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ziro4ka [17]
3 years ago
9

Explain why a car driving around a curve at a constant speed is accelerating.

Physics
1 answer:
GREYUIT [131]3 years ago
7 0
But the fact is that an accelerating object is an object that is changing it’s velocity.. for this reason , it can be safely concluded that an object moving in a circle at constant speed is indeed accelerating. It is accelerating because the direction of the velocity vector is changing .
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A circuit contains two light bulbs connected in parallel. What would happen to the brightness of each light bulb if two more lig
laila [671]

Answer;

C. The brightness of each bulb would remain the same even though the total resistance of the circuit would decrease.

Explanation;

-If light bulbs are connected in parallel to a voltage source, the brightness of the individual bulbs remains more-or-less constant as more and more bulbs are added to the circuit.

-The current increases as more bulbs are added to the circuit and the overall resistance decreases. In addition, if one bulb is removed from the circuit the other bulbs do not go out. Each bulb is independently linked to the voltage source

6 0
3 years ago
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А
bagirrra123 [75]

Explanation:

The reading on the scale is

W = m(g + a)

= (77 kg)(9.8 m/s^2 + 2 m/s^2)

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7 0
3 years ago
Calculate the electric field at the center of a square
pantera1 [17]

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
Can someone help me with the exercise 3 only c,d,e please!!!
saul85 [17]
I'm not sure...
I feel ya.
4 0
3 years ago
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What is a satellite
g100num [7]

Answer:

an artificial body placed in orbit around the earth or moon or another planet in order to collect information or for communication.

Explanation:

Look it up on google

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3 years ago
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