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Neporo4naja [7]
3 years ago
14

I need help with this, please help thank you

Physics
1 answer:
Alecsey [184]3 years ago
6 0
B,a current flows through the wire
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We know that charged objects with the same sign repel each other and objects charged with opposite signs attract. What happens w
choli [55]

solution:

When an uncharged conducting object brought near to a charged insulating object there is a force on the conducting object to move the electrons within it to opposite sides of the conductor. That means there is a separation of charges in the conducting object in the presence of the charged insulating object near to it but the charge on the conducting object is neutral.

Thus, the conducting object is uncharged.

There is a force of attraction between the uncharged conducting object and the insulating object when it brought near to the insulating object.

Thus, there is a force on the conducting object.

The conductor remains uncharged and a force is exerted on it.


3 0
3 years ago
Spidermans nemesis electro delivers 4kj of electrical energy in half a second how much power does it draw from the mains?
Elodia [21]

Power delivered = (energy delivered) / (time to deliver the energy)

Power delivered = (4,000 J) / (0.5 sec)

Power delivered = 8,000 watts

I'm a little surprised to learn that Electro draws his power from the mains.  This is VERY good news for Spiderman !  It means that Spiderman can always avoid tangling with Electro ... all he has to do is stay farther away from Electro than the length of Electro's extension cord.

But OK.  Let's assume that Electro draws it all from the mains.  Then inevitably, there must be some loss in Electro's conversion process, between the outlet and his fingertips (or wherever he shoots his bolts from).

The efficiency of Electro's internal process is

<em>(power he shoots out) / (power he draws from the mains) </em>.

So, if he delivers energy toward his target at the rate of 8,000 watts, he must draw power from the mains at the rate of

<em>(8,000 watts) / (his internal efficiency) .  </em>

4 0
4 years ago
Read 2 more answers
HELP NEEDED! Determine the mechanical energy of this object: a 1-kg ball rolls on the ground at 2 m/s.
valentina_108 [34]
The mechanical energy of an object is the sum of its potential and kinetic energy.

Because the ball is on the ground, its potential energy is 0.
Its kinetic energy is given by:

K.E = 1/2 mv²
K.E = 1/2 x 1 x 2²
K.E = 2 J

Mechanical energy = 2 + 0
Mechanical energy = 2 J

The answer is B.
4 0
4 years ago
Read 2 more answers
The maximum speed of a mass m on an oscillating spring is vmax . what is the speed of the mass at the instant when the kinetic a
umka21 [38]
Let
A =  the amplitude of vibration
k =  the spring constant
m =  the mass of the object

The displacement at time, t, is of the form
x(t) = A cos(ωt)
where
ω =  the circular frequency.

The velocity is
v(t) = -ωA sin(ωt)

The maximum velocity occurs when the sin function is either 1 or -1.
Therefore
v_{max} = \omega A
Therefore
v(t) = -V_{max} sin(\omega t)

The KE (kinetic energy) is given by
KE = \frac{m}{2}v^{2} = \frac{m}{2} V_{max}^{2} sin^{2} (\omega t)

The PE (potential energy) is given by
PE = \frac{k}{2} x^{2} = \frac{k}{2} A^{2} cos^{2} (\omega t)

When the KE and PE are equal, then
v^{2} = \frac{k}{m} A^{2} cos^{2} (\omega t)

For the oscillating spring,
\omega ^{2} =  \frac{k}{m} \\ V_{max} = \omega A =  \sqrt{ \frac{k}{m} } A
Therefore
v^{2} =  \frac{k}{m}  \frac{m}{k} V_{max}^{2} cos^{2} ( \sqrt{ \frac{k}{m} t} ) \\ v = V_{max}  \,cos( \sqrt{ \frac{k}{m} t} )

Answer: v(t) = V_{max} cos( \sqrt{ \frac{k}{m} t} )

3 0
4 years ago
If the heat of combustion for a specific compound is -1320.0 kJ/mol and its molar mass is 71.83 g/mol, how many grams of this co
natka813 [3]
Heat released from combustion is calculated by multiplying the amount of substance being burned with its heat of combustion. From this, we can calculate the mass needed from the system above. The calculation is as follows:

-543 kJ = x x -1320 kJ/mol
x = 0.411 mol 

Mass of x = 0.411 x 71.83 g/mol
Mass of x = 29.55 g 
4 0
4 years ago
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