All categories

3 years ago

Classical physics is a good approximation to modern physics under certain circumstances. What are they?

Physics

1 answer:

Aleks [24]3 years ago

4 0

Answer and Explanation:

In classical physics we study about macroscopic objects and in modern physics we study about microscopic objects but under some circumstances classical physics is a approximation to modern physics such as

The object must be microscopic but they should be large enough so that we can see that through a microscope
The speed of the object should be very less it should be about 1 % of the speed of the light
They should be involved with very week gravitational field

You might be interested in

What is the audible range of human hearing

Phantasy [73]

Answer:

on an average, 20Hz to 20kHz

8 0

3 years ago

If 300. mL of water are poured into the measuring cup, the volume reading is 10.1 oz . This indicates that 300. mL and 10.1 oz a

tigry1 [53]

Answer:

Milliliters to Ounces Conversions

some results rounded

mL - fl oz

200.00 6.7628

200.01 6.7631

200.02 6.7635

200.03 6.7638

200.04 6.7642

200.05 6.7645

200.06 6.7648

200.07 6.7652

200.08 6.7655

200.09 6.7658

200.10 6.7662

200.11 6.7665

200.12 6.7669

200.13 6.7672

200.14 6.7675

200.15 6.7679

200.16 6.7682

200.17 6.7686

200.18 6.7689

200.19 6.7692

200.20 6.7696

200.21 6.7699

200.22 6.7702

200.23 6.7706

200.24 6.7709

mL fl oz

200.25 6.7713

200.26 6.7716

200.27 6.7719

200.28 6.7723

200.29 6.7726

200.30 6.7729

200.31 6.7733

200.32 6.7736

200.33 6.7740

200.34 6.7743

200.35 6.7746

200.36 6.7750

200.37 6.7753

200.38 6.7757

200.39 6.7760

200.40 6.7763

200.41 6.7767

200.42 6.7770

200.43 6.7773

200.44 6.7777

200.45 6.7780

200.46 6.7784

200.47 6.7787

200.48 6.7790

200.49 6.7794

mL fl oz

200.50 6.7797

200.51 6.7800

200.52 6.7804

200.53 6.7807

200.54 6.7811

200.55 6.7814

200.56 6.7817

200.57 6.7821

200.58 6.7824

200.59 6.7828

200.60 6.7831

200.61 6.7834

200.62 6.7838

200.63 6.7841

200.64 6.7844

200.65 6.7848

200.66 6.7851

200.67 6.7855

200.68 6.7858

200.69 6.7861

200.70 6.7865

200.71 6.7868

200.72 6.7872

200.73 6.7875

200.74 6.7878

mL fl oz

200.75 6.7882

200.76 6.7885

200.77 6.7888

200.78 6.7892

200.79 6.7895

200.80 6.7899

200.81 6.7902

200.82 6.7905

200.83 6.7909

200.84 6.7912

200.85 6.7915

200.86 6.7919

200.87 6.7922

200.88 6.7926

200.89 6.7929

200.90 6.7932

200.91 6.7936

200.92 6.7939

200.93 6.7943

200.94 6.7946

200.95 6.7949

200.96 6.7953

200.97 6.7956

200.98 6.7959

200.99 6.7963

Explanation:

5 0

3 years ago

Read 2 more answers

Two astronauts, each having a mass of 74.3 kg are connected by a 13.1 m rope of negligible mass. They are isolated in space, orb

murzikaleks [220]

Answer:

L = 5076.5 kg m² / s

Explanation:

The angular momentum of a particle is given by

L = r xp

L = r m v sin θ

the bold are vectors, where the angle is between the position vector and the velocity, in this case it is 90º therefore the sine is 1

as we have two bodies

L = 2 r m v

let's find the distance from the center of mass, let's place a reference frame on one of the masses

$x_{cm}$ = $\frac{1}{M} \sum x_{i} m_{i}$ i

x_{cm} = $\frac{1}{m+m} ( 0 + l m)$

x_{cm} = $\frac{1}{2m} lm$

x_{cm} = $\frac{1}{2}$

x_{cm} = 13.1 / 2 = 6.05 m

let's calculate

L = 2 6.05 74.3 5.65

L = 5076.5 kg m² / s

4 0

3 years ago

Read 2 more answers

A proton moving at 3.0 × 10^4 m/s is projected at an angle of 30° above a horizontal plane. If an electric field of 400 N/C is a

GuDViN [60]

Answer:

The time it takes the proton to return to the horizontal plane is 7.83 X10⁻⁷ s

Explanation:

From Newton's second law, F = mg and also from coulomb's law F= Eq

Dividing both equations by mass;

F/m = Eq/m = mg/m, then

g = Eq/m --------equation 1

Again, in a projectile motion, the time of flight (T) is given as

T = (2usinθ/g) ---------equation 2

Substitute in the value of g into equation 2

$T = \frac{2usin \theta}{\frac{Eq}{m}} =\frac{m* 2usin \theta}{Eq}$

Charge of proton = 1.6 X 10⁻¹⁹ C

Mass of proton = 1.67 X 10⁻²⁷ kg

E is given as 400 N/C, u = 3.0 × 10⁴ m/s and θ = 30°

Solving for T;

$T = \frac{(1.67X10^{-27}* 2*3X10^4sin 30}{400*1.6X10^{-19}}$

T = 7.83 X10⁻⁷ s

6 0

3 years ago

ASAP

scoundrel [369]

Answer:

Explanation:

Hooke's law! F(spring)=-kx

There's no tricky square law here. The spring constant doesn't change, only x (distance stretched) changes. Therefore, if distance is halved, Force will be halved.

5 0

3 years ago