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Elena L [17]
3 years ago
14

The image on the left shows the charges on a balloon after it’s been rubbed with a wool cloth. The image on the right is a piece

of tissue paper. What will happen to the tissue paper as it approaches the balloon and why? A. The tissue paper will stick to the balloon because the tissue paper is negatively charged. B. The tissue paper will stick to the balloon because the tissue paper is positively charged. C. The tissue paper will move away from the balloon because the tissue paper is negatively charged. D. The tissue paper will move away from the balloon because the tissue paper is positively charged.
Physics
1 answer:
Leto [7]3 years ago
6 0

Answer:

The tissue paper will stick to the balloon because it is positively charged.

Explanation:

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Amy has a mass of 30 kg, and she is riding a skateboard traveling 5 m/s. What is her momentum
soldi70 [24.7K]
We Know, P = m*v
Here, m = 30 Kg
v = 5 m/s

Substitute it into the expression, 
P = 30*5 Kgm/s
P = 150 Kgm/s

So, your final answer is 150 Kg.m/s

Hope this helps!
3 0
3 years ago
Read 2 more answers
Consider four point charges arranged in a square with sides of length L. Three of the point charges have charge q and one of the
nydimaria [60]

Answer:F_{net}=\frac{kq^2}{(L)^2}\left [ \frac{1}{2}+\sqrt{2}\right ]

Explanation:

Given

Three charges of magnitude q is placed at three corners and fourth charge is placed at last corner with -q charge

Force due to the charge placed at diagonally opposite end on -q charge

F_1=\frac{kq(-q)}{(L\sqrt{2})^2}

where  L\sqrt{2}=Distance between the two charges

F_1=-\frac{kq^2}{2L^2}

negative sign indicates that it is an attraction force

Now remaining two charges will apply the same amount of force as they are equally spaced from -q charge

F_2=\frac{kq(-q)}{(L)^2}

The magnitude of force by both the  charge is same but at an angle of 90^{\circ}

thus combination of two forces at 2 and 3 will be

F'=\sqrt{2}\frac{kq^2}{2L^2}

Now it will add with force due to 1 charge

Thus net force will be

F_{net}=\frac{kq^2}{(L)^2}\left [ \frac{1}{2}+\sqrt{2}\right ]

6 0
3 years ago
The current and the potential difference in an inductor are in phase. B. The current lags the potential difference by π/2 in an
slava [35]

Answer:

The current lags the potential difference by π/2 in an inductor

Explanation:

The potential difference leads to the current by  \frac{\pi}{2}. Alternate signals such as current and voltage -in this case- are periodic, this means that this signals are repeated at fixed spaces of time. Thus, In an inductor the current lags the potential difference by \frac{\pi}{2}.

6 0
3 years ago
The pressure on a fluid at rest in a pipe increases by 20 Pa. How does this change in pressure affect the pressure on the fluid
timofeeve [1]

Answer:The change in pressure can affect the pressure on the fluid through the radius and diameter of the pipe.

r^² x Pressure (pa).

Therefore the narrower the other part of the pile, the greater the pressure on the fluid at such part, the wider in other part the lesser the pressure on the fluid at this part.

Explanation:

4 0
3 years ago
If the Earth and distant stars were stationary (motionless) in space, what would we observe about the wavelength from these star
dangina [55]
1) In the first case, the correct answer is
<span>A.Wavelengths measured would match the actual wavelengths emitted.
In fact, the stars are not moving relative to Earth, so there is no shift in the measured wavelength.

2) In this second case, the correct answer is
</span><span>A.Wavelengths measured would be shorter than the actual wavelengths emitted.
</span>in fact, since the stars in this case are moving towards the Earth, then apparent frequency of their emitted light will be larger than the actual frequency, because of the Doppler effect, according to the formula:
f'= \frac{c}{c+v_s} f_0
where f0 is the actual frequency, f' the apparent frequency, c the speed of light and vs the velocity of the source (the stars) relative to the obsever (Earth). Vs is negative when the source is moving towards the observer, so the apparent frequency f' is larger than the actual frequency f0. But the wavelength is inversely proportional to the frequency, so the apparent wavelength will be shorter than the actual wavelength.
6 0
3 years ago
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