Four friends push on the same block in different directions. Allie pushes on the block to the north with a force of 18 N. Bill pushes on the block to the east with a force of 14 N. Chris pushes on the block to south with a force of 23 N. Debra pushes on the block to the west with a force of 20 N. Assuming it does not move vertically, in which directions will the block move? north and west south and east south and west north and east

Answers

Answer 1

Answer:

South and West

Explanation:

Those people are pushing the hardest. It will move south faster than it moves west.


Related Questions

Which scientist proposed a mathematical solution for the wave nature of light?

Answers

Answer:

Explanation:

Christian Huygens

Light Is a Wave!

Then, in 1678, Dutch physicist Christian Huygens (1629 to 1695) established the wave theory of light and announced the Huygens' principle.

Consider two parallel plate capacitors. The plates on Capacitor B have half the area as the plates on Capacitor A, and the plates in Capacitor B are separated by twice the separation of the plates of Capacitor A. If Capacitor A has a capacitance of CA-17.8nF, what is the capacitance of Capacitor? .

Answers

Answer:

CB = 4.45 x 10⁻⁹ F = 4.45 nF

Explanation:

The capacitance of a parallel plate capacitor is given by the following formula:

C = ε₀A/d

where,

C = Capacitance

ε₀ = Permeability of free space

A = Area of plates

d = Distance between plates

FOR CAPACITOR A:

C = CA = 17.8 nF = 17.8 x 10⁻⁹ F

A = A₁

d = d₁

Therefore,

CA = ε₀A₁/d₁ = 17.8 x 10⁻⁹ F   ----------------- equation 1

FOR CAPACITOR B:

C = CB = ?

A = A₁/2

d = 2 d₁

Therefore,

CB = ε₀(A₁/2)/2d₁

CB = (1/4)(ε₀A₁/d₁)

using equation 1:

CB = (1/4)(17.8 X 10⁻⁹ F)

CB = 4.45 x 10⁻⁹ F = 4.45 nF

At what temperature (degrees Fahrenheit) is the Fahrenheit scale reading equal to:_____
(a) 3 times that of the Celsius and
(b) 1/5 times that of the Celsius

Answers

Answer:

C = 26.67° and F = 80°C = -20° and F = -4°

Explanation:

Find:

3 times that of the Celsius and 1/5 times that of the Celsius

Computation:

F = (9/5)C + 32

3 times that of the Celsius

If C = x

So F = 3x

So,

3x = (9/5)x + 32

15x = 9x +160

6x = 160

x = 26.67

So, C = 26.67° and F = 80°

1/5 times that of the Celsius

If C = x

So F = x/5

So,

x/5 = (9/5)x + 32

x = 9x + 160

x = -20

So, C = -20° and F = -4°

A 17.0 g bullet traveling horizontally at 785 m/s passes through a tank containing 13.5 kg of water and emerges with a speed of 534 m/s.
What is the maximum temperature increase that the water could have as a result of this event? (in degrees)

Answers

Answer:

The maximum temperature increase is [tex]\Delta T = 0.0497 \ ^oC[/tex]

Explanation:

From the question we are told that

    The mass of the bullet is [tex]m = 17.0 \ g =0.017 \ kg[/tex]

     The  speed is  [tex]v_1 = 785 \ m/s[/tex]

     The mass of the water is  [tex]m_w = 13.5 \ kg[/tex]

     The velocity it emerged with is  [tex]v_2 = 534 \ m/s[/tex]

Generally due to the fact that energy can nether be created nor destroyed but transferred from one form to another then  

the change in kinetic energy of the bullet =  the heat gained by the water

 So

 The change in kinetic energy of the water is  

          [tex]\Delta KE = \frac{1}{2} m (v_1^2 - v_2 ^2 )[/tex]

substituting values  

        [tex]\Delta KE =0.5 * 0.017 * (( 785)^2 - (534) ^2 )[/tex]

        [tex]\Delta KE = 2814.1 \ J[/tex]

Now the heat gained by the water is

     [tex]Q = m_w* c_w * \Delta T[/tex]

Here [tex]c_w[/tex] is the specific heat of water which has a value  [tex]c_w = 4190 J/kg \cdot K[/tex]

So  since   [tex]\Delta KE = Q[/tex]  

we have that

          [tex]2814.1 = 13.5 * 4190 * \Delta T[/tex]

          [tex]\Delta T = 0.0497 \ ^oC[/tex]

   

What is the separation in meters between two slits for which 594 nm orange light has its first maximum at an angle of 32.8°?

