All categories

3 years ago

What is the elevation at point A?

Physics

1 answer:

horsena [70]3 years ago

6 0

Answer:

i think that says 4800

its the same as the number at the bottom

You might be interested in

An object traveling in a circular path is accelerating because its

Anton [14]

It is accelerating because the direction of the velocity vector is changing

8 0

4 years ago

Two lasers are shining on a double slit, with slit separation d. Laser 1 has a wavelength of d/20, whereas laser 2 has a wavelen

xxTIMURxx [149]

Answer:

A) first laser

B) 0.08m

C) 0.64m

Explanation:

To find the position of the maximum you use the following formula:

$y=\frac{m\lambda D}{d}$

m: order of the maximum

λ: wavelength

D: distance to the screen = 4.80m

d: distance between slits

A) for the first laser you use:

$y_1=\frac{(1)(d/20)(4.80m)}{d}=0.24m\\$

for the second laser:

$y_2=\frac{(1)(d/15)(4.80m)}{d}=0.32m$

hence, the first maximum of the first laser is closer to the central maximum.

B) The difference between the first maximum:

$\Delta y=y_2-y_1=0.32m-0.24m=0.08m=8cm$

hence, the distance between the first maximum is 0.08m

C) you calculate the second maximum of laser 1:

$y_{m=2}=\frac{(2)(d/20)(4.80m)}{d}=0.48m$

and for the third minimum of laser 2:

$y_{minimum}=\frac{(m+\frac{1}{2})(\lambda)(D)}{d}\\\\y_{m=3}=\frac{(3+\frac{1}{2})(d/15)(4.80m)}{d}=1.12m$

Finally, you take the difference:

$1.12m-0.48m=0.64m$

hence, the distance is 0.64m

3 0

3 years ago

A car that experiences no frictional force is started and caused to move. For the car to continue in that motion, the gas pedal

astraxan [27]

Explanation:

Gas pedal is not required to used on friction less surface. On a friction less surface once the car is started and caused to move, car continues to move with same velocity as there is no opposing force. Therefore, no pressing of gas pedal required. However, this situation is not possible in real life, as friction is always present.

8 0

3 years ago

How many particles are there in 111.6 g of iron?

masya89 [10]

1 mol Fe contains 6.022*10^23 atoms or particles

2mol Fe will contain 2*6.022*10^23 = 1.20*10^24 atoms or particles.

4 0

4 years ago

A conducting sphere with radius R carries total charge Q. What is the magnitude of the electric field at distances R/2 and 2R fr

RideAnS [48]

Hello!

Distance of R/2:

Since a conducting sphere is referenced in this situation, all of its charge will be distributed along its SURFACE. Therefore, there is NO enclosed at a distance of R/2 from the center.

Using Gauss's Law:
$\oint E \cdot dA = E\cdot A = \frac{Q_{encl}}{\epsilon_0}$

E = Electric field strength (N/C)
A = Area of Gaussian surface (m²)

Q = Enclosed charge (C)
ε₀ = Permittivity of free space C²/Nm²)

If the enclosed charge is 0, then:
$E \cdot A = \frac{0}{\epsilon_0}\\\\\boxed{E = 0 \frac{N}{C}}$

Distance of '2R':

We can once again use Gauss's Law to solve. This time, however, a surface of radius '2R' encloses ALL of the charge of the sphere.

$E \cdot A = \frac{Q_{encl}}{\epsilon_0}$

'A' is equivalent to the surface area of a sphere of radius '2R', or:
$A = 4\pi (2R)^2\\\\A = 4\pi (4R^2)\\\\A = 16\pi R^2$

Substituting this expression back into Gauss's Law:
$E \cdot 16\pi R^2 = \frac{Q}{\epsilon_0}\\\\E = \frac{Q}{16\pi R^2\epsilon_0}$

To simplify:
$E = \frac{1}{4\pi \epsilon_0 } * \frac{Q}{4R^2}\\$

OR using k = 1/4πε₀:
$\boxed{E = \frac{kQ}{4R^2}}$

4 0

2 years ago