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My name is Ann [436]
3 years ago
11

What is laplace law???????

Physics
1 answer:
cestrela7 [59]3 years ago
8 0

{\texttt{\huge{\red{ANSWER :-}}}}

<h3><u>L</u><u>A</u><u>P</u><u>L</u><u>A</u><u>C</u><u>E</u><u> </u><u>L</u><u>A</u><u>W</u><u> </u><u>:</u><u>-</u></h3>

Laplace's law for the gauge pressure inside a cylindrical membrane is given by

<h3>ΔP = γ/r</h3>

<u>W</u><u>h</u><u>e</u><u>r</u><u>e</u>

γ => surface tension

r => the radius of the cylinder

Note : there is an inverse relation between the pressure and the radius.

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a car advertisement states that a certain car can accelerate from rest to 70km/h in 7 seconds. find the cars acceleration
ELEN [110]
<span>a = (v2 - v1)/t= acceleration formula

a = (70 - 0)/7
a = 10 km/hr/sec
-----
It's better to use as few units as possible.
10 km/hr = 10 km*1000 m/km/(1 hr*3600 sec/hr) = 25/9 m/sec
a= 25/9 m/sec/se</span>
4 0
4 years ago
If forces acting on an object are unbalanced, which factor may result from an unbalanced force? The net force is negative. There
Oksi-84 [34.3K]

Answer:the Forces cancel out each other

Explanation:the forces cancel out each other

6 0
3 years ago
Read 2 more answers
The 60.0 kg skier shown below is skiing down a 35.0 degree incline where the magnitude of the friction force is 38.5N
Sunny_sXe [5.5K]

Answer:

a) 4.98m/s²

b) 481.66N

Explanation:

a) Using the Newtons second law of motion

\sum F_x = ma_x\\F_m - F_f = ma_x\\Wsin \theta - F_f = ma_x\\mgsin \theta - F_f = ma_x\\

m is the mass of the object

g is the acceleration due to gravity

Fm is the moving force acting along the plane

Ff is the frictional force opposing the moving froce

a is the acceleration of the skier

Given

m = 60kg

g = 9.8m/s²

\theta = 35°

Ff = 38.5N

Required

acceleration of the skier a

Substituting into the formula;

60(9.8)sin 35^0 - 38.5 = 60a\\588sin35^0 - 38.5 = 60a\\337.26 - 38.5 = 60a\\298.76 = 60a\\a = 298.76/60\\a = 4.98m/s^2\\

Hence the acceleration of the skier is 4.98m/s²

b) The normal force on the skier is expressed as;

N = Wcosθ

N = mgcosθ

N = 60(9.8)cos 35°

N = 588cos 35°

N = 481.66N

Hence the normal force on the skier is 481.66N

5 0
3 years ago
if a bowling ball hits a wall a force of 6 N, the wall exerts a force of how much back. on the bowling ball
grigory [225]
It would exert the same back right?
3 0
3 years ago
A 52 kg and a 95 kg skydiver jump from an airplane at an altitude of 4750 m, both falling in the pike position. Assume all value
Scilla [17]

Answer: 52 kg skydiver: 9.09 m/s and 522.55 s

              95 kg skydiver: 12.3 m/s and 386.2 s

Explanation: <u>Drag</u> <u>Force</u> is an opposite force when an object is moving in a fluid.

For skydivers, when falling through the air, the forces acting on it are gravitational and drag forces. At a certain point, drag force equals gravitational force, which is constant on any part of the planet, producing a net force that is zero. Since there is no net force, there is no acceleration and, consequently, velocity is constant. When that happens, the person reached the <u>Terminal</u> <u>Velocity</u>.

Drag Force and Velocity are proportional to the squared speed. So, terminal velocity is given by:

F_{G}=F_{D}

mg=\frac{1}{2}C \rho Av_{T}^{2}

v_{T}=\sqrt{\frac{2mg}{\rho CA} }

where

m is mass in kg

g is acceleration due to gravitational force in m/s²

ρ is density of the fluid in kg/m³

C is drag coefficient

A is area of the object in the fluid in m²

Calculating:

The 52kg skydiver has terminal velocity of:

v_{T}=\sqrt{\frac{2(52)(9.8)}{(1.21)(0.7)(0.14)} }

v_{T}= 9.09

The 95kg skydiver's terminal velocity is

v_{T}=\sqrt{\frac{2(95)(9.8)}{(1.21)(0.7)(0.14)} }

v_{T}= 12.3

The 52 kg and 95kg skydivers' terminal velocity are 9.09m/s and 12.3m/s, respectively.

The time each one will reach the floor will be:

52 kg at 9.09 m/s:

t=\frac{4750}{9.09}

t = 522.5

95 kg at 12.3 m/s:

t=\frac{4750}{12.3}

t = 386.2

The 52 kg and 95kg skydivers' time to reach the floor are 522.5 s and 386.2 s, respectively.

3 0
3 years ago
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