A standing wave is the result of: refraction diffraction reflection resonance

frozen [14]

Answer:

Force

Explanation:

The mass of an object is the quantity of matter it contains. It is measured in kilograms.

Acceleration is the ratio of the change in the velocity of an object to the change in time. It is measured in m/ $s^{2}$ .

When the mass of an object is multiplied with its acceleration, this gives the average force applied on the object. As force is defined as agent that can change the state of an object.

i.e F = m × a

where F is the force, m is the mass of the object and a its acceleration.

The two major classes of force are; contact force and field force.

6 0

3 years ago

3. A ray of light consisting of blue light (wavelength 480 nm) and red light (wavelength 670 nm) is incident on a thick piece of

Alex Ar [27]

Answer:

The angular separation between the refracted red and refracted blue beams while they are in the glass is 42.555 - 42.283 = 0.272 degrees.

Explanation:

Given that,

The respective indices of refraction for the blue light and the red light are 1.4636 and 1.4561.

A ray of light consisting of blue light (wavelength 480 nm) and red light (wavelength 670 nm) is incident on a thick piece of glass at 80 degrees.

We need to find the angular separation between the refracted red and refracted blue beams while they are in the glass.

Using Snell's law for red light as :

$n_1\sin\theta_1=n_2\sin\theta_2\\\\\theta_2=\sin^{-1}((\dfrac{n_2}{n_1})\sin\theta_1)\\\\\theta_2=\sin^{-1}((\dfrac{1}{1.4561})\sin(80))\\\\\theta_2=42.555$

Again using Snell's law for blue light as :

$n_1\sin\theta_1=n_2\sin\theta'_2\\\\\theta'_2=\sin^{-1}((\dfrac{n_2}{n_1})\sin\theta_1)\\\\\theta'_2=\sin^{-1}((\dfrac{1}{1.4636 })\sin(80))\\\\\theta'_2=42.283$

The angular separation between the refracted red and refracted blue beams while they are in the glass is 42.555 - 42.283 = 0.272 degrees.

7 0

4 years ago

Astronauts use a centrifuge to simulate the acceleration of a rocket launch. The centrifuge takes 40.0 s to speed up from rest t

Vinvika [58]

Answer

Time period T = 1.50 s

time t = 40 s

r = 6.2 m

a)

Angular speed ω = 2π/T

= $\dfrac{2\pi }{1.5}$

= 4.189 rad/s

Angular acceleration α = $\dfrac{\omega}{t}$

= $\dfrac{4.189}{40}$

= 0.105 rad/s²

Tangential acceleration a = r α = 6.2 x 0.105 = 0.651 m/s²

b)The maximum speed.

v = 2πr/T

= $\dfrac{2\pi \times 6.2}{1.5}$

= 25.97 m/s

So centripetal acceleration.

a = $\dfrac{v^2}{r}$

= $\dfrac{25.97^2}{6.2}$

= 108.781 m/s^2

= 11.1 g

in combination with the gravitation acceleration.

$a_{total} = \sqrt{(11.1g)^2+g^2}$

$a_{total}= 11.145 g$

6 0

4 years ago

The earth's radius is about 4000 miles. kampala, the capital of uganda, and singapore are both nearly on the equator. the distan

Y_Kistochka [10]

The solution is: Earth circumference is 2*pi*r=25,130 miles 25,130 miles corresponds to 360Âş, so 5,000 miles corresponds to 360*5000/25130 = 71,6Âş 71,6Âş is equivalent to 71,6*2pi/360=1,25 radians Angular velocity is 1,25/9=0,139 rad/h

4 0

3 years ago

What average force is required to stop a 1400 kg car in 6.0 s if the car is traveling at 90 km/h ? Express your answer to two si

devlian [24]

Answer:

F=5833.3 N N

Explanation:

Newton's second law applied to the car

F= m*a Formula (1)

F: Force in Newtons (N)

m : mass in kg

a: acceleration ( m/s²)

kinematics car

vf= v₀ + a*t Formula (2)

vf : final velocity (m/s)

v₀ : final velocity (m/s)

a : acceleration ( m/s²)

t : time t

Equivalences

1 km= 1000m

1 h = 3600 s

Data

m= 1000kg

v₀ = 90 km/h = 90*1000/3600 m/s = 25 m/s

vf= 0

t= 6 s

Problem Development

We calculate the acceleration replacing the data in the formula (2) :

0 = 25 + a*6

a= -25/6 = -4.16 m/s² ( The negative sign indicates that the car is braking)

We calculate the force is required to stop the car replacing the data in the formula (1)

-F = 1400 kg*(-4.16 m/s²)

F=5833.3 N

8 0

3 years ago