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alex41 [277]
2 years ago
8

You are sitting in a car that is travelling at a constant 10 m/s. How fast is your drink, sitting in the cup holder, traveling r

elative to the car?
O 0 m/s

O 10 m/s

O speed of light

100 m/s
Physics
1 answer:
Ksenya-84 [330]2 years ago
7 0

Here, we are required to determine how fast is you drink, sitting in the cup holder, travelling relative to the car.

  • The speed of the drink, sitting in the cup holder, relative to the car is; 0m/s

From the laws of relative motion,

  • <em>when object A and Object B are travelling with speed a and b respectively in the same direction, the speed of Object A relative to B is;. (a - b)</em>

  • <em>when object A and Object B are travelling with speed a and b respectively in the same direction, the speed of Object A relative to B is;. (a - b)when object A and Object B are travelling with speed a and b respectively in opposite directions, the speed of Object A relative to B is; (a+b)</em>

  • <em>when object A and Object B are travelling with speed a and b respectively in the same direction, the speed of Object A relative to B is;. (a - b)when object A and Object B are travelling with speed a and b respectively in opposite directions, the speed of Object A relative to B is; (a+b)when object A and Object B are travelling with speed a and b respectively in the same direction, where speed a = speed b, then the speed of object A relative to object B is; zero(0).</em>

Evidently, the scenario in the question is similar to the third scenario above. The cup, sitting in the cup holder is travelling with the car at the same constant speed 10m/s.

Therefore, the speed of the drink relative to the car is zero(0).

Read more:

brainly.com/question/20549055

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A balloon will stick to a wooden wall if the balloon is charged
grigory [225]
<h2>Answer: either way</h2>

The balloon contains neutral charge atoms, that is, it has the same number of electrons (negative charge), protons (positive charge) and neutrons (no charge).

Then, when two objects come into contact, the electrons of one of them can become part of the other.

Thus, by bringing the balloon closer to the wall, the wall, which is also made up of atoms, will reorder its charges in such a way that its electrons or protons become part of the balloon, charging it.

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3 years ago
A negative charge of - 8.0 x 10^-6 C exerts an attractive force of 12 N on a second charge that is
Umnica [9.8K]

Answer:

<h2>Magnitude of the second charge is -4.17*10^{-7}C</h2>

Explanation:

According to columbs law;

F = kq1q2/r^{2}

F is the attractive or repulsive force between the charges = 12N

q1 and q2 are the charges

let q1 = - 8.0 x 10^-6 C

q2=?

r is the distance between the charges = 0.050m

k is the coulumbs constant =9*10⁹ kg⋅m³⋅s⁻⁴⋅A⁻²

On substituting the given values

12 = 9*10⁹*( - 8.0 x 10^-6)q2/0.050²

Cross multiplying

0.03=9*10^{9}*  -8.0*10^{-6} q2\\0.03 = -72*10^{3} q2\\q2 = \frac{0.03}{ -72*10^{3}} \\q2 = -4.17*10^{-7}C

6 0
4 years ago
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How much water will flow in 30 secs through 200 mm of capillary tube of 1.50 mm in diameter, if the pressure difference across t
Paladinen [302]

The water outflow in 30 secs through 200 mm of the capillary tube is mathematically given as

Qo=1.6 \times 10^{2} \mathrm{~mL}

<h3>What is the water outflow in 30 secs through 200 mm of the capillary tube?</h3>

\begin{aligned}\Delta P &=6660 \mathrm{~m} / \mathrm{m}^{2} \\\mu &=8.01 \times 10^{-4} \text { Pas } \\t &=30 \mathrm{~s} \\L &=200 \mathrm{~mm}=200 \times 10^{-3} \mathrm{~m} \\D &=1.5 \mathrm{~mm}=1.5 \times 10^{-3} \mathrm{~m} \Rightarrow \gamma=\frac{1.5 \times 10^{-3}}{2} \mathrm{~m}\end{aligned}

Generally, the equation for Rate of flow of Liquid is  mathematically given as

\\$$Q=\frac{\pi r^{4} \times \Delta P}{8 \mu L}

$$

Where dP is pressure difference r is the radius

\mu is the viscosity of water

L is the length of the pipe

Q=\frac{\pi \times\left(\frac{1.5 \times 10^{-3}}{2}\right)^{4} \times 6660}{8 \times 8.01 \times 10^{-4} \times 200 \times 10^{-3}}

Q=5.2 \mathrm{~mL} / \mathrm{s}

In $30s the quantity that flows out of the tube

&Qo=5.2 \times 30 \\&Qo=1.6 \times 10^{2} \mathrm{~mL}

In conclusion, the quantity that flows out of the tube

Qo=1.6 \times 10^{2} \mathrm{~mL}

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Paha777 [63]

Explanation:

It is given that,

Voltage of the battery, V = 12 V

Current, I = 100 ampere-hours

Energy stored is given by the product of power and time taken. So,

E=P\times t

P is the power, P=V\times I

P=12\times 100

P = 1200 watts

This power can be used for 1 hour or 3600 seconds

Energy, E=1200\times 3600

E = 4320000 J

So, the energy stored in this battery is 4320000 J. Hence, this is the required solution.

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