Answers

Answer:

1.1micro meter

Explanation:

Given that

Constructive interference is

ma = alpha x sin theta

Alpha = 1 x 594 x10^ -9/ sin 32.8°

= 1.1 x 10^ -6m

Explanation:

A 10-cm-long thin glass rod uniformly charged to 6.00 nC and a 10-cm-long thin plastic rod uniformly charged to - 6.00 nC are placed side by side, 4.4 cm apart. What are the electric field strengths E1 to E3 at distances 1.0 cm, 2.0 cm, and 3.0 cm from the glass rod along the line connecting the midpoints of the two rods?
A. Specify the electric field strength E1
B. Specify the electric field strength E2
C. Specify the electric field strength E3

Answers

Answer:

A) E(r) = 1.3957 × 10^(5) N/C

B) E(r) = 9.8864 × 10⁴ N/C

C) E(r) = 1.13 × 10^(5) N/C

Explanation:

We are given;

q = 6 nc = 6 × 10^(-9) C

L = 10 cm = 0.1 m

d = 4.4 cm = 0.044 m

r1 = 1 cm = 0.01 m

r2 = 2 cm = 0.02 m

r3 = 3 cm = 0.03 m

Formula for the electric field strength in this question is given as;

E(r) = q/(2π(ε_o)rL) + q/(2π(ε_o)(d - r)L)

When factorized, we have;

E(r) = q/(2π(ε_o)L) × [(1/r) + (1/(d - r))]

Plugging in the relevant values for q/(2π(ε_o)L)

We know that (ε_o) has a constant value of 8.854 × 10^(−12) C²/N².m

Thus; q/(2π(ε_o)L) = (6 × 10^(-9))/(2π(8.854 × 10^(−12)0.1) = 1078.53

Thus;

E(r) = 1078.52 [1/r + 1/(d - r)]

A) E1 is at r = 1 cm = 0.01m

Thus;

E(r) = 1078.52 (1/0.01 + (1/(0.044 - 0.01))

E(r) = 1.3957 × 10^(5) N/C

B) E2 is at r = 2 cm = 0.02 m

Thus;

E(r) = 1078.52 (1/0.02 + (1/(0.044 - 0.02))

E(r) = 9.8864 × 10⁴ N/C

C) E2 is at r = 3 cm = 0.03 m

Thus;

E(r) = 1078.52 (1/0.03 + (1/(0.044 - 0.03))

E(r) = 1.13 × 10^(5) N/C

Proposed Exercises: Strength and Acceleration in Circular Movement In the situation illustrated below, a 7kg sphere is connected to a rope so that it can rotate in a vertical plane around an O axis perpendicular to the plane of the figure. When the sphere is in position A, it has a speed of 3m/s. Determine for this position the modulus of tension on the string and the rate at which the tangential velocity is increased.

Answers

Answer:

81 N

7.1 m/s²

Explanation:

Draw a free body diagram of the sphere.  There are two forces:

Weight force mg pulling straight down,

and tension force T pulling up along the rope.

Sum of forces in the centripetal direction:

∑F = ma

T − mg sin 45° = m v² / r

T = m (g sin 45° + v² / r)

T = (7 kg) (10 m/s² sin 45° + (3 m/s)² / 2 m)

T = 81 N

Sum of forces in the tangential direction:

mg cos 45° = ma

a = g cos 45°

a = (10 m/s²) cos 45°

a = 7.1 m/s²

Suppose you drop paperclips into an open cart rolling along a straight horizontal track with negligible friction. As a result of the accumulating paper clips, explain whether the momentum and kinetic energy increase, decrease, or stay the same.

Answers

Answer:

Stay the same

Explanation:

Since, friction is negligible:

Initial Momentum = Final Momentum

Initial KE = Final KE

m1 * v1 = m2 * v2

When m increases v decreases.

