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gizmo_the_mogwai [7]
4 years ago
12

Which describes static electricity?

Physics
1 answer:
HACTEHA [7]4 years ago
6 0
Static electricity is like a balloon is being rubbed on your head making electricity.
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skateboarder, starting from rest, rolls down a 13.5 m ramp. When she arrives at the bottom of the ramp her speed is 7.37 m/s. If
scZoUnD [109]

Answer:

1.7 m/s²

Explanation:

d = length of the ramp = 13.5 m

v₀ = initial speed of the skateboarder = 0 m/s

v = final speed of the skateboarder = 7.37 m/s

a = acceleration

Using the equation

v² = v₀² + 2 a d

7.37² = 0² + 2 a (13.5)

a = 2.01 m/s²

θ = angle of the incline relative to ground = 29.9

a' = Component of acceleration parallel to the ground

Component of acceleration parallel to the ground is given as

a' = a Cosθ

a' = 2.01 Cos29.9

a' = 1.7 m/s²

7 0
3 years ago
The graph below shows the displacement of an object as a constant
SSSSS [86.1K]

Answer:help

Explanation:  b.3.6 j

4 0
3 years ago
Propose an experiment that could be used to demonstrate that friction is a contact force?
vesna_86 [32]
Maybe push or pull an object with a large amount of mass? you are force a (pushing through object) aka making contact. i hope i helped not good with physics :)
3 0
3 years ago
Read 2 more answers
A wire 6.60 m long with diameter of 2.05 mm has a resistance of 0.0310 Ω.
Alex73 [517]

Answer:

1.551×10^-8 Ωm

Explanation:

Resistivity of a material is expressed as shown;.

Resistivity = RA/l

R is the resistance of the material

A is the cross sectional area

l is the length of the wire.

Given;

R = 0.0310 Ω

A = πd²/4

A = π(2.05×10^-3)²/4

A = 0.000013204255/4

A = 0.00000330106375

A = 3.30×10^-6m

l = 6.60m

Substituting this values into the formula for calculating resistivity.

rho = 0.0310× 3.30×10^-6/6.60

rho = 1.023×10^-7/6.60

rho = 1.551×10^-8 Ωm

Hence the resistivity of the material is 1.551×10^-8 Ωm

6 0
3 years ago
A point charge with charge q1 = 3.40 μC is held stationary at the origin. A second point charge with charge q2 = -4.90 μC moves
expeople1 [14]

Answer:

-0.79 J

Explanation:

We are given that

q_1=3.4\mu C=3.4\times 10^{-6} C

1\mu C=10^{-6} C

q_2=-4.9\mu C=-4.9\times 10^{-6} C

x_1=0.125,y_1=0

x_2=0.280,y_2=0.235

We have to find the work done by the electric force on the moving point charge.

r_1=\sqrt{x^2_1+y^2_1}=\sqrt{(0.125)^2+0}=0.125

r_2=\sqrt{(0.280)^2+(0.235)^2}=0.366

Work done,W=kq_1q_2(\frac{1}{r_1}-\frac{1}{r_2})

Where k=9\times 10^9

Using the formula

W=9\times 10^9\times 3.4\times 10^{-6}\times(-4.9\times 10^{-6})(\frac{1}{0.125}-\frac{1}{0.366})

W=-0.79 J

5 0
4 years ago
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