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Ganezh [65]
3 years ago
9

Two small, positively charged spheres have a combined charge of 5.0 x 10 -5 C. If each sphere is repelled from the other by an e

lectrostatic force of 1.0 N when the spheres are 2.0 m apart, what is the charge, in micro-coulomb, on the sphere with the smaller charge?
Physics
1 answer:
geniusboy [140]3 years ago
4 0

Explanation:

It is given that,

Let q₁ and q₂ are two small positively charged spheres such that,

q_1+q_2=5\times 10^{-5}\ C.............(1)

Force of repulsion between the spheres, F = 1 N

Distance between spheres, d = 2 m

We need to find the charge on the sphere with the smaller charge. The force is given by :

F=k\dfrac{q_1q_2}{d^2}

q_1q_2=\dfrac{F.d^2}{k}

q_1q_2=\dfrac{1\ N\times (2\ m)^2}{9\times 10^9}

q_1q_2=4.45\times 10^{-10}\ C............(2)

On solving the system of equation (1) and (2) using graph we get,

q_1=0.0000384\ C=38.4\ \mu C

q_2=0.0000116\ C=11.6\ \mu C

So, the charge on the smaller sphere is 11.6 micro coulombs. Hence, this is the required solution.

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A roller coaster car moving at a velocity of 30 meters/second has a momentum of 2.5 × 104 kilogram meters/second. what is its ma
frez [133]

The mass of a rollercoaster car moving at a velocity of 30 meters/second and has a momentum of 2.5 × 104 kilogram meters/second is 8.3 × 10²kg.

<h3>How to calculate mass?</h3>

The mass of the roller coaster car can be calculated using the following formula:

P = m × v

Where;

  • P = momentum
  • m = mass
  • v = velocity

m = 2.5 × 10⁴ ÷ 30

m = 8.3 × 10²kg

Therefore, the mass of a rollercoaster car moving at a velocity of 30 meters/second and has a momentum of 2.5 × 104 kilogram meters/second is 8.3 × 10²kg.

Learn more about mass at: brainly.com/question/19694949

#SPJ1

8 0
2 years ago
1. Which mathematical representation correctly identifies impulse?
horsena [70]

Answer:

1. B. Impulse = Force × Time

2. A. The momentum of each ball changes, and the total momentum stays the same

3. -55 kg·m/s

4. B. 3.5 kg

5. C. 6.3 m/s

Explanation:

1. The impulse is the momentum change of an object due to a force applied for a given period

2. Given that the objects collide, and the force of the 3 kg mass moving with 24 kg·m/s acts on the 1 kg mass while the total momentum is conserved;

The stationary ball of mass 1 kg begins to moves at certain velocity after collision and therefore changes momentum, while the velocity of the ball of mass 3.0 kg reduces and the total combined momentum of the two balls in the closed system remains the same

3. By the principle of conservation of linear momentum, we have;

The sum of the momentum before the collision = The sum of the momentum after collision

Given that the objects move together after the collision, the total momentum is therefore;

Total momentum = 110 kg·m/s + -65 kg·m/s + -100 kg·m/s = 110 kg·m/s - 65 kg·m/s - 100 kg·m/s  = -55kg·m/s

4. Given that the final velocity of the two objects (m₁ + m₂) combined = 50 m/s

Where;

m₁ = The mass of the first object

m₂ = The mass of the second object

The total momentum of the system = 250 kg·m/s

From momentum = Mass × Velocity, we have;

Mass = Momentum/Velocity = 250 kg·m/s/(50 m/s) = 5.0 kg

The mass (m₁ + m₂) = 5.0 kg

Given that m₁ = 1.5 kg, we have;

m₂ = 5.0 kg - m₁ = 5.0 kg - 1.5 kg = 3.5 kg

The mass of the second object = 3.5 kg

5. The mass of the cue stick = 0.5 kg

The velocity of the cue stick = 2.5 m/s

The mass of the ball = 0.2 kg

The initial velocity of the ball = 0 m/s

Given that total initial momentum = Total final momentum, we have;

