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Darya [45]
3 years ago
7

When white light is viewed through sodium vapor in a spectroscope the spectrum is continuous?

Physics
2 answers:
morpeh [17]3 years ago
5 0

Answer:

It's continuous except for a pair of dark lines.

They are very close together in the yellow/orange section of the spectrum.

elena-14-01-66 [18.8K]3 years ago
4 0
<span>It's continuous except for a pair of dark lines.
They are very close together in the yellow/orange section of the spectrum.</span>
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On what factor does speed of sound depend​
velikii [3]

Answer:

For sound waves to travel, there is a requirement of medium and density of the medium is considered to be one of the factors on which the speed of sound depends. When the medium is dense, the molecules in the medium are closely packed which means that the sound travels faster.

Explanation:

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3 years ago
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WINSTONCH [101]
Check the attached file for the answer.

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Jennifer, who has a mass of 50.0 kg, is riding at 35.0 m/s in her red sports car when she must suddenly slam on the brakes to av
Assoli18 [71]

Answer:

Average force = 3.5 kN

Explanation:

Given:

Mass of Jennifer (m) = 50 kg

Initial velocity = 35 m/s

Time taken to stop body = 0.5 s

Find:

Average force

Computation:

v = u + at

0 = 35 + a(0.5)

Acceleration (a) =  - 70 m/s² = 70 m/s²

Average force = ma

Average force = (50(70)

Average force = 3500 N

Average force = 3.5 kN

6 0
3 years ago
A 115 g hockey puck sent sliding over ice is stopped in 15.1 m by the frictional force on it from the ice.
Hoochie [10]

Answer:

(a) Ff = 0.128 N

(b μk = 0.1135

Explanation:

kinematic analysis

Because the hockey puck  moves with uniformly accelerated movement we apply the following formulas:

vf=v₀+a*t Formula (1)

d= v₀t+ (1/2)*a*t² Formula (2)

Where:  

d:displacement in meters (m)  

t : time in seconds (s)

v₀: initial speed in m/s  

vf: final speed in m/s  

a: acceleration in m/s

Calculation of the acceleration of the  hockey puck

We apply the Formula (1)

vf=v₀+a*t      v₀=5.8 m/s ,  vf=0

0=5.8+a*t

-5.8 = a*t

a= -5.8/t   Equation (1)

We replace a= -5.8/t in the Formula (2)

d= v₀*t+ (1/2)*a*t²   ,  d=15.1 m ,  v₀=5.8 m/s

15.1 = 5.8*t+ (1/2)*(-5.8/t)*t²  

15.1= 5.8*t-2.9*t

15.1= 2.9*t

t = 15.1 / 2.9

t= 5.2 s

We replace t= 5.2 s in the equation (1)

a= -5.8/5.2

a= -1.115 m/s²

(a) Calculation of the  frictional force (Ff)

We apply Newton's second law

∑F = m*a    Formula (3)

∑F : algebraic sum of the forces in Newton (N)

m : mass in kilograms (kg)

a : acceleration in meters over second square (m/s²)

Look at the free body diagram of the  hockey puck in the attached graphic

∑Fx = m*a     m= 115g * 10⁻³ Kg/g = 0.115g    ,  a= -1.12 m/s²

-Ff = 0.115*(-1.115)  We multiply by (-1 ) on both sides of the equation

Ff = 0.128 N

(b) Calculation of the coefficient of friction (μk)

N: Normal Force (N)

W=m*g= 0.115*9.8= 1.127 N : hockey puck  Weight

g: acceleration due to gravity =9.8 m/s²

∑Fy = 0

N-W=0

N = W

N =  1.127 N

μk = Ff/N

μk = 0.128/1.127

μk = 0.1135

8 0
3 years ago
Determine the amount of work done on an ideal gas as it is heated in an enclosed thermally insulated cylinder topped with a free
sp2606 [1]

Answer:

W = 3/2 n (T₁- T₂)

Explanation:

Let's use the first law of thermodynamics

           ΔE = Q + W

in this case the cylinder is insulated, so there is no heat transfer

           ΔE = W

internal energy can be related to the change in temperature

            ΔE = 3/2 n K ΔT

we substitute

           3/2 n (T₂-T₁) = W

as the work is on the gas it is negative

            W = 3/2 n (T₁- T₂)

3 0
3 years ago
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