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Oksi-84 [34.3K]
4 years ago
6

A small metallic bob is suspended from the ceiling by a thread of negligible mass. The ball is then set in motion in a horizonta

l circle so that the thread describes a cone. Calculate the magnitude of the angular momentum of the bob about a vertical axis through the supporting point. The acceleration of gravity is 9.8 m/s2. Answer in units of kg · m2/s
Physics
1 answer:
shusha [124]4 years ago
8 0

Answer:

19.99 kg m²/s

Explanation:

Angular Momentum (L) is defined as the product of the moment of Inertia (I) and angular velocity (w)

L = m r × v.

r and  v are perpendicular to each other,

where r = lsinθ.

l = 2.4 m

θ= 34°

g = 9.8 m/s²  and m = 5 kg

resolving using newtons second law in the vertical and horizontal components.

T cos θ − m g = 0

T sin θ − mw² lsin θ = 0

where T is the force with which the wire acts on the bob

w = √g / lcosθ

= √ 9.8 / 2.4 ×cos 34

= 2.2193 rad/s

the angular momentum  L = mr× v

= mw (lsin θ)²

= 5 × 2.2193 (2.4 ×sin 34°)²

=19.99 kg m²/s

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Answer:

P=9.58 W

Explanation:

According to Newton's second law, and assuming friction force as zero:

F_m=m.a\\F_m=0.875kg*a

The acceleration is given by:

a=\frac{\Delta v}{t}\\a=\frac{0.685m/s}{21.5*10^{-3}s}\\\\a=31.9m/s^2

So the force exerted  by the motor is:

F_m=0.875kg*31.9m/s^2\\F_m=27.9N

The work done by the motor is given by:

W_m=F_m*d\\\\d=\frac{1}{2}*a*t^2\\d=\frac{1}{2}*31.9m/s^2*(21.5*10^{-3}s)^2\\\\d=7.37*10^{-3}m

W_m=27.9N*7.37*10^{-3}m\\W_m=0.206J

And finally, the power is given by:

P=\frac{W_m}{t}\\P=\frac{0.206J}{21.5*10^{-3}s}\\\\P=9.58W

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3 years ago
The relationship among speed, distance, and time is
alexdok [17]
Distance= speed (multiplied by) time
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3 years ago
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After a party the host pours the remnants of several bottles of wine into a jug. He then inserts a cork with a 2.15-cm diameter
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Answer:

3893.99675 N

Explanation:

F_1  = 125 N

A_1 = \pi 1.075^2

A_2 = \pi 6^2

From Pascal's law

\dfrac{F_1}{A_1}=\dfrac{F_2}{A_2}\\\Rightarrow F_2=\dfrac{F_1\times A_2}{A_1}\\\Rightarrow F_2=\dfrac{125\times 6^2}{1.075^2}\\\Rightarrow F_2=3893.99675\ N

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WILL GIVE BRAINLIEST!
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Rom4ik [11]

F = 1440 N. The repulsion force between two identical charges, each -8.00x10⁻⁵C separated by a distance of 20.0 cm is 1440 N.

The easiest way to solve this problem is using Coulomb's Law given by the equation F=k\frac{|q_{1}*q_{2}|}{r^{2} }, where k is the constant of proportionality or Coulomb's constant, q₁ and q₂ are the charges magnitude, and r is the distance between them.

We have to identical charges of -8.00x10⁻⁵C, are separated by a distance of 20.0 cm, and we need to know the force of repulsion between the charges.

First, we have to convert 20.0 cm to meters.

(20.0 cm x 1m)/100cm = 0.20 m

Using the Coulomb's Law equation:

F = 9.00x10^{9}\frac{Nm^{2}}{C^{2}} \frac{|-8.00x10^{-5}C*-8.00x10^{-5}C|}{(0.20m)^{2} }

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