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Yuki888 [10]
2 years ago
15

A resistor has a resistance of 120 12. How much current is in the resistor if there is a potential difference of 12.0V across th

e resistor?
A. 160 A
B. 0.43 A
C. 0.10 A
D. 2.3 A
Physics
1 answer:
Vaselesa [24]2 years ago
4 0

Answer:

c 0.10

Explanation:

V=IR

12.0V=I x 120.12ohms

I=0.099

I= 0.10

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A car driving at 20 m/s accelerates continuously 2m/s2​ ​ for 3 seconds. What is its final velocity
Lesechka [4]

Answer: V= u+ at

V= final velocity

u=initial velocity

a=acceleration

t=time taken

V= 20 + 2*3

V= 26m/s

Explanation:

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2 years ago
We can model a pine tree in the forest as having a compact canopy at the top of a relatively bare trunk. Wind blowing on the top
frez [133]

We need to consider for this exercise the concept Drag Force and Torque. The equation of Drag force is

F_D = c_D A \frac{\rho V^2}{2}

Where,

F_D = Drag Force

c_D = Drag coefficient

A = Area

\rho= Density

V = Velocity

Our values are given by,

c_D = 0.5 (That is proper of a cone-shape)

A = 9m^2

\rho = 1.2Kg/m^3

V = 6.5m/s

Part A ) Replacing our values,

F_D = 0.5*9*\frac{1.2*6.5^2}{2}

F_D = 114.075N

Part B ) To find the torque we apply the equation as follow,

\tau = F*d

\tau = (114.075N)(7)

\tau = 798.525N.m

3 0
3 years ago
Un soldado de 1.80 m de altura ha sido herido por una bala, cuya masa es de 100 g. Deciden colocar al soldado en una centrifugad
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3 0
3 years ago
The membrane that surrounds a certain type of living cell has a surface area of 4.3 x 10-9 m2 and a thickness of 1.1 x 10-8 m. A
Nadusha1986 [10]

Answer:

The charge resides on the outer surface = 1.245 \times 10^{-12} C

Explanation:

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Separation between two plate  (d) = 1.1 \times 10^{-8}  m  

Dielectric constant (k) = 4.2

Potential difference (\Delta V) = 85.7 \times 10^{-3} V

The capacitance of parallel plate capacitor in free space is given by,

           C = \frac{\epsilon_{o} A }{d}

Where \epsilon_{o}  = permittivity of free space = 8.85 \times 10^{-12}

The Capacitance of capacitor is increase by k times when it placed in dielectric medium.

C_{dielectric}  = \frac{k \epsilon_{o} A }{d}

And we know that, C = \frac{Q}{ \Delta V}

So charge on the outer surface is given by,

      Q = \frac{k \epsilon A \Delta V }{d}

      Q = \frac{4.2 \times 4.3 \times 10^{-9} \times 8.85 \times 10^{-12} \times 85.7\times 10^{-3}   }{1.1 \times 10^{-8} }

      Q = 1.245 \times 10^{-12}

3 0
3 years ago
A particle that carries a net charge of -41.8 μc is held in a region of constant, uniform electric field. the electric field vec
miss Akunina [59]
The total work done by the electric field on the charge is given by the scalar product between the electric force acting on the charge and the displacement of the charge:
W=F d cos \theta
where the force is F=qE, d=0.556 and \theta=55.2^{\circ}. Using the value of q and E given by the problem, we find
W=qEdcos\theta = 6.39\cdot10^{-5}J
3 0
3 years ago
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