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3 years ago

Which theory was first proposed by Albert Einstein

Physics

2 answers:

pishuonlain [190]3 years ago

7 0

He proposed the theory of relativity. =)

sveta [45]3 years ago

4 0

The theory of special relativity

You might be interested in

How much energy is needed to heat and melt 3.0 kg of copper initially at 83°C?

Ne4ueva [31]

As we know that in order to melt the copper we need to take the temperature of copper to its melting point

So here heat required to raise the temperature of copper is given as

$Q = ms\Delta T$

We know that

melting temperature of copper = 1085 degree C

Specific heat capacity of copper = 385 J/kg C

now we have

$Q = 3(385)(1085 - 83)$

$Q = 1157310 J$

$Q = 1157.3 kJ$

now in order to melt the copper we know the heat required is

$Q = mL$

here we know that

L = 205 kJ/kg

now from above formula

$Q = 3(205) kJ$

$Q = 615 kJ$

now total heat required will be

$Q = 1157.3 kJ + 615 kJ$

$Q = 1772.3 kJ$

As we know that

$1 Cal = 4.18 kJ$

now we have

$Q = \frac{1772.3}{4.18} = 430 KCal$

6 0

2 years ago

Light of wavelength 630 nm is incident on a long, narrow slit. Determine the angular deflection of the first diffraction minimum

asambeis [7]

Answer:

a) 1.8°

b) 0.18°

c) 0.018°

Explanation:

Wavelength (λ) = 630nm = 630 *10^-9m

The equation that describes the angular deflection of a dark band is

Wsin(βm) = mλ

w = width of the single slit

λ = wavelength of the light

βm = angular deflection of the mth dark band.

a) In order to get the angular deflection of the first dark band for a slit with 0.02mm width, substitute w = 0.02mm = 0.02*10^-3 , m = 1 , λ= 630*10^-9

0.02*10^-3 sin(β1) = 1 * 630*10^-9

Sin(β1) = 630 * 10^-9 / 0.02*10^-3

Sin(β1) = 0.0315

β1 = Sin^-1(0.0315)

= 1.8°

b) substitute w = 0.2mm = 0.2*10^-3 , m = 1 , λ= 630*10^-9

0.2*10^-3 sin(β1) = 1 * 630*10^-9

Sin(β1) = 630 * 10^-9 / 0.2*10^-3

Sin(β1) = 0.00315

β1 = Sin^-1(0.00315)

= 0.18°

c) substitute w = 2mm = 2*10^-3 , m = 1 , λ= 630*10^-9

2*10^-3 sin(β1) = 1 * 630*10^-9

Sin(β1) = 630 * 10^-9 / 2*10^-3

Sin(β1) = 3.15*10^-4

β1 = Sin^-1(3.15*10^-4)

= 0.018°

6 0

3 years ago

E) the time it takes the car to reach 20.0ms-1

Galina-37 [17]

4.2 thats the answer

8 0

2 years ago

A jet accelerates from rest down a runway at 1.75m/s² for a distance of 1500 m before takeoff.

babunello [35]

A. Using the third equation of motion:
v2 = u2 + 2as
from the question;
the jet was initially at rest
hence u = 0
a = 1.75m/s2
s = 1500m
v2 = 02 + 2(1.75)(1500)
v2 = 5250
v = √5250
v = 72.46m/s
hence it moves with a velocity of 72.46m/s.
b. s = ut + 1/2at2
1500 = 0(t) + 1/2(1.75)t2
1500 × 2 = 2× 1/2(1.75)t2
3000 = 1.75t2
1714.29 = t2
41.4 = t
hence the time taken for the plane to down the runway is 41.4s.

Read more on Brainly.com - brainly.com/question/18743384#readmore

8 0

3 years ago

Select all the correct answers.

sdas [7]

I believe there are two correct answers, and those answers are A and D

5 0

2 years ago