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3 years ago

What is the name for a quantity that has both magnitude and direction

Physics

1 answer:

matrenka [14]3 years ago

8 0

Answer:

Vector

Explanation:

Vector, in physics, a quantity that has both magnitude and direction. It is typically represented by an arrow whose direction is the same as that of the quantity and whose length is proportional to the quantity's magnitude.

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After school, several students skateboard at the nearby park. Your best friend, Erwin, is a good skateboarder and happens to fal

Alchen [17]

Answer:

A person who never gives up.

Explanation:

due to his passion for skateboarding he try's his never gives up until he will finally learns the trick.

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3 years ago

Unit Test Review

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Answer:

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2 years ago

A clothes dryer in a home draws a current of 10 amps when connected on a special 220-volts household circuit.what is the resista

aliya0001 [1]

Answer:

22Ω

Explanation:

if V ⇒ voltage

I ⇒ current

R ⇒ resistance

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2 years ago

Mrs. LaCross leaves school and accidentally leaves her coffee mug on the roof of her car as shown in the picture below.

aleksandrvk [35]

Answer:

When she stops at a fast pace the energy and wind will take the cup forward and it will most likeley brake

Explanation:

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3 years ago

Use the Bohr model to address this question. When a hydrogen atom makes a transition from the 5 th energy level to the 2nd, coun

iris [78.8K]

Answer:

A. 2.82 eV

B. 439nm

C. 59.5 angstroms

Explanation:

A. To calculate the energy of the photon emitted you use the following formula:

$E_{n1,n2}=-13.4(\frac{1}{n_2^2}-\frac{1}{n_1^2})$ (1)

n1: final state = 5

n2: initial state = 2

Where the energy is electron volts. You replace the values of n1 and n2 in the equation (1):

$E_{5,2}=-13.6(\frac{1}{5^2}-\frac{1}{2^2})=2.82eV$

B. The energy of the emitted photon is given by the following formula:

$E=h\frac{c}{\lambda}$ (2)

h: Planck's constant = 6.62*10^{-34} kgm^2/s

c: speed of light = 3*10^8 m/s

λ: wavelength of the photon

You first convert the energy from eV to J:

$2.82eV*\frac{1J}{6.242*10^{18}eV}=4.517*10^{-19}J$

Next, you use the equation (2) and solve for λ:

$\lambda=\frac{hc}{E}=\frac{(6.62*10^{-34} kg m^2/s)(3*10^8m/s)}{4.517*10^{-19}J}=4.39*10^{-7}m=439*10^{-9}m=439nm$

C. The radius of the orbit is given by:

$r_n=n^2a_o$ (3)

where ao is the Bohr's radius = 2.380 Angstroms

You use the equation (3) with n=5:

$r_5=5^2(2.380)=59.5$

hence, the radius of the atom in its 5-th state is 59.5 anstrongs

8 0

3 years ago