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Y_Kistochka [10]

3 years ago

5

An object is located 51 millimeters from a diverging lens. The object has a height of 13 millimeters and the image height is 3.5

millimeters. How far in front of the lens is the image located? A. 189 millimeters B. 13.7 millimeters C. 1.12 millimeters D. 51 millimeters

2 answers:

Umnica [9.8K]3 years ago

5 0

Answer is B 13.7 millimeters

12345 [234]3 years ago

3 0

You would get 13.7
mi/51mm=3.5mm/13mm
by solving it you will B13.7mm

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A uniformly charged ring of radius 10.0 cm has a total charge of 75.0 mC. Find the electric field on the axis of the ring at (a)

wlad13 [49]

Answer:

(a) 6650246.305 N/C

(b) 24150268.34 N/C

(c) 6408227.848 N/C

(d) 665024.6305 N/C

Explanation:

Given:

Radius of the ring (r) = 10.0 cm = 0.10 m [1 cm = 0.01 m]

Total charge of the ring (Q) = 75.0 μC = $75\times 10^{-6}\ \mu C$ [1 μC = 10⁻⁶ C]

Electric field on the axis of the ring of radius 'r' at a distance of 'x' from the center of the ring is given as:

$E_x=\dfrac{kQx}{(x^2+r^2)^\frac{3}{2}}$

Plug in the given values for each point and solve.

(a)

Given:

$Q=75\times 10^{-6}\ \mu C$ , $r=0.01\ m, a=1.00\ cm=0.01\ m,k=9\times 10^{9}\ Nm^2/C^2$

Electric field is given as:

$E_x=\dfrac{(9\times 10^{9})(75\times 10^{-6})(0.01)}{((0.01)^2+(0.1)^2)^\frac{3}{2}}\\\\E_x=\dfrac{6750}{1.015\times 10^{-3}}\\\\E_x=6650246. 305\ N/C$

(b)

Given:

$Q=75\times 10^{-6}\ \mu C$ , $r=0.01\ m, a=5.00\ cm=0.05\ m,k=9\times 10^{9}\ Nm^2/C^2$

Electric field is given as:

$E_x=\dfrac{(9\times 10^{9})(75\times 10^{-6})(0.05)}{((0.05)^2+(0.1)^2)^\frac{3}{2}}\\\\E_x=\dfrac{33750}{1.3975\times 10^{-3}}\\\\E_x=24150268.34\ N/C$

(c)

Given:

$Q=75\times 10^{-6}\ \mu C$ , $r=0.01\ m, a=30.0\ cm=0.30\ m,k=9\times 10^{9}\ Nm^2/C^2$

Electric field is given as:

$E_x=\dfrac{(9\times 10^{9})(75\times 10^{-6})(0.30)}{((0.30)^2+(0.1)^2)^\frac{3}{2}}\\\\E_x=\dfrac{202500}{0.0316}\\\\E_x=6408227.848\ N/C$

(d)

Given:

$Q=75\times 10^{-6}\ \mu C$ , $r=0.01\ m, a=100\ cm=1\ m,k=9\times 10^{9}\ Nm^2/C^2$

Electric field is given as:

$E_x=\dfrac{(9\times 10^{9})(75\times 10^{-6})(1)}{((1)^2+(0.1)^2)^\frac{3}{2}}\\\\E_x=\dfrac{675000}{1.015}\\\\E_x=665024.6305\ N/C$

7 0

3 years ago

A sound wave with a frequency of 400 Hz is moving through a solid object. If the wavelength of the sound wave is 8 m, what is th

Paraphin [41]

Answer:

v = 3200 m/s

Explanation:

As we know that the frequency of the sound wave is given as

$f = 400 Hz$

wavelength of the sound wave is given as

$\lambda = 8 m$

so now we have

$speed = wavelength \times frequency$

so we will have

$v = (8m) \times (400 Hz)$

$v = 3200 m/s$

4 0

3 years ago

An attack helicopter is equipped with a 20- mm cannon that fires 88.7 g shells in the forward direction with a muzzle speed of 9

Svetradugi [14.3K]

The impulse imparted to the shells equals the change in the momentum:
Fav*(Delta t)= Delta m*v.
The mass change is
Delta m= n*m= (89.9shells)*(88.7g)=7.97Kg
So the average force is
F=((v)*(Delta m))/t= ((929)*(7.97))/4.84=1529.78 N
Since the velocity of the shells is much greater than the velocity of the helicopter, there is no need to use relative velocity.

4 0

3 years ago

2. A 1.30-m long gas column that is open at one end and closed at the other end has a fundamental resonant frequency 80.0 Hz. Wh

Elina [12.6K]

To solve this problem, it will be necessary to apply the concepts related to the fundamental resonance frequency in a closed organ pipe.

This is mathematically given as

$f_n (2n+1)(\frac{v}{4L})$

For fundamental frequency n is 0, then,

$f_0 = \frac{v}{4L}$

When,

v = Velocity of sound

L = Length,

Rearranging to find the velocity,

$v = f_0 (4L)$

$v = (80Hz)(4)(1.3m)$

$v = 416m/s$

Therefore the speed of sound in this gas is 416m/s

7 0

4 years ago

The view that time is a scarce resource that must be rationed and controlled through the use of schedules and appointments, and

Diano4ka-milaya [45]

Answer:

Monochronic.

Explanation:

This view is called as Monochronic.

The dictionary definition of Mono-chronic is that Monochronic people just like to do one thing at a time. They respect a certain uniformity and sense of being a suitable place and time for different activities. They don't esteem interruptions. It has been generally observe that their focus and productivity remains higher than the polychronic.( Those we are comfortable with performing multiple task at a time)

6 0

4 years ago