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4 years ago

Spring tides have higher high tides and lower low tides when the moon, sun, and earth are aligned in a row. Why does this occur?

1 answer:

8 0

Spring tides have higher high tides and lower low tides when the moon, sun, and earth are aligned in a row. Why does this occur?

A.gravitational pull is stronger

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A 0.00275 kg air‑inflated balloon is given an excess negative charge q1 =−3.50×10−8 C by rubbing it with a blanket. It is found

kati45 [8]

Answer:

1) $\rm q_2$ is positive.

2) $\rm q_2=4.56\times 10^{-10}\ C.$

Explanation:

The charged rod is held above the balloon and the weight of the balloon acts in downwards direction. To balance the weight of the balloon, the force on the balloon due to the rod must be directed along the upwards direction, which is only possible when the rod exerts an attractive force on the balloon and the electrostatic force on the balloon due to the rod is attractive when the polarities of the charge on the two are different.

Thus, In order for this to occur, the polarity of charge on the rod must be positive, i.e., $\rm q_2$ is positive.

Given:

Mass of the balloon, m = 0.00275 kg.
Charge on the balloon, $\rm q_1 = -3.50\times 10^{-8}\ C.$
Distance between the rod and the balloon, d = 0.0640 m.
Acceleration due to gravity, $\rm g = 9.81\ m/s^2.$

In order to balloon to be float in air, the weight of the balloom must be balanced with the electrostatic force on the balloon due to rod.

Weight of the balloon, $\rm W = mg = 0.00275\times 9.81=2.70\times 10^{-2}\ N.$

The magnitude of the electrostatic force on the balloon due to the rod is given by

$\rm F_e = \dfrac{1}{4\pi \epsilon_o}\dfrac{|q_1||q_2|}{d^2}.$

$\rm \dfrac{1}{4\pi \epsilon_o}$ is the Coulomb's constant.

For the elecric force and the weight to be balanced,

$\rm F_e = W\\\dfrac{1}{4\pi \epsilon_o}\dfrac{|q_1||q_2|}{d^2}=W\\8.99\times 10^9\times \dfrac{3.50\times10^{-8}\times |q_2| }{0.0640^2}=2.70\times 10^{-2}\\|q_2| = \dfrac{2.70\times 10^{-2}\times 0.00640^2}{8.99\times 10^9\times 2.70\times 10^{-7}}=4.56\times 10^{-10}\ C.$

3 0

3 years ago

2. What is the kinetic energy of a 7.26 kg bowling ball that is rolling at a speed of 2 m/s?

telo118 [61]

Kinetic energy is the energy possessed by an object when that object is moving in space. The higher the mass of an object or higher the speed of an object the higher the kinetic energy will be.

So to calculate the Kinetic Energy we can use the following formula

K.E=(1/2)*m*v^2

Inserting the values in formula gives:

K.E=1/2*7.26*2^2

14.52J

This is the final answer which gives the kinetic energy of the ball.

6 0

3 years ago

Read 2 more answers

As the population in a small town grows, a new water treatment plant is built. The plant is built on a local river where it can

abruzzese [7]

Treatment water from the plant would affects the communities present in the downstream of the river.

Answer: Option (a)

Explanation:

Wastewater treatment plant is a process used to treat the water that flows from the rivers, streams, and lakes.This plant removes the dust, sand, sediments present in the river water.

When this treated water is released into the downriver stream, it affects the health of aquatic organisms present in that stream.Use of chemicals in treating the wastewater may cause genetic problems to the communities present in that region.

It also causes air pollution and more energy is needed for the treatment process and thus affects the ecosystem.

6 0

3 years ago

Read 2 more answers

A vertical spring gun is used to launch balls into the air. If the spring is compressed by 4.9 cm, the ball of mass 5.5 g is lau

AleksandrR [38]

We know, by conservation of energy :

$\dfrac{kx^2}{2}=mgh$

Therefore,

$\dfrac{x_1^2}{x_2^2}=\dfrac{h_1}{h_2}$

Putting given values, we get :

$\dfrac{x_1^2}{x_2^2}=\dfrac{h_1}{h_2}\\\\\dfrac{4.9^2}{x_2^2}=\dfrac{50.2}{2\times 50.2}\\\\x_2^2=2\times 4.9^2\\\\x_2 = 4.9\times \sqrt{2}\\\\x_2=6.93\ cm$

Therefore, the spring be compressed to 6.93 cm to send the ball twice as high.

Hence, this is the required solution.

6 0

3 years ago

Calculate the speed of a gamma ray with a frequency of 3.0 x 10^19 Hz and a wavelength of 1.0 x 10^-11 m.

Ulleksa [173]

Answer:

Speed of gamma rays = 3 x 10⁸ m/s

Explanation:

Given:

Frequency of gamma ray = 3 x 10¹⁹ Hz

Wavelength of gamma rays = 1 x 10⁻¹¹ meter

Find:

Speed of gamma rays

Computation:

Velocity = Frequency x wavelength

Speed of gamma rays = Frequency of gamma ray x Wavelength of gamma rays

Speed of gamma rays = [3 x 10¹⁹][1 x 10⁻¹¹]

Speed of gamma rays = 3 x [10¹⁹⁻¹¹]

Speed of gamma rays = 3 x [10⁸]

Speed of gamma rays = 3 x 10⁸ m/s

6 0

3 years ago