A baseball with a mass of 142 g is thrown horizontally with a speed of 38.6 m/s (86 mi/h) at a bat. The ball is in contact with the bat for 1.00 ms and then travels straight back at a speed of 46.1 m/s (103 mi/h). Determine the average force (in N) exerted on the ball by the bat. Neglect the weight of the ball (it is much smaller than the force of the bat) and choose the direction of the incoming ball to be positive. (Indicate the direction with the sign of your answer.)

Answers

Answer 1

Answer:

-11.600N

Explanation:

lol


Related Questions

A camper hears her echo off a
hillside 1.52 s after yelling. How
far is she from the hill?

Answers

Answer:

261

Explanation:

Acellus

A ceiling fan with 90-cm-diameter blades is turning at 64 rpm . Suppose the fan coasts to a stop 28 s after being turned off. What is the speed of the tip of a blade 10 s after the fan is turned off

Answers

Answer:

the speed of the tip of a blade 10 s after the fan is turned off is 16.889 m/s.

Explanation:

Given;

diameter of the ceiling fan, d = 90 cm = 0.9 m

angular speed of the fan, ω = 64 rpm

time taken for the fan to stop, t = 28 s

The distance traveled by the ceiling fan when it comes to a stop is calculated as;

[tex]d = vt = \omega r\times t= ( \frac{64 \ rev}{\min} \times \frac{2 \pi \ rad}{rev} \times \frac{1 \min}{60 \ s} \times 0.9 \ m) \times 28 \ s\\\\d = 168.89 \ m[/tex]

The speed of the tip of a blade 10 s after the fan is turned off is calculated as;

[tex]v = \frac{d}{t} \\\\v = \frac{168.89}{10} \\\\v = 16.889 \ m/s[/tex]

Therefore, the speed of the tip of a blade 10 s after the fan is turned off is 16.889 m/s.

A force of 12 N changes the momentum of a toy car from 3kgm/s t0 10kgm/s. Calculate the time the force took to produce this change in momentum.

Answers

Answer:

Time = 0.58 seconds

Explanation:

Given the following data;

Initial momentum = 3 kgm/s

Final momentum = 10 kgm/s

Force = 12 N

To find the time required for the change in momentum;

First of all, we would determine the change in momentum.

[tex] Change \; in \; momentum = final \; momentum - initial \; momentum [/tex]

[tex] Change \; in \; momentum = 10 - 3 [/tex]

Change in momentum = 7 kgm/s

Now, we can find the time required;

Note: the impulse of an object is equal to the change in momentum experienced by the object.

Mathematically, impulse (change in momentum) is given by the formula;

[tex] Impulse = force * time [/tex]

Making "time" the subject of formula, we have;

[tex] Time = \frac {impulse}{force} [/tex]

Substituting into the formula, we have;

[tex] Time = \frac {7}{12} [/tex]

Time = 0.58 seconds

What is diffraction of light

Answers

Answer:

According to "http://ww2010.atmos.uiuc.edu"  Diffraction is the slight bending of light as it passes around the edge of an object.

Some examples of Light Defraction would be..

-CD reflecting rainbow colours

-Sun appears red during sunset

-From the shadow of an object

PLEASE HELP
Section 1 - Question 6
Wave Movement Through Media
What could be happening to the wave as it travels from left to right?
A
It's moving through a medium whose density stays the same
B
It's moving from a low density medium to a high density medium.
С
It's moving from a high density medium to a low density medium.
D
It's moving from a low density medium, to a high density medium, and then back to a low density medium

Answers

Answer: B

Explanation:

Two 20kg spheres are placed with their
Centres 50cm apart. What is the magnitude of
gravitational force each exerts on the other?​

Answers

Answer:

F = 1.07 x 10⁻⁷ N

Explanation:

The gravitational force of attraction between two objects can be found by the use of Newton's Gravitational Law:

[tex]F = \frac{Gm_{1}m_{2}}{r^2}\\\\[/tex]

where,

F = Gravitational Force of attraction = ?

