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drek231 [11]
3 years ago
9

As you stand on a floor, the floor exerts an upward force on you. why are you not moved upward by this force? 1. the upward forc

e is less than your weight, so you do not move up. 2. the upward force is greater than your weight, but air pressure pushes you down. without the air, you would be lifted up. 3. the upward force is equal to your weight and the two forces cancel each other. 4. none of these 5. the upward force is negligible
Physics
1 answer:
dolphi86 [110]3 years ago
6 0
Newton's third law of motion states that for any action, there is equal and opposite force. For a person standing on a floor, the action force is the weight and thus the floor must exert an opposite and equal reaction force equivalent in magnitude to the weight of the person.

In this regard, statement 3. is correct.
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An arrow of mass 415 g is shot at a target with a speed of 68.5m/s. The target, which has a mass of 3.3 kg, is moving toward the
Murrr4er [49]

The velocity of the target and arrow after collision is 6.67m/s

<u>Explanation:</u>

Given:

Mass of arrow, mₐ = 415g

Speed of arrow, vₐ = 68.5m/s

Mass of the target, mₓ = 3.3kg = 3300g

speed of the target, vₓ = -1.1m/s (Because the target moves in opposite direction

Velocity of the target and arrow after collision, vₙ = ?

Applying the conservation of momentum,

mₐvₐ + mₓvₓ = (mₐ+mₓ) vₙ

415 X 68.5 + 3300 X -1.1 = (415+3300) X vₙ

28427.5 - 3630 = 3715 X vₙ

24797.5 = 3715 X vₙ

vₙ = 6.67m/s

Therefore, the velocity of the target and arrow after collision is 6.67m/s

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3 years ago
TRUE of FALSE: The human body responds to stressors by activating the nervous
melamori03 [73]
That is false I think sorry if its not correct
8 0
3 years ago
The magnetic field of an electromagnetic wave in a vacuum is Bz =(2.4μT)sin((1.05×107)x−ωt), where x is in m and t is in s. You
tatiyna

Answer:

Explanation:

Given

B_z=(2.4\mu T)\sin (1.05\times 10^7x-\omega t)

Em wave is in the form of

B=B_0\sin (kx-\omega t)

where \omega =frequency\ of\ oscillation

k=wave\ constant

B_0=Maximum\ value\ of\ Magnetic\ Field

Wave constant for EM wave k is

k=1.05\times 10^7 m^{-1}

Wavelength of wave \lambda =\frac{2\pi }{k}

\lambda =\frac{2\pi }{1.05\times 10^7}

\lambda =5.98\times 10^{-7} m

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3 years ago
An object's potential energy is set it cannot change is this true
zepelin [54]
The energy achieved I think
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3 years ago
Read 2 more answers
A Ferris wheel starts at rest and builds up to a final angular speed of 0.70 rad/s while rotating through an angular displacemen
PilotLPTM [1.2K]

Answer:

The average angular acceleration is 0.05 radians per square second.

Explanation:

Let suppose that Ferris wheel accelerates at constant rate, the angular acceleration as a function of change in angular position and the squared final and initial angular velocities can be clear from the following expression:

\omega^{2} = \omega_{o}^{2} + 2 \cdot \alpha\cdot (\theta-\theta_{o})

Where:

\omega_{o}, \omega - Initial and final angular velocities, measured in radians per second.

\alpha - Angular acceleration, measured in radians per square second.

\theta_{o}, \theta - Initial and final angular position, measured in radians.

Then,

\alpha = \frac{\omega^{2}-\omega_{o}^{2}}{2\cdot (\theta-\theta_{o})}

Given that \omega_{o} = 0\,\frac{rad}{s}, \omega = 0.70\,\frac{rad}{s} and \theta-\theta_{o} = 4.9\,rad, the angular acceleration is:

\alpha = \frac{\left(0.70\,\frac{rad}{s} \right)^{2}-\left(0\,\frac{rad}{s} \right)^{2}}{2\cdot \left(4.9\,rad\right)}

\alpha = 0.05\,\frac{rad}{s^{2}}

Now, the time needed to accelerate the Ferris wheel uniformly is described by this kinematic equation:

\omega = \omega_{o} + \alpha \cdot t

Where t is the time measured in seconds.

The time is cleared and obtain after replacing every value:

t = \frac{\omega-\omega_{o}}{\alpha}

If \omega_{o} = 0\,\frac{rad}{s},  \omega = 0.70\,\frac{rad}{s} and \alpha = 0.05\,\frac{rad}{s^{2}}, the required time is:

t = \frac{0.70\,\frac{rad}{s} - 0\,\frac{rad}{s} }{0.05\,\frac{rad}{s^{2}} }

t = 14\,s

Average angular acceleration is obtained by dividing the difference between final and initial angular velocities by the time found in the previous step. That is:

\bar \alpha = \frac{\omega-\omega_{o}}{t}

If \omega_{o} = 0\,\frac{rad}{s},  \omega = 0.70\,\frac{rad}{s} and t = 14\,s, the average angular acceleration is:

\bar \alpha = \frac{0.70\,\frac{rad}{s} - 0\,\frac{rad}{s} }{14\,s}

\bar \alpha = 0.05\,\frac{rad}{s^{2}}

The average angular acceleration is 0.05 radians per square second.

4 0
3 years ago
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