The momentum and kinetic energy remain the same if you drop paper clips into an open cart rolling along a straight horizontal track with negligible friction.

What is friction?

Between two surfaces that are sliding or attempting to slide over one another, there is a force called friction. For instance, friction makes it challenging to push a book down the floor. Friction always moves an object in a direction that is counter to the direction that it is traveling or attempting to move.

Given:

The paperclips into an open cart rolling along a straight horizontal track with negligible friction,

Calculate the momentum, Since friction is negligible,

Initial Momentum = Final Momentum

Initial Kinetic Energy = Final Kinetic Energy

m₁ × v₁ = m₁  × v₂

When m increases, v decreases,

Thus, momentum will remain the same.

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You add 500 mL of water at 10°C to 100 mL of water at 70°C. What is the
most likely final temperature of the mixture?
O A. 80°C
OB. 10-C
OC. 20°C
O D. 60°C

Answers

Answer:

Option (c) : 20°C

Explanation:

[tex]t(final) = \frac{w1 \times t1 + w2 \times t2}{w1 + w2} [/tex]

T(final) = 500* 10 + 100*70/600 = 20°C

A long solenoid consists of 1700 turns and has a length of 0.75 m.The current in the wire is 0.48 A. What is the magnitude of the magnetic field inside the solenoid

Answers

Answer:

1.37 ×10^-3 T

Explanation:

From;

B= μnI

μ = 4π x 10-7 N/A2

n= number of turns /length of wire = 1700/0.75 = 2266.67

I= 0.48 A

Hence;

B= 4π x 10^-7 × 2266.67 ×0.48

B= 1.37 ×10^-3 T

Consider two parallel wires where the magnitude of the left currentis 2 I0(io) and that of the right current is I0(io). Point A is midway between the wires,and B is an equal distance on the other side of the wires.
The ratio ofthe magnitude of the magnetic field at point A to that at point Bis________

Answers

Answer:

Explanation:

At the point midway between wires

magnetic field due to wire having current 2I₀

= 10⁻⁷ x 2 x2I₀ / r     where 2r is the distance between wires .

magnetic field due to wire having current I₀

= 10⁻⁷ x 4 I₀ / r

magnetic field due to wire having current I₀

= 10⁻⁷ x 2I₀ / r    

= 10⁻⁷ x 2 I₀ / r     where 2r is the distance between wires .

these fields are in opposite direction as direction of current is same in both .

net magnetic field = (4 - 2 )x 10⁻⁷ x I₀ / r

= 2 x 10⁻⁷ x  I₀ / r

At point A net magnetic field = 2 x 10⁻⁷ x  I₀ / r

At point B , we shall calculate magnetic field

magnetic field due to nearer wire having current  2 I₀ = 10⁻⁷ x 4 I₀ / r

magnetic field due to wire far away = 10⁻⁷ x 2 I₀ / 3r

These magnetic fields act in the same direction so they will add up

net magnetic field = [ (4 I₀ / r)  + (2 I₀ / 3r) ] x 10⁻⁷

= (14 I₀ / 3r ) x 10⁻⁷

Magnetic field at point B = (14 I₀ / 3r ) x 10⁻⁷

Ratio of field at A and B

= 3 / 7 . Ans

The ratio of the magnitude of the magnetic field at point A to point B is :

3 / 7

Given data :

Magnitude of the left current is  2I₀

Magnitude of the right current is  I₀

First step : Determine the magnetic field at point A  

The magnetic field due to the left current ( 2I₀ )

10⁻⁷ * 2 * 2I₀ / r       ( 2r = distance between wires )

The magnetic field due to the right current ( I₀ )

10⁻⁷ * 2 I₀ / r

From the expressions above the magnetic fields are in  opposite direction

∴ Net magnetic field = (4 - 2 )* 10⁻⁷ * I₀ / r =   2 * 10⁻⁷ *  I₀ / r

Hence The magnetic field at point A = 2 * 10⁻⁷ *  I₀ / r

Next step : determine the magnetic field at point B

Magnetic field due to the closest wire to point B ( i.e.2I₀ ) = 10⁻⁷ * 4 I₀ / r

Magnetic field due to the wire away from point A = 10⁻⁷ * 2 I₀ / 3r

Since the fields acts in the same directions

The net magnetic field =  (4 I₀ / r)  + (2 I₀ / 3r) ] * 10⁻⁷ = ( 14 I₀ / 3r ) * 10⁻⁷

