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bagirrra123 [75]
3 years ago
13

What is an example of Deceleration?

Physics
2 answers:
Elina [12.6K]3 years ago
5 0
That's a word often used to mean "slowing down". An example is what happens
when the brake is applied in a moving car.

But 'deceleration' is a very unofficial, unscientific word.  If the forward speed
is decreasing, then a much more useful concept is "negative acceleration".
Luden [163]3 years ago
4 0
When the body starts moving with a slower speed than the initial speed and it slows down continuously, the this is called deceleration. Examples of deceleration - 

When a ball is thrown upwards, its speed goes on decreasing as its height increases.
When a car applies brakes.
You might be interested in
9) A balloon is charged with 3.4 μC (microcoulombs) of charge. A second balloon 23 cm away is charged with -5.1 μC of charge. Th
lina2011 [118]

Answer:

9. The force is a force of attraction and it is 2.95N

10. The magnitude of acceleration 35.12m/s^2 and the direction of this acceleration is away from the other balloon.

Explanation:

Parameters given:

Q1 = 3.4 * 10^-6C

Q2 = - 5.1 * 10^-6C

Distance between the two balloons = 23cm = 0.23m

9. Force acting between the two balloons is a force of attraction because they are unlike charges. Hence, the force between them is:

F = kQ1Q2/r^2

F = (9 *10^9 * 3.4 * 10^-6 * -5.1 * 10^-6)/(2.3 * 10^-1)^2

F = (1.56 * 10^-1)/(5.29 * 10^-2)

F = - 2.95N

10. Assuming that Balloon A has a mass, m, of 0.084kg, then:

F = ma

Where a = acceleration

a = F/m

a = -2.95/0.084

a = - 35.12m/s^2

The acceleration has a magnitude of 35.12m/s^2 and its direction is away from balloon B.

The negative sign shows that the balloon A is slowing down as it moves towards balloon B. Hence, it's velocity is reducing slowly.

4 0
4 years ago
You drive from your house to the dry cleaners, which is 12 miles east of your house. You then drive to the grocery store from th
Leviafan [203]

Answer:

15 miles.

Explanation:

Displacement is the position you are in now relative to the position you were in when you started. If you had stopped at the dry cleaner's, your displacement would be 12 miles. However, you traveled back to your house and then 15 miles West of that, meaning you are now 15 miles away from your starting position.

3 0
4 years ago
Any girls come to talk on insta or here<br>​
xxMikexx [17]

Answer:

No

Explanation:

Sorry hun, I'm a lesbian lol

7 0
3 years ago
Electric devices use electric _____ to function
Rina8888 [55]

AC or DC

someone use active current or Direct current

6 0
3 years ago
A cannon shoots a ball at an angle θ above the horizontal ground. (a) Neglecting air resistance, use Newton's second law to find
nikitadnepr [17]

Answer:

a)  x = v₀ₓ t ,  y = v_{oy} t - ½ g t²

Explanation:

This is a projectile launch problem, let's use Newton's second law on each axis

X axis

       F = m a

Since there is no acceleration on the x axis, the force on this axis is zero

Y Axis  

       -W = m a_{y}

       -m g = m a_{y}  

       a_{y}  = -g

In this axis the acceleration is the acceleration of gravity

Now we can use science to find the position of the body on each axis

X axis

       x = v₀ₓ t + ½ a  t²

As the acceleration on this axis is zero

      x = v₀ₓ t

Y Axis

       y = v_{oy} t + ½ a_{y}  t²

The acceleration on this axis is –g

        y = v_{oy} t - ½ g t²

B) to find the maximum value of distance r

        r =√ x² + y²

        r = √( v₀ₓ² t² + (v_{oy} t + ½ g t²)²

We can find the maximum value of r using time respect derivatives

      dr / dt = 0

      0 = ½ 1/√( v₀ₓ² t² + (v_{oy} t + ½ g t²)²    (v₀ₓ² 2t + 2 (v_{oy} t - ½ g t²)(v_{oy} - ½ g2t)

We simplify this expression

         0 = v₀ₓ² 2t + 2 (v_{oy} t - ½ g t²)  (v_{oy} - ½ g2t)

       -v₀ₓ² t =( v_{oy} t - ½ g t²)  (v_{oy} - ½ g2t)  

      -v₀ₓ² t = v_{oy}² t -3/2 gt² v_{oy} + 1/2 g² t³  

       ½ g² t²- 3/2 g v_{oy} t  = v_{oy}² + v₀ₓ²

Let's use trigonometry to find go and vox

         sin θ = v_{oy} / v₀

         cos θ = vox / v₀

         v_{oy} = v₀ sin θ

        v₀ₓ = v₀ cos θ

We replace

         ½ g² t² -3/2 g v_{oy} t = v₀ (sin² θ +  cos² θ)

          g t² - 3v₀ sin θ t = 2 v₀/g

 

The time is maximum for the angle is zero

         

8 0
3 years ago
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