PHYSICS COLLEGE

Light of wavelength 608.0 nm is incident on a narrow slit. The diffraction pattern is viewed on a screen 88.5 cm from the slit. The distance on the screen between the fifth order minimum and the central maximum is 1.61 cm. What is the width of the slit?

Answers

Answer 1

Answer:

Answer:

The width of the slit is 0.167 mm

Explanation:

Wavelength of light, $\lambda=608\ nm=608* 10^(-9)\ m$

Distance from screen to slit, D = 88.5 cm = 0.885 m

The distance on the screen between the fifth order minimum and the central maximum is 1.61 cm, y = 1.61 cm = 0.0161 m

We need to find the width of the slit. The formula for the distance on the screen between the fifth order minimum and the central maximum is :

$y=(mD\lambda)/(a)$

where

a = width of the slit

$a=(mD\lambda)/(y)$

$a=(5* 0.885\ m* 608* 10^(-9)\ m)/(0.0161\ m)$

a = 0.000167 m

$a=1.67* 10^(-4)\ m$

a = 0.167 mm

So, the width of the slit is 0.167 mm. Hence, this is the required solution.

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1. If a net force of 412 N is required to accelerate an object at 5.82 m/s2, what must theobject's mass be?

Answers

Answer:

The mass of the object is approximately 70.79 kilograms

Explanation:

We use Newton's second law to solve this problem. This law states that the net force on an object equals the product of its mass times the acceleration:

$F_(net)=m\,a$

Therefore, for this case, since the net force on the object and its acceleration are given, we can use the equation above to solve for the unknown mass:

$F_(net)=m\,a\n412\,N=m\,(5.82\,(m)/(s^2) )\nm=(412\,N)/(5.82\,(m)/(s^2) ) \nm=70.79\.kg$

What minimum distance would you have to hit a baseball from the center of the earth so that it would eventually reach the moon? Assume you can hit the ball directly along the line that connects the centers of the earth and moon. The distance between the centers of the earth and moon is ???? = 3.82 × 108 m.

Answers

Answer:

d = 3.44 x 10⁸ m

Explanation:

The minimum distance required will be the distance from the centre of the earth to a point where gravitational intensity due to both earth and moon becomes equal . Once this point is reached , moon will attract the baseball on its own .

Let this distance be d from the centre of the earth

So GM / d² = G m / ( 3.82 x 10⁸ - d )²

M is mass of the earth , m is mass of the moon

M / m = ( d / 3.82 x 10⁸ - d )²

5.972 x 10²⁴ / 7.34 x 10²² = ( d / 3.82 x 10⁸ - d )²

81.36 = ( d / 3.82 x 10⁸ - d )²

9.02 = d / 3.82 x 10⁸ - d

34.45 x 10⁸ - 9.02 d = d

34.45 x 10⁸ = 10.02 d

d = 3.44 x 10⁸ m

An object propelled upwards with an acceleration of 2.0 m / s ^ 2 is launched from rest. After 6 seconds the fuel runs out. Determine the speed at this time and the maximum height at which it reaches.

Answers

Answer:43.34 m

Explanation:

Given

acceleration(a) $=2 m/s^2$

Initial Velocity(u)=0 m/s

After 6 s fuel runs out

Velocity after 6 s

v=u+at

$v=0+2* 6=12 m/s$

After this object will start moving under gravity

height reached in first 6 s

$s=ut+(at^2)/(2)$

$s=0+(2* 6^2)/(2)$

s=36 m

After fuel run out distance traveled in upward direction is

$v^2-u^2=2as_0$

here v=0

u=12 m/s

$a=9.8 m/s^2$

$0-12^2=2(-9.8)(s)$

$s_0=(144)/(2* 9.8)=7.34 m$

$s+s_0=36+7.34=43.34 m$

The best rebounders in basketball have a vertical leap (that is, the vertical movement of a fixed point on their body) of about 100 cm . a) What is their initial "launch" speed off the ground?b)How long are they in the air?

Answers

Answer:

a) 4.45 m/s

b) 0.9 seconds

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.81 m/s²

$v^2-u^2=2as\n\Rightarrow -u^2=2as-v^2\n\Rightarrow u=√(v^2-2as)\n\Rightarrow u=√(0^2-2* -9.81* 1)\n\Rightarrow u=4.45\ m/s$

a) The vertical speed when the player leaves the ground is 4.45 m/s

$v=u+at\n\Rightarrow t=(v-u)/(a)\n\Rightarrow t=(0-4.45)/(-9.81)\n\Rightarrow t=0.45\ s$

Time taken to reach the maximum height is 0.45 seconds

$s=ut+(1)/(2)at^2\n\Rightarrow 1=0t+(1)/(2)* 9.81* t^2\n\Rightarrow t=\sqrt{(1* 2)/(9.81)}\n\Rightarrow t=0.45\ s$

Time taken to reach the ground from the maximum height is 0.45 seconds

b) Time the player stayed in the air is 0.45+0.45 = 0.9 seconds

As you go up the y-axis, what happens to the number of sprouted bean seeds?A. Sprouted bean seeds decrease.

B. Sprouted bean seeds increase.

C. Sprouted bean seeds remain constant.

D. None of the above

Answers

As we go up the y-axis, the number of sproutedbean seeds increase (option B).

What is a graph?

Graph is a data chart intended to illustrate the relationship between a set (or sets) of numbers (quantities, measurements or indicative numbers) and a reference set.

In a graph, there are two axes as follows;

Y-axis or vertical axis
X-axis or horizontal axis

According to this question, a graph of number of sprouted bean seeds on the y-axis is plotted against temperature on the x-axis.

We can observe that as we go up the y-axis, the number of sprouted bean seeds increase.

Learn more about graphs at: brainly.com/question/2938738

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15.Restore the battery setting to 10 V. Now change the number of loops from 4 to 3. Explain what happens to the magnitude and direction of the magnetic field. Now change to 2 loops, then to 1 loop. What do you observe the relationship to be between the magnitude of the magnetic field and the number of loops for the same current

Answers

Answer:

we see it is a linear relationship.

Explanation:

The magnetic flux is u solenoid is

B = μ₀ N/L I

where N is the number of loops, L the length and I the current

By applying this expression to our case we have that the current is the same in all cases and we can assume the constant length. Consequently we see that the magnitude of the magnetic field decreases with the number of loops

B = (μ₀ I / L) N

the amount between paracentesis constant, in the case of 4 loop the field is worth

B = cte 4

N B

4 4 cte

3 3 cte

2 2 cte

1 1 cte

as we see it is a linear relationship.

In addition, this effect for such a small number of turns the direction of the field that is parallel to the normal of the lines will oscillate,

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