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Contact [7]
3 years ago
8

Calculate the value of two equal charges if they repel one another with a force of 10 N having 10 m distance apart in a vacuum.

What would be the size of the charges if they were situated in an insulting liquid whose permittivity was ten times that of a vacuum.
​
Physics
1 answer:
MAVERICK [17]3 years ago
4 0

Explanation:

a) From Coulombs law,

F = (kq1q2)/r²...............1

F = 10 N

r = 10m

k = 9 × 10^9 m/F

q1 = q2

from equation 1, make q the subject

q = (Fr²)/2k

q = (10×10²)/(2×9×10^9)

q = 1000/1.8×10^10

q = 5.556×10^-8 C

b) K = k/ko

k = Kko = 10times permittivity of free-space

Goodluck

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Ede4ka [16]

Answer:

72 m

Explanation:

Given:

v₀ = 0 m/s

v = 60 m/s

a = 25 m/s²

Find: Δx

v² = v₀² + 2aΔx

(60 m/s)² = (0 m/s)² + 2 (25 m/s²) Δx

Δx = 72 m

6 0
3 years ago
A graduated cylinder contains 62 ml of water. When a small metal block is added to the water, the volume of the water increases
lawyer [7]

Answer:

The density of the block is 7.4g/ml.

Explanation:

We can determine the volume of the metal block by taking the difference between the volumes measured in the graduated cylinder:

V_{block}=65.5ml-62ml\\\\V_{block}=3.5ml

Now, as we know that the average density of an object is calculated dividing its mass by its volume, we can calculate the density ρ of the metal block using the expression:

\rho_{block}=\frac{m_{block}}{V_{block}}\\\\\rho_{block}=\frac{26g}{3.5ml}\\\\\rho_{block}=7.4\frac{g}{ml}

Finally, it means that the density of the metal block is 7.4g/ml.

3 0
3 years ago
Read 2 more answers
A civil engineer wishes to redesign the curved roadway in the example What is the Maximum Speed of the Car? in such a way that a
vlabodo [156]

Answer:

24.3 degrees

Explanation:

A car traveling in circular motion at linear speed v = 12.8 m/s around a circle of radius r = 37 m is subjected to a centripetal acceleration:

a_c = \frac{v^2}{r} = \frac{12.8^2}{37} = 4.43 m/s^2

Let α be the banked angle, as α > 0, the outward centripetal acceleration vector is split into 2 components, 1 parallel and the other perpendicular to the road. The one that is parallel has a magnitude of 4.43cosα and is the one that would make the car slip.

Similarly, gravitational acceleration g is split into 2 component, one parallel and the other perpendicular to the road surface. The one that is parallel has a magnitude of gsinα and is the one that keeps the car from slipping outward.

So gsin\alpha = 4.43cos\alpha

\frac{sin\alpha}{cos\alpha} = \frac{4.43}{g} = \frac{4.43}{9.81} = 0.451

tan\alpha = 0.451

\alpha = tan^{-1}0.451 = 0.424 rad = 0.424*180/\pi \approx 24.3^0

3 0
4 years ago
An electron accelerated from rest through a voltage of 780 v enters a region of constant magnetic field. part a part complete if
maxonik [38]
The electron is accelerated through a potential difference of \Delta V=780 V, so the kinetic energy gained by the electron is equal to its variation of electrical potential energy:
\frac{1}{2}mv^2 =  e \Delta V
where
m is the electron mass
v is the final speed of the electron
e is the electron charge
\Delta V is the potential difference

Re-arranging this equation, we can find the speed of the electron before entering the magnetic field:
v= \sqrt{ \frac{2 e \Delta V}{m} } = \sqrt{ \frac{2(1.6 \cdot 10^{-19}C)(780 V)}{9.1 \cdot 10^{-31} kg} }=1.66 \cdot 10^7 m/s


Now the electron enters the magnetic field. The Lorentz force provides the centripetal force that keeps the electron in circular orbit:
evB=m \frac{v^2}{r}
where B is the intensity of the magnetic field and r is the orbital radius. Since the radius is r=25 cm=0.25 m, we can re-arrange this equation to find B:
B= \frac{mv}{er}= \frac{(9.1 \cdot 10^{-31}kg)(1.66 \cdot 10^7 m/s)}{(1.6 \cdot 10^{-19}C)(0.25 m)} =3.8 \cdot 10^{-4} T
3 0
4 years ago
10 points
JulijaS [17]

Answer:

I think the answer might be the 2nd one or 3rd one.

8 0
3 years ago
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