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3 years ago

a model rocket climbs 200 m in 4 seconds. if was moving 10 m/s to begin with, what is it’s final velocity?

1 answer:

8 0

So first Identify all the given Varibales so u can choose which Eqauton to use

D=200m
T=4s
Vi=10m/s
Vf=?

You should this equation

D= 0.50(Vf+Vi)T

Plug in the values

200= 0.50 (Vf+10) 4

Divide the 4 out of the right side and if you do sumthing to one side you gotta do it to the other

200 divided by 4= 0.50(Vf+10)
50= 0.50(Vf+10)

Now expand the 0.50

So 50= 0.5Vf + 5 (because 0.5 times 10 is 5)

Now get rid of the 5

50-5= 0.5Vf

45 =0.5Vf now Divide the 0.5 out

45 divided by 0.5 = Vf

And 45/0.5 is 90

So 90=Vf

Therefore the final Velocity is 90m/s

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Define limitations in an experiment

laila [671]

Answer:

The limitations of the experiment are those characteristics of design or methodology that impacted or influenced the application or interpretation of the results of your study.

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2 years ago

A particle with mass 1.09 kg oscillates horizontally at the end of a horizontal spring. A student measures an amplitude of 0.985

e-lub [12.9K]

Answer:

a) $f = 0.598\,hz$ , b) $v_{max} = 3.701\,\frac{m}{s}$ , c) $k = 15.385\,\frac{N}{m}$ , d) $U = 1.081\,J$ , e) $K = 6.382\,J$ , f) $v\approx 3.422\,\frac{m}{s}$

Explanation:

a) The frequency of oscillation is:

$f = \frac{76}{127\,hz}$

$f = 0.598\,hz$

b) The angular frequency is:

$\omega = 2\pi \cdot f$

$\omega = 2\pi \cdot (0.598\,hz)$

$\omega = 3.757\,\frac{rad}{s}$

Lastly, the speed at the equilibrium position is:

$v_{max} = \omega \cdot A$

$v_{max} = (3.757\,\frac{rad}{s} )\cdot (0.985\,m)$

$v_{max} = 3.701\,\frac{m}{s}$

c) The spring constant is:

$\omega = \sqrt{\frac{k}{m}}$

$k = \omega^{2}\cdot m$

$k = (3.757\,\frac{rad}{s} )^{2}\cdot (1.09\,kg)$

$k = 15.385\,\frac{N}{m}$

d) The potential energy when the particle is located 38.1 % of the amplitude away from the equilibrium position is:

$U = \frac{1}{2}\cdot (15.385\,\frac{N}{m} )\cdot (0.375\,m)^{2}$

$U = 1.081\,J$

e) The maximum potential energy is:

$U_{max} = \frac{1}{2}\cdot (15.385\,\frac{N}{m} )\cdot (0.985\,m)^{2}$

$U_{max} = 7.463\,J$

The kinetic energy when the particle is located 38.1 % of the amplitude away from the equilibrium position is:

$K = U_{max} - U$

$K = 7.463\,J - 1.081\,J$

$K = 6.382\,J$

f) The speed when the particle is located 38.1 % of the amplitude away from the equilibrium position is:

$K = \frac{1}{2}\cdot m \cdot v^{2}$

$v = \sqrt{\frac{2\cdot K}{m} }$

$v = \sqrt{\frac{2\cdot (6.382\,J)}{1.09\,kg} }$

$v\approx 3.422\,\frac{m}{s}$

4 0

3 years ago

Verify that for values of n less than 8, the system goes to a stable equilibrium, but as n passes 8, the equilibrium point becom

irakobra [83]

Answer:

Biological system is one of the major causes of oscillation due to sensitive negative feedback loops. For instance, imagine a father teaching his son how to drive, the teen is trying to keep the car in the centre lane and his father tell him to go right or go left as the case may be. This is a example of a negative feedback loop of a biological system. If the father's sensitivity to the car's position on the road is reasonable, the car will travel in a fairly straight line down the centre of the road. On the other hand, what happens if the father raise his voice at the son "go right" or when the car drifts a bit to the left? The startled the son will over correct, taking the car too far to the right. The father will then starts yelling "go left" then the boy will over correct again and the car will definitely oscillate back and forth. A scenario that indicates the behavior of a car driver under a very steep feedback control mechanism. Since the driver over corrects in each direction. Therefore causes oscillations.

Explanation:

5 0

3 years ago

PLEASE HELPPPPPPPPPP

tekilochka [14]

Answer:

sorry me computer is blocking the picture but i will try to unblock so i can answer

Explanation:

7 0

3 years ago

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What is the electric current measured in

cluponka [151]

An ampere (AM-pir), or amp

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2 years ago