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Rina8888 [55]
2 years ago
12

HELP ASAP!! The voltage from a power supply to a light bulb is increased by 2 volts and the current is recorded. A graph of this

data is
shown. Explain in detail the relationship between voltage and current

Physics
1 answer:
uysha [10]2 years ago
3 0

Answer:

It is a resistor since as u increases v increases so they are directly propotional

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Select the correct answer.
r-ruslan [8.4K]

Answer:

That would be B. Hope this helps!

Explanation:

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3 years ago
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Out of aluminum,copper,steel,and glass.Which material do you think will be the best thermal conductor?
Nastasia [14]
I believe it is copper

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3 years ago
What is the half-reaction that occurs at the cathode during the electrolysis of molten potassium bromide?
polet [3.4K]

The complete ionization of KBr into its constituents is:<span>
<span>KBr (s)  --->  K+ (aq)  +  Br- (aq)</span></span>

<span>
During electrolysis, oxidation takes place at the anode electrode. This means that an ion is stripped off its electron hence becoming more positive:
<span>2 Br- (aq)  --->  Br2 (g) + 2e- </span></span>

We can see that Bromine gas Br2 is evolved at the anode. 

<span>
<span>Meanwhile at the cathode, the reduction reaction occurs. Which means that the electron from the anode electrode is used to make an ion more negative:
<span>2K+ (aq)  +  2e-  --->  2K (s) </span></span>
Hence, through reduction, solid potassium is deposited on the plate.</span>

 

 

Half reactions:

<span>Anode: 2 Br- (aq)  --->  Br2 (g) + 2e- </span>                       

<span>Cathode: 2K+ (aq)  +  2e-  --->  2K (s) </span>

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2 years ago
What does the Greek word gymno mean?
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The Green work gymno means Naked
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A pitcher throws a 0.140 kg baseball, and it approaches the bat at a speed of 35.0 m/s. The bat does Wnc = 75.0 J of work on the
Eva8 [605]

Answer:

The speed of the ball is 42.5 m/s

Explanation:

The initial kinetic energy of the ball is:

K_1=\frac{1}{2} m v_0^2=\frac{1}{2}*0.140 kg*(35.0 m/s)^2= 85.75 J

The speed of the ball after leaving the bat is:

K_2=K_1+W_{nc}\\ \frac{1}{2}mV^2= 85.75 J + 75 J\\ (\frac{1}{2}mV^2)2=( 160.75 J)2\\ mV^2= 321.5 J\\ V^2= \frac{321.5 J}{0.140kg} \\ V=\sqrt{\frac{321.5 J}{0.140kg}}

V=47.92 m/s

Using kinematic equation we can find the speed of the ball after being 25 m above the point of collision:

V_f^2-V^2=-2gh

V_f^2-(47.92 m/s)^2=-2*9.81m/s^2*25m

V_f^2=-2*9.81m/s^2*25m+(47.92 m/s)^2

V_f=\sqrt{-2*9.81m/s^2*25m+(47.92 m/s)^2}

V_f=42.5m/s

3 0
2 years ago
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