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ipn [44]
3 years ago
9

How does increasing the amount of charge on an object affect the electric force it exerts on another charged

Physics
2 answers:
Setler [38]3 years ago
6 0

Answer:

the electric force increases because the amount of charge has a direct relationship to the force

Goshia [24]3 years ago
4 0

Answer: The electric force increases because the amount of charge has a direct relationship to the force.

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A laser beam of unknown wavelength passes through adiffraction grating having 5510 lines/cm after striking itperpendicularly. Ta
Ivahew [28]

Answer:

Explanation:

distance between two slit  d = \frac{1 \times 10^{-2}}{5510}

d = 18.15 x 10⁻⁷ m

Let wave length of light λ

formula for  position of  first pair of bright spot

Tanθ = λ / d , λ is wave length of light and d is distance between two slit .

Tan 15.4 = \frac{\lambda}{18.15\times10^{-7}}

λ = Tan 15.4 x 18.15 x 10⁻⁷

=5 x 10⁻⁷ m

If θ be the position of next bright spot

Tanθ = 2 λ / d

= \frac{2 \times\ 5\times\ 10^ {-7}}{18.15\times10^{-7}}

=\frac{2\times5}{18.5}

θ = 28.4 degree .

7 0
3 years ago
What is the cause of space
krek1111 [17]

Answer:

this depends on religion

Explanation:

some people believe in the big bang and some people believe in Jesus Christ and other gods.

big bang- Gravity

God- spoke

5 0
4 years ago
a 4.60 kg mass is placed on top of a vertical spring, which compresses a distance of 2.31 cm. calculate the force constant (in n
Lerok [7]

The restoring force of the spring cancels the weight of the mass, so by Newton's second law

∑ F = F[spring] - mg = 0   ⇒   F[spring] ≈ 45.1 N

where m = 4.60 kg and g = 9.80 m/s². Then the spring constant is k such that by Hooke's law,

F[spring] = k x

where x = 0.0231 m. Then the spring constant is

k = F[spring]/x ≈ 1950 N/m

4 0
2 years ago
A car is traveling at the posted speed limit of 6.70 m/s (15 miles/h) in a school zone. The car passing a school bus when a chil
Margarita [4]

Answer: 5.6m and 22.4m

Explanation: from the 3rd equation of motion

v²=u²+2as

v²-u²=2as  where : v=0 u=6.7m/s a=4m/s² s=?

0-6.7²=2×4×s

44.89=8s

s=44.89÷8

s=5.6m

for the second part, we shall still make use of the same equation.

v²-u²=2as         where: a=4m/s² u=13.4m/s v=0 s=?

0-13.4²=2×4×s

179.56=8s

s=179.56/8

22.4m

8 0
4 years ago
An astronomical telescope has an objective of diameter 20 cm with a focal length of 180 cm. the telescope is used with an eyepie
Bingel [31]
By definition;
M = fo/fe

Where,
M = Angular magnification
fo = Focal length of objective lens
fe = Focal length of eyepiece lens

From the information given;
M = 180/30 = 6
8 0
3 years ago
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