G = Universal Gravitational Constant = 6.67 x 10⁻¹¹ N.m²/kg²

m₁ = m₂ = mass of spheres = 20 kg

r = distance between the objects = 50 cm = 0.5 m

Therefore,

[tex]F = \frac{(6.67\ x\ 10^{-11}\ N.m^2/kg^2)(20\ kg)(20\ kg)}{(0.5\ m)^2}\\\\[/tex]

F = 1.07 x 10⁻⁷ N

If matter cannot be created or destroyed, then how do you end up with
rust? Below is the equation for rust.
4Fe + 302 → 2Fe203
oxygen from the air
water in the atmosphere
oxygen from in the metal
there shouldn't be any oxygen

Answers

Your question is a "non sequitur", which means "it doesn't follow".

Your "then" doesn't contradict your "If", so no mystery is implied.

Maybe you're trying to say that matter is somehow not conserved in the equation . . . 4Fe + 302 → 2Fe203 . But it is.  There are 4 Irons and 6 Oxygens on each side, so conservation is not violated here.

I looked up "rust" on Floogle, and got slapped with pages and pages of chemistry that I don't completely understand.  But what it's saying is that rusting is a very complex chemical process, AND it doesn't happen unless there's some water involved.

So the bottom line is that there's a lot more going on than simply

4Fe + 302 → 2Fe203 ,

there's water going in and out of the process at every stage, and when it's all over, you have rusty iron, and mass has been conserved.

Time of the day when the Sun does not shine (___time)
N____N

Answers

I-
nighttime? So put night in the blank

The amount of light that enters the pupil is controlled by the:
retina.
lens.
inis.

Answers

Answer: The amount of light that enters the pupil is controlled by the Iris

Explanation:

Which of these is the BEST answer for why science is important?
Science can take us to other planets, even if it’s only through a telescope.
Science is part of human nature; it helps answer questions about how the world works.
Science helps us learn to think more critically and weigh evidence better.
Science gives us better tablet computers and games.

QQQUUUUCCCKKKK!!!!!!

Answers

3) science helps us learn to think more critically and weigh evidence better ( I guess this the answer) :)

A physics professor wants to demonstrate the large size of the henry unit. On the outside of a 16-cm-diameter plastic hollow tube, she wants to wind an air-filled solenoid with self-inductance of 1.0 H using copper wire with a 0.79-mm diameter. The solenoid is to be tightly wound with each turn touching its neighbor (the wire has a thin insulating layer on its surface so the neighboring turns are not in electrical contact).

Required:
a. How long will the plastic tube need to be?
b. How many kilometers of copper wire will be required?
c. What will be the resistance of this solenoid?

Answers

Answer:

a) the plastic tube need to be 24.7 m long

b) the kilometer of copper wire required is 15.7

c) the resistance of this solenoid is 5538.8 2 ohms

Explanation:

Given the data in the question;

we determine the length of the plastic tube. assuming the solenoid is long.

the self inductance of a long solenoid is;

L = μ₀n²πr²l

μ₀ = 4π × 10⁻⁷ T-m/A

where

n = number of turns per unit length

r = radius of the solenoid = 8cm (as the diameter of the plastic hollow tube is 16 cm)

l = length of the solenoid or the length of the plastic tube

we find n = number of turns per unit length

given that, the copper wire to be wound around the solenoid is 0.79 mm in diameter

number of turns per meter = n = 1 / ( 0.79 × 10⁻³ m ) = 1265.8 turns/meter

So from our previous formula, we find l

L = μ₀n²πr²l

we substitute

1.0 H = (4π × 10⁻⁷ T-m/A)( 1265.8 )²(3.14)(0.08)² ( l)

1 = 0.04048 × l

l = 1 / 0.04048

l = 24.7 m

Therefore, the plastic tube need to be 24.7 m long

b)  

n = number of turns per unit length = 1265.8 turns/metre

so, the length of the plastic tube over which the copper wire is to be wound,

number of turns of copper wire required = n × l

= 1265.8 turns/meter × 24.7 m

= 31,265.26 turns

Now each turn of the copper wire is to be wound across the 18cm diameter of the plastic tube.

so for each turn length of copper wire required = 2π × r

= 2π × 0.08 m

= 0.5026548 m

So copper wire required for 31,265.26 turns will be;

