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2 years ago

What is the acceleration of a 4,000 kg car pushed with a force of 12,000 N?

Physics

2 answers:

jek_recluse [69]2 years ago

8 0

Answer:

3 m/s

Explanation:

A= F/m

12,000/ 4000 = 3

Lady bird [3.3K]2 years ago

7 0

Answer:

3 m/s^2

Explanation:

The equation you have to use is F=ma because the problem is a Newton's 2nd law problem.

Our known values are:

F ( Force ) = 12,000 N

m ( mass ) = 4,000 kg

a ( acceleration ) = ?

Now we plug in the known values into the equation and solve

F=ma

12,000=4,000a

We have to divide 4,000 by both sides to isolate the a value

12,000/4,000=4,000/4,000a

The 4,000s on the right of the equation cancel.

And 12,000 divided by 4,000 equals 3

The acceleration (a) is 3 meters per second squared (m/s^2)

Next, check to make sure 3 does work by plugging it back into the equation.

12,000=4,000*3

12,000=12,000 ✔

As you can see, the acceleration will be 3 m/s^2

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3 years ago

A small electronic device is rated at 0.16 W when connected to 115 V. What is the resistance of this device?

Ierofanga [76]

Answer:

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Explanation:

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P = V²/R................ equation 1

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8 0

2 years ago

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Two 13 cm -long thin glass rods uniformly charged to +11nC are placed side by side, 4.0 cm apart. What are the electric field st

DedPeter [7]

Answer:

E1 = 10.15 * 10^4 N/C

E2 = 0

E3 = 10.15 *10^4 N/C

Explanation:

Given data:

Two 13 cm-long thin glass rods ( L ) = 0.13 m

charge (Q) = +11nC

distance between thin glass rods = 4 cm .

<u>Calculate the electric field strengths </u>

electric charge due to a single glass rod in the question ( E ) = $\frac{Q}{2\pi e_{0}rL }$

equation 1 can be used to determine E1, E2 and E3 because the points lie within the two rods hence the net electric field produced will be equal to the difference in electric fields produced

applying equation 1 to determine E1

E1 = $\frac{Q}{2\pi e_{0}rL }$ $( \frac{1}{0.01} - \frac{1}{0.03} )$ ( distance from 1 rod is 0.01 m and from the other rod is 0.03 )

= $\frac{11*10^{-9} }{2*3.14*8.85*10^{-12}*0.13 } ( 66.67 )$

= 10.15 * 10^4 N/C

applying equation 1 to determine E2

E2 = $\frac{Q}{2\pi e_{0}rL }$ $( \frac{1}{0.02} - \frac{1}{0.02} )$

therefore E2 = 0

E1 = E3

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2 years ago

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iren2701 [21]

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