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3 years ago

Help asap will mark brainliest

Physics

1 answer:

Mila [183]3 years ago

8 0

Answer:

Average speed = total distance/total time

= 10/15

= 0.67 km/min

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Precision measurements of the acceleration due to gravity show that the acceleration is slightly different in different location

DochEvi [55]

Gravity largely depends on the comparison of two objects; it's why you have the equation F= (GMm)/r^2. On Earth, you have different altitudes that, with the formula, will give different results for gravity because the radius is different everywhere. This difference on calculations, however, are seen to be miniscule. We know gravity as 9.81 m/s^2 but it might be different by thousandths or hundreds of thousandths of a decimal.

6 0

4 years ago

Read 2 more answers

11. A cyclist accelerates from 0 m/s to 10 m/s in 3 seconds. What is his acceleration ? Is this acceleration higher than that of

Marat540 [252]

$a = \frac{v - u}{t}$

v = final velocity

u = initial velocity

t = time taken

the acceleration of the cyclist is

$\frac{10 - 0}{3} = 3.333333....$

approximately 3.33 m/s^2

the acceleration of the car is

$\frac{40 - 0 }{8} = 5.0$

5.0 m/s^2

$5.0 > 3.33 \\ so \: the \: answer \: is \: no$

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3 years ago

A man pushes on a trunk with a force of 250 newtons. The trunk does not move. How much positive work is done on the trunk?

Digiron [165]

Answer:

F is 250 N

d is 0 m

F x d

=250 x 0

The answer is 0.0 J.

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4 years ago

A baseball player leads off the game and hits a long home run. The ball leaves the bat at an angle of 70.0 from the horizontal w

professor190 [17]

Answer: 211.059 m

Explanation:

We have the following data:

$\theta=70\°$ The angle at which the ball leaves the bat

$V_{o}=55 m/s$ The initial velocity of the ball

$g=-9.8 m/s^{2}$ The acceleration due gravity

We need to find how far (horizontally) the ball travels in the air: $x$

Firstly we need to know this velocity has two components:

<u>Horizontally:</u>

$V_{ox}=V_{o}cos \theta$ (1)

$V_{ox}=55 m/s cos(70\°)=18.811 m/s$ (2)

<u>Vertically:</u>

$V_{oy}=V_{o}sin \theta$ (3)

$V_{oy}=55 m/s sin(70\°)=51.683 m/s$ (4)

On the other hand, when we talk about parabolic movement (as in this situation) the ball reaches its maximum height just in the middle of this parabola, when $V=0$ and the time $t$ is half the time it takes the complete parabolic path.

So, if we use the following equation, we will find $t$ :

$V=V_{o}+gt=0$ (5)

Isolating $t$ :

$t=\frac{-V_{o}}{g}$ (6)

$t=\frac{-55 m/s}{-9.8 m/s^{2}}$ (7)

$t=5.61 s$ (8)

Now that we have the time it takes to the ball to travel half of is path, we can find the total time $T$ it takes the complete parabolic path, which is twice $t$ :