What is the required force to bring a 1550 kg vehicle moving at 32 m/s to a stop in a distance of 45 m?
1 answer:
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We have here what is known as parallel combination of resistors.
Using the relation:
And then we can turn take the inverse to get the effective resistance.
Where r is the magnitude of the resistance offered by each resistor.
In this case we have,
(every term has an mho in the end)
To ger effective resistance take the inverse:
we get,
The potential difference is of 9V.
So the current flowing using ohm's law,
V = IR
will be, 0.0139 Amperes.
Using the relation:
And then we can turn take the inverse to get the effective resistance.
Where r is the magnitude of the resistance offered by each resistor.
In this case we have,
(every term has an mho in the end)
To ger effective resistance take the inverse:
we get,
The potential difference is of 9V.
So the current flowing using ohm's law,
V = IR
will be, 0.0139 Amperes.
Answer:
Explanation:
Given that,
The mass of the motorbike and rider is 286 kg.
We need to find the acceleration of the motorbike at this moment in time.
Net force acting on the bike is given by :
F = 2000 N - 845 N = 1155 N
Let a be the acceleration of the motorbike. So,
So, the acceleration of the motorbike is .
