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Lapatulllka [165]
3 years ago
6

An arrow is shot vertically upward from a platform 33ft high at a rate of 214ft/sec. when will the arrow hit the ground? use the

formula: h=−16t2+v0t+h0. (round your answer to the nearest tenth.)
Physics
1 answer:
Setler [38]3 years ago
8 0

Answer:

Time = 11.1 seconds

Explanation:

h = -16t^2 + v(0)t + h(0)

but we are given;

v(0) = 174 ft/sec

h(0) = 33 ft

Find h = 0:

Therefore;

0 = -16t^2 + 174t + 33

16t^2 - 174t - 33 = 0

Use the quadratic equation:

a = 16

b = -174

c = -33

t = (174 +/- sqrt((-174)^2 - 4(16)(-33)))/2(16)

t = (174 +/- sqrt(30276 + 2112))/32

t = (174 +/- sqrt(32388))/32

t = (174 +/- 179.97)/32

t = 353.97/32 or -5.97/32

t = 11.1 or -.2

Therefore;

The arrow will hit the ground at:  

t = 11.1 sec

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Current Flow and Ohm's Law

Ohm's law is the most important, basic law of electricity. It defines the relationship between the three fundamental electrical quantities: current, voltage, and resistance. When a voltage is applied to a circuit containing only resistive elements (i.e. no coils), current flows according to Ohm's Law, which is shown below.

I = V / R 

Where: 

I =

Electrical Current (Amperes)

V =

Voltage (Voltage)

R =

Resistance (Ohms)

    

Ohm's law states that the electrical current (I) flowing in an circuit is proportional to the voltage (V) and inversely proportional to the resistance (R). Therefore, if the voltage is increased, the current will increase provided the resistance of the circuit does not change. Similarly, increasing the resistance of the circuit will lower the current flow if the voltage is not changed. The formula can be reorganized so that the relationship can easily be seen for all of the three variables.

The Java applet below allows the user to vary each of these three parameters in Ohm's Law and see the effect on the other two parameters. Values may be input into the dialog boxes, or the resistance and voltage may also be varied by moving the arrows in the applet. Current and voltage are shown as they would be displayed on an oscilloscope with the X-axis being time and the Y-axis being the amplitude of the current or voltage. Ohm's Law is valid for both direct current (DC) and alternating current (AC). Note that in AC circuits consisting of purely resistive elements, the current and voltage are always in phase with each other.

Exercise: Use the interactive applet below to investigate the relationship of the variables in Ohm's law. Vary the voltage in the circuit by clicking and dragging the head of the arrow, which is marked with the V. The resistance in the circuit can be increased by dragging the arrow head under the variable resister, which is marked R. Please note that the vertical scale of the oscilloscope screen automatically adjusts to reflect the value of the current.

See what happens to the voltage and current as the resistance in the circuit is increased. What happens if there is not enough resistance in a circuit? If the resistance is increased, what must happen in order to maintain the same level of current flow?


4 0
4 years ago
In an old-fashioned amusement park ride, passengers stand inside a 3.0-m-tall, 5.0-m-diameter hollow steel cylinder with their b
lesya692 [45]

Answer:2.55 rad/s

Explanation:

Given

Diameter of ride=5 m

radius(r)=2.5 m

Static friction coefficient range=0.60-1

Here Frictional force will balance weight

And limiting  frictional force is provided by Centripetal force

f=\mu N=\mu m\omega ^2\cdot r

weight of object=mg

Equating two

f=mg

\mu m\omega ^2\cdot r=mg

\omega ^2=\frac{g}{\mu r}

\omega =\sqrt{\frac{g}{\mu r}}

\omega =2.55 rad/sec

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Natali [406]

Answer:

The potential energy of the hiker is 1.13\times 10^6\ J.

Explanation:

Given that,

Mass of the hiker, m = 61 kg

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g is the acceleration due to gravity

E=61\ kg\times 9.8\ m/s^2\times 1900\ m\\\\E=1.13\times 10^6\ J

So, the potential energy of the hiker is 1.13\times 10^6\ J. Hence, this is the required solution.

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max2010maxim [7]

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