Muscle surround your body

A stationary bomb explodes in space breaking into a number of small fragments. At the location of the explosion, the net force d

Kaylis [27]

Answer:

.D)The Vector sum of the linear momenta of the fragments must be zero.

Explanation:

.D)The Vector sum of the linear momenta of the fragments must be zero.

This statement is true. This is so because no external force is acting on the masses. The motion is created by internal force so momentum of fragments will be conserved.

A) this statement is false because kinetic energy was zero in the beginning ( the bomb was stationary in the beginning )

B ) This statement is false because it violates the law of conservation of momentum .( it does not violates only when all the fragments have equal mass )

C ) This statement is zero because kinetic energy is not a vector quantity so two kinetic energy when added can not sum up to zero.

6 0

3 years ago

A beam of light strikes a sheet of glass at an angle of 57.0° with the normal in air. You observe that red light makes an angle

Contact [7]

<h2>Answers: </h2>

1) 1.359, 1.403

2) $2.207(10)^{8}m/s$ , $2.138(10)^{8}m/s$

Explanation:

The described situation is known as Refraction.

Refraction is a phenomenon in which a wave (the light in this case) bends or changes it direction when passing through a medium with a refractive index different from the other medium.

In this context, the Refractive index $n$ is a number that describes how fast light propagates through a medium or material, and is defined as the relation between the speed of light in vacuum ( $c=3(10)^{8}m/s$ ) and the speed of light $v$ in the second medium:

$n=\frac{c}{v}$ (1)

On the other hand we have the Snell’s Law:

$n_{1}sin(\theta_{1})=n_{2}sin(\theta_{2})$ (2)

Where:

$n_{1}$ is the first medium refractive index . We are told is the air, hence $n_{1}\approx 1$

$n_{2}$ is the second medium refractive index

$\theta_{1}$ is the angle of the incident ray

$\theta_{2}$ is the angle of the refracted ray

Knowing this, let's begin with the answers:

<h2><u>1) Indexes of refraction for red and violet light</u></h2><h2 /><h2>1a) Red light</h2>

Using equation (2) according to Snell's Law and $\theta_{1}=57.0\º$ $\theta_{2}=38.1\º$ :

$(1)sin(57.0\º)=n_{2}sin(38.1\º)$

Finding $n_{2}$ :

$n_{2}=\frac{sin(57.0\º)}{sin(38.1\º)}$

$n_{2}=1.359$ (3)>>>Index of Refraction for red light

<h2>1b) Violet light</h2>

Again, using equation (2) according to Snell's Law and $\theta_{1}=57.0\º$ $\theta_{2}=36.7\º$ :

$(1)sin(57.0\º)=n_{2}sin(36.7\º)$

Finding $n_{2}$ :

$n_{2}=\frac{sin(57.0\º)}{sin(36.7\º)}$

$n_{2}=1.403$ (4) >>>Index of Refraction for violet light

<h2><u>2) Speeds of red and violet light</u></h2><h2 /><h2>1a) Red light</h2>

Here we are going to use equation (1):

$n_{red}=\frac{c}{v_{red}}$

$v_{red}=\frac{c}{n_{red}}$

Substituting (3) in this equation:

$v_{red}=\frac{3(10)^{8}m/s}{1.359}$

$v_{red}=2.207(10)^{8}m/s$ >>>>Speed of red light

<h2>1a) Violet light</h2>

Using again equation (1):

$n_{violet}=\frac{c}{v_{violet}}$

$v_{violet}=\frac{c}{n_{violet}}$

Substituting (4) in this equation:

$v_{violet}=\frac{3(10)^{8}m/s}{1.403}$

$v_{red}=2.138(10)^{8}m/s$ >>>>Speed of violet light

3 0

3 years ago

A vibrating tuning fork is held over a water column with one end closed and the other open. As the water level is allowed to fal

DiKsa [7]

Answer:

630 Hz.

Explanation:

As we are considering the one end open pipe. So for the sound wave there will be a pressure node at the open end of the tube as at that place the molecules can not move back and forth. However on the closed end there will be a flow node as the water molecules their are moving back and forth. So it will produces the resonance at the positions 1/4, 3/4.......

we can find the wavelength by multiplying the levels distance by 2.

λ = 2 × 0.27 m = 0.54

f = Vs/λ

= 340/0.54

= 630 Hz

6 0

3 years ago

An electric field of 2.09 kV/m and a magnetic field of 0.358 T act on a moving electron to produce no net force. If the fields a

My name is Ann [436]

Answer:

The velocity is $v = 5838 \ m/s$