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4 years ago

What would happen if there were no action-reaction forces in effect and you tried walking down a sidewalk

Physics

1 answer:

Tema [17]4 years ago

7 0

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A single strain gage has a nominal resistance of 120 ohm. For a quarter bridge with 120 ohm fixed resistors, strain gauge factor

natita [175]

Answer:

Output voltage is 1.507 mV

Solution:

As per the question:

Nominal resistance, R = $120\Omega$

Fixed resistance, R = $120\Omega$

Gauge Factor, G.F = 2.01

Supply Voltage, $V_{s} = 3\ V$

Strain, $\epsilon = 1000\times 10^{-6}\ strain$

Now,

To calculate the output voltage, $V_{o}$ :

WE know that strain is given by:

$\epsilon = \frac{(R + R')^{2}V_{o}}{RR'V_{s}\times G.F}$

Thus

$V_{o} = \frac{RR'V_{s}\epsilon \times G.F}{(R + R')^{2}}$

Now, substituting the suitable values in the above eqn:

$V_{o} = \frac{120\times 120\times 3\times 1000\times 10^{-6}\times 2.01}{(120 + 120)^{2}}$

$V_{o} = 1.507\ mV$

6 0

4 years ago

6. Kinetic energy depends upon and the of the object

Musya8 [376]

I don’t understand your question?

5 0

3 years ago

A satellite is launched to orbit the Earth at an altitude of 2.90 x10^7 m for use in the Global Positioning System (GPS). Take t

marin [14]

Answer:

$T=66262.4s$

Explanation:

From the question we are told that:

Altitude $A=2.90 *10^7$

Mass $m=5.97 * 10^{24} kg$

Radius $r=6.38 *10^6 m.$

Generally the equation for Satellite Speed is mathematically given by

$V=(\frac{GM}{d} )^{0.5}$

$V=(\frac{6.67*10^{-11}*5.97 * 10^{24}}{6.38 *10^6+2.90 *10^7} )^{0.5}$

$V=3354.83m/s$

Therefore

Period T is Given as

$T=\frac{2 \pi *a}{V}$

$T=\frac{2 \pi *(6.38 *10^6+2.90 *10^7}{3354.83}$

$T=66262.4s$

4 0

3 years ago

A fly sits on a potter's wheel 0.30 m from its axle. The wheel's rotational speed decreases from 4.0 rad/s to 2.0 rad/s in 5.0 s

Likurg_2 [28]

Answer: The wheel's average rotational acceleration is -0.4 radians per second squared (rad/s^2)

Explanation: Please see the attachments below

7 0

3 years ago

Read 2 more answers

The severity of a fall depends on your speed when you strike the ground. All factors but the acceleration from gravity being the

Diano4ka-milaya [45]

Answer:

The object could fall from six times the original height and still be safe

Explanation:

Free Falling

When an object is released from rest in free air (no friction), the motion is completely dependant on the acceleration of gravity g.

If we drop an object of mass m near the Earth surface from a height h, it has initial mechanical energy of

$U=m.g.h$

When the object strikes the ground, all the mechanical energy (only potential energy) becomes into kinetic energy

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

Where v is the speed just before hitting the ground

If we know the speed v is safe for the integrity of the object, then we can know the height it was dropped from

$\displaystyle m.g.h=\frac{1}{2}m.v^2$

Solving for h

$\displaystyle h=\frac{m.v^2}{2mg}=\frac{v^2}{2g}$

If the drop had occurred in the Moon, then

$\displaystyle h_M=\frac{v_M^2}{2g_M}$

Where hM, vM and gM are the corresponding parameters on the Moon. We know v is the safe hitting speed and the gravitational acceleration on the Moon is g_M=1/6 g

$\displaystyle h_M=\frac{v^2}{2\frac{1}{6}g}$

$\displaystyle h_M=6\frac{v^2}{2g}=6h$

This means the object could fall from six times the original height and still be safe

6 0

4 years ago