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rusak2 [61]
3 years ago
8

Two spherical objects are separated by a distance that is 9.00 X 10m The objects are initially electrically neutral and are very

small compared to the stance between them. Each objectures the same negative charge due to the addition of electrons. As a result, each experiences an lectr ic force that has a magnitude of 1.137 10 . How many electrons did t o produce the charge on one of the objects?
Physics
1 answer:
romanna [79]3 years ago
5 0

Answer: 3.2 * 10^-3 C

Explanation: In order to solve this problem we have to use the Coulomb force given by:

F=k*q^2/ d^2 where he consider the same charge for each point

so, we have

q^2= F*d^2/k= 1.137* (10 C*9*10m)^2/9*10^9 N*m^2/C^2)=3.2 * 10^-3 C.

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A baseball has a mass of 145 g. a bat exerts a force of 18,400 n on the ball. what is the acceleration of the ball? 1.27 x 102 m
fomenos

The acceleration of the ball is 1.27 X 10⁵ m/s²

The Newtons second law states that the acceleration of an object is dependent upon two variables - the net force acting upon the object and the mass of the object.

Given,

Mass = 145 kg

Force = 18,400 N

We need to calculate the acceleration

Using formula of acceleration(a)

a = F/m  where F= force and m= mass

Put the values in the formula,

 a = 18,400/(145 X 10⁻³) = 1.27 X 10⁵ m/s²

Therefor, the acceleration of the ball is 1.27 X 10⁵ m/s².

Learn more about the Newton's second law with the help of the given link:

brainly.com/question/27573481

#SPJ4

7 0
2 years ago
Suppose a spring has a relaxed length of 28.3 cm. The simulation refers to this as the natural length. This is the length of the
den301095 [7]

Answer:

Explanation:

Normal length of spring = 28.3 cm

stretched length of spring = 38.2 cm

length of extension = 38.2 - 28.3 = 9.9 cm

= 9.9 x 10⁻² m

force applied to stretch = .55 x 9.8 ( mg )

= 5.39 N

Force constant = force applied / extension

= 5.39 / 9.9 x 10⁻²

= .5444 x 10² N /m

= 54.44 N/m

4 0
3 years ago
What type of stress is based on folds
Ipatiy [6.2K]
Life stress is bad on the fold

3 0
4 years ago
Fluid flows at 2.0 m/s through a pipe of diameter 3.0 cm. What is the volume flow rate of the fluid in m^3/
fenix001 [56]

Answer:

Volume flow rate = 1.41 \times 10^{-3}m^3/s

Explanation:

The volume flow rate through a channel can be gotten by multiplying its area and its velocity.

The channel under consideration is a circular channel. Hence, the cross-sectional area can be calculated by using the relation for calculating the area of a circle

Cross-sectional area = \pi \times d^{2}/4

A = \pi \times (3\times 10^-2)^2  /4 = 7.07 \times 10 ^-4 m^2

volume flow rate= A =  7.07 \times 10 ^-4 m^2 X 2.0 = 1.41 \times 10^{-3}m^3/s

Volume flow rate = 1.41 \times 10^{-3}m^3/s

7 0
4 years ago
A ping-pong ball is traveling at 10m/s in air, spinning 100 revolutions per second in the clockwise direction. Calculate the lif
allsm [11]

Answer:

lift force = 0.213 N

Drag force =2*10^-3 N

Explanation:

Given: velocity v = 10 m/s

w = 100 rev/sec

diameter d = 3cm

density D = 1.2kg/m3

lift force =16/3*(pi^2*d^3*w*D*v)

Substituting the values into the equation, we obtain

lift force = 0.213 N

Drag force = C*D*A*v/2

where C = 0.5

substituting the values into the equation again, we have

Drag force =2*10^-3 N

5 0
3 years ago
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