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Alenkasestr [34]
4 years ago
5

8. Simple machines make a job easier to perform by

Physics
1 answer:
WINSTONCH [101]4 years ago
8 0

Answer:

C

Explanation:

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Determine the magnitude as well as direction of the electric field at point A, shown in the above figure. Given the value of k =
Pavlova-9 [17]

Answer:

The magnitude of the electric field is 8.99*10^{12}N/C in the r direction.

Explanation:

The equation of the electric field is given by:

|\vec{E}|=k\frac{q}{r^{2}}

Where:

  • k is the Coulomb constant is 8.99 *10^{9}\: Nm^{2}C^{-2}
  • q is the charge
  • r is the distance from A to q

|\vec{E}|=8.99*10^{9}\frac{12.5}{0.11^{2}}    

\vec{E}=9.29*10^{12} \vac{r} \: N/C

Therefore, the magnitude of the electric field is 8.99*10^{12}N/C in the r direction.

I hope it helps you!

4 0
3 years ago
A loop of radius r = 3 cm is placed parallel to the xy-plane in a uniform magnetic field = (0.75 T) . The resistance of the loop
matrenka [14]

Answer:

i = 0.00077A

Explanation:

Given:

loop radius, r = 3.0 cm = 0.03 m

Area, A = π x r² = π x 0.03² = 0.0028 m²

Magnetic Field, B = 0.75 T

Loop resistance, R = 18 Ω

time, t = 0.15 seconds

Now,

the induced emf is given as:

EMF = - BA/t .......1

Likewise,

EMF = iR.......2

Equate 1 and 2

iR = - BA/t

i = - BA/tR

i = 0.75×0.0028/0.15×18

i = 0.0021/2.7

i = 0.00077A

8 0
3 years ago
Is there a unit of measure called potential
Umnica [9.8K]
No. Potentional (other wordy here) are a few of though.
4 0
3 years ago
What is the approximate weight of a 20-kg cannonball on Earth? 2 N 20 N 196 N 1,960 N
Margaret [11]
Since weight is the force an object is exerting on another object, and the formula to calculate force is Force = Mass * Acceleration, the answer to your question is 196 N, since the mass of the cannonball times Earth's gravitational pull equals 196 N.
5 0
4 years ago
Read 2 more answers
A businesswoman is rushing out of a hotel through a revolving door with a force of 80 N applied at the edge of the 3 m wide door
mel-nik [20]

Answer:

Explanation:

Given

Force applied F=80\ N

Door is d=3\ m wide

for gate to revolve Properly Pivot Point must be at center i.e. 1.5 from either end

Torque applied is T=force\times distance

Maximum torque

T_{max}=F\times \frac{d}{2}

T_{max}=80\times \frac{3}{2}

T_{max}=120\ N-m

7 0
3 years ago
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