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Harlamova29_29 [7]
2 years ago
12

Two 10-N weights are hanging from two Springs A and B. Spring A has a greater spring constant and therefore will ____. a. have a

greater displacement compared to Spring B b. have a lower displacement compared to Spring B c. have the same displacement as Spring B d. none of the above
Physics
1 answer:
arsen [322]2 years ago
8 0
The spring should have a greater displacement .
Greater spring constant = the more difficult it is to stretch a spring , due to stiffness.
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A hockey player uses her hockey stick to exert a force of 6.81 N on a stationary hockey puck. The hockey puck has a mass of 165
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Answer:

41.3 m/s^2 option (e)

Explanation:

force, F = 6.81 N

mass, m = 165 g = 0.165 kg

Let a be the acceleration of the puck.

Use newtons' second law

Force = mass x acceleration

6.81 = 0.165 x a

a = 41.27 m/s^2

a = 41.3 m/s^2

Thus, the acceleration of the puck is 41.3 m/s^2.

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A student constructed a series circuit consisting of a 12.0-volt battery, a 10.0-ohm lamp, and a
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The power dissipated across a component can be calculated through the formula P=I^2xR

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1) A pendulum is configured to have a period of 2 seconds.
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Calculate the electric field at the center of a square
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Answer:

E_y=1175510.2\ N.C^{-1}

The Magnitude of electric field is in the upward direction as shown directly towards the charge q_1.

Explanation:

Given:

  • side of a square, a=52.5\ cm
  • charge on one corner of the square, q_1=+45\times 10^{-6}\ C
  • charge on the remaining 3 corners of the square,q_2=q_3=q_4=-27\times 10^{-6}\ C

<u>Distance of the center from each corners</u>=\frac{1}{2} \times diagonals

diagonal=\sqrt{52.5^2+52.5^2}

diagonal=74.25\cm=0.7425\ m

∴Distance of center from corners, b=0.3712\ m

Now, electric field due to charges is given as:

E=\frac{1}{4\pi\epsilon_0}\times \frac{q}{b^2}

<u>For charge q_1 we have the field lines emerging out of the charge since it is positively charged:</u>

E_1=9\times 10^9\times \frac{45\times 10^{-6}}{0.3712^2}

  • E_1=2938775.5\ N.C^{-1}

<u>Force by each of the charges at the remaining corners:</u>

E_2=E_3=E_4=9\times 10^9\times \frac{27\times 10^{-6}}{0.3712^2}

  • E_2=E_3=E_4=1763265.3\ N.C^{-1}

<u> Now, net electric field in the vertical direction:</u>

E_y=E_1-E_4

E_y=1175510.2\ N.C^{-1}

<u>Now, net electric field in the horizontal direction:</u>

E_y=E_2-E_3

E_y=0\ N.C^{-1}

So the Magnitude of electric field is in the upward direction as shown directly towards the charge q_1.

8 0
3 years ago
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