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2 years ago

a car travels east with a velocity of 80 km/h for 90 minutes how far did the car travel (show your work)

Physics

1 answer:

olga55 [171]2 years ago

3 0

$▪▪▪▪▪▪▪▪▪▪▪▪▪ {\huge\mathfrak{Answer}}▪▪▪▪▪▪▪▪▪▪▪▪▪▪$

Given terms :

velocity (v) = 80 km/h

time (t) = 90 minutes = 1.5 hour

As we know, Distance covered is equal to :

$velocity \times time$

$80 \times 1.5$

$120 \: \: km$

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Question 1 (1 point)

MatroZZZ [7]

Answer:

The work done by the frictional force is 600J.

Explanation:

The work $W$ done by the frictional force is

$W= Fd$ .

Now, $F = 60N$ and $d =10m$ ; therefore,

$W= (60N)(10m)$

$\boxed{W = 600J.}$

Hence, the work done by friction is 660J.

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2 years ago

WILL GIVE BRAINLY!! PLEASE HELPP

LekaFEV [45]

The purpose of the machine is to leverage its mechanical advantage such that the force it outputs to move the heavy object is greater than the force required for you to input.

But there's no such thing as a free lunch! When you apply the conservation of energy, the work the machine does on the object will always be equal to (in an ideal machine) or less than the work you input to the machine.

This means that you will apply a lesser force for a longer distance so that the machine can supply a greater force on the object to push it a smaller distance. That is the trade-off of using the machine: it enables you to use a smaller force but at the cost of having to apply that smaller force for a greater distance.

The answer is: The work input required will equal the work output.

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3 years ago

7. Starting at rest, a car accelerates at 5.5m/s/s for 12s. What is its

Musya8 [376]

Answer:

66 m/s

Explanation:

v=u+at

= 0 + 5.5 * 12

= 66 m/s

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2 years ago

Unpolarized light with intensity 370 W/m2 passes first through a polarizing filter with its axis vertical, then through a second

vampirchik [111]

To solve this problem it is necessary to apply the concepts given by Malus regarding the Intensity of light.

From the law of Malus intensity can be defined as

$I' = \frac{I_0}{2} cos^2 \theta$

Where

$\theta =$ Angle From vertical of the axis of the polarizing filter

$I_0 =$ Intensity of the unpolarized light

The expression for the intensity of the light after passing through the first filter is given by

$I = \frac{I_0}{2}$

Replacing we have that

$I = \frac{370}{2}$

$I = 185W/m^2$

Re-arrange the equation,

$I'= \frac{I_0}{2}cos^2\theta$

Re-arrange to find \theta

$cos^2\theta = \frac{2I'}{I_0}$

$cos^2\theta = \frac{2*138}{370}$

$\theta = cos^{-1}(\sqrt{\frac{2*138}{370}})$

$\theta = 0.5282rad$

$\theta = 30.27\°$

The value of the angle from vertical of the axis of the second polarizing filter is equal to 30.2°

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2 years ago

A man is standing on the edge of a 20.0 m high cliff. He throws a rock horizontally with an initial velocity of 10.0 m/s. How lo

Arlecino [84]

Answer:

t = 4.08 s

R = 40.8 m

Explanation:

The question is asking us to solve for the time of flight and the range of the rock.

Let's start by finding the total time it takes for the rock to land on the ground. We can use this constant acceleration kinematic equation to solve for the displacement in the y-direction:

Δx = v_0 t + 1/2at²

We have these known variables:

(v_0)_y = 0 m/s
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Δx_y = -20 m

And we are trying to solve for t (time). Therefore, we can plug these values into the equation and solve for t.

-20 = 0t + 1/2(-9.8)t²
-20 = 1/2(-9.8)t²
-20 = -4.9t²
t = 4.08 sec

The time it takes for the rock to reach the ground is 4.08 seconds.

Now we can use this time in order to solve for the displacement in the x-direction. We will be using the same equation, but this time it will be in terms of the x-direction.

List out known variables:

v_0 = 10 m/s
t = 4.08 s
a_x = 0 m/s

We are trying to solve for:

Δx_x = ?

By using the same equation, we can plug these known values into it and solve for Δx.

Δx = 10 * 4.08 + 1/2(0)(4.08)²
Δx = 10 * 4.08
Δx = 40.8 m

The rock lands 40.8 m from the base of the cliff.

7 0

2 years ago