Hence The magnetic field at point A = ( 14 I₀ / 3r ) * 10⁻⁷

Therefore the ratio of the magnitude of the magnetic field at point A to point B  =  3/ 7

Hence we can conclude that the ratio of the magnitude of the magnetic field at point A to point B  = 3 / 7

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Suppose that a sound source is emitting waves uniformly in all directions. If you move to a point twice as far away from the source, the frequency of the sound will be:________.
a. one-fourth as great.
b. half as great.
c. twice as great.
d. unchanged.

Answers

Answer:

d. unchanged.

Explanation:

The frequency of a wave is dependent on the speed of the wave and the wavelength of the wave. The frequency is characteristic for a wave, and does not change with distance. This is unlike the amplitude which determines the intensity, which decreases with distance.

In a wave, the velocity of propagation of a wave is the product of its wavelength and its frequency. The speed of sound does not change with distance, except when entering from one medium to another, and we can see from

v = fλ

that the frequency is tied to the wave, and does not change throughout the waveform.

where v is the speed of the sound wave

f is the frequency

λ is the wavelength of the sound wave.

CAN SOMEONE HELP ME PLEASE ITS INTEGRATED SCIENCE AND I AM STUCK

Answers

Answer:

[tex]\huge \boxed{\mathrm{Option \ D}}[/tex]

Explanation:

Two forces are acting on the object.

Subtracting 2 N from both forces.

2 N → Object ← 5 N

- 2 N                 - 2N

0 N → Object ← 3 N

The force 3 N is pushing the object to the left side.

The mass of the object is 10 kg.

Applying formula for acceleration (Newton’s Second Law of Motion).

a = F/m

a = 3/10

a = 0.3

Two ice skaters, Paula and Ricardo, initially at rest, push off from each other. Ricardo weighs more than Paula.
A. Which skater, if either, has the greater momentum after the push-off? Explain.
B. Which skater, if either, has the greater speed after the push-off? Explain.

Answers

Answer:

the two ice skater have the same momentum but the are in different directions.

Paula will have a greater speed than Ricardo after the push-off.

Explanation:

Given that:

Two ice skaters, Paula and Ricardo, initially at rest, push off from each other. Ricardo weighs more than Paula.

A. Which skater, if either, has the greater momentum after the push-off? Explain.

The law of conservation of can be applied here in order to determine the skater that possess a greater momentum after the push -off

The law of conservation of momentum states that the total momentum of two  or more objects acting upon one another will not change, provided there are no external forces acting on them.

So if two objects in motion collide, their total momentum before the collision will be the same as the total momentum after the collision.

Momentum is the product of mass and velocity.

SO, from the information given:

Let represent the mass of Paula with [tex]m_{Pa}[/tex] and its initial velocity with [tex]u_{Pa}[/tex]

Let represent the mass of Ricardo with [tex]m_{Ri}[/tex] and its initial velocity with [tex]u_{Ri}[/tex]

At rest ;

their velocities will be zero, i.e

[tex]u_{Pa}[/tex] = [tex]u_{Ri}[/tex] = 0

The initial momentum for this process can be represented as :

[tex]m_{Pa}[/tex][tex]u_{Pa}[/tex] +  [tex]m_{Ri}[/tex][tex]u_{Ri}[/tex] = 0

after push off from each other then their final velocity will be [tex]v_{Pa}[/tex] and [tex]v_{Ri}[/tex]

The we can say their final momentum is:

[tex]m_{Pa}[/tex][tex]v_{Pa}[/tex] +   [tex]m_{Ri}[/tex][tex]v_{Ri}[/tex] = 0

Using the law of conservation of momentum as states earlier.

