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Leya [2.2K]
3 years ago
7

Which formula represents Charles’s law? P1V1 = P2V2 V1T1 = V2T2

Physics
2 answers:
Karo-lina-s [1.5K]3 years ago
7 0

or it could be V1/T1=V2/T2

Lynna [10]3 years ago
3 0

Charles's law is best represent by the formula, V1T2= V2T1.

Charles's law state that for an ideal gas, at constant pressure, the absolute volume of a gas is directly proportional to the temperature.

that is

V α T

Or

V1T2= V2T1.

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A 40 g ball rolls around a 30 cm -diameter L-shaped track, shown in the figure, (Figure 1)at 60 rpm . What is the magnitude of t
levacccp [35]

Answer:

0.47 N

Explanation:

Here we have a ball in motion along a circular track.

For an object in circular motion, there is a force that "pulls" the object towards the centre of the circle, and this force is responsible for keeping the object in circular motion.

This force is called centripetal force, and its magnitude is given by:

F=m\omega^2 r

where

m is the mass of the object

\omega is the angular velocity

r is the radius of the circle

For the ball in this problem we have:

m = 40 g = 0.04 kg is the mass of the ball

\omega =60 rpm \cdot \frac{2\pi rad/rev}{60 s/min}=6.28 rad/s is the angular velocity

r = 30 cm = 0.30 m is the radius of the circle

Substituting, we find the force:

F=(0.040)(6.28)^2(0.30)=0.47 N

3 0
3 years ago
The terminal velocity of a person falling in air depends upon the weight and the area of the person facing the fluid. Find the t
AVprozaik [17]

Answer:

v=115 m/s

or

v=414 km/h

Explanation:

Given data

A_{area}=0.140m^{2}\\  p_{air}=1.21 kg/m^{3}\\  m_{mass}=80kg

To find

Terminal velocity (in meters per second and kilometers per hour)

Solution

At terminal speed the weight equal the drag force

mg=1/2*C*p_{air}*v^{2}*A_{area}\\   v=\sqrt{\frac{2*m*g}{C**p_{air}*A_{area}} }\\ Where C=0.7\\v=\sqrt{\frac{2*9.8*80}{1.21*0.14*0.7} }\\ v=115m/s

For speed in km/h(kilometers per hour)

To convert m/s to km/h you need to multiply the speed value by 3.6

v=(115*3.6)km/h\\v=414km/h

5 0
3 years ago
A solid metal ball of radius 1.5 cm bearing a charge of -15 nC is located near a hollow plastic ball of radius 1.9 cm bearing a
blsea [12.9K]

We have that the electric field at the center of the metal ball due only to the charges on the surface of the metal ball is

E=7*10^{9}N/C

From the question we are told that

A solid metal ball of radius 1.5 cm

bearing a charge of -15 nC is located near a hollow plastic ball of radius 1.9 cm bearing

uniformly distributed charge of -7 nC

The distance between the centers of the balls is 9 cm

Generally the equation for the electric field  is mathematically given as

E=\frac{kq_2}{d^2}\\\\E=\frac{(9*10^9)7*10^{-2}}{9*10^{-2}}\\\\

E=7*10^{9}N/C

For more information on this visit

brainly.com/question/21811998

4 0
3 years ago
Which of the following is the best definition of gravity?
IRISSAK [1]

In physics, gravity is the natural force that causes things to fall toward the earth. The noun gravity can also mean seriousness or solemnity. Someone who conducts themselves with an air of gravity is someone who takes what they are doing seriously.

8 0
3 years ago
Read 2 more answers
A yet-to-be-built spacecraft starts from Earth moving at constant speed to the yet-tobe-discovered planet Retah, which is 20 lig
Gre4nikov [31]

Answer:

The  time elapsed at the spacecraft’s frame is less that the time elapsed at earth's  frame

Explanation:

From the question we are told that

The distance between earth and Retah is  d = 20 \ light \ hours =  20 * 3600 *  c =  72000c \ m

Here c is the peed of light with value c =  3.0*10^8 m/s

The time taken to reach Retah from earth is  t =  25 \ hours  =  25 * 3600 =90000 \ sec

The velocity of the spacecraft is mathematically evaluated  as

     v_s =  \frac{d }{t}

substituting values

   v_s =  \frac{72000 * 3.0*10^{8} }{90000}

    v_s =  2.40*10^{8} \ m/s

The time elapsed in the spacecraft’s frame is mathematically evaluated as

      T  =  t *  \sqrt{ 1 -  \frac{v^2}{c^2} }

substituting value

       T  =  90000 *  \sqrt{ 1 -  \frac{[2.4*10^{8}]^2}{[3.0*10^{8}]^2} }

        T = 54000 \ s

=>    T  = 15 \ hours

So  The  time elapsed at the spacecraft’s frame is less that the time elapsed at earth's  frame

       

7 0
3 years ago
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