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Savatey [412]
2 years ago
6

Q1: An object with a charge of 1.2 C is located 4.5 m away from a second object that has a charge of 0.36 C. Find the electrical

force that the objects exert on each other. Use 8.988100 N·m2/C2 for the value of Coulomb's constant. Give your answer to two decimal places.
Physics
1 answer:
gizmo_the_mogwai [7]2 years ago
4 0

Answer:

a) F= 0,19  [N]   according to problem statement

b) F = 0,19*10⁹ [N]  using the right value of K

Explanation:

The force between two electric charges is according to Coulomb´s law is:

F = K * q₁*q₂ / d²    where  q₁  and q₂ are the charges on body one and body 2 respectively, d is the distance between the two bodies and K is a constant  K = 8,988100*10⁹ N.m²/C². The problem establishes to use        K = 8,988100 N.m²/C².

NOTE: To value of is :  K = 8,988100*10⁹ N.m²/C². I am going to solve the problem using K = 8,988100 N.m²/C² if that information was an error, all we need to get the right answer is multiply the result by 10⁹

Then:

F = 8,988100 * 1,2* 0,36 / (4,5)²     [ N*m²/C² ] * [ C*C*/m²]

F = 3,882859/ 20,25  [N]

F= 0,19  [N]

The force is of repulsion since the two charges are positive and in the direction of the straight line which passes through the centers of the bodies

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babunello [35]

Answer:

v= 2413.5 m/s

Explanation:

maximum change of speed of rocket

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let initial mass= m

final mass = m-m(4/5) = m/5

[since the 80% of mass which is fuel is exhausted]

V-0 = 1500 ln (1/0.2)

V= 1500×1.609 = 2413.5 m/s

therefore, its exhaust speed v= 2413.5 m/s

4 0
2 years ago
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a child hits a ball with a force of 350 N. (a) If the ball and bat are in contact for 0.12 is, what impulse does the ball receiv
Lina20 [59]

Explanation:

Given that,

Force with which a child hits a ball is 350 N

Time of contact is 0.12 s

We need to find the impulse received by the ball. The impulse delivered is given by :

J=F\times t\\\\J=350\times 0.12\\\\J=42\ N-m

So, the impulse is 42 N-m..

We know that he change in momentum is also equal to the impulse delivered.

So, impulse = 42 N-m and change in momentum =42 N-m.

7 0
2 years ago
A rocket in deep space has an exhaust-gas speed of 2000 m/s. When the rocket is fully loaded, the mass of the fuel is five times
notka56 [123]

Answer:

 v_{f} = 1,386 m / s

Explanation:

Rocket propulsion is a moment process that described by the expression

       v_{f} - v₀ =  v_{r} ln (M₀ / Mf)

Where v are the velocities, final, initial and relative and M the masses

The data they give are the relative velocity (see = 2000 m / s) and the initial mass the mass of the loaded rocket (M₀ = 5Mf)

We consider that the rocket starts from rest (v₀ = 0)

At the time of burning half of the fuel the mass ratio is that the current mass is    

       M = 2.5 Mf

       v_{f} - 0 = 2000 ln (5Mf / 2.5 Mf) = 2000 ln 2

       v_{f} = 1,386 m / s

3 0
2 years ago
A car is traveling at 20.0 m/s on tires with a diameter of 70.0 cm. The car slows down to a rest after traveling 300.0 m. If the
cupoosta [38]

Answer: deceleration of 1.904\ rad/s^2

Explanation:

Given

Car is traveling at a speed of u=20 m/s

The diameter of the car is d=70 cm

It slows down to rest in 300 m

If the car rolls without slipping, then it must be experiencing pure rolling i.e. a=\alpha \cdot r

Using the equation of motion

v^2-u^2=2as\\

Insert v=0,u=20,s=300

0-(20)^2=2\times a\times 300\\\\a=\dfrac{-400}{600}\\\\a=-\dfrac{2}{3}\ m/s^2

Write acceleration as a=\alpha \cdot r

-\dfrac{2}{3}=\alpha \times 0.35\\\\\alpha =-\dfrac{2}{1.05}\\\\\alpha =-1.904\ rad/s^2

So, the car must be experiencing the deceleration of 1.904\ rad/s^2.

4 0
2 years ago
A satellite, orbiting the earth at the equator at an altitude of 400 km, has an antenna that can be modeled as a 1.76-m-long rod
ivann1987 [24]

Answer:

The inducerd emf is 1.08 V

Solution:

As per the question:

Altitude of the satellite, H = 400 km

Length of the antenna, l = 1.76 m

Magnetic field, B = 8.0\times 10^{- 5}\ T

Now,

When a conducting rod moves in a uniform magnetic field linearly with velocity, v, then the potential difference due to its motion is given by:

e = - l(vec{v}\times \vec{B})

Here, velocity v is perpendicular to the rod

Thus

e = lvB           (1)

For the orbital velocity of the satellite at an altitude, H:

v = \sqrt{\frac{Gm_{E}}{R_{E}} + H}

where

G = Gravitational constant

m_{e} = 5.972\times 10^{24}\ kg = mass of earth

R_{E} = 6371\ km = radius of earth

v = \sqrt{\frac{6.67\times 10^{- 11}\times 5.972\times 10^{24}}{6371\times 1000 + 400\times 1000} = 7670.018\ m/s

Using this value value in eqn (1):

e = 1.76\times 7670.018\times 8.0\times 10^{- 5} = 1.08\ V

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2 years ago
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