Initial momentum = final momentum = 0

[tex]m_{Pa}[/tex][tex]u_{Pa}[/tex] +  [tex]m_{Ri}[/tex][tex]u_{Ri}[/tex] =  [tex]m_{Pa}[/tex][tex]v_{Pa}[/tex] +   [tex]m_{Ri}[/tex][tex]v_{Ri}[/tex]

Since the initial velocities are stating at rest then ; u = 0

[tex]m_{Pa}[/tex](0) + [tex]m_{Pa}[/tex](0) = [tex]m_{Pa}[/tex][tex]v_{Pa}[/tex] +   [tex]m_{Ri}[/tex][tex]v_{Ri}[/tex]

[tex]m_{Pa}[/tex][tex]v_{Pa}[/tex] +   [tex]m_{Ri}[/tex][tex]v_{Ri}[/tex]  = 0

[tex]m_{Pa}[/tex][tex]v_{Pa}[/tex] = - [tex]m_{Ri}[/tex][tex]v_{Ri}[/tex]

Hence, we can conclude that the two ice skater have the same momentum but the are in different directions.

 B. Which skater, if either, has the greater speed after the push-off? Explain.

Given that Ricardo weighs more than Paula

So [tex]m_{Ri} > m_{Pa}[/tex] ;

Then [tex]\mathsf{\dfrac{{m_{Ri}}}{m_{Pa} }= 1}[/tex]

The magnitude of their momentum which is a product of mass and velocity can now be expressed as:

[tex]m_{Pa}[/tex][tex]v_{Pa}[/tex] =  [tex]m_{Ri}[/tex][tex]v_{Ri}[/tex]

The ratio is

[tex]\dfrac{v_{Pa}}{v_{Ri}} =\dfrac{m_{Ri}}{m_{Pa}} = 1[/tex]

[tex]v_{Pa} >v_{Ri}[/tex]

Therefore, Paula will have a greater speed than Ricardo after the push-off.

(A) Both the skaters have the same magnitude of momentum.

(B) Paula has greater speed after push-off.

Conservation of momentum:

Given that two skaters Paula and Ricardo are initially at rest.

Ricardo weighs more than Paula.

Let us assume that the mass of Ricardo is M, and the mass of Paula is m.

Let their final velocities be V and v respectively.

(A) Initially, both are at rest.

So the initial momentum of Paula and Ricardo is zero.

According to the law of conservation of momentum, the final momentum of the system must be equal to the initial momentum of the system.

Initial momentum = final momentum

0 = MV + mv

MV = -mv

So, both of them have the same magnitude of momentum, but in opposite directions.

(B) If we compare the magnitude of the momentum of Paula and Ricardo, then:

MV = mv

M/m = v/V

Now, we know that M>m

so, M/m > 1

therefore:

v/V > 1

v > V

So, Paula has greater speed.

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A ball travels with velocity given by [21] [ 2 1 ​ ], with wind blowing in the direction given by [3−4] [ 3 −4 ​ ] with respect to some co-ordinate axes. What is the size of the velocity of the ball in the direction of the wind?

Answers

Answer:

2/5 m/s

Explanation:

There are two vectors  v and w . Let θ be angle b/w the two vector.

[tex]cos\theta =\frac{\overleftarrow{v}\cdot \overleftarrow{w}}{\left | v \right |\left | w \right |}\\=\frac{6-4}{\sqrt(2^2+1^2)\sqrt(3^2+4^2)} =\frac{2}{5\sqrt(5)}[/tex]

velocity of the ball in direction of the the wind

[tex]\left | vcos\theta \right |\\\left | v \right |cos\theta\\\sqrt(2^2+1^2)\frac{2}{5\sqrt(5)} = \frac{2}{5}[/tex]

The size of the velocity of the ball in the direction of the wind is 2/5 ms.

Calculation of the size of velocity:

Since there are two vectors v and w

Also, here we assume θ be angle b/w the two vector.

So

Cos θ = 6-4 / √(2^2 + 1^2) √(3^2 + 4^2)

= 2/5√5

Now the velocity of the ball should be

= √(2^2 + 1^2) 2 ÷ 5√(5)

= 2 /5

hence, The size of the velocity of the ball in the direction of the wind is 2/5 ms.

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Niobium metal becomes a superconductor when cooled below 9 K. Its superconductivity is destroyed when the surface magnetic field exceeds 0.100 T. In the absence of any external magnetic field, determine the maximum current a 5.68-mm-diameter niobium wire can carry and remain superconducting.

