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3 years ago

7

Debbie touches the metal doorknob to her bedroom at 7 a.m. and feels a static electric shock. In the afternoon, Debbie comes bac

k and touches the doorknob again, but she experiences no shock.

1 answer:

AleksandrR [38]3 years ago

5 0

Answer:

she had socks one and was shuffeling so she had static

Explanation:

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Can A positively charged body attract another positively charged body

andriy [413]

Like charges repel, unlike charges attract

Two protons will also tend to repel each other because they both have a positive charge. On the other hand, electrons and protons will be attracted to each other because of their unlike charges.

So I would say no, unless the two bodies are placed close to each other where one has much more charge than the other, then due to induction, force of attraction becomes more than the force of repulsion.

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3 years ago

A go-cart is traveling at a rate of 25 m/sec for 20 seconds. How far will the go cart travel?

notsponge [240]

Answer:

Distance travel by go-cart = 500 meter

Explanation:

Given:

Speed of go cart = 25 m/s

Time travel = 20 seconds

Find:

Distance travel by go-cart

Computation:

Distance = Speed x time

Distance travel by go-cart = Speed of go cart x Time travel

Distance travel by go-cart = 25 x 20

Distance travel by go-cart = 500 meter

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3 years ago

How much bigger is the sun than the earth?

Zinaida [17]

The sun is about 109 times larger than the earth in diameter. <span>1,300,000 Earths can fit in the sun.</span>

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3 years ago

Read 2 more answers

local AM radio station broadcasts at a frequency of 685.9 kHz. Calculate the wavelength at which they are broadcasting. Waveleng

natta225 [31]

Answer:

λ = 437 m.

Explanation:

Since a radio wave is a electromagnetic type of wave, it propagates approximately at the same speed of light in vacuum, 3*10⁸ m/s.
As in any wave, there exist a fixed relationship between the propagation speed, the wavelength and the frequency of the radio wave, as follows:

$c = \lambda* f (1)$

Replacing by the values of the speed and the frequency, and solving for the wavelength λ, we get:

$\lambda = \frac{c}{f} =\frac{3e8 m/s}{685.9e31/s} =437 m (2)$

8 0

2 years ago

In 2005 astronomers announced the discovery of a large black hole in the galaxy Markarian 766 having clumps of matter orbiting a

IRISSAK [1]

A. $4.64\cdot 10^{11}m$

The orbital speed of the clumps of matter around the black hole is equal to the ratio between the circumference of the orbit and the period of revolution:

$v=\frac{2\pi r}{T}$

where we have:

$v=30,000 km/s = 3\cdot 10^7 m/s$ is the orbital speed

r is the orbital radius

$T=27 h \cdot 3600 =97,200 s$ is the orbital period

Solving for r, we find the distance of the clumps of matter from the centre of the black hole:

$r=\frac{vT}{2\pi}=\frac{(3\cdot 10^7 m/s)(97200 s)}{2\pi}=4.64\cdot 10^{11}m$

B. $6.26\cdot 10^{36}kg, 3.13\cdot 10^6 M_s$

The gravitational force between the black hole and the clumps of matter provides the centripetal force that keeps the matter in circular motion:

$m\frac{v^2}{r}=\frac{GMm}{r^2}$

where

m is the mass of the clumps of matter

G is the gravitational constant

M is the mass of the black hole

Solving the formula for M, we find the mass of the black hole:

$M=\frac{v^2 r}{G}=\frac{(3\cdot 10^7 m/s)^2(4.64\cdot 10^{11} m)}{6.67\cdot 10^{-11}}=6.26\cdot 10^{36}kg$

and considering the value of the solar mass

$M_s = 2\cdot 10^{30}kg$

the mass of the black hole as a multiple of our sun's mass is

$M=\frac{6.26\cdot 10^{36} kg}{2\cdot 10^{30} kg}=3.13\cdot 10^6 M_s$

C. $9.28\cdot 10^9 m$

The radius of the event horizon is equal to the Schwarzschild radius of the black hole, which is given by

$R=\frac{2MG}{c^2}$

where M is the mass of the black hole and c is the speed of light.

Substituting numbers into the formula, we find

$R=\frac{6.26\cdot 10^{36} kg)(6.67\cdot 10^{-11})}{(3\cdot 10^8 m/s)^2}=9.28\cdot 10^9 m$

8 0

3 years ago