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3 years ago

15

A 0.40-μF capacitor is connected to a 5.0-V battery. How much charge is on each plate of the capacitor? Express your answer usin

g two significant figures.

1 answer:

weqwewe [10]3 years ago

7 0

Answer:

$Q=2.0\times10^{-6}C$

Explanation:

The equation of charge on the capacitor given its capacitance and a voltage applied to it is Q=CV.

Our capacitance is expressed in <em>microfarads</em>, where micro stands for $\times10^{-6}$ , so in S.I. we sill have:

$Q=CV=(0.4\times10^{-6}F)(5V)=0.000002C=2.0\times10^{-6}C$

Where the results has been expresed with <em>two significant figures</em>.

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Two blocks with masses 1 and 2 are connected by a massless string that passes over a massless pulley as shown. 1 has a mass of 2

Bess [88]

Answer:

The acceleration of $M_2$ is $a = 0.7156 m/s^2$

Explanation:

From the question we are told that

The mass of first block is $M_1 = 2.25 \ kg$

The angle of inclination of first block is $\theta _1 = 43.5^o$

The coefficient of kinetic friction of the first block is $\mu_1 = 0.205$

The mass of the second block is $M_2 = 5.45 \ kg$

The angle of inclination of the second block is $\theta _2 = 32.5^o$

The coefficient of kinetic friction of the second block is $\mu _2 = 0.105$

The acceleration of $M_1 \ and\ M_2$ are same

The force acting on the mass $M_1$ is mathematically represented as

$F_1 = T - M_1gsin \theta_1 - \mu_1 M_1 g cos\theta_1$

=> $M_1 a = T - M_1gsin \theta_1 - \mu_1 M_1 g cos\theta_1$

Where T is the tension on the rope

The force acting on the mass $M_2$ is mathematically represented as

$F_2 = M_2gsin \theta_2 - T -\mu_2 M_2 g cos\theta_2$

$M_2 a = M_2gsin \theta_2 - T -\mu_2 M_2 g cos\theta_2$

At equilibrium

$F_1 = F_2$

So

$T - M_1gsin \theta_1 - \mu_1 M_1 g cos\theta_1 =M_2gsin \theta_2 - T -\mu_2 M_2 g cos\theta_2$

making a the subject of the formula

$a = \frac{M_2 g sin \theta_2 - M_1 g sin \theta_1 - \mu_1 M_1g cos \theta - \mu_2 M_2 g cos \theta_2 }{M_1 +M_2}$

substituting values $a = \frac{(5.45) (9.8) sin (32.5) - (2.25) (9.8) sin (43.5) - (0.205)*(2.25) *9.8cos (43.5) - (0.105)*(5.45) *(9.8) cos(32.5) }{2.25 +5.45}$

=> $a = 0.7156 m/s^2$

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4 years ago

What is the relationship between sunspots and geomagnetic storms?

shutvik [7]

Sunspots are spot like structure on the surface of sun, they are created due to geomagnetic flux which inhibits the convection of heat waves thus reducing the temperature of spot with some factor

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3 years ago

On a distant planet, a rock falls in 48.4 s from the top of a 1.10e+02 m cliff to the planet surface below. What is the accelera

pav-90 [236]

this can be solve using the formala of free fall

t = sqrt( 2y/ g)

where t is the time of fall

y is the height

g is the acceleration due to gravity

48.4 s = sqrt (2 (1.10e+02 m)/ g)

G = 0.0930 m/s2

The velocity at impact

V = sqrt(2gy)

= sqrt( 2 ( 0.0930 m/s2)( 1.10e+02 m)

V = 4.523 m/s

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If revenue is $3,000 and operating expenses are $4,000, cash flow equals _____. -$1,000 $1,000 -$7,000 $7,000

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Answer:

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Explanation:

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jarptica [38.1K]

Answer:

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Explanation:

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3 years ago