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Pavel [41]
3 years ago
6

A pendulum consists of a 0.5 kg mass attached to the end of 1-meter-long rod of negligible mass. When the rod makes an angle of

60° with the vertical, find the magnitude of the torque about the pivot.
Physics
1 answer:
Alenkinab [10]3 years ago
7 0

Answer:

T=4.24 N.m

Explanation:

Torque is equal to force for distance for sinus of the angle between the direction of the force and the distance, the distance between the mass and the pivot is 1 m, and to obtain the force that is the mass for the gravity in this case, we need to know the component that produces a torque in the pivot

F=0.5 kg* 9.8 m/s^{2}= 4.9 N

and we decompose the force in parallel direction to the rod and perpendicular direction to the rod, the magnitude that produces torque is the perpendicular component, because the torque is in function of the sinus

so, we obtain -> Fy= 4.9 N*sin(60)= 4.24 N

and, T= (4.24 N)*(1 m)*(Sin(90))= 4.24 N.m

anothe way to do it is,

T= (4.9 N)*(1 m)*(Sin(60))= 4.24 N.m, and we obtain the same result

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Question 5 At 12:00 pm, a spaceship is at position ⎡⎣324⎤⎦ km ⎣ ⎢ ⎡ ​ 3 2 4 ​ ⎦ ⎥ ⎤ ​ km away from the origin with respect to so
Anettt [7]

Answer:

[1, 6, -2]

Explanation:

Given the following :

Initial Position of spaceship : [3 2 4] km

Velocity of spaceship : [-1 2 - 3] km/hr

Location of ship after two hours have passed :

Distance moved by spaceship :

Velocity × time

[-1 2 -3] × 2 = [-2 4 -6]

Location of ship after two hours :

Initial position + distance moved

[3 2 4] + [-2 4 -6] = [3 + (-2)], [2 + 4], [4 + (-6)]

= [3-2, 2+4, 4-6] = [1, 6, -2]

4 0
3 years ago
Whiat is quantum numbers describes the size and energy of an orbital?
Levart [38]

Answer:

The answer is the principal Quantum number (n)

Explanation:

The principal quantum number is one of the four quantum numbers associated with an atom.

It is denoted by a number n=1,2,3,4 etc

It tells both size (directly) and energy (indirectly) of an orbital.

When n=1 means it is the closest to the nucleus and is the smallest orbital and with increase in principal quantum number, it depicts that size of the orbital is increasing.

It tells the energy of the orbital as well as smaller number means less distance from nucleus and having less energy. Since electrons requires to absorb energy to jump into higher orbitals making n=2,3,4 etc. Thus electrons in the orbitals with higher n number indicates higher energy orbitals.

7 0
3 years ago
what is the acceleration of an object if the object has an initial speed of 230 m/s and speeds up to 650 m/s. The time it takes
adell [148]

Answer:

explain this

Explanation:

7 0
2 years ago
A 11,000-watt radio station transmits at 880 kHz. Determine the number of joules transmitted per second.
NARA [144]

1 watt = 1 joule per sec

11,000 Watts = 11,000 joules per sec

The frequency doesn't matter.

3 0
3 years ago
A circular loop of wire with a radius of 15.0cm and oriented in the horizontal xy-plane is located in a region of uniform magnet
Drupady [299]

Answer:

See answer

Explanation:

The area of the circular loop is given by:

A = \pi r^2

The magnetic flux is given by:

\phi = \int \vec{B} \cdot d\vec{A}

d\vec{A} is parallel to \vec{B} and \vec{B} is constant in magnitude and direction therefore:

\phi = \int \vec{B} \cdot d\vec{A}= \int BdAcos(0)= B\int dA= B*(\pi r^2)= \pi Br^2

Part A)

initially the flux is \phi =\pi B r^2

after the interval \Delta t= 2.4 [m/s]

the flux is

\phi = 0

now, the EMF is defined as:

\epsilon =- \frac{d \phi}{dt},

if we consider \Delta t= 2.4 [m/s] very small then we can re-write it as:

\epsilon =- \frac{\Delta \phi}{\Delta t}

\Delta \phi = 0 - \pi B r^2=-\pi (1.7) (0.15)^2=-0.12

then:

\epsilon =- \frac{-0.12}{0.0024} = 50 [V]

Part B)

When looked down from above, the current flows counter clockwise, according to the right hand rule, if you place your thumb upwards (the direction of the magnetic field) and close your fingers, then the current will flow in the direction of your fingers.

3 0
3 years ago
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