Answers

Answer:

The current is  [tex]I = 1420 \ A[/tex]

Explanation:

From the question we are told that

   The  diameter of the wire is  [tex]d = 5.68 \ mm = 0.00568 \ m[/tex]

    The  magnetic field is  [tex]B = 0.100 \ T[/tex]

   

Generally the radius of the wire is mathematically evaluated as

       [tex]r = \frac{d}{2}[/tex]

substituting values

     [tex]r = \frac{ 0.00568}{2}[/tex]

     [tex]r = 0.00284 \ m[/tex]

Generally the magnetic field is mathematically represented as

       [tex]B = \frac{\mu_o * I}{ 2 \pi r }[/tex]

=>    [tex]I =\frac{ B * 2 \pi r }{\mu_o}[/tex]

Here [tex]\mu_o[/tex] is the permeability of free space  with value [tex]\mu_o = 4 \pi *10^{-7} N/A^2[/tex]

substituting values

=>     [tex]I =\frac{ 0.100 * 2 * 3.142 * 0.00284 }{ 4 \pi * 10^{-7}}[/tex]

=>     [tex]I = 1420 \ A[/tex]

The frequency of light emitted from hydrogen present in the Andromeda galaxy has been found to be 0.10% higher than that from hydrogen measured on Earth.
Is this galaxy approaching or receding from the Earth, and at what speed?

Answers

Answer:

3x10^5m/s

Explanation:

See attached file

Explanation:

The speed of the light emitted from the earth is approaching the galaxy at [tex]3\times 10^5\;\rm m/s[/tex].

Doppler's Effect

According to the Doppler effect, the difference between the frequency at which light wave leave a source and reaches an observer is caused by the relative motion of the observer and the wave source.

Given that the difference in the frequency is 0.10 %. The speed of light emitted from the galaxy can be calculated by the Doppler effect.

[tex]\dfrac {\Delta f}{f} = \dfrac {v}{c}[/tex]

Where f is the frequency of the light, v is the speed of light emitted from the galaxy and c is the speed of light emitted from the earth.

[tex]\dfrac {0.10 f}{100 f} = \dfrac {v}{3\times 10^8}[/tex]

[tex]v = 3\times 10^5\;\rm m/s[/tex]

Hence we can conclude that the speed of the light emitted from the earth is approaching the galaxy at [tex]3\times 10^5\;\rm m/s[/tex].

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A wave travelling along the positive x-axis side with a
frequency of 8 Hz. Find its period, velocity and the distance covered
along this axis when its wavelength and amplitude are 40 and 15 cm
respectively.​

Answers

Explanation:

The frequency is given to be f = 8 Hz.

Period is the inverse of frequency.

T = 1/f = 0.125 s

Velocity is wavelength times frequency.

v = λf = (0.40 m) (8 Hz) = 3.2 m/s

The wave travels 3.2 meters every second.

An organ pipe open at both ends is 1.5 m long. A second organ pipe that is closed at one end and open at the other is 0.75 m long. The speed of sound in the room is 330 m/s. Which of the following sets of frequencies consists of frequencies which can be produced by both pipes?

a. 110Hz,220Hz, 330 Hz
b. 220Hz 440Hz 66 Hz
c. 110Hz, 330Hz, 550Hz
d. 330 Hz, 550Hz, 440Hz
e. 660Hz, 1100Hz, 220Hz

Answers

Answer:

A. 110Hz,220Hz, 330 Hz

Explanation:

for organ open at open both ends;

the length of the organ for the fundamental frequency, L = A---->N + N----->A

A---->N  = λ /4 and N----->A = λ /4

L = λ /4 + λ /4 = λ /2

[tex]L = \frac{\lambda}{2} \\\\\lambda = 2L[/tex]

λ  = 2 x 1.5m = 3.0 m

Wave equation is given by;

V = Fλ

Where;

V is the speed of sound

F is the frequency of the wave

F = V/ λ

F₀ = V / 2L

Where;

F₀  is the fundamental frequency

F₀ = 330 / 2(1.5)

F₀ = 330 / 3

F₀ = 110 Hz

the length of the organ for the first overtone, L = A---->N + N----->A + A----->N +  N----->A

