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Murrr4er [49]
2 years ago
9

You have been training and can run a mile in 6.5 minutes. How long would it take to run a lap around the outside track, which is

400 meters?
Physics
1 answer:
Paha777 [63]2 years ago
5 0

Answer:

1 minute 36.85 seconds

Explanation:

First we need to convert the miles into meters, as the demanded result should be in meters.

1 mile = 1,609.34 meters

Also, 6.5 minutes should be converted into seconds.

1 minute = 60 seconds

6.5 x 60 = 390 seconds

Now we need to divide the miles with the seconds to see how much meters have been run in a second.

1,609.34 / 390 = 4.13 meters

The suggested meters now should be divided with the distance run in one second.

400 / 4.13 = 96.85 seconds

So we get a result of 96.85 seconds, or 1 minute 36.85 seconds.

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vazorg [7]
We make use of the equation: v^2=v0^2+2a Δd. We substitute v^2 equals to zero since the final state is halting the truck. Hence we get the equation           -<span>v0^2/2a = Δd. F = m a from the second law of motion. Rearranging, a = F/m
</span>F = μ Fn where the force to stop the truck is the force perpendicular or normal force multiplied by the static coefficient of friction. We substitute, -v0^2/2<span>μ Fn/m</span> = Δd. This is equal to 
3 0
3 years ago
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An object is placed 4.0 cm to the left of a convex lens with a focal length of +8.0 cm . Where is the image of the object?
Serjik [45]

The image of the object is 8cm to the left of the lens (D)

<h3></h3>

What is the image of an object?

The image of an object is said to be the location where light rays from that object intersect with a mirror by reflection.

It is calculated thus:

1÷v = 1÷f - 1÷u

<h3>How to calculate the image of an object</h3>

From the formula

1÷v = 1÷f - 1÷u

<h3>Where </h3>

V = image distance fromthe object

U = object

f = focal length

Substitute the values

1÷v = 1÷8 - 1÷ 4

1÷v = - 1÷8

Make v the subject of formula

v = -8cm

Therefore, the image of the object is 8cm to the left of the lens (D)

Learn more on focal length here:

brainly.com/question/25779311

#SPJ1

6 0
2 years ago
A shoe and a shirt are released from the same height. They take different amounts of time to fall to the ground. How can this be
Law Incorporation [45]

The best explanation for the difference in time is: A. The difference in weight doesn't affect the time, but they are affected differently by air resistance.

<h3>What is weight?</h3>

Weight can be defined as the force acting on an object or a physical body due to the effect of gravity. Also, the weight of an object (body) is typically measured in Newton.

<h3>The factors that affect weight.</h3>

Some of the factors that affect the weight that is possessed by an object or a physical body include the following:

  • Mass
  • Distance
  • Air resistance

In conclusion, the weight possessed by the shoe and shirt has no effect on time but would be affected differently by air resistance.

Read more on weight here: brainly.com/question/13833323

4 0
2 years ago
If f(x) = -x2+x-1 then value of f(f(2)) is
qaws [65]

Answer:

<u>-13</u>

Explanation:

<u>Function</u>

  • f(x) = -x² + x - 1

<u>Solving</u>

  • Substitute 2 in place of x
  • f(2) = -(2)² + 2 - 1
  • f(2) = -4 + 2 - 1
  • f(2) = -3
  • f(f(2)) = f(-3)
  • f(-3) = -(-3)² - 3 - 1
  • f(-3) = -9 - 3 - 1
  • f(f(2)) = <u>-13</u>
6 0
1 year ago
Find an expression for the electric field E⃗ at the center of the semicircle. Hint: A small piece of arc length Δs spans a small
Hatshy [7]

Answer:

Electric Field a the centreE=\dfrac{Q}{2\pi^2\epsilon_0 R^2}(-\vec j)

Explanation:

<u>Given:</u>

Total charge on the semicircle =Q

Radius of the semicircle=R

Let consider a elemental charge on the semicircle at an angle \theta\\ with the horizontal. Since the The charge distribution is symmetrical about y axis so the there will symmetrical charge on other side. The horizontal component will cancel out each other and vertical component will get added so let dE be the electric Field due to elemental charge

Let \lambda be the charge per unit length such that\lambda=\dfrac{Q}{\pi R}

k=\dfrac{1}{4\pi \epsilon_0}

Total Electric Field at the centre

=2dE\sin\theta\\=2\dfrac{k\lambda }{R}\int \sin \theta d\theta\\

integrating 0 to \dfrac{\pi}{2}

E=\dfrac{2k\lambda}{R}(-\vec j)

E=\dfrac{Q}{2\pi^2\epsilon_0 R^2}(-\vec j)

So the Electric field at the centre is calculated.

7 0
3 years ago
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