⇒ 31,265.26 × 0.5026548 = 15715.63m = 15.7 km

Therefore, the kilometer of copper wire required is 15.7

c)

p = resistivity of copper = 1.68 × 10⁻⁸ ohm-m

Resistance = pl/a

where l is length of copper wire, a is cross sectional area;

diameter of copper wire is 0.79-mm

radius of copper wire is 0.79/2 = 0.395 mm = 0.000395 m

area of cross section of copper wire a = πr² = π( 0.00395)² = 4.9 × 10⁻⁷ m²

Resistance = pl/a

we substitute

Resistance = [(1.68 × 10⁻⁸ ohm-m)( 15715.63m )] / [ 4.9 × 10⁻⁷ m² ]

Resistance =  5538.8 2 ohms

Therefore,  the resistance of this solenoid is 5538.8 2 ohms

Energy is transferred through a solid
by if there is a
between the ends of it.
Liquids and solids transfer energy
by because the particles can
Energy is transferred much
through an insulator
than it is through a conductor.
Tore

Answers

Answer:

Conduction is the transfer of thermal energy through direct contact between particles of a substance, without moving the particles to a new location. Usually occurs in solids. When heat is supplied to one end, molecules at that end start to move more quickly.

Explanation:

Conduction is the transfer of heat and electricity through a material or surface. Generally, metals are good conductors of heat and electricity.

What is Conduction?

Conduction is the process through which heat energy is transmitted through the collisions which occur between neighboring atoms or molecules in a system. Conduction occurs more readily in the solids and liquid substances, where the particles are closer together than that of the gases, where the particles are further apart from each other.

Metals are good conductors of heat and electricity as they contain valence electrons and charge whereas non-metals are bad or poor conductors of heat and electricity as these elements does not contain valence electrons.

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An astronaut has a mass of 75 kg and is floating in space 500 m from his 125,000 kg spacecraft. What will be the force of gravitational attraction between the two? Since there is no force opposing him, he will accelerate toward the ship. Find his acceleration.

Answers

Answer:

1. 2.5×10¯⁹ N

2. 3.33×10¯¹¹ m/s²

Explanation:

1. Determination of the force of attraction.

Mass of astronaut (M₁) = 75 Kg

Mass of spacecraft (M₂) = 125000 Kg

Distance apart (r) = 500 m

Gravitational constant (G) = 6.67×10¯¹¹ Nm²/Kg²

Force of attraction (F) =?

The force of attraction between the astronaut and his spacecraft can be obtained as follow:

F = GM₁M₂ /r²

F = 6.67×10¯¹¹ × 75 × 125000 / 500²

F = 2.5×10¯⁹ N

Thus, the force of attraction between the astronaut and his spacecraft is 2.5×10¯⁹ N

2. Determination of the acceleration of the astronaut.

Mass of astronaut (m) = 75 Kg

Force (F) = 2.5×10¯⁹ N

Acceleration (a) of astronaut =?

The acceleration of the astronaut can be obtained as follow:

F = ma

2.5×10¯⁹ N = 75 × a

Divide both side by 75

a = 2.5×10¯⁹ / 75

a = 3.33×10¯¹¹ m/s²

Thus, the acceleration the astronaut is 3.33×10¯¹¹ m/s²

A force of 10 N is applied at right angles to the handle of a spanner, 0.5 m from the centre of a nut. The
moment on the nut is:
20 Nm
50 Nm
5 Nm

Answers

Explanation:

the movement of the nut is 20Nm

Discuss how the pressure cooker is designed to achieve temperatures above 100°C.​

Answers

With rising heat, the steam pressure inside the pot builds up beyond atmospheric pressure, allowing the temperatures to rise well above boiling point. This design enables to save time, energy, and resources. The temperature inside a pressure cooker can well go beyond 110° C, which reduces the time needed to cook food.

When6-2 He He-6 undergoes beta decay, the daughter is?​

Answers

Answer: The daughter is named Susie.

Explanation: LIL SUSIE!!!

                      HUH? DIDN'T UNDERSTAND THE QUESTION!

                                        HAVE A GREAT DAY!!!!!

Answer:6/3 Li

Explanation:

I’m not sure what the person under me is talking about but yeah

Cells use nutrients and oxygen to supply the body with the energy it needs. What three-body systems are working together in this situation?
A
nervous, digestive, and circulatory systems

B
digestive, circulatory, and excretory systems

C
circulatory, immune, and respiratory systems

D
digestive, respiratory, and circulatory systems

Answers

Answer:

The respiratory system provides oxygen for cells, while the circulatory system transports oxygen to cells.

