While a bug fly in the air the weight of bug is counter balanced by the buoyancy force that is applied by air on the bug
This buoyancy force is net upthrust on the bug due to air which will help the bug to float in the air
This buoyancy force is given by the formula
here we know that
= density of air
V = volume of bug
so here we can say that upthrust of air is the force that bug will experience during his uplift
<h2>
Hey There!</h2><h2>
_____________________________________</h2><h2>
Answer:</h2>
<h2>_____________________________________</h2><h3>DATA:</h3>
Radius of Mercury = =
Mass of Mercury =
Distance Satellite above the surface of the Mercury = d = 265,000m
Gravitational Constant =
<h2>_____________________________________</h2><h3>SOLUTION:</h3>
Since the Satellite is orbiting around the Planet Mercury, due to the centripetal force, and Centripetal force is the force that acts towards the center of the circle, Whereas The gravitational force also acts towards the center of the circle thus we can say that Centripetal force is equal or same as centripetal force. So,
Fg is Given by,
Fc is Given by,
Where,
G is Gravitational Constant
is mass of Planet Mercury
is Mass of Satellite
r(small letter) is the distance between the center of the Planet Mercury and the satellite.
V is velocity of satellite
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Now,
r can also be written as,
Substitute the variables,
Simplify the equation,
V = 2814
Approximately,
V = 2800
<h2>_____________________________________</h2><h2>Best Regards,</h2><h2>'Borz'</h2><h3 />
Answer:
Explanation:
The gravitational force between two objects is given by
where
G is the gravitational constant
m1, m2 are the two masses
r is the separation between the two masses
In this problem, we have
the mass of each truck
r = 2.0 m is their separation
Substituting,
Answer:
The final speed will be "".
Explanation:
The given values are:
Potential difference,
Δv = 400 v
Radius,
r = 0.5580 cm
As we know,
⇒
and,
⇒
then,
⇒
On substituting the values, we get
⇒
⇒
⇒
Answer:
Hence lowest (nonzero) frequency that gives destructive interference in this case = 3400 Hz
Explanation:
Since, the two are in out of phase,
their path difference is
d= nλ
Given d1= 2.75 m
D= 3.90 m
d_2= 2.76 m
2.76-2.75= 1×λ
λ= 0.01 m
0.01= 1*λ
λ =0.01
frequency ν = v/λ = 340/0.01
f= 3400 Hz
Hence lowest (nonzero) frequency that gives destructive interference in this case = 3400 Hz