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Over [174]
3 years ago
15

80PTS

Physics
1 answer:
Nitella [24]3 years ago
6 0
A) The magnitude of the electric field generated by a single point charge is given by
E=k \frac{q}{r^2}
where k is the Coulomb's constant, q is the charge and r is the distance at which we calculate the field.
Since we want to know the intensity of the field at distance of r=4.52 m from the charge of q=8.15 C, the intensity of the electric field is
E=k \frac{q}{r^2}=(8.99 \cdot 10^9 Nm^2C^{-2}) \frac{8.15 C}{4.52 m}=3.6 \cdot 10^9 N/C

b) Direction of the field
The charge that produces the field is negative, and we know that the electric field generated by a negative charge points toward the charge. Therefore, the direction of the electric field is towards the charge.
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Is it possible to be moving but not be in motion?
GREYUIT [131]

No. Motion is the thing that when you're moving, you're in it.

But it IS possible for one person to say you're moving and another person to say you're not moving, both at the same time, and both of them are correct !

4 0
3 years ago
Read 2 more answers
ow long must a simple pendulum be if it is to make exactly ten swings per second? (That is, one complete vibration takes exactly
Igoryamba
The period T of a pendulum is given by:
T=2 \pi  \sqrt{ \frac{L}{g} }
where L is the length of the pendulum while g=9.81 m/s^2 is the gravitational acceleration.

In the pendulum of the problem, one complete vibration takes exactly 0.200 s, this means its period is T=0.200 s. Using this data, we can solve the previous formula to find L:
L=g ( \frac{T}{2\pi} )^2=(9.81 m/s^2)( \frac{0.2 s}{2 \pi} )^2=1 \cdot 10^{-3} m=1 mm
4 0
3 years ago
Physics Homework MathPhys homie if you see this pls help
cluponka [151]

Answer:

1. -8.20 m/s²

2. 73.4 m

3. 19.4 m

Explanation:

1. Apply Newton's second law to the car in the y direction.

∑F = ma

N − mg = 0

N = mg

Apply Newton's second law to the car in the x direction.

∑F = ma

-F = ma

-Nμ = ma

-mgμ = ma

a = -gμ

Given μ = 0.837:

a = -(9.8 m/s²) (0.837)

a = -8.20 m/s²

2. Given:

v₀ = 34.7 m/s

v = 0 m/s

a = -8.20 m/s²

Find: Δx

v² = v₀² + 2aΔx

(0 m/s)² = (34.7 m/s)² + 2 (-8.20 m/s²) Δx

Δx = 73.4 m

3. Since your braking distance is the same as the car in front of you, the minimum safe following distance is the distance you travel during your reaction time.

d = v₀t

d = (34.7 m/s) (0.56 s)

d = 19.4 m

6 0
3 years ago
I need help plz and thank you this is do
love history [14]

Answer:

Oh I am sorry this is my first time on brainly i dont how to exit and sorry but dont know the answer

Explanation:

4 0
3 years ago
A car stopped at a red light, not moving?
timofeeve [1]

I think it's a) 1st Newton's law... so sorry if it's wrong...

5 0
3 years ago
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