Explanation:

so the answer is D

A particle with an initial linear momentum of 2.00 kg-m/s directed along the positive x-axis collides with a second particle, which has an initial linear momentum of4.00 kg-m/s, directed along the positive y-axis. The final momentum of the first particle is 3.00 kg-m/s, directed 45.0 above the positive x-axis.

a. the magnitude and direction (angle expressed counter-clockwise with respect to the positive x-axis) of the final momentum for the second particle
b. assuming that these particles have the same mass, % loss of their total kinetic energy after they collided

Answers

Answer:

a) p₂ = 1.88 kg*m/s

   θ = 273.4 º

b)  Kf = 37% of Ko

Explanation:

a)

Assuming no external forces acting during the collision, total momentum must be conserved.Since momentum is a vector, their components (projected along two axes perpendicular each other, x- and y- in this case) must be conserved too.The initial momenta of both particles are directed one along the x-axis, and the other one along the y-axis.So for the particle moving along the positive x-axis, we can write the following equations for its initial momentum:

       [tex]p_{o1x} = 2.00 kg*m/s (1)[/tex]

       [tex]p_{o1y} = 0 (2)[/tex]

We can do the same for the particle moving along the positive y-axis:

        [tex]p_{o2x} = 0 (3)[/tex]

        [tex]p_{o2y} = 4.00 kg*m/s (4)[/tex]

Now, we know the value of magnitude of the final momentum p1, and the angle that makes with the positive x-axis.Applying the definition of cosine and sine of an angle, we can find the x- and y- components of the final momentum of the first particle, as follows:

       [tex]p_{f1x} = 3.00 kg*m/s * cos 45 = 2.12 kg*m/s (5)[/tex]

      [tex]p_{f1y} = 3.00 kg*m/s sin 45 = 2.12 kg*m/s (6)[/tex]

Now, the total initial momentum, along these directions, must be equal to the total final momentum.We can write the equation for the x- axis as follows:

       [tex]p_{o1x} + p_{o2x} = p_{f1x} + p_{f2x} (7)[/tex]

We know from (3) that p₀₂ₓ = 0, and we have the values of p₀1ₓ from (1) and pf₁ₓ from (5) so we can solve (7) for pf₂ₓ, as follows:

       [tex]p_{f2x} = p_{o1x} - p_{f1x} = 2.00kg*m*/s - 2.12 kg*m/s = -0.12 kg*m/s (8)[/tex]

Now, we can repeat exactly the same process for the y- axis, as follows:

       [tex]p_{o1y} + p_{o2y} = p_{f1y} + p_{f2y} (9)[/tex]

We know from (2) that p₀1y = 0, and we have the values of p₀₂y from (4) and pf₁y from (6) so we can solve (9) for pf₂y, as follows:

       [tex]p_{f2y} = p_{o1y} - p_{f1y} = 4.00kg*m*/s - 2.12 kg*m/s = 1.88 kg*m/s (10)[/tex]

Since we have the x- and y- components of the final momentum of  the second particle, we can find its magnitude applying the Pythagorean Theorem, as follows:

       [tex]p_{f2} = \sqrt{p_{f2x} ^{2} + p_{f2y} ^{2} } = \sqrt{(-0.12m/s)^{2} +(1.88m/s)^{2}} = 1.88 kg*m/s (11)[/tex]

We can find the angle that this vector makes with the positive x- axis, applying the definition of tangent of an angle, as follows:

       [tex]tg \theta = \frac{p_{2fy} }{p_{2fx} } = \frac{1.88m/s}{(-0.12m/s} = -15.7 (12)[/tex]

The angle that we are looking for is just the arc tg of (12) which measured in a counter-clockwise direction from the positive x- axis, is just 273.4º.

b)

Assuming that both masses are equal each other, we find that the momenta are proportional to the speeds, so we find that the relationship from the final kinetic energy and the initial one can be expressed as follows:

       [tex]\frac{K_{f}}{K_{o} } = \frac{v_{f1}^{2} + v_{f2} ^{2}}{v_{o1}^{2} + v_{o2} ^{2} } = \frac{12.5}{20} = 0.63 (13)[/tex]

So, the final kinetic energy has lost a 37% of the initial one.

