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ch4aika [34]
3 years ago
8

PLEASE HELPPPPPP <333​

Physics
2 answers:
Alexus [3.1K]3 years ago
8 0
I think the answer is c
zloy xaker [14]3 years ago
4 0

Answer:

Explanation:

The answer is C  i think

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How many atoms of carbon would two molecules of glucose (C6H12O6) have?
Sauron [17]
Well according to the molecular formula of glucose, one molecule would have 6 carbon atoms, and thus 2 molecules of glucose would have 12 carbon atoms.

The correct response would be B. 12.
4 0
3 years ago
Read 2 more answers
A light wave travels through air at a speed of 3.0x108 m/s. Green light has a wavelength of about 5.76x1014Hz. What is the wavel
solong [7]

Answer:

521 nm

Explanation:

Given the values and units we are given, I'm assuming  5.76*10^14 Hz is frequency.

The formula to use here is λ * υ = c, where λ is wavelength, υ is frequency, and c is the speed of light.

λ = \frac{3*10^8\frac{m}{s} }{5.76*10^{14}Hz} = {5.20833*10^{-7} m}\approx{521 *10^{-9}m}={521 nm}

4 0
3 years ago
PLEASE HELP ILL MARK BRAINLIEST
ikadub [295]
I think the answer is A?
7 0
3 years ago
A 60-W, 120-V light bulb and a 200-W, 120-V light bulb are connected in series across a 240-V line. Assume that the resistance o
gulaghasi [49]

A. 0.77 A

Using the relationship:

P=\frac{V^2}{R}

where P is the power, V is the voltage, and R the resistance, we can find the resistance of each bulb.

For the first light bulb, P = 60 W and V = 120 V, so the resistance is

R_1=\frac{V^2}{P}=\frac{(120 V)^2}{60 W}=240 \Omega

For the second light bulb, P = 200 W and V = 120 V, so the resistance is

R_1=\frac{V^2}{P}=\frac{(120 V)^2}{200 W}=72 \Omega

The two light bulbs are connected in series, so their equivalent resistance is

R=R_1 + R_2 = 240 \Omega + 72 \Omega =312 \Omega

The two light bulbs are connected to a voltage of

V  = 240 V

So we can find the current through the two bulbs by using Ohm's law:

I=\frac{V}{R}=\frac{240 V}{312 \Omega}=0.77 A

B. 142.3 W

The power dissipated in the first bulb is given by:

P_1=I^2 R_1

where

I = 0.77 A is the current

R_1 = 240 \Omega is the resistance of the bulb

Substituting numbers, we get

P_1 = (0.77 A)^2 (240 \Omega)=142.3 W

C. 42.7 W

The power dissipated in the second bulb is given by:

P_2=I^2 R_2

where

I = 0.77 A is the current

R_2 = 72 \Omega is the resistance of the bulb

Substituting numbers, we get

P_2 = (0.77 A)^2 (72 \Omega)=42.7 W

D. The 60-W bulb burns out very quickly

The power dissipated by the resistance of each light bulb is equal to:

P=\frac{E}{t}

where

E is the amount of energy dissipated

t is the time interval

From part B and C we see that the 60 W bulb dissipates more power (142.3 W) than the 200-W bulb (42.7 W). This means that the first bulb dissipates energy faster than the second bulb, so it also burns out faster.

7 0
3 years ago
4. A metal fork has a density of 7.8 g/cm ³. Is this fork made up of aluminum?
Morgarella [4.7K]
No, aluminum has a density near 2.7 g/cm^3 
<span>7.8 g/cm^3 is near the density of iron (or in the case of a fork, steel).
this is it

</span>
5 0
4 years ago
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