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3 years ago

14

You have a 78.7 mF capacitor initially charged to a potential difference of 11.5 V. You discharge the capacitor through a 3.03 Ω

resistor. What is the time constant?

1 answer:

Aloiza [94]3 years ago

8 0

Answer:

$\tau=0.23\;second$

Explanation:

Given,

$C=78.7\;mF\\V=11.5\;V\\R=3.03\;\Omega\\$

Time constant

$\tau=RC\\\tau=78.7\times10^{-3}\times3.03\\\tau=238.461\times10^{-3}\;second\\\tau=0.23\;second$

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What is 300N divided by 25 kg?

snow_lady [41]

Answer:

Explanation:

it is acceleration because we know that

F=ma

a=F/m

a=300N/25 kg

a=12 m/s^2

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2 years ago

When an object is placed 110 cm from a diverging thin lens, its image is found to be 55 cm from the lens. The lens is removed, a

Kazeer [188]

Explanation:

Given that,

Object distance u= -110 cm

Image distance v= 55 cm

We need to calculate the focal length for diverging lens

Using formula of lens

$\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}$

Put the value into the formula

$\dfrac{1}{-f}=\dfrac{1}{55}-\dfrac{1}{-110}$

$\dfrac{1}{f}=-\dfrac{3}{110}$

$f=-36.6\ cm$

The focal length of the diverging lens is 36.6 cm.

Now given a thin lens with same magnitude of focal length 36.6 cm is replaced.

Here, The object distance is again the same.

We need to calculate the image distance for converging lens

Using formula of lens

$\dfrac{1}{36.6}=\dfrac{1}{v}-\dfrac{1}{-110}$

Here, focal length is positive for converging lens

$\dfrac{1}{v}=\dfrac{1}{36.6}-\dfrac{1}{110}$

$\dfrac{1}{v}=\dfrac{367}{20130}$

$v=54.85\ cm$

The distance of the image is 54.85 cm from converging lens.

Hence, This is the required solution.

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2 years ago

Reflection off of a smooth surface like a mirror is an example of diffuse reflection.

Akimi4 [234]

Answer:

true

Explanation:

Reflection is when light bounces off an object. If the surface is smooth and shiny, like glass, water or polished metal, the light will reflect at the same angle as it hit the surface. ... This is called diffuse reflection. Diffuse reflection is when light hits an object and reflects in lots of different directions.

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2 years ago

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Answer the following questions

zaharov [31]

Answer:

9 - 10N to the left

10 - There is no change on the object

Explanation:

Can I have brainliest answer pls?

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2 years ago

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Last night Mookie Betts hit a baseball at 32.5 m/s at a 45° angle. Betts

podryga [215]

Answer:

a) Since the height of the baseball at 99 m was 8.93 m and the fence at that distance is 3m tall, the hit was a home run.

b) The total distance traveled by the baseball was 108.7 m.

Explanation:

a) To know if the hit was a home run we need to calculate the height of the ball at 99 m:

$y_{f} = y_{0} + v_{0_{y}}t - \frac{1}{2}gt^{2}$

Where:

$y_{f}$ : is the final height =?

$y_{0}$ : is the initial height = 1 m

$v_{0_{y}$ : is the initial vertical velocity = v₀sin(45)

v₀: is the initial velocity = 32.5 m/s

g: is the gravity = 9.81 m/s²

t: is the time

First, we need to find the time by using the following equation:

$t = \frac{x}{v_{0_{x}}} = \frac{99 m}{32.5 m/s*cos(45)} = 4.31 s$

Now, the height is:

$y_{f} = y_{0} + v_{0_{y}}t - \frac{1}{2}gt^{2} = 1m + 32.5 m/s*sin(45)*4.31 s - \frac{1}{2}9.81 m/s^{2}*(4.31 s)^{2} = 8.93 m$

Since the height of the baseball at 99 m was 8.93 m and the fence at that distance is 3m tall, the hit was a home run.

b) To find the distance traveled by the baseball first we need to find the time of flight:

$y_{f} = y_{0} + v_{0_{y}}t - \frac{1}{2}gt^{2}$

$0 = 1 m + 32.5m/s*sin(45)t - \frac{1}{2}9.81 m/s^{2}t^{2}$

$1 m + 32.5m/s*sin(45)t - \frac{1}{2}9.81 m/s^{2}t^{2} = 0$

By solving the above quadratic equation we have:

t = 4.73 s

Finally, with that time we can find the distance traveled by the baseball:

$x = v_{0_{x}}*t = 32.5 m/s*cos(45)*4.73 s = 108.7 m$

Hence, the total distance traveled by the baseball was 108.7 m.

I hope it helps you!

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2 years ago