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3 years ago

10

A car starts from rest and accelerates uniformly over a time of 10.21 seconds for a distance of 210 m. Determine the acceleratio

n of the car.

1 answer:

uysha [10]3 years ago

6 0

Answer:

4.03 m/s/s

Explanation:

d=1/2 a t^2

210 = 1/2 a (10.21)^2

210 = 52.1 a

a = 4.03 m/s/s

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In one sentence describe how the electricity in an action potential is generated

vova2212 [387]

Due to influx of potassium ions, electricity is generated in axon of a neuron.

<u>Explanation:</u>

Axon membrane is the semi permeable membrane that is full of potassium and sodium channels. There’s also Sodium Potassium ATPase pumps. When there’s an impulse coming through the synapse, the potassium channels open. This leads to influx of sodium from outside the membrane to inside it. Then the membrane becomes positive.

Then the electricity is generated and its conducted from one part to another. After the impulse is conducted, the sodium potassium pumps come in action which transports 3 sodium inside and 2 potassium outside in consumption to an ATP.

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3 years ago

A 20-kg boy slides down a smooth, snow-covered hill on a plastic disk. The hill is at a 10° angle to the horizontal, and the slo

zvonat [6]

Answer:

13 m/s

Explanation:

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2 years ago

Does someone know how to do math with that equation

mestny [16]

Answer:

f = 5 cm

Explanation:

using the thin lens equation, given as follows:

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do = the distance of object from lens = 20 cm

di = the distance of image from lens = 6.6667 cm

Therefore,

$\frac{1}{f} = \frac{1}{20\ cm}+\frac{1}{6.6667\ cm}\\\\\frac{1}{f} = 0.199999\ cm^{-1}\\\\f = \frac{1}{0.199999\ cm^{-1}}\\\\$

<u>f = 5 cm</u>

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3 years ago

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wlad13 [49]

Answer:

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3 years ago

Read 2 more answers

In Example 2.7, we investigated a jet landing on an aircraft carrier. In a later maneuver, the jet comes in for a landing on sol

igor_vitrenko [27]

Answer:

a) 20 seconds

b) No.

Explanation:

t = Time taken for jet to stop

u = Initial velocity = 100 m/s (given in the question)

v = Final velocity = 0 (because the jet will stop at the end)

s = Displacement of the jet (Distance between the moment the jet touches the ground to the point the point it stops)

a = Acceleration = -5.00 m/s² (slowing down, so it is negative)

a) Equation of motion

$v=u+at\\\Rightarrow t=\frac{v-u}{a}\\\Rightarrow t=\frac{0-100}{-5}\\\Rightarrow t=20\ s$

The time required for the plane to slow down from the moment it touches the ground is 20 seconds.

$s=ut+\frac{1}{2}at^2\\\Rightarrow s=100\times 20+\frac{1}{2}\times -5\times 20^2\\\Rightarrow s=1000\ m$

The distance it requires for the jet to stop is 1000 m so in a small tropical island airport where the runway is 0.800 km long the plane would not be able to land. The runway needs to be atleast 1000 m long here the runway on the island is 1000-800 = 200 m short.

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3 years ago