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4 years ago

Compare the force of air resistance and the force of gravity on an object falling at its terminal velocity.

1 answer:

7 0

The viscous force on an object moving through air is proportional to its velocity.
The only forces acting on an object when falling are air resistance and its weight itself. The weight acts vertically downwards whereas air resistance acts vertically upward.
Let F be the viscous force due to air molecules, B be buoyant force due to air and W be the weight of falling object. Initially, the velocity of falling object and hence the viscous force F is zero and the object is accelerated due to force
(W-B). Because of the acceleration the velocity increases and accordingly the viscous force also increases. At a certain instant, the viscous force becomes equal to W-B. The net force then becomes zero and the object falls with constant velocity. This constant velocity is called terminal velocity.
Thus at terminal velocity, air resistance and force of gravity becomes equal.

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Three point charges are positioned on the x axis. If the charges and corresponding positions are +32 µC at x = 0, +20 µC at x =

Crank

Answer:

Fnet = 12 N

Explanation:

Force on a point charge due to another point charge = kq1q2 / d^2

Force on +32uC = due to + 20uC + due to -60uC

where uC = 1 x 10^-6 C and k = 9 x 10^9 N m^2 / C^2

Net Force =

$= \frac{1}{4\pi \epsilon _0} [\frac{32 \times 10^-^6 \times60\times10^-^6}{(60/100)^2}-\frac{32 \times 10^-^6 \times20\times10^-^6}{ (40/100)^2} ]$

$F_{net}=9 \times10^9\times 10^-^1^2[\frac{32\times60\times10^4}{60\times60} -\frac{32\times20\times10^4}{40\times40} ]$

$=90[32(\frac{80-60}{60\times 80} )]\\\\=90\times32\times0.004167\\\\=12N$

Fnet = 12 N

8 0

4 years ago

If a machine has an efficiency of 50% and an input of 3 J, what is the output?

uranmaximum [27]

Answer:

150J

Explanation:

work output/work input=100%

so just make work output the subject

6 0

3 years ago

A 4.00 µf capacitor is connected to a 12.0 v battery.

atroni [7]

(a) The capacitance of the capacitor is:
$C=4 \mu F=4 \cdot 10^{-6}F$
and the voltage applied across its plates is
$V=12.0 V$

The relationship between the charge Q on each plate of the capacitor, the capacitance and the voltage is:
$C= \frac{Q}{V}$
and re-arranging it we find the charge stored in the capacitor:
$Q=CV=(4 \cdot 10^{-6} F)(12.0 V)=4.8 \cdot 10^{-5} C$

(b) The electrical potential energy stored in a capacitor is given by
$U= \frac{1}{2}CV^2$
where C is the capacitance and V is the voltage. The new voltage is
$V=1.50 V$
so the energy stored in the capacitor is
$U= \frac{1}{2}(4 \cdot 10^{-6} F)(1.50 V)^2=4.5 \cdot 10^{-6} J$

3 0

3 years ago

Read 2 more answers

Four equal masses m are located at the corners of a square of side L, connected by essentially massless rods. Find the rotationa

skelet666 [1.2K]

Answer:

Explanation:

a )

Moment of inertial of four masses about axis that coincides with one side :

Out of four masses . location of two masses will lie on the axis so their moment of inertia will be zero .

Moment of inertia of the two remaining masses

= m L² + m L²

= 2 mL²

b )

Axis that bisects two opposite sides

Each of the four masses will lie at a distance of L / 2 from this axis so moment of inertia of the four masses

= 4 x m x ( L/2 )²

= 4 x mL² / 4

= m L² .

3 0

3 years ago

A circular 10-turn coil with a radius of 5.0 cm carries a current of 5 A. It lies in the xy plane in a uniform magnetic field =

Tju [1.3M]

Answer:

U = – 0.12J

Explanation:

Given N = 10 turns, I = 5A, r = 5×10-²m

B^ = 0.05 T iˆ+ 0.3 T kˆ

Magnitude of the magnetic field vector B = √(0.05²+0.3²) = 0.304T

Area = πr² = π(5×10-²)² = 7.85×10-³m²

Magnetic moment μ = NIA

μ = 10×5×7.85×10-³ = 0.3925Am²

U = -μ•B = –0.3925×0.304 = –0.12J

The sign is negative because the magnetic moment is aligned with the magnetic field.

3 0

3 years ago