What is the acceleration of a 4,000 kg car pushed with a
force of 12,000 N?

Answers

Answer:

3 m/s

Explanation:

A= F/m

12,000/ 4000 = 3

Answer:

3 m/s^2

Explanation:

The equation you have to use is F=ma because the problem is a Newton's 2nd law problem.

Our known values are:

F ( Force ) = 12,000 N

m ( mass ) = 4,000 kg

a ( acceleration ) = ?

Now we plug in the known values into the equation and solve

F=ma

12,000=4,000a

We have to divide 4,000 by both sides to isolate the a value

12,000/4,000=4,000/4,000a

The 4,000s on the right of the equation cancel.

And 12,000 divided by 4,000 equals 3

The acceleration (a) is 3 meters per second squared (m/s^2)

Next, check to make sure 3 does work by plugging it back into the equation.

12,000=4,000*3

12,000=12,000 ✔

As you can see, the acceleration will be 3 m/s^2

The speed of a wave is its wavelength multiplied by its

vibration

frequency

reflection

amplitude

Answers

The speed of a wave is its wavelength multiplied by its Frequency.

The speed of a wave is its wavelength multiplied by its frequency.

What is frequency ?

The frequency referred to the  number of oscillation per unit time and it is used for defining the cyclic process like rotation, oscillation, wave etc.

The SI unit of the frequency is represented as  Hertz and the symbol λ represents it where one hertz refers to  the wave completed per cycle in one second.

The frequency explains the process of oscillatory and vibration like the mechanical vibration, sound signals, light, frequency waves etc.

If the flashes, then the period is the time between the two flashes and the frequency is the total number of flashes per second.

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5. How much heat is needed to warm .052 kg of gold from 30°C to 120°C? Note: Gold has a specific heat of 136

J/kg °C

Answers

Answer:

Q = 636.48 J

Explanation:

Given that,

The mass of gold, m = 0.052 kg

The temperature increase from 30°C to 120°C.

The specific heat of gold is 136  J/kg °C.

We need to find the heat needed to warm the gold. The formula for heat needed is given by :

[tex]Q=mc\Delta T\\\\Q=0.052\times 136\times (120-30)\\\\Q=636.48\ J[/tex]

So, 636.48 J of heat is needed to warm gold.

Soap bubbles can display impressive colors, which are the result of the enhanced reflection of light of particular wavelengths from the bubbles' walls. For a soap solution with an index of refraction of 1.21, find the minimum wall thickness that will enhance the reflection of light of wavelength 711 nm in air.

Answers

Answer:

the minimum wall thickness that will enhance the reflection of light is 146.9 nm

Explanation:

Given the data in the question;

At the first interface, a phase shift occurs as the incident light is in air that has less refractive index compare to the thin film of soap bubble.

At the second interface, no shift occurs,

condition for constructive interference;

t = ( m + 1/2) × λ/2n

where m = 0, 1, 2, 3 . . . . . .

now, the condition for the constructive interference;

t = mλ/2n

where t is the thickness of the soap bubble,  λ is the wavelength of light and n is the refractive index of soap bubble.

so the minimum thickness of the film which will enhance reflection of light will be;

t[tex]_{min[/tex] =  ( m + 1/2) × λ/2n

we substitute

t[tex]_{min[/tex] =  ( 0 + 1/2) × 711 /2(1.21)

t[tex]_{min[/tex] = 0.5 × 711/2.42

t[tex]_{min[/tex] = 0.5 × 293.80165

t[tex]_{min[/tex] = 146.9 nm

Therefore,  the minimum wall thickness that will enhance the reflection of light is 146.9 nm

A small glider is placed against a compressed spring at the bottom of an air track that slopes upward at an angle of 38.0 ∘ above the horizontal. The glider has mass 9.00×10−2 kg. The spring has 590 N/m and negligible mass. When the spring is released, the glider travels a maximum distance of 1.70 m along the air track before sliding back down. Before reaching this maximum distance, the glider loses contact with the spring.

