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juin [17]
3 years ago
11

Do these examples represent unbalanced or balanced forces?

Physics
1 answer:
neonofarm [45]3 years ago
8 0
I  belive that  they are undalanced
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An object initially at rest experiences an acceleration of 1.90 ­m/s² for 6.60 s then travels at that constant velocity for anot
borishaifa [10]

Answer:

The magnitude of the object's average velocity is 9.74 m/s (9.80 m/s without any intermediate rounding).

Explanation:

Hi there!

The average velocity is calculated as the displacement of the object divided by the time it takes the object to do that displacement.

The displacement is calculated as the distance between the final position of the object and the initial position. <u>In this problem</u>, the displacement is equal to the traveled distance because the object travels only in one direction:

a.v = Δx/t

Where:

a.v = average velocity.

Δx = displacement = final position - initial position

t = time

So, let's find the distance traveled while the object was accelerating. For that, we will use this equation:

x = x0 + v0 · t + 1/2 · a · t²

Where:

x = position of the object at time t.

x0 = initial position.

v0 = initial velocity.

t = time.

a = acceleration.

In this case, since the object is initially at rest, v0 = 0. If we place the origin of the frame of reference at the point where the object starts moving, then x0 = 0. So, the equation of the position of the object after a time t will be:

x = 1/2 · a · t²

x = 1/2 · 1.90 m/s² · (6.60 s)²

x = 41.4 m

The object traveled 41.4 m during the first 6.60 s.

Now, let's find the rest of the traveled distance.

When the velocity is constant, a = 0. Then, the equation of position will be:

x = x0 + v · t

Let's place now the origin of the frame of reference at the point where the object starts traveling at constant velocity so that x0 = 0:

x = v · t

The velocity reached by the object during the acceleration phase is calculated as follows:

v = v0 + a · t   (v0 = 0 because the object started from rest)

v = 1.90 m/s² · 6.60 s

v = 12.5 m/s

Then, the distance traveled by the object at a constant velocity will be:

x = 12.5 m/s · 8.50 s

x = 106 m

The total traveled distance in 15.1 s is (106 m + 41.4 m) 147 m.

Then the displacement will be:

Δx = final position - initial position

Δx = 147 m - 0 = 147 m

and the average velocity will be:

a.v = Δx/t

a.v = 147 m / 15.1 s

a.v = 9.74 m/s

The magnitude of the object's average velocity is 9.74 m/s (9.80 m/s without any intermediate rounding).

8 0
3 years ago
A spaceship accelerates from 0m/s to 40m/s in 5 seconds. What is the acceleration of the spaceship
nadya68 [22]
Acceleration = change in velocity/time
= 40/5
=8m/s^2
6 0
3 years ago
Which of these is a type of mass movement that is likely to happen after a large amount of rain
Gala2k [10]
The answer is landslide
3 0
3 years ago
Read 2 more answers
A rock is thrown horizontally from a high building at 33.8 m/s. What is the magnitude of its velocity 4.25 s later?
Alex17521 [72]
<h2>Answer:53.63ms^{-2}</h2>

Explanation:

The equations of motion used in this question is v=u+at

When a object is projected horizontally from a sufficiently height,the x-component of acceleration remains zero because there is no force that drags the object in x direction.

But,due to gravity,the object accelerates downward at a rate of 9.8ms^{-2}.

In X-Direction,

Given that initial velocity=u_{x}=33.8ms^{-1}

Using v=u+at,

v_{x}=33.8+(0)4.25=33.8ms^{-1}

In Y-Direction,

Given that initial velocity=u_{x}=0ms^{-1}

Using v=u+at,

v_{y}=0+(9.8)4.25=41.65ms^{-1}

v=\sqrt{v_{x}^{2}+v_{y}^{2}}

v=\sqrt{1142.44+1734.72}=\sqrt{2877.163}=53.63ms^{-1}

7 0
3 years ago
A taxi is travelling at 15m/s. Its driver accelerates with acceleration 3m/s^2 for 4 s. What would be the new velocity?..... pls
Aleksandr-060686 [28]

Answer:

27 m/s

Explanation:

Given:

v₀ = 15 m/s

a = 3 m/s²

t = 4 s

Find: v

v = at + v₀

v = (3 m/s²) (4 s) + (15 m/s)

v = 27 m/s

5 0
3 years ago
Read 2 more answers
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