Magnetic Field: is a region around the charged particle where the electric force is exerted on other charged particles.

If the work put into a lever is 25.0 joules and the work done by the lever is 20.0 joules, what is the efficiency of the lever?

Liula [17]

The efficiency is Wout/Win = 25/20

5 0

3 years ago

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A CD is spinning on a CD player. In 12 radians, the cd has reached an angular speed of 17 r a d s by accelerating with a constan

mote1985 [20]

Answer:

The initial angular speed of the CD is equal to 14.73 rad/s.

Explanation:

Given that,

Angular displacement, $\theta=12\ rad$

Final angular speed, $\omega_f=17\ rad/s$

The acceleration of the CD, $\alpha =3\ rad/s^2$

We need to find the initial angular speed of the CD. Using third equation of kinematics to find it such that,

$\omega_f^2=\omega_i^2+2\alpha \theta\\\\\omega_i^2=\omega_f^2-2\alpha \theta$

Put all the values,

$\omega_i^2=(17)^2-2\times 3\times 12\\\\\omega_i=\sqrt{217}\\\\\omega_i=14.73\ rad/s$

So, the initial angular speed of the CD is equal to 14.73 rad/s.

3 0

3 years ago

A hunter aims at a deer which is 40 yards away. Her cross- bow is at a height of 5ft, and she aims for a spot on the deer 4ft ab

shutvik [7]

Answer:

a) θ₁ = 0.487º , b) t = 0.400 s , x = 11.73 ft

Explanation:

For this exercise let's use the projectile launch relationships.

The initial height is I = 5 ft and the final height y = 4 ft

y = y₀ + $v_{oy}$ t - ½ g t²

The distance to the band is x = 40 yard (3 ft / 1 yard) = 120 ft

x = v₀ₓ t

t = x / v₀ₓ

We replace

y –y₀ = v_{oy} x / v₀ₓ - ½ g x² / v₀ₓ²

v_{oy} = v₀ sin θ

v₀ₓ = vo cos θ

y –y₀ = x tan θ - ½ g x² / v₀² cos² θ

5-4 = 120 tan θ - ½ 32 120 / (300 2 cos2 θ)

1 = 120 tan θ - 0.0213 sec² θ

Let's use the trigonometry relationship

Sec² θ = 1 - tan² θ

1 = 120 tan θ - 0.0213 (1 –tan²θ)

0.0213 tan²θ + 120 tanθ -1.0213 = 0

We change variables

u = tan θ

u² + 5633.8 u - 48.03 = 0

We solve the second degree equation

u = [-5633.8 ±√(5633.8 2 + 4 48.03)] / 2

u = [- 5633.8 ± 5633.82] / 2

u₁ = 0.0085

u₂= -5633.81

u = tan θ

θ = tan⁻¹ u

For u₁

θ₁ = tan⁻¹ 0.0085

θ₁ = 0.487º

For u₂

θ₂ = -89.99º

The launch angle must be 0.487º

b) let's look for the time it takes for the arrow to arrive

x = v₀ₓ t

t = x / v₀ cos θ

t = 120 / (300 cos 0.487)

t = 0.400 s

The deer must be at a distance of

v = 20 mph (5280 ft / 1 mi) (1 h / 3600s) = 29.33 ft / s

x = v t

x = 29.33 0.4

x = 11.73 ft

3 0

3 years ago

Determine the wavelengths for Potassium (violet, λ = 400 nm) and Strontium (red, λ = 700 nm) light emissions. Calculate the freq

TEA [102]

Answer:

For Potassium:

Frequency = 7.5 x 10¹⁴ Hz; E (energy) = 8.83 x 10⁻²¹ J

For Strontium:

Frequency = 4.3 x 0¹⁴ Hz

E (energy) = 2.85 x 10⁻¹⁹ J

Explanation:

Wavelength is represented by λ, and Frequency is represented by ν .

E (energy) = hν = hc/λ, where ν = frequency; c = speed of light = 3 x 10⁸ m/s; 1 s-1 = 1 Hz

h = planck's constant = 6.62 x 10⁻³⁴ J.s; 1 nm = 10⁻⁹ m

1. Potassium λ (wavelength) = 400 nm, Frequency, ν is given by :

ν = c/λ = (3 x 10⁸ m/s) / 400 nm

= (3 x 10⁸ m/s) / 400 x 10⁻⁹ s-1

= 0.0075 x 10¹⁷s-1

= 7.5 x 10¹⁴ s-1

Frequency = 7.5 x 10¹⁴ Hz

E (energy) = hν = (6.62 x 10⁻³⁴ J.s) x (7.5 x 10¹⁴s-1)

E (energy) = 8.83 x 10⁻²¹ J

2. Strontium λ (wavelength) = 700 nm ,Frequency, ν is given by :

ν = c/λ = (3 x 10⁸ m/s) / 700 nm

= (3 x 10⁸ m/s) / 700 x 10⁻⁹s-1

= 0.00428571428 x 10¹⁷s-1

= 4.3 x 10¹⁴ s-1

Frequency = 4.3 x 0¹⁴ Hz

E (energy) = hν = (6.62 x 10⁻³⁴ J.s) x (4.3 x 0¹⁴s-1)

E (energy) = 2.85 x 10⁻¹⁹ J

4 0

3 years ago

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What are igneous intrusions and extrusions

spayn [35]

Igneous intrusions form when magma cools and solidifies before it can reach the surface. An extrusion consists of extrusive rock; which forms above the surface of the crust.

3 0

3 years ago