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3 years ago

7

Let theta denote the angular displacement of a simple pendulum oscillating in a vertical plane. If the mass of the Bob is m, the

n the tension in string is

2 answers:

posledela3 years ago

8 0

The tension has to hold the part of the weight in the direction of the string:

T = mg*cos(theta)

Theta=0, whole weight, theta=90, T=0, if the pendulum is horizontal, the string will be loose! Yeah

Harrizon [31]3 years ago

6 0

Answer:

So the tension in the string T =mg cosθ

Explanation:

Now given that

mass of bob = m

Angle from vertical plane =θ

Lets T is the tension in the string

From the diagram

Weight of bob have two component ,first one(mg cosθ) balance the tension in the string and other one(mg sinθ) make the bob to move in the plane.

So the tension in the string T =mg cosθ

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A wire of length L is wound into a square coil with 167 turns and used in a generator that operates at 60.0 Hz and 120 V rms val

Mrac [35]

Answer:

$171.43m$

Explanation:

First, define $emf$ ( electromotive force )-is the unit electric charge imparted by an energy source. In this case the generator.

The peak emf is:

$E_p_e_a_k=\sqrt2(E_m_a_x)$ = $\sqrt(2\times 120V)=170V$

Substituting $w=2\pi f$ and the value for peaf $E$ gives:

Total length= $4\sqrt {\frac{NE_p_e_a_k}{Bw}$ = $4\sqrt{\frac{167\times 170}{0.041T\times 2pi \times 60.0Hz}$

= $171.43m$

Hence, wire's length is 171.43m

8 0

3 years ago

How much force must a locomotive exert on a 12840-kg boxcar to make it accelerate forward at 0.490 m/s2?

ddd [48]

Data given:

m=12840kg

a=0.490m/s²

Formula:

F=ma

Solution:

F= 12840kg×0.490m/s²

F=6291.6N

6 0

3 years ago

A wave is propagating from left to right in a medium. The particles in the medium are also vibrating from left to right. What ki

sattari [20]

Answer:

In this case, the particles of the medium move parallel to the direction that the pulse moves. This type of wave is a longitudinal wave.

6 0

3 years ago

Calculati rezistenta filamentului unui bec al lanternei de buzunar daca la tensiunea de 3,5 v el este parcurs de un curent cu in

Gennadij [26K]

Answer:

Resistencia = 12.5 Ohmios

Explanation:

Dados los siguientes parámetros;

Voltaje = 3.5 Volts

Corriente = 0.28 Amps

Para encontrar el resistencia;

La ley de Ohm establece que a temperatura constante, la corriente que fluye en un circuito eléctrico es directamente proporcional al voltaje aplicado en los dos puntos e inversamente proporcional a la resistencia en el circuito eléctrico.

Matemáticamente, La ley de Ohm viene dada por la fórmula;

Voltaje = corriente * resistencia

Resistencia = voltaje/corriente

Resistencia = 3.5/0.28

Resistencia = 12.5 Ohmios

Por tanto, la resistencia del filamento de una linterna de bolsillo es de 12,5 ohmios.

3 0

3 years ago

Calculate the unit cell edge length for an 79 wt% Ag- 21 wt% Pd alloy. All of the palladium is in solid solution, and the crysta

ankoles [38]

Answer:

The edge length is 0.4036 nm

Solution:

As per the question:

Density of Ag, $\rho = 10.49 g/cm^{3}$

Density of Pd, $\rho = 12.02 g/cm^{3}$

Atomic weight of Ag, A = 107.87 g/mol

Atomic weight of Pd, A' = 106.4 g/mol

Now,

The average density, $\rho_{a} = \frac{n A_{avg}} {V_{c}\times N_{A}}$

where

$V_{c} = a^{3}$ = Volume of crystal lattice

a = edge length

n = 4 = no. of atoms in FCC

Therefore,

$\rho_{a} = = \frac{n A_{avg}} {V_{c}\times N_{A}}$

Therefore, the length of the unit cell is given as:

$a = (\frac{nA_{avg}}{\rho_{a}\times N_{a}})^{1/3}$ (1)

Average atomic weight is given as:

$A_{avg} = \frac{100}{\frac{C_{Ag}}{A_{Ag}} + \frac{C_{Pd}}{A_{Pd}}}$

where

$C_{Ag}$ = 79 %

$A_{Ag}$ = 107

$C_{Pd}$ = 21%

$A_{Pd}$ = 106

Therefore,

$A_{avg} = \frac{100}{\frac{79}{107} + \frac{21}{106}} = 106.78$

In the similar way, average density is given as:

$\rho_{a} = \frac{100}{\frac{C_{Ag}}{\rho_{Ag}} + \frac{C_{Pd}}{\rho_{Pd}}}$

$\rho_{a} = \frac{100}{\frac{79}{10.49} + \frac{21}{12.02}} = 10.78 g/cm^{3}$

Therefore, edge length is given by eqn (1) as:

$a = (\frac{4\times 106.78}{10.78\times 6.023 X 10^23})^{1/3} = 4.036\times 10^{- 8} cm = 0.4036\times 10^{- 9} m = 0.4036 nm$

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3 years ago