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pogonyaev
3 years ago
5

On a playground, there is a merry‑go‑round. In order to get it moving, Bonnie applies a force of 31 N31 N . The merry‑go‑round m

easures 8.5 m8.5 m from the axis of rotation to the edge where Bonnie applies her force. Assuming she applies her force perpendicularly to a line drawn from the axis of rotation, what is the magnitude of the torque ????τ Bonnie imparts to the merry‑go‑round?
Physics
1 answer:
nika2105 [10]3 years ago
4 0

Answer:

The magnitude of the torque is 263.5 N.

Explanation:

Given that,

Applied force = 31 N

Distance from the axis = 8.5 m

She applies her force perpendicularly to a line drawn from the axis of rotation

So, The angle is 90°

We need to calculate the torque

Using formula of torque

\tau=Fd\sin\theta

Where, F = force

d = distance

Put the value into the formula

\tau=31\times8.5\sin90

\tau= 263.5\ N

Hence, The magnitude of the torque is 263.5 N.

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A series RC circuit contains two resistors and two capacitors. The resistors are 39 ohms and 56 ohms. The capacitors have capaci
Nataly [62]

Answer:

The  correct option is b

Explanation:

From the question we are told that

    The resistance of the first resistor is  R_1 =  39 \ \Omega

     The resistance of the second resistor is  R_2 =  56 \  \Omega

      The capacitive reactance of the first capacitor is jX_{c_1 }  =  80 \ \Omega

       The capacitive reactance of the first capacitor is jX_{c_2 }  =  40 \ \Omega

Generally given that the resistors are connected in parallel , their equivalent resistance is  

            R_e =  R_1 +R_2

=>         R_e = 39 + 56

=>         R_e = 95 \  \Omega

Generally given that the capacitors are connected in parallel , their equivalent capacitive reactance  is  

          jX_e =  jX_{c_1} +  jX_{c_2}

=>       jX_e = 80 + 40

=>       jX_e = 120

Hence the impedance of the circuit is  

         Z =  R_e - jX_e        

=>      Z = 95  -  j120          

Generally from the impedance equation , the total used resistance is  

       R_e =  95 \  \Omega

3 0
3 years ago
Take one jar with salt water and another with pure water .put an egg into both jars. In which the jar egg floated over the liqui
DanielleElmas [232]

Answer:

if the which coduct salt has more density so the egg floats .

Explanation:

8 0
3 years ago
Rank the pressures from highest to lowest:
n200080 [17]

Answer:

P₃ > P₁ > P₂

Explanation:

To rank pressure of the given situation

a) we know

  Pressure at height h below

      P = ρ g h

density of salt water, ρ = 1029 kg/m³

      P₁ = 1029 x 10 x 0.2

      P₁ = 2058 Pa

b) density of fresh water, ρ = 1000 kg/m³

      P₂ = 1000 x 10 x 0.2

      P₂ = 2000 Pa

c) density of mercury, ρ = 13593 kg/m³

      P₃ = 13593 x 10 x 0.05

      P₃ = 6796.5 Pa

Rank of Pressures from highest to lowest

        P₃ > P₁ > P₂

3 0
3 years ago
Need help to get this question right!!
Charra [1.4K]

Answer:

16 times as strong

Explanation:

From the question given above, the following assumptions were made:

Initial Force (F₁) = F

Initial distance apart (r₁) = r

Final distance apart (r₂) = ¼r

Final force (F₂) =?

Next, we shall obtain a relationship between the force and the distance apart. This can be obtained as follow:

F = GM₁M₂ / r²

Cross multiply

Fr² = GM₁M₂

If G, M₁ and M₂ are kept constant, then,

F₁r₁² = F₂r₂²

Finally, we determine the new force as follow:

Initial Force (F₁) = F

Initial distance apart (r₁) = r

Final distance apart (r₂) = ¼r

Final force (F₂) =?

Fr² = F₂ × (¼r)²

Fr² = F₂ × r²/16

Fr² = F₂r² / 16

Cross multiply

16Fr² = F₂r²

Divide both side by r²

F₂ = 16Fr² / r²

F₂ = 16F

From the calculations made above, we can see that the new force is 16 times the original force.

Thus, the new force is 16 times stronger.

8 0
3 years ago
Sean, after being so happy for two full days that he reported he "never needed much sleep," now is stating he is so sad that he
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D or A i'm not that sure
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