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Travka [436]
2 years ago
7

Someone help me please

Physics
1 answer:
Fiesta28 [93]2 years ago
5 0

Explanation:

the unknown resistance is 9ohm

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An object moving with a constant
jeka57 [31]

Answer:

Acceleration:

{ \tt{a =  \frac{v - u}{t} }} \\ { \tt{a =  \frac{20 - 10}{5} }} \\ { \tt{a = 2 \: m {s}^{ - 2} }}

From third equation:

{ \bf{ {v}^{2}  =  {u}^{2}  + 2as}} \\ { \tt{s =  \frac{ {20}^{2}  -  {10}^{2} }{2 \times 2} }} \\   = { \tt{s = 75 \: m}}

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2 years ago
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Collect information about working of Geyser and prepare a report
ikadub [295]
A geyser is actually a devise that coverts electrical energy into heat energy for heating up water. The heating element that is inside the geyser actually gets heated up and then in turn it heats the water in contact with it within the geyser. There is also a thermostat device within the geyser that cuts off the heating when the water temperature reaches the desired level. This helps in stopping of electrical energy loss. One inlet brings in cold water while another outlet gets rid of the hot water. When the temperature of the water falls below the desired level the heating is again started by the thermostat.



6 0
3 years ago
Constants A capacitor is connected across an ac source that has voltage amplitude 59 0 V and frequency 77 0 Hz Part C What is th
Vladimir79 [104]

Answer:

C = 1.77 \times 10^{-4} F

Explanation:

As we know that in AC circuit we have

V = i x_c

here we have

V = 59 V

i = 5.05 A

so we will have

x_c = \frac{59}{5.05}

x_c = 11.68 ohm

also we know that

x_c = \frac{1}{\omega C}

here we will have

11.68 = \frac{1}{(2\pi 77)C}

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2 years ago
A crate is placed on an adjustable, incline board. the coefficient of static friction between the crate and the board is 0.29.
sasho [114]

Let the angle be Θ (theta)

Let the mass of the crate be m.

a) When the crate just begins to slip. At that moment the net force will be equal to zero and the static friction will be at the maximum vale.

Normal force (N) = mg CosΘ

μ (coefficient of static friction) = 0.29

Static friction = μN = μmg CosΘ

Now, along the ramp, the equation of net force will be:

mg SinΘ - μmg CosΘ = 0

mg SinΘ = μmg CosΘ

tan Θ = μ

tan Θ = 0.29

Θ = 16.17°

b) Let the acceleration be a.

Coefficient of kinetic friction = μ = 0.26

Now, the equation of net force will be:

mg sinΘ - μ mg CosΘ = ma

a = g SinΘ - μg CosΘ

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a = 9.8 × 0.278 - 0.26 × 9.8 × 0.96

a = 2.7244 - 2.44608

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7 0
3 years ago
Write short letters.
Svetradugi [14.3K]

Answer:

No. 67

Peter Street

12th Road

Chennai

24th June 201_

Dear Amrish

I have come to know that since your school has closed for the Autumn Break you have plenty of free time at your disposal at the moment. I would like to tell you that even I am having holidays now.

It has been a long time since we have spent some time together. If you are free, I would welcome to have your company this weekend. Why don’t you come over to my house and spend a day or so with me?

I am anxiously waiting for your reply.

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