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JulijaS [17]
2 years ago
9

(ASAP) would it be 125 m/s2 to calculate for her speeding up?

Physics
1 answer:
serg [7]2 years ago
8 0

Answer:

0\:\mathrm{ m/s^2}

Explanation:

Recall the formula for acceleration:

\displaystyle\\a=\frac{v_f-v_i}{\Delta t}, where v_f is final velocity, v_i is initial velocity, and \Delta t is elapsed time (change in velocity over this amount of time).

Let's look at our time vs velocity graph. At t=0 seconds, V=25 m/s. So her initial velocity is 25 m/s.

We want to find the acceleration during the first 5 seconds of motion. Well, looking at our graph, at t=5 seconds, isn't our velocity still 25 m/s? Therefore, final velocity is 25 m/s (for this period of 5 seconds).

We are only looking from t=0 seconds to t=5 seconds which is a total period of 5 seconds. Therefore, elapsed time is 5 seconds.

Substituting values in our formula, we have:

\displaystyle a=\frac{25-25}{5}=\frac{0}{5}=\boxed{0\:\mathrm{m/s^2}}

Alternative:

Without even worrying about plugging in numbers, let's think about what acceleration actually is! Acceleration is the change in velocity over a certain period of time. If we are not changing our velocity at all, we aren't accelerating! In the graph, we can see that we have a straight line from t=0 seconds to t=5 seconds, the interval we are worried about. This indicates that our velocity is staying the same! At t=0 seconds, we have a velocity of 25 m/s and that velocity stays the same until t=5 seconds. Even though we are moving, we haven't changed velocity, which means our average acceleration is zero!

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You can use any coordinate system you like in order to solve a projectile motion problem. To demonstrate the truth of this state
posledela

Answer:

a)  y₂ = 49.1 m ,    t = 1.02 s , b)   y = 49.1 m , t= 1.02 s

Explanation:

a) We will solve this problem with the missile launch kinematic equations, to find the maximum height, at this point the vertical speed is zero

            v_{y}² = v_{oy}² - 2 g (y –yo)

The origin of the coordinate system is on the floor and the ball is thrown from a height

           y-yo = v_{oy}² /2 g
            y- 0 = 10.0²/2 9.8
            y - 0 = 5.10 m
            
The height from the ground is the height that rises from the reference system plus the depth of the ground from the reference system
             y₂ = 5.1 + 44
             y₂ = 49.1 m
Let's use the other equation to find the time
              [tex]v_{y} = v_{oy} - g t

              t = v_{oy} / g

              t = 10 / 9.8

              t = 1.02 s

b) the maximum height

            y- 44.0 = v_{y}² / 2 g

            y - 44.0 = 5.1

            y = 5.1 +44.0

            y = 49.1 m

The time is the same because it does not depend on the initial height

              t = 1.02 s

7 0
3 years ago
A driver notices an upcoming speed limit change from 45 mi/h (20 m/s) to 25 mi/h (11 m/s). If she estimates
zloy xaker [14]

Answer:

-2.79 m/s²

Explanation:

Given:

v₀ = 20 m/s

v = 11 m/s

Δx = 50 m

Find: a

v² = v₀² + 2aΔx

(11 m/s)² = (20 m/s)² + 2a (50 m)

a = -2.79 m/s²

Round as needed.

8 0
3 years ago
A 120-kg object and a 420-kg object are separated by 3.00 m At what position (other than an infinitely remote one) can the 51.0-
djverab [1.8K]

Answer:

1.045 m from 120 kg

Explanation:

m1 = 120 kg

m2 = 420 kg

m = 51 kg

d = 3 m

Let m is placed at a distance y from 120 kg so that the net force on 51 kg is zero.

By use of the gravitational force

Force on m due to m1 is equal to the force on m due to m2.

\frac{Gm_{1}m}{y^{2}}=\frac{Gm_{2}m}{\left ( d-y \right )^{2}}

\frac{m_{1}}{y^{2}}=\frac{m_{2}}{\left ( d-y \right )^{2}}

\frac{3-y}{y}=\sqrt{\frac{7}{2}}

3 - y = 1.87 y

3 = 2.87 y

y = 1.045 m

Thus, the net force on 51 kg is zero if it is placed at a distance of 1.045 m from 120 kg.

6 0
3 years ago
As mass increases what happens to the kinetic energy
AnnyKZ [126]
As mass increases kinetic energy also increases; kinetic energy is directly proportional to mass so whatever is done to either affects the other one the same. i hope this helps :)
5 0
3 years ago
A man has been guarding a house for one hour . Why is it not considered work in science
leva [86]

Answer: because there is no displacement or movement in the watchman's work. according to science when displacement or movement take place it is said to be work. hope this helps you.

8 0
2 years ago
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