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Svetach [21]
3 years ago
15

2. A car accelerates down the road. What is the "reaction" to the tires pushing on the road?

Physics
1 answer:
Artist 52 [7]3 years ago
7 0
The answer would be friction
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Please<br>help<br>me<br>with this problem​
uranmaximum [27]

Answer:

speeding up

Explanation:

because its speeding up, theres going to be more newtons in the back

i really hope this is right, tell me if so

3 0
2 years ago
Which game represents a wave​
alisha [4.7K]
Dominoes does because when you hit one, it knocks over the next one, and so on, so forth. The same type of pattern happens in a wave.
6 0
3 years ago
10)A car is moving from rest and attained a velocity of 80 m/s. Calculate the
Mashcka [7]

Explanation:

the velocity graph of a ball mass 20mg moving along a straight line

5 0
3 years ago
Read 2 more answers
Hansel pushes a 2 lb. box 5 feet in 20 seconds. How much work has he done?
Licemer1 [7]
Work = force x distance

You can see time doesn’t matter (if we were talking about power, which is the RATE at which work is performed, that would be a different story).

W = 2 x 5 = 10 foot-pounds of work

Foot-pounds are gross units. Better to work in SI units when you can!

8 0
3 years ago
A very long uniform line of charge has charge per unit length λ1 = 4.68 μC/m and lies along the x-axis. A second long uniform li
Kitty [74]

Answer:

E_{net} = 6.44 \times 10^5 N/C

Explanation:

As we know that electric field due to infinite line charge distribution at some distance from it is given as

E = \frac{2k \lambda}{r}

now we need to find the electric field at mid point of two wires

So here we need to add the field due to two wires as they are oppositely charged

Now we will have

E_{net} = \frac{2k\lambda_1}{r} + \frac{2k\lambda_2}{r}

now plug in all data

\lambda_1 = 4.68 \muC/m

\lambda_2 = 2.48 \mu C/m

r = 0.200 m

now we have

E_{net} = \frac{2k}{r}(4.68 + 2.48)

E_{net} = \frac{2(9\times 10^9)}{0.200}(7.16 \times 10^{-6})

E_{net} = 6.44 \times 10^5 N/C

8 0
3 years ago
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