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Len [333]
3 years ago
8

Katniss, an alien from another planet, shoots an arrow straight into the air 2.4 meters from the ground at 36 meters per second.

According to the laws of physics on her planet, the gravity is around 2.78 meters per second squared.
Which of the quadratic equations correctly models the height in meters of the object, h, in relation to the amount of time in seconds after it is thrown, t?
Physics
1 answer:
kirill [66]3 years ago
7 0
Y(t)=2.4+36t+1.39t^2
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Help with all of them plsss
MakcuM [25]
I):final velocity=initial velocity+acceleration due to gravity*time of travel

Therefore,
v=0+9.8*3.1
=30.38 m/s

If by floor you meant water bed then i think there isn't enough info in the question.
8 0
3 years ago
A wave has a frequency of 775 Hz. What is its period?
Leto [7]

The period is simply the inverse of the frequency, therefore:

T = 1 / f

T = 1 / 775 Hz

T = 0.001290 s    → possible answer

T = 1.29 × 10⁻³ s   → possible answer

T = 1.29 ms          → possible answer

4 0
2 years ago
The energy processed and used by living beings is <br>​
Molodets [167]

Answer:

Nutrition , heat , wind

Explanation:

6 0
2 years ago
La energía cinética es la energía que presentan los cuerpos que se encuentran en movimiento.
myrzilka [38]
Do you speak a little English cuz I can’t help you if a can’t understand you
6 0
2 years ago
A 50 kg runner runs up a flight of stairs. The runner starts out covering 3 steps every second. At the end the runner stops. Thi
nadezda [96]

To solve the problem it is necessary to take into account the concepts of kinematic equations of motion and the work done by a body.

In the case of work, we know that it is defined by,

W = F * d

Where,

F= Force

d = Distance

The distance in this case is a composition between number of steps and the height. Then,

d=h*N, for h as the height of each step and N number of steps.

On the other hand we have the speed changes, depending on the displacement and acceleration (omitting time)

V_f^2-V_i^2 = 2a\Delta X

Where,

V_f = Final velocity

V_i = Initial Velocity

a = Acceleration

\Delta X = Displacement

PART A) For the particular case of work we know then that,

W = F*d

W = m*g*(h*N)

W = 50*9.8*(0.3*30)

W = 4.41kJ

Therefore the Work to do that activity is 4.41kJ

PART B) To find the acceleration (from which we can later find the time) we start from the previously given equation,

V_f^2-V_i^2 = 2a\Delta X

Here,

V_i = \frac{0.3*3}{1} = 0.90m/s\rightarrow3 steps in one second

v_f = 0

Replacing,

V_f^2-V_i^2 = 2a\Delta X

0-0.9^2=2a(30*0.3)

Re-arrange for a,

a = -\frac{0.9^2}{2*30*0.3}

a = -45*10^{-3}m/s^2

At this point we can calculate the time, which is,

t = \frac{\Delta V}{a}

t = \frac{0-0.9}{-45*10^{-3}}

t = 20s

With time and work we can finally calculate the power

P = \frac{W}{t} = \frac{4.41}{20}

P = 0.2205kW

6 0
3 years ago
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