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Tems11 [23]
3 years ago
12

A point charge of 4 µC is located at the center of a sphere with a radius of 25 cm. Find the electric flux through the surface o

f the sphere. The Coulomb constant is 8.98755 × 109 N · m2 /C 2 and the acceleration due to gravity is 9.8 m/s 2 .Answer in units of N · m2 /C.
Physics
1 answer:
givi [52]3 years ago
5 0

Answer:

Ф = 4.5176x10⁵ N . m² / C

Explanation:

In this case, we need to use two expressions in order to calculate the electric flux of the sphere.

First the Electric flux is calculated using this expression:

Ф = E * A  (1)

Where:

E: Electric field

A: Area of the sphere

To get the electric field E, we use this expression:

E = K * q / r²    (2)

If we replace (2) into (1) we have the following:

Ф = K * q * A / r²    (3)

Finally, we need to know the expression to get the area of a sphere which is the following:

A = 4πr²   (4)

Replacing into (3):}

Ф = K * Q * 4πr² / r²       discarting r²:

Ф = K * Q * 4π   (5)

Now, all we need to do is replace the given values and solve for the electric flux of the sphere:

Ф = 8.98755x10⁹ * 4x10⁻⁶ * 4 * π

<h2>Ф = 4.5176x10⁵ N . m² / C</h2>

Hope this helps

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Two speedboats are traveling at the same speed relative to the water in opposite directions in a moving river. An observer on th
DENIUS [597]

Answer:

a) vboat = 5.95 m/s  b) vriver= 1.05 m/s

Explanation:

a) As observed from the shore, the speed of the boats can be expressed as the vector sum, of the boat speed relative to the water and the river speed relative to the shore, as follows:

vb₁s = vb₁w + vrs

In one case, the boat is moving in the same direction as the water:

vb₁s = vb₁w + vrs = 7.0 m/s (1)

For the other boat, it is clear that is moving in an opposite direction:

vb₂s = vb₂w - vrs = 4.9 m/s (2)

As  we know that vb₁w = vb₂w, adding both sides, we can remove the river speed from the equation, as follows:

vb₁w = vb₂w =  \frac{7.0 m/s + 4.9 m/s}{2} =5.95 m/s

b) Replacing this value in (1) and solving for vriver, we have:

vriver = 7.0 m/s - 5.95 m/s = 1.05 m/s

(we could have arrived to the same result subtracting both sides in (1), and (2))

3 0
3 years ago
An object is dropped from a bridge. A second object is thrown downwards 1.00 s later. They both reach the water 20.0 m below at
Deffense [45]

Answer:

v_{o}=-14.60m/s

Explanation:

<u>Kinematics equation for first Object:</u>

y(t)=y_{o}+v_{o}t-1/2*g*t^{2}

but:

v_{o}=0m/s       The initial velocity is zero

y_{o}=20m

it reach the water at in instant, t1, y(t)=0:

0=y_{o}-1/2*g*t_{1}^{2}

t_{1}=\sqrt{2y_{o}/g}=\sqrt{2*20/9.81}=2.02s

<u>Kinematics equation for the second Object:</u>

The initial velocity is zero

y(t)=y_{o}+v_{o}t-1/2*g*t^{2}

but:

y_{o}=20m

it reach the water at in instant, t2, y(t)=0. If the second object is thrown 1s later, t2=t1-1=1.02s

0=y_{o}+v_{o}t_{2}-1/2*g*t_{2}^{2}

v_{o}=1/(t_{2})*(1/2*g*t_{2}^{2}-y_{o})=(1/1.02)*(1/2*9.81*1.02^{2}-20)=-14.60m/s

The velocity is negative, because the object is thrown downwards.

6 0
2 years ago
The weight of an object _____.
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The weight of an object is the force with which it is attracted to earth. The gravity of an object or body of an object is high on earth than at the atmosphere. It has an average of gravitational constant equal to 9.8066 or 9.8 meters per second. In truth, the acceleration of the object depend upon its location, the latitude and altitude, on earth.     

<span> </span>

8 0
3 years ago
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Answer: 5.944\times 10^{23}\ kg

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Orbital speed v=3.40\times 10^3\ m/s

mass of GS m_{GS}=930\ kg

Radius of Mars r=3.43\times 10^6\ m

Consider the mass of mars is M

Here, Gravitational pull will provide the centripetal force

F_G=F_c

\dfrac{GMm_{GS}}{r^2}=\dfrac{m_{GS}v^2}{r}\\M=\dfrac{v^2\cdot r}{G}\\M=\dfrac{(3.43\times 10^3)^2\cdot 3.43\times 10^6}{6.67\times 10^{-11}}

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Answer: C

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