CAN SOMEONE HELP ME PLEASE! SOMEONE HAS YET TO ANSWER MY QUESTIONS! I WILL GREATLY APPRECIATE IT! (Anxious)

A student measures the speed of sound by echo destiny classes hands and then measures the time to hear the echo his distance to

777dan777 [17]

Explanation:

∆x=300 m×2

∆t=1.5 s

v=∆x/∆t → v=2×300/1.5 = 400 m/s

6 0

3 years ago

If an object has a mass of 38 kg, what is its approximate weight on earth?

klasskru [66]

38*10=380 N
To be more exact, 38 should be multiplied by 9.8 instead of 10.

3 0

3 years ago

Light of wavelength 650 nm is normally incident on the rear of a grating. The first bright fringe (other than the central one) i

garik1379 [7]

Answer

given,

wavelength (λ) = 650 nm

angle = 5°

using bragg's law

$sin \theta = \dfrac{n \lambda}{d}$

$d= \dfrac{n \lambda}{sin \theta}$

$d= \dfrac{1 \times 650 \times 10^{-9}\ m}{sin5^0}$

d = 7.46 x 10⁻⁴ cm

number of slits per centimeter

= $\dfrac{1}{d}\\\Rightarrow \dfrac{1}{7.46\times 10^{-4}}\\\Rightarrow 1340 split per centimeter.$

b) wavelength of two rays 650 nm and 420 nm

$d = \dfrac{1}{5000}$

d = 2 x 10⁻6 m

we now,

$sin \theta = \dfrac{n \lambda}{d}$

for 650 nm

$sin \theta = \dfrac{2\times 650\times 10^{-9}}{2\times 10^{-6}}$

$\theta =sin^{-1}(0.65)$

θ = 40.54°

for 450 nm

$sin \theta = \dfrac{2\times 450\times 10^{-9}}{2\times 10^{-6}}$

$\theta =sin^{-1}(0.45)$

θ = 24.83°

now, difference

|θ_{650} -θ_{420}| =40.54°-24.83°

|θ_{650} -θ_{420}| =19.71°

8 0

4 years ago

Two small spheres with mass 10 gm and charge q, are suspended from a point by threads of length L=0.22m. What is the charge on e

Nataliya [291]

Answer:

$q=1.95*10^{-7}C$

Explanation:

According to the free-body diagram of the system, we have:

$\sum F_y: Tcos(15^\circ)-mg=0(1)\\\sum F_x: Tsin(15^\circ)-F_e=0(2)$

So, we can solve for T from (1):

$T=\frac{mg}{cos(15^\circ)}(3)$

Replacing (3) in (2):

$(\frac{mg}{cos(15^\circ)})sin(15^\circ)=F_e$

The electric force ( $F_e$ ) is given by the Coulomb's law. Recall that the charge q is the same in both spheres:

$(\frac{mg}{cos(15^\circ)})sin(15^\circ)=\frac{kq^2}{r^2}(4)$

According to pythagoras theorem, the distance of separation (r) of the spheres are given by:

$sin(15^\circ)=\frac{\frac{r}{2}}{L}\\r=2Lsin(15^\circ)(5)$

Finally, we replace (5) in (4) and solving for q:

$mgtan(15^\circ)=\frac{kq^2}{(2Lsin(15^\circ))^2}\\q=\sqrt{\frac{mgtan(15^\circ)(2Lsin(15^\circ))^2}{k}}\\q=\sqrt{\frac{10*10^{-3}kg(9.8\frac{m}{s^2})tan(15^\circ)(2(0.22m)sin(15^\circ))^2}{8.98*10^{9}\frac{N\cdot m^2}{C^2}}}\\q=1.95*10^{-7}C$

5 0

4 years ago

The wavelengths of terrestrial radiation from earth's surface are concentrated in _____.

Lady bird [3.3K]

The wavelengths of terrestrial radiation from earth's surface are concentrated in thermal infrared spectrum.

<u>Explanation:</u>

The wavelengths used in thermal infrared are in the range of micrometer ranging from 8 to 15. Terrestrial radiation is defined as the radiation that is originating from earth due to radioactive materials like uranium, thorium e.t.c.

Increase in terrestrial radiation by the use of radioactive materials will consequently increase global warming as it releases green house gases thereby degrading the quality of earth's atmosphere.

4 0

3 years ago