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3241004551 [841]
2 years ago
15

Tim is pushing a box with a 7 N force. The weight of the box is 5 N. A frictional force

Physics
1 answer:
Vesnalui [34]2 years ago
5 0

Friction and the applied force are not balanced. The 7 N push is countered by the frictional force of 3 N to push the box with a net force of 4 N.

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"How did your current and voltage measurements differ between the series and parallel circuits you created
irakobra [83]

Answer:

Series circuit:

The voltage that is measured across the circuit is different.

The current measured in a series circuit remains the same at all points in the circuit.

Parallel circuit:

The current measured across each resistor varies

The voltage measured across a parallel circuit will remain the same

Explanation:

Series and parallel circuits behave differently when it comes to the circulation of current and the interaction with a potential difference.

In a series circuit, the resistances are connected end to end. As a result, the voltage that is measured across the circuit is different once resistance is encountered. However, the current measured in a series circuit remains the same at all points in the circuit.

A parallel circuit behaves in an exactly opposite manner to the series circuit. In a parallel circuit, the resistances are connected side by side. As a result of this, the current measured across each resistor varies as there are circuit branches through which electric current can flow into. On the other hand, the voltage measured across a parallel circuit will remain the same

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3 years ago
Water is falling on the blades of a turbine at a rate of 100 kg/s from a certain spring. If the height of spring be 100m, then t
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2 years ago
True or False:<br><br> It is safe to touch an electric current while soaking wet.
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Answer:

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3 years ago
A 3.0-kg object moves to the right with a speed of 2.0 m/s. It collides in a perfectly elastic collision with a 6.0-kg object mo
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Answer:

The kinetic energy of the system after the collision is 9 J.

Explanation:

It is given that,

Mass of object 1, m₁ = 3 kg

Speed of object 1, v₁ = 2 m/s

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Speed of object 2, v₂ = -1 m/s (it is moving in left)

Since, the collision is elastic. The kinetic energy of the system before the collision is equal to the kinetic energy of the system after the collision. Let it is E. So,

E=\dfrac{1}{2}m_1v_1^2+\dfrac{1}{2}m_2v_1^2

E=\dfrac{1}{2}\times 3\ kg\times (2\ m/s)^2+\dfrac{1}{2}\times 6\ kg\times (-1\ m/s)^2

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