Required:
a. What distance was the spring originally compressed?
b. When the glider has traveled along the air track 0.80 m from its initial position against the compressed spring, is it still in contact with the spring? What is the kinetic energy of the glider at this point?

Answers

Answer:

x = 0.056 m

ΔKE = 0.489 J

Explanation:

Given that

Angle, θ = 38°

Length, L = 1.7 m

Mass, m = 0.09 kg

Spring constant, K = 590 N/m

If we use the Work-Energy theorem, then we know that Potential Energy, PE = Kinetic Energy, KE

This is mathematically written as

1/2kx² = mgH

The height, H we can get by using the relation

H = L.Sinθ

H = 1.7 * Sin 38

H = 1.7 * 0.6157

H = 1.047 m

Next, we use the Work-Energy theorem

1/2kx² = mgH

1/2 * 590 * x² = 0.09 * 9.8 * 1.047

295 * x² = 0.9234

x² = 0.9235 / 295

x² = 0.00313

x = √0.00313

x = 0.056 m

If the spring loses contact at x = 0.056, definitely, it will also lose contact at x = 0.8

Then we use the formula

ΔKE = mg(H - H1)

ΔKE = mg(xsinθ - x2.sinθ)

Where, x = 1.7 , x2 = 0.8

ΔKE = 0.09 * 9.8 (1.7 * sin 38 - 0.8 * sin 38)

ΔKE = 0.882(1.047 - 0.493)

ΔKE = 0.882 * 0.554

ΔKE = 0.489 J

If you swing an object on the end of a string around a circle, the string pulls on the object to keep it moving in a circle. What is the name of this force?
A. inertial
B. centripetal
C. resistance
D. gravitational

Answers

Answer:

B

Explanation:

The centripetal force keeps an object moving in a circular path. Therefore option (B) is correct.

What is centripetal force?

A centripetal force can be described as a force that makes a body follow a curved path and its direction is orthogonal to the motion of the body. Gravity offers the centripetal force causing astronomical orbits.

The centripetal force is directed perpendicular to the direction of the displacement of an object. It always acts towards the center of the circle on an object moving in a circular path. For example, When spinning a ball on a string, the tension on the rope pulls the object toward the center.

The Centripetal Force can be described as the product of mass and velocity squared, divided by the radius.

F = mv²/r

Where F is the Centripetal force, m is the mass, r is the radius of the circle and v is the velocity of the object.

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If you push with a power of 20 Watts
on a 150 Newton object, how long would
it take to push it over the 4.3 m?

Answers

Answer:

32.25 s

Explanation:

From the question,

P = W/t.............. Equation 1

Where P = Power, W = work done, t = time.

But

W = F×d................. Equation 2

Where F = force and d = distance

Substitute equation 2 into equation 1

P = F×d/t............... Equation 3

make t the subject of euqation 3

t = (F×d)/P............. Equation 4

Givn: F = 150 N, d = 4.3 m, P = 20 watts.

Substitute these values into equation 4

t = (150×4.3)/20

t = 32.25 s

A 1500-kg car travelling 90 km/h[N] collides with a 1200-kg minivan travelling 40 km/h[S]. After the collision, the two vehicles stick together.

a. Calculate the initial momentum of the car and the minivan.

b. Using the law of conservation of momentum, determine the total momentum of the two vehicles after the collision.

c. Calculate the final velocity of the two vehicles after the collision in metres per second.

Answers

Answer:

A) car - 37500 kg*m/s, minivan - 13332 kg*m/s

B) 50832 kg*m/s

C) 18.83 m/s

Explanation:

Realize that sticky collisions are modeled by: m1v1+m2v2=(m1+m2) vf

conevert to m/s....car going 25 m/s, minivan going 11.11 m/s

A) p=mv

p(car)=(1500)(25)

p(car)=37500 kg*m/s

p(minivan)=(1200)(11.11)

p(minivan)=13332 kg*m/s

B) 37500+13332=50832 kg*m/s

C) 37500+13332=(1500+1200) vf

50832=2700(vf)

18.83 m/s = vf

Calculate the first and second order angles for light of wavelength 400. nm and 700. nm of the grating contains 1.00 x 104 lines/cm.

