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2 years ago

A 3.0 cm object is 12.0 cm from a convex mirror that has a focal length of 20.0 cm.

Physics

2 answers:

poizon [28]2 years ago

7 0

Given conditions for question 2:

height of object = 3.0 cm

distance of object from mirror = 12 cm

focus length = 20 cm

image distance = ?

Using mirror formula:

1/(focus length) = 1/(object distance) + 1/(image distance)

or, 1/20 = 1/12 + 1/(image distance)

or, image distance = -30cm (the image formed is a virtual image)

Also, magnification of image is:

image height /height of object<span> = - image distance </span>/object distance

or, image height = - image distance / object distance * height of object

or, image height = -(-30) / 12 * 3 = 7.5

user100 [1]2 years ago

3 0

Distance of the image from the mirror: -30

Height of the image: 7.5

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If the voltage amplitude across an 8.50-nF capacitor is equal to 12.0 V when the current amplitude through it is 3.33 mA, the fr

Dmitriy789 [7]

Answer:

Frequency will be equal to 5.20 kHz

So option (c) will be correct answer

Explanation:

We have given value of capacitance $C=8.5nF=8.5\times 10^{-9}f$

Potential difference across capacitor V = 12 volt

Current through capacitor $i=3.33mA=3.33\times 10^{-3}A$

Capacitive reactance will be equal to $X_c=\frac{V}{i}=\frac{12}{3.33\times 10^{-3}A}=3603.60ohm$

Capacitive reactance is equal to $X_c=\frac{1}{\omega C}$

$3603.60=\frac{1}{\omega\times 8.5\times 10^{-9}}$

$\omega =32647.091rad/sec$

$2\pi f=32647.091$

$f=5198.98Hz$

f = 5.20 kHz

So frequency will be equal to 5.20 kHz

So option (c) will be correct answer

3 0

2 years ago

an iron rod of length 100m at 10 degree Celsius is used to measure a distance of 2km on a day when the temperature is 40 degree

german

Answer:

0.68 m

Explanation:

α = dL / L1*(dT)

dL = L1(dT) * α

Initial length, L1 = 100

Chang in Length =dL

α linear expansivity ; dL = change in length ; dT = change in temperature ; L1 = initial length

α of iron rod = 1.13 * 10^-5 k

dL = 100(40 - 10) * 1.13 * 10^-5

dL = 100(30) * 1.13 * 10^-5

dL = 3000 * 1.13 * 10^-5

dL = 3390 * 10^-5

dL = 0.0339 m

Error :

Distance measured = 2km = (2 * 1000) = 2000m

[Distance measured / (initial length + change in length)] × change in length

Error = (2000 / (100 + 0.0339)) * 0.0339

Error = (2000 / 100.0339) * 0.0339

Error = 19.993222 * 0.0339

Error = 0.6777702

Error = 0.68 m

4 0

2 years ago

Un tubo de acero de 40000 kilómetros forma un anillo que se ajusta bien a la circunferencia de la tierra. Imagine que las person

Darina [25.2K]

Answer:

82.76m

Explanation:

In order to find the distance of the steel ring to the ground, when its temperature has raised by 1°C, you first calculate the radius of the steel tube before its temperature increases.

You use the formula for the circumference of the steel ring:

$C=2\pi r$ (1)

C: circumference of the ring = 40000 km = 4*10^7m (you assume the circumference is the length of the steel tube)

you solve for r in the equation (1):

$r=\frac{C}{2\pi}=\frac{4*10^7m}{2\pi}=6,366,197.724m$

Next, you use the following formula to calculate the change in the length of the tube, when its temperature increases by 1°C:

$L=Lo[1+\alpha \Delta T]$ (2)

L: final length of the tube = ?

Lo: initial length of the tube = 4*10^7m

ΔT = change in the temperature of the steel tube = 1°C

α: thermal coefficient expansion of steel = 13*10^-6 /°C

You replace the values of the parameters in the equation (2):

$L=(4*10^7m)(1+(13*10^{-6}/ \°C)(1\°C))=40,000,520m$

With the new length of the tube, you can calculate the radius of a ring formed with the tube. You again solve the equation (1) for r:

$r'=\frac{C}{2\pi}=\frac{40,000,520m}{2\pi}=6,366,280.484m$

Finally, you compare both r and r' radius:

r' - r = 6,366,280.484m - 6,366,197.724m = 82.76m

Hence, the distance to the ring from the ground is 82.76m

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2 years ago

A 4.50-kg wheel that is 34.5 cm in diameter rotates through an angle of 13.8 rad as it slows down uniformly from 22.0 rad/s to 1

Mila [183]

Answer:

-10.9 rad/s²

Explanation:

ω² = ω₀² + 2α(θ - θ₀)

Given:

ω = 13.5 rad/s

ω₀ = 22.0 rad/s

θ - θ₀ = 13.8 rad

(13.5)² = (22.0)² + 2α (13.8)

α = -10.9 rad/s²

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2 years ago

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Rotational dynamics about a fixed axis: A person pushes on a small doorknob with a force of 5.00 N perpendicular to the surface

FrozenT [24]

Answer:

I = 2 kgm^2

Explanation:

In order to calculate the moment of inertia of the door, about the hinges, you use the following formula:

$\tau=I\alpha$ (1)

I: moment of inertia of the door

α: angular acceleration of the door = 2.00 rad/s^2

τ: torque exerted on the door

You can calculate the torque by using the information about the Force exerted on the door, and the distance to the hinges. You use the following formula:

$\tau=Fd$ (2)

F: force = 5.00 N

d: distance to the hinges = 0.800 m

You replace the equation (2) into the equation (1), and you solve for α:

$Fd=I\alpha\\\\I=\frac{Fd}{\alpha}$

Finally, you replace the values of all parameters in the previous equation for I:

$I=\frac{(5.00N)(0.800m)}{2.00rad/s^2}=2kgm^2$

The moment of inertia of the door around the hinges is 2 kgm^2

3 0

2 years ago