All categories

3 years ago

Can someone help please

Physics

2 answers:

sesenic [268]3 years ago

7 0

Answer:

Third Side = 24 units.

Explanation:

Let the third side be 'm'.

Use Pythagoras theorem,

x² + 10² = 26²

x² = 676 - 100

x² = 576

x = 24 units

love history [14]3 years ago

4 0

Answer:

The correct answer is 24.

Explanation:

You might be interested in

Glass is transparent to visibile light under normal conditions; however, at extremely high intensities, glass will absorb most o

8_murik_8 [283]

Answer:

3 photons

Explanation:

The energy of a photon E can be calculated using this formula:

$E=\frac{hc}{\lambda}$

Where $h$ corresponds to Plank constant (6.626070x10^-34Js), $c$ is the speed of light in the vacuum (299792458m/s) and $\lambda$ is the wavelength of the photon(in this case 800nm).

$E=\frac{hc}{\lambda}=\frac{(6.626070\times10^{-34})(299792458)}{800\times10^{-9}}=\frac{1.986445812\times10^-25}{800}=2.483057265\times10^{-19}J$

Tranform the units

$1eV=1.602176634\times10^{-19}J\\2.483057265\times10^{-19}J(\frac{1eV}{1.602176634\times10^{-19}J})=1.549802445eV$

The band Gap is 4eV, divide the band gap between the energy of the photon:

$\frac{4ev}{1.549802445eV}=2.508974118$

Rounding to the next integrer: 3.

Three photons are the minimum to equal or exceed the band gap.

4 0

3 years ago

Which is a characteristic of the part of the atom marked "A"?Which is a characteristic of the part of the atom marked "A"?Which

WITCHER [35]

Here's link to the answer:

tinyurl.com/wpazsebu

4 0

3 years ago

Read 2 more answers

Select the correct answer. Which healthy snack can provide protein after physical activity? A. an orange slice B. pretzels C. yo

givi [52]

Answer:

From the given choices, the healthy snack that can provide protein after physical activity is the third option, yogurt. Yogurts are derived from fermentation of raw materials by the healthy bacteria, usually of lactobacillus origin. Other items in the choices are not able to provide protein.

Explanation:

4 0

3 years ago

A helium nucleus (charge = 2e, mass = 6.63 10-27 kg) traveling at 6.20 105 m/s enters an electric field, traveling from point ci

MA_775_DIABLO [31]

Answer:

$v_B=3.78\times 10^5\ m/s$

Explanation:

It is given that,

Charge on helium nucleus is 2e and its mass is $6.63\times 10^{-27}\ kg$

Speed of nucleus at A is $v_A=6.2\times 10^5\ m/s$

Potential at point A, $V_A=1.5\times 10^3\ V$

Potential at point B, $V_B=4\times 10^3\ V$

We need to find the speed at point B on the circle. It is based on the concept of conservation of energy such that :

increase in kinetic energy = increase in potential×charge

$\dfrac{1}{2}m(v_A^2-v_B^2)=(V_B-V_A)q\\\\\dfrac{1}{2}m(v_A^2-v_B^2)={(4\times 10^3-1.5\times 10^3)}\times 2\times 1.6\times 10^{-19}=8\times 10^{-16}\\\\v_A^2-v_B^2=\dfrac{2\times 8\times 10^{-16}}{6.63\times 10^{-27}}\\\\v_A^2-v_B^2=2.41\times 10^{11}\\\\v_B^2=(6.2\times 10^5)^2-2.41\times 10^{11}\\\\v_B=3.78\times 10^5\ m/s$

So, the speed at point B is $3.78\times 10^5\ m/s$ .

7 0

3 years ago

What are the conditions to increase the stability of a body?

cricket20 [7]

Training everyday
..........

5 0

3 years ago