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3 years ago

A race car starts at rest and speeds up to 40 m/s in a distance of 100 m. Determine the acceleration of the car.

Physics

1 answer:

Finger [1]3 years ago

8 0

Answer:

Acceleration of the car, $a=8\ m/s^2$

Explanation:

Given that,

Initial speed of the car, u = 0 (at rest)

Final speed of the car, v = 40 m/s

Distance covered, s = 100 m

We need to find the acceleration of the car. It can be calculated using third equation of motion as :

$v^2-u^2=2as$

a is the acceleration of the car

$a=\dfrac{v^2}{2s}$

$a=\dfrac{(40)^2}{2\times 100}$

$a=8\ m/s^2$

So, the acceleration of the car is $8\ m/s^2$ . Hence, this is the required solution

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3 years ago

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The near point of an eye is 56.0 cm. A corrective lens is to be used to allow this eye to focus clearly on objects at the distan

irakobra [83]

Answer:

Explanation:

Near point = 56 cm .

near point of healthy person = 25 cm

person suffers from long sightedness

convex lens will be required .

object distance u = 25 cm

image distance v = 56 cm

both will be negative as both are in front of the lens.

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I/v - 1 / u = 1/f

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3 years ago

When a roller coaster gets to the bottom of a descent, describe the energy transfers and changes to energy stores that happen if

Feliz [49]

Answer:

its B

Explanation:

It's B & A at the same time because A. a roller coaster uses brakes to slow down and stop. B is the most reasonable answer. Because all roller coasters go up and over a second time over the hill, but they also slow down. But go with B.

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2 years ago

A 39-foot ladder is leaning against a vertical wall. If the bottom of the ladder is being pulled away from the wall at the rate

Viefleur [7K]

Answer:

The rate of change of the area when the bottom of the ladder (denoted by $b$ ) is at 36 ft. from the wall is the following:

$\frac{dA}{dt}|_{b=36}=-571.2\, ft^2/s$

Explanation:

The Area of the triangle is given by $A=h\times b$ where $h=\sqrt{l^2-b^2}$ (by using the Pythagoras' Theorem) and $b$ is the length of the base of the triangle or the distance between the bottom of the ladder and the wall.

The area is then

$A=\sqrt{l^2-b^2}b$

The rate of change of the area is given by its time derivative

$\frac{dA}{dt}=\frac{d}{dt}\left(\sqrt{l^2-b^2}\cdot b\right)$

$\implies \frac{dA}{dt}=\frac{d}{dt}\left(\sqrt{l^2-b^2}\right)\cdot b+\frac{db}{dt}\cdot\sqrt{l^2-b^2}$

$\implies\frac{dA}{dt}=\frac{1}{2\sqrt{l^2-b^2}}\frac{d}{dt}(l^2-b^2)\cdot b+\sqrt{l^2-b^2}}\cdot \frac{db}{dt}$ Product rule

$\implies\frac{dA}{dt}=-\frac{1}{2\sqrt{l^2-b^2}}\cdot 2\cdot b^2\cdot \frac{db}{dt}+\sqrt{l^2-b^2}}\cdot \frac{db}{dt}$ Chain rule

$\implies\frac{dA}{dt}=-\frac{1}{\sqrt{l^2-b^2}}\cdot b^2\cdot \frac{db}{dt}+\sqrt{l^2-b^2}}\cdot \frac{db}{dt}$

$\implies\frac{dA}{dt}=\frac{db}{dt}\left(-\frac{1}{\sqrt{l^2-b^2}}\cdot b^2+\sqrt{l^2-b^2}}\right)$

In here we can identify $b=36\, ft$ , $l=39$ and $\frac{db}{dt}=8\,ft/s$ .

The result is then

$\frac{dA}{dt}=8\left(-\frac{1}{\sqrt{39^2-36^2}}\cdot 36^2+\sqrt{39^2-36^2}}\right)=-571.2\, ft^2/s$

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2 years ago

An elevator is being lowered at steadily decreasing speed by a steel cable attached to an electric motor. There is no air resist

shtirl [24]

Answer:

F = W + ma a> 0

Explanation:

For this exercise let's use Newton's second law

we assume the upward direction as positive

F - W = m a

F = W + ma

F = m (g + a)

In this case they indicate that the speed is less and less as it goes down, therefore the acceleration must be opposite to the speed, that is, the acceleration is upwards, consequently it is positive

We can see that since a> 0 the force F must have greater than the weight of the elevator

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2 years ago

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