PHYSICS COLLEGE

7) Straws work on the principle of the outside atmospheric pressure pushing the fluid (for example water) up the straw after you have lowered the pressure at the top of the straw (in your mouth). Assuming you could create a perfect vacuum in your mouth, what is the longest vertical straw you could drink water from?

Answers

Answer 1

Answer:

Answer:

The longest straw will be 10.328 meters long.

Explanation:

The water will rise up to a height pressure due to which will balance the atmospheric pressure.

We know

$P_(atm)=101325N/m^(2)$

Pressure due to water column of height 'h'

$P_(water)=1000* 9.81* h$

Equating both the values we get the value of height 'h' as

$h=(101325)/(1000* 9.81)\n\nh=10.328m$

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You observe three carts moving to the right. Cart A moves to the right at nearly constant speed. Cart B moves to the right, gradually speeding up. Cart C moves to the right, gradually slowing down. Which cart or carts, if any, experience a net force to the right

Answers

Answer:

Explanation:

Cart A is moving to the right with constant speed i.e. net acceleration is zero

because acceleration is change in velocity in given time

Cart B is moving towards right with gradually speed up so there is net acceleration which helps to increase the velocity s

This indicates the net force acting on the cart towards right

For cart C there is gradual slow down of cart which indicates cart is decelerating and a net force is acting towards which opposes its motion.

Help!!! Need answer ASAP.

Answers

Answer:

Hey

Explanation:

The volume control on a stereo is designed so that three clicks of the dial increase the output by 10 dB. How many clicks are required to increase the power output of the loudspeakers by a factor of 100?

Answers

Answer:

300 clicks...

Explanation:

Output on 3 clicks = 10 dB

Increasing 10 by a factor of 100 equals 1000 dB so,

Its simple math, clicks will also increase in the same ratio and it shall take 300 clicks to increase the volume by a factor of 100.

Point charge 3.5 μC is located at x = 0, y = 0.30 m, point charge -3.5 μC is located at x = 0 y = -0.30 m. What are (a)the magnitude and (b)direction of the total electric force that these charges exert on a third point charge Q = 4.0 μC at x = 0.40 m, y = 0?

Answers

Hi, thank you for posting your question here at Brainly.

To solve this problem, we use Coulomb's Law:

F = kQ1Q2/d^2, where k = 9x10^9

Q1 = 3.5 uC
Q2 = -3.5 uC
Q3 = 4.0 uC

But first, we find the distance between Q1 and Q3 and between Q2 and Q3.

d between Q1 and Q2:
d = sqrt[(0-0.4)^2+(0.3-0)^2]
d = 0.5 m

d between Q1 and Q3:
d = sqrt[(0-0.4)^2+(-0.3-0)^2]
d = 0.5 m

Through force balance, F between Q2 and Q3 - F between Q1 and Q3:

$F_(net) = ((9x 10^(9))(-3.5)(4) )/( 0.5^(2) ) -((9x 10^(9))(3.5)(4) )/( 0.5^(2) )=-1.008* 10^(12)$

Thus, the net force is -1 x 10^-12 C

Final answer:

The total electric force exerted by point charges -3.5 μC and 3.5 μC on a point charge 4.0 μC is zero. This is because the forces due to each of these charges on the third charge are equal in magnitude but opposite in direction, hence they cancel each other completely.

Explanation:

The question asks for the magnitude and direction of the total electric force exerted by point charges -3.5 μC and 3.5 μC on a point charge 4.0 μC. This is related to Coulomb's Law, which describes the force between charged objects. Specifically, Coulomb's Law states that the force (F) between two point charges is directly proportional to the product of their charges (q1*q2) and inversely proportional to the square of the distance (r) between them. It also depends on the permittivity of free space (ε₀).

First, you would determine the force between each of the point charges and the third charge separately, and then superpose these forces to find the total force. The force in each case can be calculated using the equation F = k*|q1*q2|/r², where k is Coulomb's constant (8.99 * 10^9 N.m²/C²). You would need to make sure you take into account the signs of the charges when deciding the directions of the forces and when superposing the separate forces.

Assume upwards to be the positive direction. The 3.5 uC charge forces and -3.5 uC charge forces on the 4 uC charge would be opposite in direction (one downwards and one upwards) and identical in magnitude. Therefore, they will cancel each other out, and hence, the total electric force on the third charge (4 uC) will be zero.

Learn more about Total Electric Force here:

brainly.com/question/30448827

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A rider on a bike with the combined mass of 100 kg attains a terminal speed of 15 m/s on a 12% slope. Assuming that the only forces affecting the speed are the weight and the drag, calculate the drag coefficient. The frontal area is 0.9 m2.

Answers

Answer:

7.84

Explanation:

Draw free body diagram and put all forces on it. Forces are

Gravity in direction of slope = $(12)/(100) *100*12$ =117.6newton
Viscous force(opposite to slope) after attaining terminal speed =k*V=k*15

As the bike+man system has attained a terminal velocity thus acceleration is zero .

Both forces are opposite then equate them

117.6=k.15

k=7.84

Here k is drag coefficient.

Answer:

0·95

Explanation:

Given the combined mass of the rider and the bike = 100 kg

Percent slope = 12%

∴ Slope = 0·12

Terminal speed = 15 m/s

Frontal area = 0·9 m²

Let the slope angle be β

tanβ = 0·12

As it attains the terminal speed, the forces acting on the combined rider and the bike must be balanced and therefore the rider must be moving download as the directions of one of the component of weight and drag force will be in opposite directions

The other component of weight will get balance by the normal reaction and you can see the figure which is in the file attached

From the diagram m × g × sinβ = drag force

Drag force = 0·5 × d × $C_(D)$ × v² × A

where d is the density of the fluid through which it flows

$C_(D)$ is the drag coefficient

v is the speed of the object relative to the fluid

A is the cross sectional area

As tanβ = 0·12

∴ sinβ = 0·119

Let the fluid in this case be air and density of air d = 1·21 kg/m³

m × g × sinβ = 0·5 × d × $C_(D)$ × v² × A

100 × 9·8 ×0·119 = 0·5 × 1·21 × $C_(D)$ × 15² × 0·9

∴ $C_(D)$ ≈ 0·95

∴ Drag coefficient is approximately 0·95

The inner and outer surfaces of a cell membrane carry a negative and positive charge, respectively. Because of these charges, a potential difference of about 0.078 V exists across the membrane. The thickness of the membrane is 7.1 x 10-9 m. What is the magnitude of the electric field in the membrane?