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3 years ago

Can someone help me answer this

Physics

2 answers:

netineya [11]3 years ago

6 0

Answer: b is sedimentary. c is metamorphic. and a is igneous.

Explanation:

bogdanovich [222]3 years ago

5 0

A is igneous B is Setitarian or whatever its called and C is the other thing im not sure tho

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Vector sum of 15 m, East and 9 m, West is 24 m, East. True False

Firlakuza [10]

Answer:

False

Explanation:

Because when you go through east

( +x axis ) then you go to west ( -x axis )

You will subtract -9 from +15

it's become +6

( I talk about the displacement not distance) ( West = - East )

I hope that it's a clear ") .

7 0

3 years ago

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The amount of heat energy required to raise the temperature of a unit mass of a material one degree is

charle [14.2K]

The amount of heat energy required to raise the temperature of a unit mass of a material to one degree is called D. its heat capacity.

The relationship of the heat when applied to the object and the change in temperature of the object when heat is being applied is directly proportional to each other. This means that when heat is applied to the object, the temperature of the object increases and when heat is not applied to the object, the temperature of the object decreases.

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2 years ago

Which county in Florida is most in need of safe rooms and hurricane ties?

Murrr4er [49]

Probably Pasco. Lol.

8 0

3 years ago

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what is the energy (in j) of a photon required to excite an electron from n = 2 to n = 8 in a he⁺ ion? submit an answer to three

grin007 [14]

Answer:

Approximately $5.11 \times 10^{-19}\; {\rm J}$ .

Explanation:

Since the result needs to be accurate to three significant figures, keep at least four significant figures in the calculations.

Look up the Rydberg constant for hydrogen: $R_{\text{H}} \approx 1.0968\times 10^{7}\; {\rm m^{-1}$ .

Look up the speed of light in vacuum: $c \approx 2.9979 \times 10^{8}\; {\rm m \cdot s^{-1}}$ .

Look up Planck's constant: $h \approx 6.6261 \times 10^{-34}\; {\rm J \cdot s}$ .

Apply the Rydberg formula to find the wavelength $\lambda$ (in vacuum) of the photon in question:

$\begin{aligned}\frac{1}{\lambda} &= R_{\text{H}} \, \left(\frac{1}{{n_{1}}^{2}} - \frac{1}{{n_{2}}^{2}}\right)\end{aligned}$ .

The frequency of that photon would be:

$\begin{aligned}f &= \frac{c}{\lambda}\end{aligned}$ .

Combine this expression with the Rydberg formula to find the frequency of this photon:

$\begin{aligned}f &= \frac{c}{\lambda} \\ &= c\, \left(\frac{1}{\lambda}\right) \\ &= c\, \left(R_{\text{H}}\, \left(\frac{1}{{n_{1}}^{2}} - \frac{1}{{n_{2}}^{2}}\right)\right) \\ &\approx (2.9979 \times 10^{8}\; {\rm m \cdot s^{-1}}) \\ &\quad \times (1.0968 \times 10^{7}\; {\rm m^{-1}}) \times \left(\frac{1}{2^{2}} - \frac{1}{8^{2}}\right)\\ &\approx 7.7065 \times 10^{14}\; {\rm s^{-1}} \end{aligned}$ .

Apply the Einstein-Planck equation to find the energy of this photon:

$\begin{aligned}E &= h\, f \\ &\approx (6.6261 \times 10^{-34}\; {\rm J \cdot s}) \times (7.7065 \times 10^{14}\; {\rm s^{-1}) \\ &\approx 5.11 \times 10^{-19}\; {\rm J}\end{aligned}$ .

(Rounded to three significant figures.)

6 0

2 years ago

A 7.00-cm-long wire is pulled along a U-shaped conducting rail in a perpendicular magnetic field. The total resistance of the wi

Anvisha [2.4K]

Answer:

(a) 1.075 N

(b) 4.254 T

Explanation:

(a)

From the question,

Power = Force×Velocity.

P = Fv................... Equation 1

Where P = Power dissipated in the circuit, F = Pulling force of the wire, v = speed of the wire.

make F the subject of the equation

F = P/v................. Equation 2

Given: P = 4.30 W, v = 4.0 m/s.

Substitute into equation 2

F = 4.30/4

F = 1.075 N.

(b)

Applying,

F = BILsinФ.............. Equation 2

Where B = Strength of the magnetic field, I = current in the wire, L = Length of the wire, Ф = angle between the wire and the magnetic field.

make B the subject of the equation

B = F/ILsinФ................... Equation 3

But,

P = I²R...................... Equation 4

Where R = resistance of the wire.

make I the subject of the equation

I = √(P/R)............... Equation 5

Given: P = 4.30 W, R = 0.330 Ω

Substitute into equation 5

I = √(4.3/0.33)

I = √13.03

I = 3.61 A.

Also given: L = 7 cm = 0.07 m, Ф = 90°

Substitute into equation 3

B = 1.075/(0.07×3.61×sin90)

B = 1.075/0.2527

B = 4.254 T.

4 0

3 years ago