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svetlana [45]
3 years ago
11

Two equal charges with magnitude Q and Q experience a force of 12.3442 when held at a distance r. What is the force between two

charges of magnitude 2*Q and 2*Q when held at a distance r/2.?
Physics
1 answer:
Pepsi [2]3 years ago
4 0

Answer:197.504 N

Explanation:

Given

Two Charges with magnitude Q experience a  force of 12.344 N

at distance r

and we know Electrostatic force is given

F=\frac{kq_1q_2}{r^2}

F=\frac{kQ\cdot Q}{r^2}

F=\frac{kQ^2}{r^2}

Now the magnitude of charge is 2Q and is at a distance of \frac{r}{2}

F'=\frac{k2Q\cdot 2Q}{\frac{r^2}{2^2}}

F'=16F

F'=197.504 N

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A singly charged positive ion has a mass of 3.46 × 10−26 kg. After being accelerated through a potential difference of 215 V the
jasenka [17]

Answer:

1.8 cm

Explanation:

m = mass of the singly charged positive ion = 3.46 x 10⁻²⁶ kg

q = charge on the singly charged positive ion = 1.6 x 10⁻¹⁹ C

\Delta V =Potential difference through which the ion is accelerated = 215 V

v = Speed of the ion

Using conservation of energy

Kinetic energy gained by ion = Electric potential energy lost

(0.5) m v^{2} = q \Delta V\\(0.5) (3.46\times10^{-26}) v^{2} = (1.6\times10^{-19}) (215)\\(1.73\times10^{-26}) v^{2} = 344\times10^{-19}\\v = 4.5\times10^{4} ms^{-1}

r = Radius of the path followed by ion

B = Magnitude of magnetic field = 0.522 T

the magnetic force on the ion provides the necessary centripetal force, hence

qvB = \frac{mv^{2} }{r} \\qB = \frac{mv}{r}\\r =\frac{mv}{qB}\\r =\frac{(3.46\times10^{-26})(4.5\times10^{4})}{(1.6\times10^{-19})(0.522)}\\r = 0.018 m \\r = 1.8 cm

5 0
3 years ago
the speed of a train is decreased in a uniform rate from 96 km/h to 48km/h through a distance of 800m. calculate the distance co
GREYUIT [131]

Answer:

1066.67 m

Explanation:

Given:

v₀ = 96 km/h = 26.67 m/s

v = 48 km/h = 13.33 m/s

Δx = 800 m

Find: a

v² = v₀² + 2aΔx

(13.33 m/s)² = (26.67 m/s)² + 2a (800 m)

a = -0.333 m/s²

Given:

v₀ = 26.67 m/s

v = 0 m/s

a = -0.333 m/s²

Find: Δx

v² = v₀² + 2aΔx

(0 m/s)² = (26.67 m/s)² + 2 (-0.333 m/s²) Δx

Δx = 1066.67 m

Round as needed.

7 0
3 years ago
The Jamaican bobsled team was moving at a velocity of 50 m/s, then they hit the brakes on their sled to decelerate at a uniform
atroni [7]

Answer:

The time it took the bobsled to come to rest is 10 s.

Explanation:

Given;

initial velocity of the bobsled, u = 50 m/s

deceleration of the bobsled, a = - 5 m/s²

distance traveled, s = 250 m

Apply the following kinematic equation to determine the time of motion of the bobsled;

s = ut + ¹/₂at²

250 = 50t + ¹/₂(-5)t²

250 = 50t - ⁵/₂t²

500 = 100t - 5t²

100 = 20t -t²

t² - 20t + 100 = 0

t² -10t - 10t + 100 = 0

t (t - 10) - 10(t - 10) = 0

(t - 10)(t - 10) = 0

t = 10 s

Therefore, the time it took the bobsled to come to rest is 10 s.

3 0
3 years ago
A car with an initial speed of 31.4 km/h accelerates at a uniform rate of 1.2 m/s2 for 1.3 s. What is the displacement of the ca
Kaylis [27]

Answer:

21.5 m

Explanation:

A car has an initial speed of 31.4 km/hr

Convert to m/s

= 31.4 × 1000/3600

= 31,400/3600

= 8.722 m/s

Acceleration = 1.2 m/s^2

Time= 1.3 seconds.

Therefore the displacement can be calculated as follows

S= 8.722 × 1.3 + 1/2 × 1.2 × 1.3^2

= 11.34 + 1/2 × 20.28

= 11.34 + 10.14

= 21.5 m

5 0
3 years ago
You are leaving for a party at your cousin’s house in a city that is 150 km away. You will travel at an average rate of 50 km/hr
mrs_skeptik [129]

The distance you should cover is: S=150 km

Your average speed is: v=50 km/h

Therefore, to calculate the time you will take to cover this distance, we can use the basic relationship between space, distance and time, and we find:

t=\frac{S}{v}=\frac{150 km}{50 km/h}=3 h

so, you will take 3 hours.

3 0
3 years ago
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