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Halogen lightbulbs allow their filaments to operate at a higher temperature than the filaments in standard incandescent bulbs. For comparison, the filament in a standard lightbulb operates at about 2900K, whereas the filament in a halogen bulb may operate at 3400K. Which bulb has the higher peak frequency? Calculate the ratio of the peak frequencies. The human eye is most sensitive to a frequency around 5.5x10^14 Hz. Which bulb produces a peak frequency close to this value?

Answers

Answer 1

Answer:

Answer:

Halogen

0.85294

Explanation:

c = Speed of light = $3* 10^8\ m/s$

b = Wien's displacement constant = $2.897* 10^(-3)\ mK$

T = Temperature

From Wien's law we have

$\lambda_m=(b)/(T)\n\Rightarrow \lambda_m=(2.897* 10^(-3))/(2900)\n\Rightarrow \lambda_m=9.98966* 10^(-7)\ m$

Frequency is given by

$\nu=(c)/(\lambda_m)\n\Rightarrow \nu=(3* 10^8)/(9.98966* 10^(-7))\n\Rightarrow \nu=3.00311* 10^(14)\ Hz$

For Halogen

$\lambda_m=(b)/(T)\n\Rightarrow \lambda_m=(2.897* 10^(-3))/(3400)\n\Rightarrow \lambda_m=8.52059* 10^(-7)\ m$

Frequency is given by

$\nu=(c)/(\lambda_m)\n\Rightarrow \nu=(3* 10^8)/(8.52059* 10^(-7))\n\Rightarrow \nu=3.52088* 10^(14)\ Hz$

The maximum frequency is produced by Halogen bulbs which is closest to the value of $5.5* 10^(14)\ Hz$

Ratio

$(3.00311* 10^(14))/(3.52088* 10^(14))=0.85294$

The ratio of Incandescent to halogen peak frequency is 0.85294

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Answers

Answer:

Explanation:

A baseball is hit with a speed of 47.24 m/s from a height of 0.42 meters. If the ball is in the air 5.73 seconds and lands 130 meters from the batters feet, (a) at what angle did the ball leave the bat? (b) with what velocity will the baseball hit the ground?

Answers

Answer:

a)the ball will leave the bat at an angle of 61.3° .

b) the velocity at which it will hit the ground will be v = 27.1 m/s

Explanation:

Given,

v = 47.24 m

h = 0.42 m

t = 5.73 s

R = 130 m

a)We know that

R = v cosθ × t

cosθ = $(R)/(v t ) = (130)/(47.24* 5.73 ) =0.4803$

θ = 61.3°

the ball will leave the bat at an angle of 61.3° .

b)Vx = v cos(θ) = 47.24 x cos 61.3 = 22.7 m/s

v = u + at

Vy = 47.24 x sin 61.3 - 9.81 x 5.73

= -14.8 m/s

v = $√(v_x^2 + v_y^2))$

v = $√(22.7^2 + -14.8^2)$

v = 27.1 m/s

the velocity at which it will hit the ground will be v = 27.1 m/s

Before 1960, people believed that the maximum attainable coefficient of static friction for an automobile tire on a roadway was ?s = 1. Around 1962, three companies independently developed racing tires with coefficients of 1.6. This problem shows that tires have improved further since then. The shortest time interval in which a piston-engine car initially at rest has covered a distance of one-quarter mile is about 4.43 s. (A) Assume the car's rear wheels lift the front wheels off the pavement as shown in the figure above. What minimum value of ?s is necessary to achieve the record time?

Answers

Answer:

4.18

Explanation:

Givens

The car's initial velocity $v_(i)$ = 0 and covering a distance Δx = 1/4 mi = 402.336 m in a time interval t = 4.43 s.

Knowns

We know that the maximum static friction force is given by:

$f_(s_max) =$ μ_s*n (1)

Where μ_s is the coefficient of static friction and n is the normal force.

Calculations

(a) First, we calculate the acceleration needed to achieve this goal by substituting the given values into a proper kinematic equation as follows:

Δx= $v_(i) +(1)/(2) at^2$

a=41 m/s

This is the acceleration provided by the engine. Applying Newton's second law on the car, so in equilibrium, when the car is about to move, we find that:

$f_(y)=n-mg=0\n n=mg\nf_(x)=F-f_(s,max) =0\n f_(s,max)=F=ma\n$

Substituting (3) into (1), we get:

$f_(s,max)=$ μ_s*m*g

Equating this equation with (4), we get:

ma= μ_s*m*g

μ_s=a/g

=4.18

A steel ball is dropped onto a thick piece of foam. The ball is released 2.5 meters above the foam. The foam compresses 3.0 cm as the ball comes to rest. What is the magnitude of the ball's acceleration as it comes to rest on the foam

Answers

Answer:

the magnitude of the ball's acceleration as it comes to rest on the foam is 817.5 m/s²

Explanation:

Given the data in the question;

initial velocity; u = 0 m/s

height; h = 2.5 m

we find the velocity of the ball just before it touches the foam.

using the equation of motion;

v² = u² + 2gh

we know that acceleration due gravity g = 9.81 m/s²

so we substitute

v² = ( 0 )² + ( 2 × 9.81 × 2.5 )

v² = 49.05

v = √49.05

v = 7.00357 m/s

Now as the ball touches the foam

final velocity v₀ = 0 m/s

compresses S = 3 cm = 0.03 m

v₀² = v² + 2as

we substitute

( 0 )² = 49.05 + 0.06a

0.06a = -49.05

a = -49.05 / 0.06

a = -817.5 m/s²

Therefore, the magnitude of the ball's acceleration as it comes to rest on the foam is 817.5 m/s²

A ball is thrown into the air with 100 J of kinetic energy, which is transformed to gravitational potential energy at the top of its trajectory.When it returns to its original level after encountering air resistance, its kinetic energy is __________.

A) more than 100 J.

B) Not enough information given.

C) less than 100 J.

D) 100 J.

Answers

To solve this problem we could apply the concepts given by the conservation of Energy.

During the launch given in terms of kinetic energy and reaching the maximum point of the object, the potential energy of the body is conserved. However, part of all this energy is lost due to the work done by the friction force due to friction with the air, therefore

$E_T = PE + KE -W_f$

The potential and kinetic energy are conserved and are the same PE = KE and this value is equivalent to 100J, therefore

$E_T = 100-W_f$

The kinetic energy will ultimately be less than 100J, so the correct answer is C.

A 60kg woman on skates throws a 3.9kg ball with a velocity of37m.s west. What is the velocity of the woman?

Answers

Answer:

2.405 m/s

Explanation:

Given that,

Mass of a women, m₁ = 60 kg

Mass of a ball, m₂ = 3.9 kg

Velocity of the ball, v₂ = 37 m/s

We need to find the velocity of the woman. It is a concept based on the conservation of linear momentum. Let v₁ is the velocity of the woman. So,

$m_1v_1=m_2v_2\n\nv_1=(m_2v_2)/(m_1)\n\nv_1=(3.9* 37)/(60)\n\nv_1=2.405\ m/s$

So, the velocity of the woman is 2.405 m/s.

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