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Natali [406]
3 years ago
7

Two very small charged particles exert an electrostatic force F on each other. If the distance between them is doubled, the forc

e
between them is

A. Doubled
B. halved
C. Quadrupled
D. Quartered
Physics
1 answer:
Solnce55 [7]3 years ago
4 0

Answer:

The answer is D

Explanation:

Your welcome

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Please help me with this question. BRAINLIEST if answered.
IceJOKER [234]

... The top branch of the 3-branched parallel block ... the 9 and 6 in series ...
is equivalent to a single resistor of 15 ohms.

... The 3-branched parallel block boils down to (30, 10, and 15) in parallel.
That's (1/30 + 1/10 + 1/15)⁻¹  =   5 ohms.

... The 5-ohm-equivalent block and the 20-ohm resistor form a
voltage divider across the battery.
The voltage across the 5-ohm-equivalent block is (5/25 x 30v) = 6v .

... The top branch of the block is equivalent to a (9 + 6) = 15-ohmer.
With 6v across its ends, the current through that branch is (6/15) = 0.4A .

... With 0.4A flowing through it, the 9-ohm resistor is dissipating

            I²R = (0.4A)² (9 ohms)  =  (0.16 A²) (9 ohms)  =  1.44 W  (choice-3)   
8 0
3 years ago
What is the strength of the electric field ep 0.90 mm from a proton?
dem82 [27]

The electric field strength is inversely related to the square of the distance.so  the strength of the electric field is given by

E=\frac{1}{4\pi \epsilon _{0}  } \frac{q}{r^{2} } = \frac{k q}{r^{2} }

Here,  \frac{1}{4\pi \epsilon _{0}  } = k  is constant depend upon medium and its value is 9.0 \times10^{9} \ N m^2/C^2 and q is charge and  r is the distance.

Given  r = 0.90 mm = 9.0 \times 10^{-4} m and we know the charge of proton, q = 1.6\times 10^{-19} \ C.

Therefore,

E=\frac{9.0 \times10^{9} \ N m^2/C^2 \times1.6\times 10^{-19} \ C  }{(9.0 \times 10^{-4} m)^2} = 0.177 \times 10^{-2} \ N/C \\\\  E= 1.77 \times 10^{-3}  N/C

4 0
3 years ago
A girl pulls her younger brother on a sled along a flat sidewalk (the total mass of the sled is 30kg). A frictional force of 50N
harkovskaia [24]

Answer:

6.13 s

219 N

Explanation:

Newton's law in the x direction:

∑F = ma

150 cos 30° N − 50 N = (30 kg) a

a = 2.66 m/s²

Δx = v₀ t + ½ at²

(50 m) = (0 m/s) t + ½ (2.66 m/s²) t²

t = 6.13 s

Newton's law in the y direction:

∑F = ma

Fn + 150 sin 30° N − (30 kg) (9.8 m/s²) = 0

Fn = 219 N

4 0
3 years ago
a roller coaster begins at the top of a hill if it acelerates at the rate of 2 m/s2 and has a mass of 2000 kg what net force is
AlladinOne [14]

Answer:

\boxed {\boxed {\sf 4000 \ Newtons }}

Explanation:

Force can be found by multiplying the mass by the acceleration.

F=m*a

The mass of the roller coaster is 2000 kilograms and the acceleration is 2 meters per second squared.

m= 2000 \ kg \\a= 2 \ m/s^2

Substitute the values into the formula.

F= 2000 \ kg * 2 \ m/s^2

Multiply.

F= 4000  \ kg*m/s^2

  • 1 kg*m/s² is equal to 1 N
  • Therefore our answer of 4000 kg*m/s² is equal to 4000 Newtons

F= 4000 \ N

The net force acting on the roller coaster is <u>4000 Newtons.</u>

7 0
3 years ago
Two identical peaches, Peach 1 and Peach 2, fall from a cliff at time t=0t=0t, equals, 0 from the same height. Peach 1 is droppe
Hunter-Best [27]

Answer: The correct answer is graph A.

Explanation:

See Khan Academy.

4 0
3 years ago
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