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3 years ago

12

PLEASE HELP ME SOMEONE

1 answer:

tensa zangetsu [6.8K]3 years ago

4 0

To convert<span> temperatures in degrees </span>Fahrenheit to Celsius<span>, subtract 32 and multiply by .5556 (or 5/9)
So your answer is D</span>

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Citrus2011 [14]

Answer:

net force is positive downward..B

6 0

2 years ago

Calculate the kinetic energy in joules of a ball of mass 40g moving at a velocity of 4 metres per second

AfilCa [17]

Given : A ball of mass 40 g moving at a velocity of 4 m/s.
To find : Calculate the kinetic energy in joules ?
Solution :
The kinetic energy formula is given by,

where, v is the velocity v=4 m/s
m is the mass m=40 g
Convert g into kg,

Substitute the values,

Therefore, the kinetic energy is 0.32 Joules.

6 0

1 year ago

Bx = -1.33 m and By = 2.81 m<br> Find the magnitude of the<br> vector.

NISA [10]

Answer:

Explanation:

The formula for the magnitude of a vector is

$B_{mag}=\sqrt{(-1.33)^2+(2.81)^2}$ and then round to the hundredths place:

3.11 m. Since we are in Q2, we can also find the direction of this vector:

$tan^{-1}(\frac{2.81}{-1.33})=-64.7$ but since we are in Q2, we add 180 degrees to the result, getting the angle to be 115.3

3 0

2 years ago

Which of the following is a<br> compound?<br> A. Sodium<br> B. Salt<br> C. Potassium<br> D. Argon

ELEN [110]

B. Salt

Hope it helpz~ uh..

7 0

2 years ago

Read 2 more answers

A 2 eV electron encounters a barrier 5.0 eV high and width a. What is the probability b) 0.5 that it will tunnel through the bar

denis23 [38]

Answer:

The tunnel probability for 0.5 nm and 1.00 nm are $5.45\times10^{-4}$ and $7.74\times10^{-8}$ respectively.

Explanation:

Given that,

Energy E = 2 eV

Barrier V₀= 5.0 eV

Width = 1.00 nm

We need to calculate the value of $\beta$

Using formula of $\beta$

$\beta=\sqrt{\dfrac{2m}{\dfrac{h}{2\pi}}(v_{0}-E)}$

Put the value into the formula

$\beta = \sqrt{\dfrac{2\times9.1\times10^{-31}}{(1.055\times10^{-34})^2}(5.0-2)\times1.6\times10^{-19}}$

$\beta=8.86\times10^{9}$

(a). We need to calculate the tunnel probability for width 0.5 nm

Using formula of tunnel barrier

$T=\dfrac{16E(V_{0}-E)}{V_{0}^2}e^{-2\beta a}$

Put the value into the formula

$T=\dfrac{16\times 2(5.0-2.0)}{5.0^2}e^{-2\times8.86\times10^{9}\times0.5\times10^{-9}}$

$T=5.45\times10^{-4}$

(b). We need to calculate the tunnel probability for width 1.00 nm

$T=\dfrac{16\times 2(5.0-2.0)}{5.0^2}e^{-2\times8.86\times10^{9}\times1.00\times10^{-9}}$

$T=7.74\times10^{-8}$

Hence, The tunnel probability for 0.5 nm and 1.00 nm are $5.45\times10^{-4}$ and $7.74\times10^{-8}$ respectively.

6 0

3 years ago