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Oliga [24]
3 years ago
12

If the charges attracting each other in the preceding problem have equal magnitudes,show that the magnitude of each charge is 2.

8
Physics
1 answer:
Schach [20]3 years ago
6 0

Answer:

The magnitude of each charge is 2.82\times10^{-6}\ C

Explanation:

Suppose the two point charges are separated by 6 cm. The attractive force between them is 20 N.

We need to calculate the magnitude of each charge

Using formula of force

F=\dfrac{kq^2}{r^2}

Where, q = charge

r = separation

Put the value into the formula

20=\dfrac{9\times10^{9}\times q^2}{(6\times10^{-2})^2}

q^2=\dfrac{(6\times 10^{-2})^2\times20}{9\times10^{9}}

q=\sqrt{\dfrac{(6\times 10^{-2})^2\times20}{9\times10^{9}}}

q=2.82\times10^{-6}\ C

Hence, The magnitude of each charge is 2.82\times10^{-6}\ C

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Explanation:

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A train moving with a velocity of 42.9 km/hour North, increases its speed with a uniform acceleration of 0.250 m/s^2 North until
ankoles [38]

Answer:

3658.24m

Explanation:

Hello!

the first thing that we must be clear about is that the train moves with constant acceleration

A body that moves with constant acceleration means that it moves in "a uniformly accelerated motion", which means that if the velocity is plotted with respect to time we will find a line and its slope will be the value of the acceleration, it determines how much it changes the speed with respect to time.

When performing a mathematical demonstration, it is found that the equations that define this movement are as follows.

\frac {Vf^{2}-Vo^2}{2.a} =X

Vf = final speed =160km/h=44.4m/s

Vo = Initial speed =42.9km/h=11.92m/s

A = acceleration =0.25m/s^2

X = displacement

solving

\frac {44.4^{2}-11.92^2}{2.(0.25)} =X\\X=3658.24m

the distance traveled by the train is 3658.24m

5 0
3 years ago
The unit of kinetic energy is the _______. The unit of kinetic energy is the _______. hertz meter watt joule radian
Musya8 [376]

Answer:

joule

Explanation:

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3 years ago
As shown in the figure below, a bullet is fired at and passes through a piece of target paper suspended by a massless string. Th
NikAS [45]

Answer:

M = 0.730*m

V = 0.663*v

Explanation:

Data Given:

v_{bullet, initial} = v\\v_{bullet, final} = 0.516*v\\v_{paper, initial} = 0\\v_{paper, final} = V\\mass_{bullet} = m\\mass_{paper} = M\\Loss Ek = 0.413 Ek

Conservation of Momentum:

P_{initial} = P_{final}\\m*v_{i} = m*0.516v_{i} + M*V\\0.484m*v_{i} = M*V .... Eq1

Energy Balance:

\frac{1}{2}*m*v^2_{i} = \frac{1}{2}*m*(0.516v_{i})^2 + \frac{1}{2}*M*V^2 + 0.413*\frac{1}{2}*m*v^2_{i}\\\\0.320744*m*v^2_{i} = M*V^2\\\\M = \frac{0.320744*m*v^2_{i} }{V^2}  ....... Eq 2

Substitute Eq 2 into Eq 1

0.484*m*v_{i} = \frac{0.320744*m*v^2_{i} }{V^2} *V  \\0.484 = 0.320744*\frac{v_{i} }{V} \\\\V = 0.663*v_{i}

Using Eq 1

0.484m*v_{i} = M* 0.663v_{i}\\\\M = 0.730*m

7 0
3 years ago
Suppose you are on a cart, initially at rest, which rides on a frictionless horizontal track. You throw a ball at a vertical sur
Len [333]

Answer:

F_c t_ c = -F_b t_b

And the forces are equal but in the opposite direction. So then we can write by general rule:

m_c \Delta V_{c} = -m_b \Delta V_b

Or equivalently:

m_c \Delta V_{c} +m_b \Delta V_b =0

Where: V_c represent the speed of the car and V_b the speed of the ball

m_c represent the mass of the car

m_b represent the mass of the ball

Since the ball is moving to the left and we assume that the total momentum not changes then the car need to move to the right in order to satisfy the equation and satisfy the balance.

By conservation of the momentum the car will move to the right since the ball is moves to the left.

So then the correct option for this case is :

A.Yes, and it moves to the right.

Explanation:

If we assume that we have the situation in the figure attached.

For this case we assume that the momentum changes are equal in magnitude and opposite in direction, so then we satisfy this:

F_c t_ c = -F_b t_b

And the forces are equal but in the opposite direction. So then we can write by general rule:

m_c \Delta V_{c} = -m_b \Delta V_b

Or equivalently:

m_c \Delta V_{c} +m_b \Delta V_b =0

Where: V_c represent the speed of the car and V_b the speed of the ball

m_c represent the mass of the car

m_b represent the mass of the ball

Since the ball is moving to the left and we assume that the total momentum not changes then the car need to move to the right in order to satisfy the equation and satisfy the balance.

By conservation of the momentum the car will move to the right since the ball is moves to the left.

So then the correct option for this case is :

A.Yes, and it moves to the right.

3 0
3 years ago
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