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LUCKY_DIMON [66]
3 years ago
10

Science: Please help due in 5 minutes...

Physics
1 answer:
AlekseyPX3 years ago
6 0

Answer:

fossil fuels are petroleum and natural gas. Well Fossils are mineralized remains of ancient plants and animals

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A solid conducting sphere with radius R = 0.390 m carries a net charge of +0.650 nC.
Tom [10]

Answer:

38.5 N/C

Explanation:

The electric field generated by a charged sphere at a point outside the sphere is equivalent to the electric field generated by a single point charge, and it is given by

E=k\frac{Q}{r^2}

where

k=9\cdot 10^9 Nm^2C^{-2} is the Coulomb's constant

Q is the net charge

r is the distance from the centre of the sphere

In this problem, we have

Q=+0.650 nC=0.65\cdot 10^{-9}C

r=0.390 m

Substituting into the equation, we find

E=(9\cdot 10^9)\frac{(0.65\cdot 10^{-9}C)}{(0.390m)^2}=38.5 N/C

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3 years ago
The wave shown below is produced in a rope.
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Answer:

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Explanation:

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3 years ago
A spring has a length of .200 m when a .300 kg hangs from it,and a length of .750 m when a 1.95 kg hangs from it.what is the for
musickatia [10]

1) 29.5 N/m

2) 0.100 m

Explanation:

1)

The force constant of the spring can be found by using the fact that the force on the spring is proportional to the extension of the spring (Hooke's Law). Therefore, we can write:

\Delta F= k \Delta x

where

\Delta F = F_2 - F_1 is the change in the force on the spring, where

F_1 = m_1 g = (0.300)(9.8)=2.94 N is the force applied when the hanging mass is

m_1 = 0.300 kg

F_2 = m_2 g = (1.95)(9.8)=19.1 N is the force applied when the hanging mass is

m_2 = 1.95 kg

\Delta x=x_2 -x_1 is the change in extension of the spring, where

x_1=0.200 m is the extension of the spring when the hanging mass is 0.300 kg

x_2=0.750 m is the extension of the spring when the hanging mass is 1.95 kg

Solving for k,

k=\frac{F_2-F_1}{x_2-x_1}=\frac{19.1-2.94}{0.750-0.200}=29.5 N/m

2)

When the first mass is hanging on the spring, we have

F_1 = k (x_1 - x_0)

where:

F_1 is the force applied on the spring (the weight of the hanging mass)

k is the spring constant

x_1 is the extension of the spring wrt its natural length

x_0 is the natural length of the spring (the unloaded length)

Here we have

F_1=2.94 N

k = 29.5 N/m

x_1=0.200 m

Solving for x_0, we find:

x_0 = x_1 - \frac{F_1}{k}=0.200 - \frac{2.94}{29.5}=0.100 m

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