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kiruha [24]
3 years ago
15

How does friction affect the distance of an object?

Physics
1 answer:
stiv31 [10]3 years ago
4 0
It slows the object down so it cannot move well and evetually the object cannot be pushed and farther
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Given: Q1 = +10 uc = 1.0 x 10-5C
ser-zykov [4K]

Answer:

-0.038 N

Explanation:

F=K Q1 Q2/r^2 by COULOMB'S LAW

F= 9×10^9×1×10^-5×-1.5×10^-5/(6)^2

F= -0.038 N

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An ion accelerated through a potential difference of 115 V experiences an increase in kinetic energy of 7.37 x 1017 J. Calculate
riadik2000 [5.3K]

Answer: 6.408(10)^{-19} C

Explanation:

This problem can be solved by the following equation:

\Delta K=q V

Where:

\Delta K=7.37(10)^{-17} J is the change in kinetic energy

V=115 V is the electric potential difference

q is the electric charge

Finding q:

q=\frac{\Delta K}{V}

q=\frac{7.37(10)^{-17} J}{115 V}

Finally:

q=6.408(10)^{-19} C

4 0
3 years ago
Which statement describes the density of the inner planets?
Eduardwww [97]

Answer:

The inner planets are less dense than the outer planets

Please brainliest my answer iam begging you for god sake

5 0
2 years ago
What will happen to the wavelength of light uf the frequency is doubled?What will happen to the wavelength of light uf the frequ
otez555 [7]

Answer:

if the frequency is double, the wavelength is only half as long

Explanation:

6 0
3 years ago
A girl rolls a ball up an incline and allows it to re- turn to her. For the angle and ball involved, the acceleration of the bal
zalisa [80]

Answer:

3.28 m

3.28 s

Explanation:

We can adopt a system of reference with an axis along the incline, the origin being at the position of the girl and the positive X axis going up slope.

Then we know that the ball is subject to a constant acceleration of 0.25*g (2.45 m/s^2) pointing down slope. Since the acceleration is constant we can use the equation for constant acceleration:

X(t) = X0 + V0 * t + 1/2 * a * t^2

X0 = 0

V0 = 4 m/s

a = -2.45 m/s^2 (because the acceleration is down slope)

Then:

X(t) = 4*t - 1.22*t^2

And the equation for speed is:

V(t) = V0 + a * t

V(t) = 4 - 2.45 * t

If we equate this to zero we can find the moment where it stops and begins rolling down, that will be the highest point:

0 = 4 - 2.45 * t

4 = 2.45 * t

t = 1.63 s

Replacing that time on the position equation:

X(1.63) = 4 * 1.63 - 1.22 * 1.63^2 = 3.28 m

To find the time it will take to return we equate the position equation to zero:

0 = 4 * t - 1.22 * t^2

Since this is a quadratic equation it will have to answers, one will be the moment the ball was released (t = 0), the other will eb the moment when it returns:

0 = t * (4 - 1.22*t)

t1 = 0

0 = 4 - 1.22*t2

1.22 * t2 = 4

t2 = 3.28 s

7 0
3 years ago
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