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den301095 [7]
3 years ago
8

F-40Nm=5kk = 0,2a=? ​

Physics
1 answer:
tester [92]3 years ago
3 0

Answer:

1......... .......... ......

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A positive point charge q is placed at the center of an uncharged metal sphere insulated from the ground. The outside of the sph
Olenka [21]

Answer:

B.The charge on A is -q; there is no charge on B.

Explanation:

We are given that

Charge=+q

We have to find the correct statement.

When positive charge is placed at center of uncharged metal sphere

insulated from the ground then negative charge(-q) induced on inner

surface  A  of sphere  and the outer  surface B  is grounded then the surface is neutral .

It means there is no charge on surface B.

Hence, option B is true .

B.The charge on A is -q; there is no charge on B.

4 0
3 years ago
An object weighs 10N on earth .what is the objects weight on a planet one tenth the earths mass and one half its radius?
earnstyle [38]
We know, weight = mass * gravity 
10 = m * 9.8
m = 10/9.8 = 1.02 Kg

Now, Let, the gravity of that planet = g'
g' = m/r²   [m,r = mass & radius of that planet ]
g' = M/10 / (1/2R)²   [M, R = mass & radius of Earth ]
g' = 4M / 10R²
g' = 2/5 * M/R²
g' = 2/5 * g   
g' = 2/5 * 9.8
g' = 3.92

Weight on that planet = planet's gravity * mass
W' = 3.92 * 1.02
W' = 4 N

In short, Your Answer would be 4 Newtons

Hope this helps!
3 0
3 years ago
What is the frequency of a photon with an energy of 4. 56 x 10^-19 j
Sauron [17]

The frequency of a photon with an energy of 4.56 x 10⁻¹⁹ J is 6.88×10¹⁴ s⁻¹.

<h3>What is a frequency?</h3>

The number of waves that travel through a particular point in a given length of time is described by frequency. So, if a wave takes half a second to pass, the frequency is 2 per second.

Given that the energy of the photon is 4.56 x 10⁻¹⁹ J. Therefore, the frequency of the photon can be written as,

\rm \gamma = \dfrac{E}{h} = \dfrac{4.56x10^{-19} J}{6.626 \times 10^{-34}\ Jsec^{-1}}\\\\\\\gamma  = 6.88 \times 10^{14}\ s^{-1}

Hence, the frequency of a photon with an energy of 4.56 x 10⁻¹⁹ J is 6.88×10¹⁴ s⁻¹.

Learn more about Frequency:

brainly.com/question/5102661

#SPJ4

5 0
2 years ago
Read 2 more answers
Two charges, one of 2.50μC and the other of -3.50μC, are placed on the x-axis, one at the origin and the other at x = 0.600 m
aev [14]

Answer:

Explanation:

Given

charge of first body q_1=2.5\ mu C

charge of second body q_2=-3.5\ mu C

Particle 1 is at origin and particle 2 is at x=0.6\ m

third Particle which charge +q must be placed left of 2.5\mu C because it will repel the q charge while -3.5\mu C will attract it

suppose it is placed at a distance of x m

F_{1q}=\frac{kq(2.5)}{x^2}

F_{2q}=\frac{kq(-3.5)}{(0.6+x)^2}

F_{1q}+F_{2q}=0

\frac{kq(2.5)}{x^2}+\frac{kq(-3.5)}{(0.6+x)^2}=0

\frac{kq(2.5)}{x^2}=\frac{kq(3.5)}{(0.6+x)^2}

\frac{0.6+x}{x}=(\frac{3.5}{2.5})^{0.5}

0.6+x=1.1832x

x=3.27\ m

5 0
3 years ago
Identify the statement below that is true about a type of stress.
Sunny_sXe [5.5K]

1-D Eustress is a positive response

2-B resistance

3-A Imagine you are doing your gymnastics

6 0
3 years ago
Read 2 more answers
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