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Oksi-84 [34.3K]
3 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]3 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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Get active. How well your body functions affects your ability to accomplish your daily activities.

Explanation:

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Two charges that are separated by one meter exert 1-N forces on each other. If the charges are pushed together so the separation
Nana76 [90]

Answer: 16N

Explanation:

According to coulombs law which states that the force of attraction between two charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. Mathematically,

F = kq1q2/r²

If two charges that are separated by one meter exert 1-N forces on each other, the equation will become

1 = kq1q2/1²

kq1q2 = 1 ... (1)

If the charges are pushed together so the separation is 25 centimeters, the force between them becomes,

F = kq1q2/0.25² (25cm converted to meters)

0.0625F = kq1q2 ... (2)

Dividing equation 1 by 2 to determine the force F on each charges, it becomes;

1/0.0625F = kq1q2/kq1q2

1/0.0625F = 1

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6 0
3 years ago
A truck’s suspension spring each have a spring constant of 769 N/m. If the potential energy of the right front spring is 250 J,
Alex Ar [27]

Answer:

x = 0.81 m

Explanation:

given,

spring constant, k = 769 N/m

Potential energy of the spring = 250 J

distance of spring compression = ?

using conservation of energy

potential energy will equal to the spring energy

PE = \dfrac{1}{2}kx^2

250= \dfrac{1}{2}\times 769\times x^2

   x² = 0.650

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Hence, the spring is compressed to  0.81 m

6 0
3 years ago
A proton is traveling horizontally to the right at 4.20×10^6m/s.Part A:Find (a)the magnitude and (b) direction of the weakest el
anygoal [31]

Answer:

2630250 N/C, horizontally left

0

1.67\times 10^{-8}\ s

1434.825 N/C, horizontally left

Explanation:

m = Mass of particle

u = Initial velocity = 4.2\times 10^6\ m/s

v = Final velocity = 0

t = Time taken

s = Displacement = 3.5 cm

q = Charge of particle = 1.6\times 10^{-19}\ C

Force is given by

F=qE

Acceleration is given by

a=\dfrac{F}{m}\\\Rightarrow a=\dfrac{qE}{m}\\\Rightarrow a=\dfrac{1.6\times 10^{-19}\times E}{1.67\times 10^{-27}}\\\Rightarrow a=95808383.23353E

v^2-u^2=2as\\\Rightarrow v^2-u^2=2\times 95808383.23353Es\\\Rightarrow E=\dfrac{0^2-(4.2\times 10^6)^2}{2\times 95808383.23353\times 0.035}\\\Rightarrow E=-2630250\ N/C

Magnitude of electric field is 2630250 N/C

Direction is horizontally to the left

The angle counterclockwise from left is zero.

a=\dfrac{F}{m}\\\Rightarrow a=\dfrac{qE}{m}\\\Rightarrow a=\dfrac{1.6\times 10^{-19}\times -2630250}{1.67\times 10^{-27}}\\\Rightarrow a=-2.52\times 10^{14}\ m/s^2

v=u+at\\\Rightarrow t=\dfrac{v-u}{a}\\\Rightarrow t=\dfrac{0-4.2\times 10^6}{-2.52\times 10^{14}}\\\Rightarrow t=1.67\times 10^{-8}\ s

The time taken is 1.67\times 10^{-8}\ s

Acceleration is given by

a=\dfrac{F}{m}\\\Rightarrow a=\dfrac{qE}{m}\\\Rightarrow a=\dfrac{1.6\times 10^{-19}\times E}{9.11\times 10^{-31}}\\\Rightarrow a=175631174533.4797E

v^2-u^2=2as\\\Rightarrow v^2-u^2=2\times 175631174533.4797Es\\\Rightarrow E=\dfrac{0^2-(4.2\times 10^6)^2}{2\times 175631174533.4797\times 0.035}\\\Rightarrow E=-1434.825\ N/C

Magnitude of electric field is 1434.825 N/C

Direction is horizontally to the left

8 0
3 years ago
A motorcycle is moving at 15 m/s when its brakes are applied, bringing the cycle to rest in 5.3 s. To the nearest meter, how far
Nonamiya [84]

Answer:

Motorcycle travel 39.75 m while coming to a stop

Explanation:

We have equation of motion

        v = u + at

   Final velocity, v = 0 m/s, u = 15 m/s, t = 5.3 s

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 Now we have the other equation of motion

        v² = u² + 2as

        0² = 15² - 2 x 2.83 x s

         s = 39.75 m

Motorcycle travel 39.75 m while coming to a stop

8 0
3 years ago
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