Answers

Answer:

[tex]23.58^{\circ}[/tex] and [tex]53.13^{\circ}[/tex]

[tex]44.43^{\circ}[/tex], second order does not exist

Explanation:

n = Number of lines grating = [tex]1\times10^4\ \text{Lines/cm}[/tex]

[tex]\lambda[/tex] = Wavelength

m = Order

Distance between slits is given by

[tex]d=\dfrac{1}{n}\\\Rightarrow d=\dfrac{1}{1\times 10^4}\\\Rightarrow d=10^{-6}\ \text{m}[/tex]

[tex]\lambda=400\ \text{nm}[/tex]

m = 1

We have the relation

[tex]d\sin\theta=m\lambda\\\Rightarrow \theta=\sin^{-1}\dfrac{m\lambda}{d}\\\Rightarrow \theta=\sin^{-1}\dfrac{1\times 400\times 10^{-9}}{10^{-6}}\\\Rightarrow \theta=23.58^{\circ}[/tex]

m = 2

[tex]\theta=\sin^{-1}\dfrac{2\times 400\times 10^{-9}}{10^{-6}}\\\Rightarrow \theta=53.13^{\circ}[/tex]

The first and second order angles for light of wavelength 400 nm are [tex]23.58^{\circ}[/tex] and [tex]53.13^{\circ}[/tex].

[tex]\lambda=700\ \text{nm}[/tex]

m = 1

[tex]\theta=\sin^{-1}\dfrac{1\times 700\times 10^{-9}}{10^{-6}}\\\Rightarrow \theta=44.43^{\circ}[/tex]

m = 2

[tex]\theta=\sin^{-1}\dfrac{2\times 700\times 10^{-9}}{10^{-6}}[/tex]

Here [tex]\dfrac{2\times 700\times 10^{-9}}{10^{-6}}=1.4>1[/tex] so there is no second order angle for this case.

The first order angle for light of wavelength 700 nm are [tex]44.43^{\circ}[/tex].

Second order angle does not exist.

What type of weather would a continental Polar air mass bring

Answers

Answer:

Continental polar ( cp):

Explanation:

Cold and dry, originating from high latitudes, typically as air flowing out of the polar highs. This air mass often brings the rattleing cold, dry and clear weather on a perfect winter day and also dry and warm weather on a pleasant day in summer.

(d) Suppose you use a spring to launch a payload horizontally from the asteroid so that the payload ends up far from the asteroid, travelling at a speed of 3 m/s. The payload has a mass of 29 kg. If the spring is to be compressed initially an amount of 1.4 m, what stiffness ks must the spring be designed to have

Answers

Answer:

ks= 133.2 N/m

Explanation:

Assuming that we can neglect the gravitational potential energy of the mass, and that no other forces acting on the payload, total mechanical energy must be conserved.This energy, at any time, is part elastic potential energy (stored in the spring) and part kinetic energy.When the spring is initially compressed, the payload is at rest, so all energy is elastic potential.Once the spring has returned to its natural state, all this elastic potential energy must have been turned into kinetic energy.If the payload is launched horizontally, and no gravity is present,this means that its final speed will be horizontal only also, according to Newton's First Law.So, we can write the following equation:

       [tex]\Delta U + \Delta K = 0 (1)[/tex]

where ΔU = -1/2*k*(Δx)²  (2)and ΔK = 1/2*m*v² (3)Replacing in (2) and (3) by the givens, and simplifying, we can find the stiffness ks as follows:

       [tex]k_{s} =\frac{m*v^{2}}{\Delta x^{2}} = \frac{29 kg*(3m/s)^{2}}{(1.4m)^{2}} = 133.2 N/M (4)[/tex]

Name one similarity and one difference between a set and a bump in volleyball??

Answers

Bump: Bumping the ball means a player uses their forearms to pass the ball to a teammate or to hit the ball back over the net to the other team. Set: Setting the ball means a player positions the ball in a way that lets a teammate spike it over the net.

One similarity is the use of physical body whereas one difference is that one is exercise and the other is a sport.

One similarity and one difference

One similarity between a set and a bump in volleyball is the movement and use of legs and hands.

Whereas one difference between a set and a bump in volleyball is that completing several reps of a particular exercise in a row is called a set while on the other hand,  the basic pass in volleyball is known as bump.

Learn more about set here: https://brainly.com/question/1090891

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