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Valentin [98]
3 years ago
11

If a particle with a charge of +4.3 × 10−18 C is attracted to another particle by a force of 6.5 × 10−8 N, what is the magnitude

of the electric field at this location?
2.2 × 1026 NC
2.8 × 10−25 NC
1.5 × 1010 N/C
6.6 × 10−11 N/C
Physics
1 answer:
AveGali [126]3 years ago
7 0

Answer: 1.5×10^10 N/C

Explanation:

E= F/q

Where E= magnitude of the electric field

F= force of attraction

q= charge of the given body

Given F= 6.5×10^-8 N

q= 4.3× 10^-18 C

Therefore, E = 6.5×10 ^-8/ 4.3×10^-18

E = 1.5×10^10 N/C

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How much power is used if a force of 54 newtons is used to pull a wagon a distance of 10 meters in 6 seconds?
AnnZ [28]
First, calculate the work done: 54 x 10 = 540J.
Now calculate power: 540 / 6 = 90.
The answer is 90W.

Hope this helps.
7 0
3 years ago
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A rock is dropped from a bridge to the water below. It takes 2.40 s for the rock to hit the water (a) find the speed in m/s of t
Artyom0805 [142]

Answer: I think Its the Height is 11.76 Meters (38.582677 Feet) between the bridge and the ground

Explanation: Supposing that where not counting air resistance in the equation, the equation D = ut+\frac{1}{2}  at^{2} states that 1/2 multiplied by earths gravitational acceleration multiplied by the amount of time to reach the bottom: 2.4 seconds equals 11.76 meters of height between the bridge and the ground.

6 0
2 years ago
Explain how soilds non metals are different from solid metals
marysya [2.9K]
Solid metal is all different types of metals or some thing that a magnet can pick up that's a full hard solid

but a non metal is everything not metal that's a solid like plastic a hot glue gun can burn throught plastic and not metal

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7 0
3 years ago
a horse began running due east and covered 25km in 4.0 hours what is the average velocity of the horse
Komok [63]

25km / 4hr = <em>6.25 km/hr east</em>

6 0
3 years ago
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To start an avalanche on a mountain slope, an artillery shell is fired with an initial velocity of 290 m/s at 57.0° above the ho
cupoosta [38]

Answer:

xf = 5.68 × 10³ m  

yf = 8.57 × 10³ m  

Explanation:

given data

vi = 290 m/s

θ = 57.0°

t = 36.0 s

solution

firsa we get here origin (0,0) to where the shell is launched

xi = 0                            yi = 0

xf = ?                            yf = ?

vxi =  vicosθ               vyi = visinθ  

ax = 0                          ay = −9.8 m/s

now we solve x motion: that is

xf = xi + vxi × t + 0.5 × ax × t²     ............1

simplfy it we get

xf = 0 + vicosθ × t + 0

put here value and we get

xf = 0 + (290 m/s) cos(57) (36.0 s)

xf = 5.68 × 10³ m  

and

now we solve for y motion: that is

yf = yi + vyi × t + 0.5 × ay × t ²     ............2

put here value and we get

yf = 0 + (290 m/s) × sin(57) × (36.0 s) + 0.5 × (−9.8 m/s2) × (36.0 s)  ²

yf = 8.57 × 10³ m  

5 0
3 years ago
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