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alexira [117]
3 years ago
10

Two 0.50 g spheres are charged equally and placed 2.5 cm apart. When released, they begin to accelerate at 170 m/s^2 .What is th

e magnitude of thecharge on each sphere?
Physics
1 answer:
vitfil [10]3 years ago
8 0

Answer:

q=7.65*10^{-8}C

Explanation:

Using Newton's second law, we calculate the magnitude of the electric force between the spheres:

F=ma\\F=0.5*10^{-3}kg(170\frac{m}{s^2})\\F=0.085N

The magnitude of the charge in both spheres is the same. So, we calculate the charge, using Coulomb's law:

F=\frac{kq^2}{d^2}\\q=\sqrt\frac{Fd^2}{k}\\q=\sqrt\frac{(0.085N)(2.5*10^{-2}m)^2}{8.99*10^9\frac{N\cdot m^2}{C^2}}\\q=7.65*10^{-8}C

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A car travels at a constant velocity of 70 mph for one hour. At the second hour, the car’s velocity was 60 mph. At the third hou
katovenus [111]

When the velocity increases, then the acceleration will be positive and when the velocity decrease then the acceleration will be negative.

During the first hour, the velocity was 70 mph and during the seconds hour the velocity was 60 mph. Hence, the velocity decrease in the seconds hour. So, the acceleration will be negative during the second hour.

Now, during the third hour the velocity increases as it is 80 mph. Hence, the acceleration will be positive during the third hour.


8 0
3 years ago
Read 2 more answers
Four power transistors, each dissipating 12 W, are mounted on a thin vertical aluminum plate 22 cm 3 22 cm in size. The heat gen
Sveta_85 [38]

Answer:

The temperature of the Aluminium plate 44.84⁰C

Explanation:

Number of transistors = 4

Since the heat dissipated by each transistor is 12W

Total heat dissipated, Q = 4 * 12 = 48 W

Q = 48 W

Cross sectional Area of the Aluminium plate, A = 2(l * b)

l = Length of the aluminium plate = 22 cm = 0.22 m

b = width of the aluminium plate = 22 cm = 0.22 m

A =2( 0.22 * 0.22 )

A = 0.0968 m²

From the heat balance equation, Q = hAΔT

h = 25 W/m²·K

A = 0.0968 m²

ΔT = T - T(air)

T(air) = 25°C

ΔT = T - 25°C

Q = 25 * 0.0968 * (  T - 25)

Q = 2.42 (T - 25)

Substitute Q = 48 into the equation above

48 =  2.42 (T - 25)

T - 25 = 19.84

T = 25 + 19.84

T = 44.84 ⁰C

6 0
3 years ago
If a truck loses 5610 J of energy as it slows down due to an external force of 425 N, what distance will it have moved after the
lara31 [8.8K]

Answer:

13.2m

Explanation:

Step one:

given data

Energy= 5610J

Force F= 425N

Required

The distance traveled

Step two:

We know that work done is given as

WD= force* distance

so

5610=425*d

divide both sides by 425

d= 5610/425

d=13.2m

3 0
3 years ago
A motorcycle is traveling up one side of a hill and down the other side. The crest of the hill is a circular arc with a radius o
Setler79 [48]

Answer:

21.8 m/s

Explanation:

At the top of the hill (crest), there are two forces acting on the motorcycle:

- The reaction force of the road, N (upward)

- The force of gravity, mg (downward)

Since the motorcycle is moving by circular motion, the resultant of these forces will give the centripetal force, so:

mg-N = m\frac{v^2}{r}

where the direction of the weight (mg) is equal to that of the centripetal force, and where

m is the mass of the cycle

g = 9.8 m/s^2 is the acceleration of gravity

v is the speed

r = 48.6 is the radius of the hill

The cycle loses contact with the road when the reaction force becomes zero:

N = 0

Substituting into the equation, we therefore find the maximum speed that is allowed for the cycle before losing constact:

mg = m\frac{v^2}{r}\\v=\sqrt{gr}=\sqrt{(9.8)(48.6)}=21.8 m/s

6 0
3 years ago
I need help!!! 60 POINTS!!!!
Scrat [10]
Correct one is b
Good luck
3 0
3 years ago
Read 2 more answers
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