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4 years ago

3

Write a hypothesis about the affect of the diffraction grating line spacing on the diffraction angle. Use the "if . . . then . .

. because . . ." format and be sure to answer the lesson question, "How does diffraction occur?" Refer to the variables.

2 answers:

pishuonlain [190]4 years ago

6 0

Answer: There is an inverse relationship between the line spacing in diffraction grating and the diffraction angle.

Explanation :

We know that the grating equation is given by :

$d\ sin\ \theta=n\lambda$

or

$d=\dfrac{n\ \lambda}{sin\ \theta}$

where,

d is the distance between two slits or obstacles.

$\theta$ is the diffraction angle.

n is the order

and $\lambda$ is the wavelength of light.

So, the line spacing inversely depends on the diffraction angle.

If the line spacing increases, the diffraction angle decreases because there is an inverse relationship between the line spacing and the diffraction angle.

Diffraction occurs when the light passes through a narrow opening or obstacles.

Lelu [443]4 years ago

3 0

<span>If the line spacing on a diffraction grating decreases, then the diffraction angle will increase because it is dependent on wavelength and line spacing. </span>

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A 20.94 g sample of an unknown metal is heated to 99.4 oC in a hot water bath. The metal sample is transferred to a calorimeter

MAXImum [283]

Answer:

1092 J

Explanation:

From the question,

Assuming no heat is lost to the surrounding

Heat lost by the metal = Heat gained by water.

H' = Cm(t₂-t₁)................. Equation 1

Where H' = Heat lost by the metal, C = specific heat capacity of water, m = mass of water, t₂ = Final temperature of the mixture, t₁ = Initial temperature of water

But,

m = D'v................... Equation 2

Where D' = Density of water, v = volume of water.

Given: D' = 1000 kg/m³, v = 100 mL = 100/1000000 = 0.0001 m²

Substitute into equation 2

m = 1000(0.0001)

m = 0.1 kg.

Also given: C = 4200 J/kg.°C, t₁ = 22 °C, 24.6 °C

Substitute into equation 1

H' = 4200×0.1×(24.6-22)

H' = 420(2.6)

H' = 1092 J.

Hence the heat (q) lost by the metal = 1092 J

3 0

4 years ago

Kisachek [45]

B. Inelastic collision.

In elastic collision , both momentum and kinetic energy are conserved while in inelastic collision only momentum is conserved. there is some loss of energy in inelastic collision during collision.

During the collision of bat with baseball, some energy gets lost to heat and sound. hence the kinetic energy is not conserved although the momentum is conserved.

3 0

3 years ago

Read 2 more answers

Recognizing the face of someone you run into at the mall is a function of the __________ hemisphere; being able to retrieve that

Alchen [17]

Answer:

right; left

Explanation:

The right hemisphere of the brain has a schema formation capacity, this can be applied to recognize faces. The left hemisphere is in charge of the memory and therefore to recall names.

I hope you find this information useful and interesting! Good luck!

5 0

4 years ago

A 54.0-kg projectile is fired at an angle of 30.0° above the horizontal with an initial speed of 126 m/s from the top of a cliff

Nikolay [14]

Answer:

a) The initial total mechanical energy of the projectile is 498556.296 joules.

b) The work done on the projectile by air friction is 125960.4 joules.

c) The speed of the projectile immediately before it hits the ground is approximately 82.475 meters per second.

Explanation:

a) The system Earth-projectile is represented by the Principle of Energy Conservation, the initial total mechanical energy ( $E$ ) of the project is equal to the sum of gravitational potential energy ( $U_{g}$ ) and translational kinetic energy ( $K$ ), all measured in joules:

$E = U_{g} + K$ (Eq. 1)

We expand this expression by using the definitions of gravitational potential energy and translational kinetic energy:

$E = m\cdot g\cdot y + \frac{1}{2}\cdot m\cdot v^{2}$ (Eq. 1b)

Where:

$m$ - Mass of the projectile, measured in kilograms.

$g$ - Gravitational acceleration, measured in meters per square second.

$y$ - Initial height of the projectile above ground, measured in meters.

$v$ - Initial speed of the projectile, measured in meters per second.

If we know that $m = 54\,kg$ , $g = 9.807\,\frac{m}{s^{2}}$ , $y = 132\,m$ and $v = 126\,\frac{m}{s}$ , the initial mechanical energy of the earth-projectile system is:

$E = (54\,kg)\cdot \left(9.807\,\frac{m}{s^{2}}\right)\cdot (132\,m)+\frac{1}{2}\cdot (54\,kg)\cdot \left(126\,\frac{m}{s} \right)^{2}$

$E = 498556.296\,J$

The initial total mechanical energy of the projectile is 498556.296 joules.

b) According to this statement, air friction diminishes the total mechanical energy of the projectile by the Work-Energy Theorem. That is:

$W_{loss} = E_{o}-E_{1}$ (Eq. 2)

Where:

$E_{o}$ - Initial total mechanical energy, measured in joules.

