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Simora [160]
3 years ago
6

A bicyclist starts at rest and speeds up to 30 m/s while accelerating at 4 m/s^2. Determine the distance traveled.

Physics
1 answer:
raketka [301]3 years ago
6 0

Answer:

Distance, d = 112.5 meters

Explanation:

Initially, the bicyclist is at rest, u = 0

Final speed of the bicyclist, v = 30 m/s

Acceleration of the bicycle, a=4\ m/s^2

Let s is the distance travelled by the bicyclist. The third equation of motion is given as :

v^2-u^2=2as

s=\dfrac{v^2-u^2}{2a}

s=\dfrac{(30)^2}{2\times 4}

s = 112.5 meters

So, the distance travelled by the bicyclist is 112.5 meters. Hence, this is the required solution.

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5 0
3 years ago
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Imagine a negative test charge sitting at the coordinate origin (0,0). Two bunches of positive charges are located on the x-axis
Oxana [17]

Answer:

the total force vector, on test charge is points from origin to point C( 1, 1 )

Explanation:

Given the data in the question, as illustrated in the image below;

from the Image, OA = 1, OB = AC = 1

so using Pythagoras theorem

a² = b² + c²

a = √( b² + c² )

so

OC = √( OB² + AC² )

we substitute

OC = √( OA² + AC² )

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5 0
3 years ago
a 5.00 kg plane is speeding up on the ground with an applied force of 706 N. If the net force is 450 N, what would be the force
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Answer:

The force due to air resistance is 256 N.

Explanation:

Given;

mass of the plane, m = 5 kg

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Let the force due to air resistance = Fr

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Net force = applied force - force due to air resistance

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Fr = Fa - ∑F

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Therefore, the force due to air resistance is 256 N.

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