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kiruha [24]
3 years ago
10

Help!!! Line B touches the circle at a single point. Line A extends through the center of the circle.

Physics
1 answer:
Ulleksa [173]3 years ago
3 0

Answer:

If I understand correctly. Line B is parallel to the circle. Also, the angle is less than 90.

  1. The size of the circle determines.
  2. The diameter should not be fixed either.
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How much total energy is dissipated in 10. seconds
noname [10]

Answer : Total energy dissipated is 10 J

Explanation :

It is given that,

Time. t = 10 s

Resistance of the resistors, R = 4-ohm

Current, I = 0.5 A

Power used is given by :

P=\dfrac{E}{t}

Where

E is the energy dissipated.

So, E = P t.............(1)

Since, P=I^2R

So equation (1) becomes :

E=I^2Rt

E=(0.5\ A)^2\times 4\Omega \times 10\ s

E=10\ J

So, the correct option is (3)

Hence, this is the required solution.

7 0
3 years ago
Read 2 more answers
A 1060-kg car moving west at 16 m/s collides with and locks onto a 1830-kg stationary car. determine the velocity of the cars ju
iren [92.7K]
M1U1 + M2V2 = (M1+M2)V, where M1 is the mass of the moving car, M2 is the mass of the stationary car, U1 is the initial velocity, and V is the common velocity after collision.
therefore; 
(1060× 16) + (1830 ×0) = (1060 +1830) V
 16960 = 2890 V
      V = 5.869 m/s
The velocity of the cars after collision will be 5.689 m/s
5 0
3 years ago
The apparent weight of a student in alift is 564N . if the mass of the student is 60.3kg, what is the acceleration of the lift ?
Yuki888 [10]

Answer:

-.457 m/s^2

Explanation:

Actual weight =   60 .3 (9.81) = 591.54 N

Accel of lift changes this to    60.3 ( 9.81 - L)     where L - accel of lift

                                           60.3 ( 9.81 - L ) = 564

                                               solve for L = .457 m/s^2  DOWNWARD

                                                        so L = - .457 m/s^2

4 0
2 years ago
What is true for two pieces of iron at the same temperature?
Lubov Fominskaja [6]
D.the average kinetic energy of their particles is the same
3 0
3 years ago
Read 2 more answers
A simple pendulum consists of a 1.00-kg bob on a string 1.00 m long. During a time interval of 25.0 s , the maximum angle this p
yuradex [85]

Answer:

The time constant is 76.92 sec.

Explanation:

Given that,

Mass of bob = 1.00 kg

String length=1.00 m

Time = 25.0 s

Suppose we need to calculate the numerical value of the time constant τ.

We need to calculate the numerical value of the time constant τ

Using formula of time constant

A=A_{0}e^{\dfrac{-t}{2\tau}}

Put the value into the formula

5.10=6.00e^{\dfrac{-25.0}{2\times\tau}}

e^{\dfrac{-25}{2\times\tau}}=\dfrac{5.10}{6.00}

e^{\dfrac{-25}{2\times\tau}}=0.85

\dfrac{25}{2\tau}=0.1625

\dfrac{1}{\tau}=\dfrac{0.1625\times2}{25}

\dfrac{1}{\tau}=0.013

\tau=\dfrac{1}{0.013}

\tau=76.92\ sec

Hence, The time constant is 76.92 sec.

7 0
3 years ago
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