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enot [183]
3 years ago
7

When parts of your body functions as lever systems, what part supplies the force?

Physics
1 answer:
mylen [45]3 years ago
5 0
Muscles supply the force.  They exert force by getting shorter (contracting).
A muscle can't get longer by itself.  There needs to be another muscle
pulling in the opposite direction.
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A student connects a small solar panel to a 40 a resistor to make a simple circuit. The solar panel produces a voltage of 2 0.
Nastasia [14]
<h3>Solution for the above question : -</h3>

Ohm's law states that :

  • v = ir

the terms used are :

  • r = resistance
  • v = potential \:  \: difference
  • i  = current

let's solve for electric current :

  • 2 = i \times 40

  • i =  \dfrac{2}{40}

  • i = 0.05 \: A

  • i = 50 \: mA

\mathfrak{good\:  \: luck \:  \: for \:  \: your \:  \: assignment}

8 0
2 years ago
A 7600 kg rocket blasts off vertically from the launch pad with a constant upward acceleration of 2.35 m/s2 and feels no appreci
ollegr [7]

Answer:

a) The rocket reaches a maximum height of 737.577 meters.

b) The rocket will come crashing down approximately 17.655 seconds after engine failure.

Explanation:

a) Let suppose that rocket accelerates uniformly in the two stages. First, rocket is accelerates due to engine and second, it is decelerated by gravity.

1st Stage - Engine

Given that initial velocity, acceleration and travelled distance are known, we determine final velocity (v), measured in meters per second, by using this kinematic equation:

v = \sqrt{v_{o}^{2} +2\cdot a\cdot \Delta s} (1)

Where:

a - Acceleration, measured in meters per square second.

\Delta s - Travelled distance, measured in meters.

v_{o} - Initial velocity, measured in meters per second.

If we know that v_{o} = 0\,\frac{m}{s}, a = 2.35\,\frac{m}{s^{2}} and \Delta s = 595\,m, the final velocity of the rocket is:

v = \sqrt{\left(0\,\frac{m}{s} \right)^{2}+2\cdot \left(2.35\,\frac{m}{s^{2}} \right)\cdot (595\,m)}

v\approx 52.882\,\frac{m}{s}

The time associated with this launch (t), measured in seconds, is:

t = \frac{v-v_{o}}{a}

t = \frac{52.882\,\frac{m}{s}-0\,\frac{m}{s}}{2.35\,\frac{m}{s} }

t = 22.503\,s

2nd Stage - Gravity

The rocket reaches its maximum height when final velocity is zero:

v^{2} = v_{o}^{2} + 2\cdot a\cdot (s-s_{o}) (2)

Where:

v_{o} - Initial speed, measured in meters per second.

v - Final speed, measured in meters per second.

a - Gravitational acceleration, measured in meters per square second.

s_{o} - Initial height, measured in meters.

s - Final height, measured in meters.

If we know that v_{o} = 52.882\,\frac{m}{s}, v = 0\,\frac{m}{s}, a = -9.807\,\frac{m}{s^{2}} and s_{o} = 595\,m, then the maximum height reached by the rocket is:

v^{2} -v_{o}^{2} = 2\cdot a\cdot (s-s_{o})

s-s_{o} = \frac{v^{2}-v_{o}^{2}}{2\cdot a}

s = s_{o} + \frac{v^{2}-v_{o}^{2}}{2\cdot a}

s = 595\,m + \frac{\left(0\,\frac{m}{s} \right)^{2}-\left(52.882\,\frac{m}{s} \right)^{2}}{2\cdot \left(-9.807\,\frac{m}{s^{2}} \right)}

s = 737.577\,m

The rocket reaches a maximum height of 737.577 meters.

b) The time needed for the rocket to crash down to the launch pad is determined by the following kinematic equation:

s = s_{o} + v_{o}\cdot t +\frac{1}{2}\cdot a \cdot t^{2} (2)

Where:

s_{o} - Initial height, measured in meters.

s - Final height, measured in meters.

v_{o} - Initial speed, measured in meters per second.

a - Gravitational acceleration, measured in meters per square second.

t - Time, measured in seconds.

If we know that s_{o} = 595\,m, v_{o} = 52.882\,\frac{m}{s}, s = 0\,m and a = -9.807\,\frac{m}{s^{2}}, then the time needed by the rocket is:

0\,m = 595\,m + \left(52.882\,\frac{m}{s} \right)\cdot t + \frac{1}{2}\cdot \left(-9.807\,\frac{m}{s^{2}} \right)\cdot t^{2}

-4.904\cdot t^{2}+52.882\cdot t +595 = 0

Then, we solve this polynomial by Quadratic Formula:

t_{1}\approx 17.655\,s, t_{2} \approx -6.872\,s

Only the first root is solution that is physically reasonable. Hence, the rocket will come crashing down approximately 17.655 seconds after engine failure.

7 0
3 years ago
An emf of 22.0 mV is induced in a 519-turn coil when the current is changing at the rate of 10.0 A/s. What is the magnetic flux
zhenek [66]

Answer:

\phi=1.56\times 10^{-5}\ Wb

Explanation:

Given that,

Emf, V = 22 mV

Number of turns in the coil us 519

Rate of change of current is 10 A/s.

We need to find the magnetic flux through each turn of the coil at an instant when the current is 3.70 A.

Let's find the inductance first. So,

L=\dfrac{\epsilon}{(dI/dt)}\\\\L=\dfrac{0.022}{10}\\\\L=0.0022\ H

We have,

L=\dfrac{N\phi}{I}, \phi is magnetic flux

\phi=\dfrac{LI}{N}\\\\\phi=\dfrac{0.0022\times3.7}{519}\\\\\phi=1.56\times 10^{-5}\ Wb

So, the magnetic flux is 1.56\times 10^{-5}\ Wb.

8 0
3 years ago
Melanie watched the path a baseball followed after a pitcher threw it. She noticed that the ball traveled horizontally away from
Alex17521 [72]

Answer:gravity

Explanation:

7 0
3 years ago
A handful of professional skaters have taken a skateboard through an inverted loop in a full pipe. For a typical pipe with a dia
Bingel [31]

Answer

given,

diameter of the pipe is  =  (14 ft)4.27 m

minimum speed of the skater must have at very top = ?

At the topmost point of the pipe the  normal force will be equal to zero.

F = mg

centripetal force acting on the skateboard

F = \dfrac{mv^2}{r}

equating both the force equation

mg = \dfrac{mv^2}{r}

v = \sqrt{gr}

r = d/2 = 14/ 2 = 7 ft

or

r = 4.27/2 = 2.135 m

g = 32 ft/s²   or g = 9.8 m/s²

v = \sqrt{32 \times 7}

v = 14.96 ft/s

or

v = \sqrt{9.8 \times 2.135}

v = 4.57 m/s

5 0
3 years ago
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