0.5 kg × 2.5 m/s + 0.2 kg × 0 = 0.2 kg × v + 0.5 kg × 0

0.5 kg × 2.5 m/s = 0.2 kg × v

v = (0.5 kg × 2.5 m/s)/(0.2 kg) = 6.25  m/s ≈ 6.3 m/s

3 0
3 years ago
Which ecosystem contains 32% of the world's producers
tangare [24]
The Answer Is D 32 % Of The worlds production is In the marine ecosystem
4 0
3 years ago
A car driver traveling at a speed of 108km per hour ,sees a traffic light and stopped after travelling for 20seconds .Find the a
navik [9.2K]

Answer:

– 2.5 m/s²

Explanation:

We have,

• Initial velocity, u = 180 km/h = 50 m/s

• Final velocity, v = 0 m/s (it stops)

• Time taken, t = 20 seconds

We have to find acceleration, a.

\longrightarrow a = (v ― u)/t

\longrightarrow a = (0 – 50)/20 m/s²

\longrightarrow a = –50/20 m/s²

\longrightarrow a = – 5/2 m/s²

\longrightarrow a = – 2.5 m/s² (Velocity is decreasing) [Answer]

6 0
2 years ago
The 1.53-kg uniform slender bar rotates freely about a horizontal axis through O. The system is released from rest when it is in
OlgaM077 [116]

Answer:

The spring constant = 104.82 N/m

The angular velocity of the bar when θ = 32° is 1.70 rad/s

Explanation:

From the diagram attached below; we use the conservation of energy to determine the spring constant by using to formula:

T_1+V_1=T_2+V_2

0+0 = \frac{1}{2} k \delta^2 - \frac{mg (a+b) sin \ \theta }{2}  \\ \\ k \delta^2 = mg (a+b) sin \ \theta \\ \\ k = \frac{mg(a+b) sin \ \theta }{\delta^2}

Also;

\delta = \sqrt{h^2 +a^2 +2ah sin \ \theta} - \sqrt{h^2 +a^2}

Thus;

k = \frac{mg(a+b) sin \ \theta }{( \sqrt{h^2 +a^2 +2ah sin \ \theta} - \sqrt{h^2 +a^2})^2}

where;

\delta = deflection in the spring

k = spring constant

b = remaining length in the rod

m = mass of the slender bar

g = acceleration due to gravity

k = \frac{(1.53*9.8)(0.6+0.2) sin \ 64 }{( \sqrt{0.6^2 +0.6^2 +2*0.6*0.6 sin \ 64} - \sqrt{0.6^2 +0.6^2})^2}

k = 104.82\ \  N/m

Thus; the spring constant = 104.82 N/m

b

The angular velocity can be calculated by also using the conservation of energy;

T_1+V_1 = T_3 +V_3  \\ \\ 0+0 = \frac{1}{2}I_o \omega_3^2+\frac{1}{2}k \delta^2 - \frac{mg(a+b)sin \theta }{2} \\ \\ \frac{1}{2} \frac{m(a+b)^2}{3}  \omega_3^2 +  \frac{1}{2} k \delta^2 - \frac{mg(a+b)sin \ \theta }{2} =0

\frac{m(a+b)^2}{3} \omega_3^2  + k(\sqrt{h^2+a^2+2ah sin \theta } - \sqrt{h^2+a^2})^2 - mg(a+b)sin \theta = 0

\frac{1.53(0.6+0.6)^2}{3} \omega_3^2  + 104.82(\sqrt{0.6^2+0.6^2+2(0.6*0.6) sin 32 } - \sqrt{0.6^2+0.6^2})^2 - (1.53*9.81)(0.6+0.2)sin \ 32 = 0

0.7344 \omega_3^2 = 2.128

\omega _3 = \sqrt{\frac{2.128}{0.7344} }

\omega _3 =1.70 \ rad/s

Thus, the angular velocity of the bar when θ = 32° is 1.70 rad/s

7 0
3 years ago
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