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3 years ago

Why are objects that fall near Earth's surface rarely in free fall?

Physics

1 answer:

sveta [45]3 years ago

6 0

Answer: Air exerts forces on falling objects near Earth's surface.

Explanation: Objects falling near the earth surface are rarely in free fall due to the force exerted by air on falling object near the earth( air resistance).

The acceleration of free fall tend to pull the body towards the earth surface while air resistance (drag) tends to act in the opposite direction.

Given that the weight of the body is always constant.

The drag acts in the upward direction, thereby negating the downward weight of the object.

From Newton's Second law:

Force = mass × acceleration due to gravity

Acceleration = force / mass

The net force acting on the object becomes :

(Downward weight - upward drag)

This hampers the free fall of the object due to gravity.

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Describe a situation that includes no less than four charges of any magnitude, but they combine so that another location, p, has

Degger [83]

Answer:

Four charges of equal magnitude sitting at the vertices of a square

Explanation:

We can arrive at such a situation by thinking of a simple example first, a configuration of two charges. The force acting on the middle point of a straight line joining the two points(charges) will be zero. That is, the net Electric field will be zero as they cancel out being equal in magnitude and opposite in direction.

Now, we can extend this idea to a square having charge q at each vertex. If we put 'p' at the geometric center, we can see that the Electric fields along the diagonals cancel out due to the charges at the diagonally opposite vertices(refer to the figure attached). Actually, the only requirement is that the diagonally opposite charges are equal.

We can further take this to 3 dimensions. Consider a cube having charges of equal magnitude at each vertex. In this case, the point 'p' will yet again be the geometric center as the Electric field due to the diagonally opposite charges will cancel out.

6 0

2 years ago

A stone is dropped from the

ICE Princess25 [194]

Height=h=500m
Acceleration=g=10m/s^2
Initial velocity=u=0
Speed of sound=c=340m/s
TIME TAKEN BY STONE TO HIT WATER=t
Time taken by sound to hear back=T

Now

$\\ \sf\longmapsto h=ut+\dfrac{1}{2}gt^2$

$\\ \sf\longmapsto h=0t+\dfrac{1}{2}10t^2$

$\\ \sf\longmapsto 500=5t^2$

$\\ \sf\longmapsto t^2=100$

$\\ \sf\longmapsto t=10s$

Now

$\\ \sf\longmapsto h=cT$

$\\ \sf\longmapsto T=\dfrac{h}{c}$

$\\ \sf\longmapsto T=\dfrac{500}{340}$

$\\ \sf\longmapsto T=1.47\approx 1.5s$

Total time:-

$\\ \sf\longmapsto T_{net}=t+T=10+1.5=11.5s$

8 0

2 years ago

A force of 20N changes the position of a body. If mass of the body is 2kg, find the acceleration produced in the body.2. A ball

shepuryov [24]

Explanation:

Hello there!!!

You just need to use simple formula for force and momentum, 

F= m.a

and momentum (p)= m.v

where m= mass

v= velocity.

a= acceleration .

And the solutions are in pictures. 

Hope it helps..

5 0

3 years ago

Show that the entire Paschen series is in the infrared part of the spectrum. To do this, you only need to calculate the shortest

mr_godi [17]

Answer and Explanation:

The computation of the shortest wavelength in the series is shown below:-

$\frac{1}{\lambda} = R(\frac{1}{n_f^2} - \frac{1}{n_i^2} )$

Where

$\lambda$ represents wavelength

R represents Rydberg's constant

$n_f$ represents Final energy states

and $n_i$ represents initial energy states

Now Substitute is

$1.097\times 10^7\ m^{-1}\ for\ R, \infty for\ n_i,\ 3 for\ n_i,\\\\\ \frac{1}{\lambda} = R(\frac{1}{n_f^2} - \frac{1}{n_i^2} )$

now we will put the values into the above formula

$= 1.097\times 10^7 m^{-1}(\frac{1}{3^2} - \frac{1}{\infty^2} )\\\\ = 1.097\times10^7\ m^{-1} (\frac{1}{9} )$

$= 1218888.889 m^{-1}$

Now we will rewrite the answer in the term of $\lambda$

$\lambda = \frac{1}{1218888.889} m\\\\ = 0.82\times 10^{-6} m$

So, the whole Paschen series is in the part of the spectrum.

8 0

3 years ago

An athlete jumping vertically on a trampoline leaves the surface with a velocity of 8.5 m/s upward. what maximum height does she

Mumz [18]

Her center of mass will rise 3.7 meters. First, let's calculate how long it takes to reach the peak. Just divide by the local gravitational acceleration, so 8.5 m / 9.8 m/s^2 = 0.867346939 s And the distance a object under constant acceleration travels is d = 0.5 A T^2 Substituting known values, gives d = 0.5 9.8 m/s^2 (0.867346939 s)^2 d = 4.9 m/s^2 * 0.752290712 s^2 d = 3.68622449 m Rounded to 2 significant figures gives 3.7 meters. Note, that 3.7 meters is how much higher her center of mass will rise after leaving the trampoline. It does not specify how far above the trampoline the lowest part of her body will reach. For instance, she could be in an upright position upon leaving the trampoline with her feet about 1 meter below her center of mass. And during the accent, she could tuck, roll, or otherwise change her orientation so she's horizontal at her peak altitude and the lowest part of her body being a decimeter or so below her center of mass. So it would look like she jumped almost a meter higher than 3.7 meters.

8 0

3 years ago