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3 years ago

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Next weekend you are going to Eleanor Rigby’s house to fix her grandfather clock that is a bit slow. Based on what you have lear

ned about pendulums, how will you fix her slow clock?

1 answer:

Bumek [7]3 years ago

8 0

Answer:

By pushing the pendulum Bob up so it moves faster

Explanation:

In pendulum physics the length of the pendulum Bob determines the speed of the clock. So since the grandfather's clock is slow it means the Bob is has moved down so to move it up you have to achieve this by adjusting the but upwards thereby making the clock move faster.

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Suppose that a steel bridge, 1000 m long, were built without any expansion joints. Suppose that only one end of the bridge was h

Stels [109]

Answer:

The difference in the length of the bridge is 0.42 m.

Explanation:

Given that,

Length = 1000 m

Winter temperature = 0°C

Summer temperature = 40°C

Coefficient of thermal expansion $\alpha= 10.5\times10^{-6}\ K^{-1}$

We need to calculate the difference in the length of the bridge

Using formula of the difference in the length

$\Delta L=L\alpha\Delta T$

Where, $\Delta T$ = temperature difference

$\alpha$ =Coefficient of thermal expansion

L= length

Put the value into the formula

$\Delta L=1000\times10.5\times10^{-6}(40^{\circ}-0^{\circ})$

$\Delta L=0.42\ m$

Hence, The difference in the length of the bridge is 0.42 m.

5 0

3 years ago

A magnet is moved in and out of a coil of wire connected to a high-resistance voltmeter. If the number of coils doubles, the ind

Varvara68 [4.7K]

Answer:

A. Doubles.

Explanation:

In an electromagnetic device such as a generator, when a wire (conductor) moves through the magnetic field between the South and North poles of a magnet, an electromotive force (e.m.f) is usually induced across a wire

The mode of operation of a generator is that a metal core with copper tightly wound to it (conductor coil) rotates rapidly between the two (2) poles of a horseshoe magnet type. Thus when the conductor coil rotates rapidly, it cuts the magnetic field existing between the poles of the horseshoe magnet and then induces the flow of current.

When a high-resistance voltmeter is connected to an electric circuit, a deflection will arise due to the flow of electricity. Moving the magnet towards the coil of wire will cause the needle of the high-resistance voltmeter to move in one direction. Also, as the magnet is moved out from the coil of wire, the needle of the high-resistance voltmeter moves in the opposite direction.

In this scenario, a magnet is moved in and out of a coil of wire connected to a high-resistance voltmeter. If the number of coils doubles, the induced voltage doubles because the number of turns (voltage) in the primary winding is directly proportional to the number of turns (voltage) in the secondary winding.

4 0

3 years ago

Two resistors, A and B, are connected in series to a 6.0 V battery. A voltmeter connected across resistor A measures a potential

mestny [16]

Answer:

Resistance of resistor A = 6.0 Ω and resistance of resistor B = 3.0 Ω

Explanation:

When the two resistors are in series, let V₁ = voltage in resistor A and R₁ = resistance of resistor A and V₂ = voltage in resistor B and R₂ = resistance of resistor B.

Given that V₁ + V₂ = 6.0 V and V₁ = 4.0 V,

V₂ = 6.0 V - V₁ = 6.0 V - 4.0 V = 2.0 V

Also, let the current in series be I.

So, V₁ = IR₁ and V₂ = IR₂

I = V₁/R₁ and I = V₂/R₂

equating both expressions, we have

V₁/R₁ = V₂/R₂

4.0 V/R₁ = 2.0 V/R₂

dividing through by 2.0 V, we have

2/R₁ = 1/R₂

taking the reciprocal, we have

R₂ = R₁/2

R₁ = 2R₂

From the parallel connection, let V₁ = voltage in resistor A and R₁ = resistance of resistor A and V₂ = voltage in resistor B and R₂ = resistance of resistor B. Since it is parallel, V₁ = V₂ = V = 6.0 V

Also, V₂ = I₂R₂ where I₂ = current in resistor B = 2.0 A and R₂ = resistance of resistor B

So, R₂ = V₂/I₂

= 6.0 V/2.0 A

= 3.0 Ω

R₁ = 2R₂

= 2(3.0 Ω)

= 6.0 Ω

So, resistance of resistor A = 6.0 Ω and resistance of resistor B = 3.0 Ω

6 0

3 years ago

One of the main factors driving improvements in the cost and complexity of integrated circuits (ICs) is improvements in photolit

nika2105 [10]

Answer:

0.000003782 m

0.000001891 m

0.000001197125 m

Explanation:

$\lambda$ = Wavelength = 248 nm

D = Diameter of beam = 1 cm

f = Focal length = 0.625 cm

The angle is given by

$\theta=\dfrac{1.22\lambda}{D}$

The width is given by

$d=2\theta f\\\Rightarrow d=2\dfrac{1.22\lambda f}{D}\\\Rightarrow d=2\dfrac{1.22\times 248\times 10^{-9}\times 6.25\times 10^{-2}}{1\times 10^{-2}}\\\Rightarrow d=0.000003782\ m$

The required width is 0.000003782 m

Minimum resolvable line separation is given by

$\dfrac{0.000003782}{2}=0.000001891\ m$

The minimum resolvable line separation between adjacent lines is 0.000001891 m

when $\lambda=157\ nm$

$d=2\dfrac{1.22\times 157\times 10^{-9}\times 6.25\times 10^{-2}}{1\times 10^{-2}}\\\Rightarrow d=0.00000239425\ m$

The new minimum resolvable line separation between adjacent lines is

$\dfrac{0.00000239425}{2}=0.000001197125\ m$

6 0

3 years ago

Four traveling waves are described by the following equations, where all quantities are measured in SI units and y represents th

agasfer [191]

Answer:

$T_1=T_3=\dfrac{2\pi}{21}$

$T_2=T_4=\dfrac{2\pi}{42}$

Explanation:

Wave 1, $y_1=0.12\ cos(3x-21t)$

Wave 2, $y_2=0.15\ sin(6x+42t)$

Wave 3, $y_3=0.13\ cos(6x+21t)$

Wave 4, $y_4=-0.27\ sin(3x-42t)$

The general equation of travelling wave is given by :

$y=A\ cos(kx\pm \omega t)$

The value of $\omega$ will remain the same if we take phase difference into account.

For first wave,

$\omega_1=21$

$\dfrac{2\pi }{T_1}=21$

$T_1=\dfrac{2\pi}{21}$

For second wave,

$\omega_2=42$

$\dfrac{2\pi }{T_2}=42$

$T_2=\dfrac{2\pi}{42}$

For the third wave,

$\omega_3=21$

$\dfrac{2\pi }{T_3}=21$

$T_3=\dfrac{2\pi}{21}$

For the fourth wave,

$\omega_4=42$

$\dfrac{2\pi }{T_4}=42$

$T_4=\dfrac{2\pi}{42}$

It is clear from above calculations that waves 1 and 3 have same time period. Also, wave 2 and 4 have same time period. Hence, this is the required solution.

3 0

3 years ago