# An electric toy car with a mass of #3 kg# is powered by a motor with a voltage of #9 V# and a current supply of #8 A#. How long will it take for the toy car to accelerate from rest to #8/3 m/s#?

The power input is the product of the voltage and the current.

The change in kinetic energy for the car is

Thus, the time needed is the energy divided by the power.

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To find the time taken for the toy car to accelerate, you can use the equation:

[ t = \frac{m \cdot v}{F} ]

where: ( t ) = time taken to accelerate (in seconds) ( m ) = mass of the car (in kilograms) ( v ) = final velocity (in meters per second) ( F ) = force acting on the car (in Newtons)

First, calculate the force using Ohm's law:

[ V = I \cdot R ] [ R = \frac{V}{I} ]

where: ( V ) = voltage (in volts) ( I ) = current (in amperes) ( R ) = resistance (in ohms)

Then, use Newton's second law to find the force:

[ F = m \cdot a ]

where: ( m ) = mass of the car (in kilograms) ( a ) = acceleration (in meters per second squared)

Finally, substitute the values into the equation for time:

[ t = \frac{m \cdot v}{F} ]

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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- A circuit with a resistance of #8 Omega# has a fuse with a capacity of #5 A#. Can a voltage of #62 V# be applied to the circuit without blowing the fuse?
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