$E_{1}$ - FInal total mechanical energy, measured in joules.

$W_{loss}$ - Work losses due to air friction, measured in joules.

We expand this expression by using the definitions of gravitational potential energy and translational kinetic energy:

$W_{loss} = E_{o}-K_{1}-U_{g,1}$

$W_{loss} = E_{o} -\frac{1}{2}\cdot m\cdot v_{1}^{2}-m\cdot g\cdot y_{1}$ (Eq. 2b)

Where:

$m$ - Mass of the projectile, measured in kilograms.

$g$ - Gravitational acceleration, measured in meters per square second.

$y_{1}$ - Maximum height of the projectile above ground, measured in meters.

$v_{1}$ - Current speed of the projectile, measured in meters per second.

If we know that $E_{o} = 498556.296\,J$ , $m = 54\,kg$ , $g = 9.807\,\frac{m}{s^{2}}$ , $y_{1} = 297\,m$ and $v_{1} = 89.3\,\frac{m}{s}$ , the work losses due to air friction are:

$W_{loss} = 498556.296\,J -\frac{1}{2}\cdot (54\,kg)\cdot \left(89.3\,\frac{m}{s} \right)^{2} -(54\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (297\,m)$

$W_{loss} = 125960.4\,J$

The work done on the projectile by air friction is 125960.4 joules.

c) From the Principle of Energy Conservation and Work-Energy Theorem, we construct the following model to calculate speed of the projectile before it hits the ground:

$E_{1} = U_{g,2}+K_{2}+1.5\cdot W_{loss}$ (Eq. 3)

$K_{2} = E_{1}-U_{g,2}-1.5\cdot W_{loss}$

Where:

$E_{1}$ - Total mechanical energy of the projectile at maximum height, measured in joules.

$U_{g,2}$ - Potential gravitational energy of the projectile, measured in joules.

$K_{2}$ - Kinetic energy of the projectile, measured in joules.

$W_{loss}$ - Work losses due to air friction during the upward movement, measured in joules.

We expand this expression by using the definitions of gravitational potential energy and translational kinetic energy:

$\frac{1}{2}\cdot m \cdot v_{2}^{2} = E_{1}-m\cdot g\cdot y_{2}-1.5\cdot W_{loss}$ (Eq. 3b)

$m\cdot v_{2}^{2} = 2\cdot E_{1}-2\cdot m \cdot g \cdot y_{2}-3\cdot W_{loss}$

$v_{2}^{2} = 2\cdot \frac{E_{1}}{m}-2\cdot g\cdot y_{2}-3\cdot \frac{W_{loss}}{m}$

$v_{2} = \sqrt{2\cdot \frac{E_{1}}{m}-2\cdot g\cdot y_{2}-3\cdot \frac{W_{loss}}{m} }$

If we know that $E_{1} = 372595.896\,J$ , $m = 54\,kg$ , $g = 9.807\,\frac{m}{s^{2}}$ , $y_{2} =0\,m$ and $W_{loss} = 125960.4\,J$ , the final speed of the projectile is:

$v_{2} =\sqrt{2\cdot \left(\frac{372595.896\,J}{54\,kg}\right)-2\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (0\,m)-3\cdot \left(\frac{125960.4\,J}{54\,kg}\right) }$

$v_{2} \approx 82.475\,\frac{m}{s}$

The speed of the projectile immediately before it hits the ground is approximately 82.475 meters per second.

7 0

3 years ago

Two 3.0 μC charges lie on the x-axis, one at the origin and the other at What is the potential (relative to infinity) due to the

miskamm [114]

Complete Question:

Two 3.0µC charges lie on the x-axis, one at the origin and the other at 2.0m. A third point is located at 6.0m. What is the potential at this third point relative to infinity? (The value of k is 9.0*10^9 N.m^2/C^2)

Answer:

The potential due to these charges is 11250 V

Explanation:

Potential V is given as;

$V =\frac{Kq}{r}$

where;

K is coulomb's constant = 9x10⁹ N.m²/C²

r is the distance of the charge

q is the magnitude of the charge

The first charge located at the origin, is 6.0 m from the third charge; the potential at this point is:

$V =\frac{9X10^9 X3X10^{-6}}{6} =4500 V$

The second charge located at 2.0 m, is 4.0 m from the third charge; the potential at this point is:

$V =\frac{9X10^9 X3X10^{-6}}{4} =6750 V$

Total potential due to this charges = 4500 V + 6750 V = 11250 V

3 0

4 years ago