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Ghella [55]
3 years ago
8

11. Maria left her house bicycling at 10 km/h. Her acceleration was 30 km/h2 . How fast was she bicycling when she arrived at sc

hool 30 minutes later?
Physics
1 answer:
weqwewe [10]3 years ago
7 0

Answer:

25km/h

Explanation:

we have the information:

initial velocity: v_{0}=10km/h

acceleration: a=30km/h^2

and the time: t=30min=0.5h

the 30 minutes must be in hours because the velocities have hour units.

and now we use the equation for velocity at constant acceleration to now the velocity when she arrived at school 30 minutes (0.5 of hour) later:

v=v_{0}+at

and we substitute known values:

v=10km/h+(30kn/h^2)(0.5h)\\v=10km/h+15km/h\\v=25km/h

the answer is 25km/h

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Calculate the time needed for a 0.600 kg hammer to reach the surface of the Earth
USPshnik [31]

The time needed for the hammer to reach the surface of the Earth is 3.54 s.

<h3>Time of motion of the hammer</h3>

The time of motion is calculated as follows;

t = √(2h/g)

where;

  • h is height of fall
  • g is acceleration due to gravity

t = √(2 x 10 / 1.6)

t = 3.54 s

Thus, the time needed for the hammer to reach the surface of the Earth is 3.54 s.

Learn more about time of motion here: brainly.com/question/2364404

#SPJ1

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2 years ago
What is the most likely outcome of increasing the number of slits per unit distance on a diffraction grating?1) lines become nar
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It's important to know that diffraction gratings can be identified by the number of lines they have per centimeter. Often, more lines per centimeter is more useful because the images separation is greater when this happens. That is, the distance between lines increases.

<h2>Therefore, the answer is 2.</h2>
8 0
1 year ago
Which type of energy is released when a bond between atoms is broken
scoray [572]

Answer:

D

Explanation:

7 0
3 years ago
As air pressure decreases, what happens
Naya [18.7K]

Answer:

I believe it is B, not 100% sure though

Explanation:

5 0
2 years ago
Rock X is released from rest at the top of a cliff that is on Earth. A short time later, Rock Y is released from rest from the s
frosja888 [35]

Answer:

C) True. S increases with time, v₁ = gt and v₂ = g (t-t₀)  we see that for the same t v₁> v₂

Explanation:

You have several statements and we must select which ones are correct. The best way to do this is to raise the problem.

Let's use the vertical launch equation. The positive sign because they indicate that the felt downward is taken as an opponent.

Stone 1

    y₁ = v₀₁ t + ½ g t²

    y₁ = 0 + ½ g t²

Rock2

It comes out a little later, let's say a second later, we can use the same stopwatch

     t ’= (t-t₀)

    y₂ = v₀₂ t ’+ ½ g t’²

    y₂ = 0 + ½ g (t-t₀)²

    y₂ = + ½ g (t-t₀)²

Let's calculate the distance between the two rocks, it should be clear that this equation is valid only for t> = to

    S = y₁ -y₂

    S = ½ g t²– ½ g (t-t₀)²

    S = ½ g [t² - (t²- 2 t to + to²)]  

    S = ½ g (2 t t₀ - t₀²)

    S = ½ g t₀ (2 t -t₀)

This is the separation of the two bodies as time passes, the amount outside the Parentheses is constant.

For t <to.  The rock y has not left and the distance increases

For t> = to.  the ratio (2t/to-1)> 1 therefore the distance increases as time

passes

Now we can analyze the different statements

A) false. The difference in height increases over time

B) False S increases

C) Certain s increases with time, v₁ = gt and V₂ = g (t-t₀) we see that for the same t   v₁> v₂

3 0
3 years ago
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