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evablogger [386]

2 years ago

13

A wrench is placed at 30 cm in front of a diverging lens with a focal length of magnitude 10 cm. What is the magnification of th

e wrench?

1 answer:

WARRIOR [948]2 years ago

4 0

Answer:

0.25

Explanation:

Magnification = image distance/object distance

mag = v/u.................. Equation 1

Given: f = -10 cm ( diverging lens) u = 30 cm.

Where can calculate for the value of v using

1/f = 1/u+1/v

make v the subject of the equation

v = fv/(u-f)..................... Equation 2

Substitute into equation 2

v = -30(10)/(30+10)

v = -300/40

v = -7.5 cm.

substituting into equation 1,

mag = 7.5/30

mag = 0.25

hence the magnification of the wretch = 0.25

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A projectile of mass 0.2 kg and an initial velocity of 50 m/s collides with the end of a blade attached to a turbine. The rotati

fgiga [73]

Answer:

5.5 rad/sec

Explanation:

8 0

2 years ago

An electric toaster is rated 1200 watts at 120 volts. what is the total electrical energy used to operate the toaster for 30 sec

Anarel [89]

Energy is calculated as power*time, so give the wattage of 1200 W (equivalent to 1200 Joules/second) and time of 30 seconds, multiplying these gives 36000 J or 36 kJ of electrical energy.

If electrical charge or current is needed: Power = voltage * current, so given the power of 1200 watts and voltage of 120 V, current is 1200 W / 120 V = 10 Amperes. Charge is calculated by multiplying 10 A*30 s = 300 C.

3 0

2 years ago

A particular string resonates in four loops at a frequency of 320 Hz . Name at least three other (smaller) frequencies at which

goldfiish [28.3K]

Answer:

160 Hz , 240 Hz , 400 Hz

Explanation:

Given that

Frequency of forth harmonic is 320 Hz.

Lets take fundamental frequency = f₁

$f_1=\dfrac{320}{4}\ Hz$

f₁=80 Hz

Frequency of first harmonic = f₂

f₂=2 f₁

f₂ =2 x 80 = 160 Hz

Frequency of second harmonic = f₃

f₃= 3 f₁=3 x 80 = 240 Hz

Frequency of fifth harmonic = f₅

f₅= 5 f₁= 5 x 80 = 400 Hz

Three frequencies are as follows

160 Hz , 240 Hz , 400 Hz

6 0

2 years ago

A convex mirror with a focal length of 0.25 m forms a 0.080 m tall image of an automobile at a distance of 0.24 m behind the mir

Semmy [17]

Answer:

The distance and height of the object is 6 m and 2 m.

The image is virtual and upright.

Explanation:

Given that,

Focal length = 0.25 m

Length of image = 0.080 m

Image distance = 0.24 m

We need to calculate the distance of the object

Using formula of lens

$\dfrac{1}{v}=\dfrac{1}{f}+\dfrac{1}{u}$

Put the value into the formula

$\dfrac{1}{0.24}=\dfrac{1}{0.25}+\dfrac{1}{u}$

$\dfrac{1}{u}=\dfrac{1}{0.24}-\dfrac{1}{0.25}$

$\dfrac{1}{u}=\dfrac{1}{6}$

$u=6\ m$

We need to calculate the magnification

Using formula of magnification

$m=-\dfrac{v}{u}$

Put the value into the formula

$m=-\dfrac{0.24}{-6}$

$m=0.04$

We need to calculate the height of the object

Using formula of magnification

$m=\dfrac{h'}{h}$

$h=\dfrac{0.080}{0.04}$

$h=2\ m$

A convex mirror produce a virtual and upright image behind the mirror.

Hence, The distance and height of the object is 6 m and 2 m.

The image is virtual and upright.

6 0

2 years ago

Read 2 more answers

A charged paint is spread in a very thin uniform layer over the surface of a plastic sphere of diameter 13.0 cm , giving it a ch

Leokris [45]

a) Electric field inside the paint layer: zero

b) Electric field just outside the paint layer: $-3.62\cdot 10^7 N/C$

c) Electric field 8.00 cm outside the paint layer: $-7.27\cdot 10^7 N/C$

Explanation:

a)

We can find the electric field inside the paint layer by applying Gauss Law: the total flux of the electric field through a gaussian surface is equal to the charge contained within the surface divided by the vacuum permittivity, mathematically:

$\int EdS = \frac{q}{\epsilon_0}$

where

E is the electric field

dS is the element of surface

q is the charge within the gaussian surface

$\epsilon_0 = 8.85\cdot 10^{-12}F/m$ is the vacuum permittivity

Here we want to find the electric field just inside the paint layer, so we take a sphere of radius $r$ as Gaussian surface, where

R = 6.5 cm = 0.065 m is the radius of the plastic sphere (half the diameter)

By taking the sphere of radius r, we note that the net charge inside this sphere is zero, therefore

$q=0$

So we have

$\int E dS=0$

which means that the electric field inside the paint layer is zero.

b)

Now we want to find the electric field just outside the paint layer: therefore, we take a Gaussian sphere of radius

$r=R=0.065 m$

The area of the surface is

$A=4\pi R^2$

And since the electric field is perpendicular to the surface at any point, Gauss Law becomes

$E\cdot 4\pi R^2 = \frac{q}{\epsilon_0}$

The charge included within the sphere in this case is the charge on the paint layer, therefore

$q=-17.0\mu C=-17.0\cdot 10^{-6}C$

So, the electric field is:

$E=\frac{q}{4\pi \epsilon_0 R^2}=\frac{-17.0\cdot 10^{-6}}{4\pi(8.85\cdot 10^{-12})(0.065)^2}=-3.62\cdot 10^7 N/C$

where the negative sign means the direction of the field is inward, since the charge is negative.

c)

Here we want to calculate the electric field 8.00 cm outside the surface of the paint layer.

Therefore, we have to take a Gaussian sphere of radius:

$r=8.00 cm + R = 8.00 + 6.50 = 14.5 cm = 0.145 m$

Gauss theorem this time becomes

$E\cdot 4\pi r^2 = \frac{q}{\epsilon_0}$

And the charge included within the sphere is again the charge on the paint layer,

$q=-17.0\mu C=-17.0\cdot 10^{-6}C$

Therefore, the electric field is

$E=\frac{q}{4\pi \epsilon_0 r^2}=\frac{-17.0\cdot 10^{-6}}{4\pi(8.85\cdot 10^{-12})(0.145)^2}=-7.27\cdot 10^7 N/C$

Learn more about electric field:

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5 0

2 years ago