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3 years ago

An airplane flies with a constant speed of 540 miles per hour. How far can it travel in 1 1/4 hours?

Physics

1 answer:

USPshnik [31]3 years ago

7 0

Answer:

Hey!

Your answer is 675 MILES!

Explanation:

Using the S = D x T formula...

We first need to convert any values... 1 1/4 hours--75 mins (1.25hrs)

So now we substitute into the formula!

<u>D (distance)=</u> 540 x 1.25 = 675 (Distance = Speed x Time)

<u>Distance Travelled=</u> 675 MILES!

<h2><u>I HOPE THIS HELPED YOU!</u></h2>

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Which term describes an image that is in the opposite orientation than the object from which it was formed? virtual erect invert

DiKsa [7]

Answer:

lateral

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3 years ago

A bullet is fired straight up from a gun with a

melamori03 [73]

Answer: 815.51 m

Explanation:

This situation is related to projectile motion or parabolic motion, in which the initial velocity of the bullet has only y-component, since it was fired straight up. In addition, we are dealing with constant acceleration (due gravity), therefore the following equations will be useful to solve this problem:

$V=V_{o}+gt$ (1)

$V^{2}=V_{o}^{2}+2gy$ (2)

Where:

$V$ is the final velocity of the bullet

$V_{o}=152 m/s$ is the initial velocity of the bullet

$g=-9.8 m/s^{2}$ is the acceleration due gravity, always directed downwards

$t=6.9 s$ is the time

$y$ is the vertical position of the bullet at $t=6.9 s$

Let's begin by finding $V$ from (1):

$V=152 m/s-9.8 m/s^{2}(6.9 s)$ (3)

$V=84.38 m/s$ (4)

Now we have to substitute (4) in (2):

$(84.38 m/s)^{2}=(152 m/s)^{2}-2(9.8 m/s^{2})y$ (5)

Isolating $y$ :

$y=815.511 m$ This is the displacement of the bullet after 6.9 s

8 0

3 years ago

During a game the same batter swings at a ball thrown by the pitcher and hits a line drive. Just before the ball is hit it is tr

sergeinik [125]

Answer:

The total work on the ball is 36.25 Joules

Explanation:

There is an important principle on classical mechanics that is the work-energy principle it states that the total work on an object is equal the change on its kinetic energy, mathematically expressed as:

$W_{net}=\Delta K = K_f -K_i$ (1)

With W net the total work, Kf the final kinetic energy and Ki the initial kinetic energy. We're going to use this principle to calculate the total work on the baseball by the force exerted by the bat.

Kinetic energy is the energy related with the movement of an object and every classical object with velocity has some kinetic energy, it is defined as:

$K=\frac{mv^2}{2}$

With m the mass of the object and v its velocity, knowing this we can use on:

$W_{net}= \frac{mv_f^2 -mv_i^2}{2}=\frac{m(v_f^2 -v_i^2)}{2}$

In our case vf is the velocity just after the hit and vi the velocity just before the hit. For an average baseball its mass is 145g that is 0.145 kg, then

$W_{net}=\frac{0.145*(30.0^2 -20.0^2)}{2}$