for nature such as pollution, global warming, and depletion of wildlife such as plants and animals.

6 0

3 years ago

Read 2 more answers

A policeman in a stationary car measures the speed of approaching cars by means of an ultrasonic device that emits a sound with

Ymorist [56]

Answer:

the frequency of the beats is 8.7687 kHz

Explanation:

Given the data in the question;

The frequency for stationary source and moving observer is;

f' = f( 1 +( v_observer/v_sound))

we know that, speed of sound in dry air = 343 m/s

so we substitute

f' = 41.2 kHz( 1 + (33.0 m/s / 343 m/s) ) = 41.2kHz( 1 + 0.0962) = 41.2kHz(1.0962)

f' = 45.1634 kHz

Now the frequency for stationary observer and moving source with frequency f' will be

f" = f'( 1 / (1 - ( v_observer/v_sound)))

45.1634 kHz( 343 / 343 - 33)

we substitute

f" = 45.1634 kHz( 1 / (1 - (33.0 m/s / 343 m/s)))

f" = 45.1634 kHz( 1 / (1 - 0.0962))

f" = 45.1634 kHz( 1 / 0.9038 )

f" = 45.1634 kHz( 1.1064 )

f" = 49.9687 kHz

Now the beat frequency will be;

f_beat = f' - f

we substitute

f_beat = 49.9687 kHz - 41.2 kHz

f_beat = 8.7687 kHz

Therefore, the frequency of the beats is 8.7687 kHz

7 0

3 years ago

An astronaut circling the earth at an altitude of 400 km is horrified to discover that a cloud of space debris is moving in the

elena-14-01-66 [18.8K]

One of the essential concepts to solve this problem is the utilization of the equations of centripetal and gravitational force.

From them it will be possible to find the speed of the body with which the estimated time can be calculated through the kinematic equations of motion. At the same time for the calculation of this speed it is necessary to clarify that this will remain twice the ship, because as we know by relativity, when moving in the same magnitude but in the opposite direction, with respect to the ship the debris will be double speed.

By equilibrium the centrifugal force and the gravitational force are equal therefore

$F_c = F_g$