PHYSICS COLLEGE

Light emitted by element X passes through a diffraction grating that has 1200 slits/mm. The interference pattern is observed on a screen 75.0 cm behind the grating. First-order maxima are observed at distances of 56.2 cm, 65.9 cm, and 93.5 cm from the central maximum. What are the wavelengths of light emitted by element X?

Answers

Answer 1

Answer:

Answer:

Explanation:

Width of slit = 10⁻³ / 1200

d = 8.3 x 10⁻⁷ m

First order maxima will be observed at

x = λD/d

D = 75 cm = 75 x 10⁻² m

56.2 x 10⁻² = λ₁D/d

= λ₁ x 75 x 10⁻² / 8.3 x 10⁻⁷

λ₁ = 56.2 x 8.3 x 10⁻⁷ / 75

= 6.219 x 10⁻⁷ m

= 6219 A

Similarly

λ₂ = 65.9 x 8.3 x 10⁻⁷ / 75

= 7293 A

λ₃ = 93.5 x 8.3 x 10⁻⁷ / 75

= 10347 A

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A small grinding wheel is attached to the shaft of an electric motor which has a rated speed of 3600 rpm. When the power is turned on, the unit reaches its rated speed in 5 s, and when the power is turned off, the unit coasts to rest in 70 s. Assuming uniformly accelerated motion, determine the number of revolutions that the motor executes (a) in reaching its rated speed, (b) in coasting to rest.

Answers

Answer:

(a) θ1 = 942.5rad, (b) θ2 = 13195 rad

Explanation:

(a) Given

ωo = 0 rad/s

ω = 3600rev/min = 3600×2(pi)/60 rad/s

ω = 377rad/s

t1 = 5s

θ1 = (ω + ωo)t/2

θ1 = (377 +0)×5/2

θ1 = 942.5 rads

(b) ωo = 377rad/s

ω = 0 rad/s

t2 = 70s

θ2 = (ω + ωo)t/2

θ2 = (377 +0)×70/2

θ2 = 13195 rad

Last night, Shirley worked on her accounting homework for one and one half hours. During that time, she completed 6 problems. What is the velocity in problems per hour? a. 6 per hour
b. 4 per hour
c. 10 per hour
d. 0.67 per hour
e. 15 per hour

Answers

Answer:

velocity in problems per hour = 4 per hour

so correct option is b. 4 per hour

Explanation:

given data

worked on homework time = 1.5 hour

completed = 6 problems

to find out

What is the velocity in problems per hour

solution

we know that Shirley solve complete 6 accounting homework problem in 1.5 hour so her velocity in problems per hour will be as

velocity in problems per hour = $(complete\ problem)/(time\ taken)$ ..................1

put here value we will get

velocity in problems per hour = $(6)/(1.5)$

velocity in problems per hour = 4 per hour

so correct option is b. 4 per hour

Answer:

option (b) is correct

Explanation:

time t = 1 and half hour = 1.5 hour

Number of problems, n = 6

So, the velocity in problems

v = n / t

v = 6 / 1.5

v = 4

So, the rate is 4 problems per hour.

A positive magnification means the image is inverted compared to the object

Answers

False

Explanation:

A positive magnification means the image is erect compared to the object. Magnifications with values greater than one represent images that are smaller than the object. A magnification of 1 (plus or minus) means that the image is the same size as the object. If m has a magnitude greater than 1 the image is larger than the object, and an m with a magnitude less than 1 means the image is smaller than the object. If the magnification is positive, the image is upright compared to the object; if m is negative, the image is inverted compared to the object.

Nichrome wire, often used for heating elements, has resistivity of 1.0 × 10-6 Ω ∙ m at room temperature. What length of No. 30 wire (of diameter 0.250 mm) is needed to wind a resistor that has 50 ohms at room temperature?

Answers

Answer:

Length = 2.453 m

Explanation:

Given:

Resistivity of the wire (ρ) = 1 × 10⁻⁶ Ω-m

Diameter of the wire (d) = 0.250 mm = 0.250 × 10⁻³ m

Resistance of the wire (R) = 50 Ω

Length of the wire (L) = ?

The area of cross section is given as:

$A=(1)/(4)\pi d^2\n\nA=(1)/(4)*\ 3.14* (0.250* 10^(-3))^2\n\nA=0.785* 6.25* 10^(-8)\n\nA=4.906* 10^(-8)\ m^2$

We know that, for a constant temperature, the resistance of a wire is directly proportional to its length and inversely proportional to its area of cross section. The constant of proportionality is called the resistivity of the wire. Therefore,

$R=\rho (L)/(A)$

Expressing the above in terms of length 'L', we get:

$L=(RA)/(\rho)$

Plug in the given values and solve for 'L'. This gives,

$L=(50* 4.906* 10^(-8))/(1* 10^(-6))\ m\n\nL=(2.453)/(1)=2.453\ m$

Therefore, length of No. 30 wire (of diameter 0.250 mm) is 2.453 m.

Which of the following are electromagnetic waves?a. Water wavesb. Radio wavesc. Sound wavesd. Seismic waves

Answers

Answer:

Radio waves

Explanation:

Radio wavs are electromagnetic waves.

Hope this helped!

The answer is Radio Waves because it is electromagnetic

A pitcher is in 85° of abduction, holding a 1.4 N baseball at point C, 65 cm from the joint axis at point O • The center of gravity of his arm is 25 cm from the joint axis of shoulder abduction at point O • The weight of the arm W is 0.06 of the pitcher’s weight of 100 N • Deltoids muscles are at an angle θ of 15° with respect to the humerus and insert 15 cm from the joint axis at point A • Determine the force applied by the Deltoid muscles and the joint reaction force at the shoulder joint and its orientation β