L = 4λ /4

L = λ

λ = 1.5 m

F₁ = 330 / 1.5

F₁ = 220 Hz

Thus, F₁ = 2F₀

For open organ at one end

the length of the organ for the fundamental frequency, L = N------A

L = λ /4

λ = 4L

F₀ = V/4L

F₀ = 330 / (4 x 0.75)

F₀ = 110 Hz

the length of the organ for the first overtone, L = N-----N + N-----A

L = λ/2 + λ / 4

L = 3λ /4

F₁ = 3F₀

F₁ = 3 x 110

F₁ = 330 Hz

Thus the fundamental frequency for both organs is 110 Hz,

The first overtone for the organ open at both ends is 220 Hz

The first overtone for the organ open at one end is 330 Hz

The correct option is "A. 110Hz,220Hz, 330 Hz"

The correct option is option (A)

the frequencies produced by the pipes are (A) 110Hz,220Hz, 330 Hz

Frequencies and overtones:

(I) For an organ pipe open at open both ends the frequency of different modes is given by:

F =  nv/2L

where

F is the frequency

L is the length of the organ pipe

v is the speed of the wave

and, n is the mode of frequency

the fundamental frequency corresponds to n = 1, given by:

F₀ = v/2L

F₀ = 330 / 2(1.5)

F₀ = 330 / 3

F₀ = 110 Hz

The first overtone corresponds to n = 2, the second overtone corresponds to n = 3, and so on...

F₁ =2v/2L

F₁ = 330 / 1.5

F₁ = 220 Hz

Thus, F₁ = 2F₀

The difference between successive overtones is F₀

(II) For an organ pipe open at one end the frequency of different modes is given by:

F =  nv/4L

where

F is the frequency

L is the length of the organ pipe

v is the speed of the wave

and, n is the mode of frequency

the fundamental frequency corresponds to n = 1, given by:

F₀ = V/4L

F₀ = 330 / (4 x 0.75)

F₀ = 110 Hz

For an organ pipe open at one end, only those overtones are present which correspond to odd n, that is n = 3,5,...so:

F₁ = 3F₀

F₁ = 3 x 110

F₁ = 330 Hz

Learn more about overtones:

https://brainly.com/question/1515875?referrer=searchResults

A radar installation operates at 9000 MHz with an antenna (dish) that is 15 meters across. Determine the maximum distance (in kilometers) for which this system can distinguish two aircraft 100 meters apart.

Answers

Answer:

R = 36.885 km

Explanation:

In order to distinguish the two planes we must use the Rayleigh criterion that establishes two distinguishable objects if in their diffraction the central maximum of one coincides with the first minimum of the other

The diffraction equation for slits is

            a sin θ = m λ

the first minimum occurs for m = 1

             sin θ = λ a

as the diffraction experiments the angles are very small, we approximate

             sin θ = θ

 

             θ = λ / a

This expression is for a slit, in the case of circular objects, when solving the system in polar coordinates, a numerical constant appears, leaving the expression of the form

            θ = 1.22 λ / a

In this problem they give us the frequency, let's find the wavelength with the relation

           c = λ f

           λ = c / f

           θ = 1.22 c/ f a

since they ask us for the distance between the planes, we can use the definition of radians

          θ = s / R

if we assume that the distance is large, we can approximate the arc to the horizontal distance

          s = x

       

we substitute

             x / R = 1.22 c / fa

             R = x f a / 1.22c

Let's reduce the magnitudes to the SI system

            f = 9000 MHz = 9 109 Hz

            a = 15 m

           x = 100 m

let's calculate

            R = 100 10⁹ 15 / (1.22 3 108)

            R = 3.6885 10⁴ m

let's reduce to km

            R = 3.6885 10¹ km

            R = 36.885 km

A train on one track moves in the same direction as a second train on the adjacent track. The first train, which is ahead of the second train and moves with a speed of 36.4 m/s , blows a horn whose frequency is 123 Hz .what is its speed?

Answers

Answer:

51. 7m/s

Explanation:

Take speed of sound in air = 340 m/s

fp = fs (V + Vp)/(V + Vs)

128 = 123 (340 + Vp)/(340 + 36.4)

Vp = 51.7m/s

Explanation:

A rectangular coil having N turns and measuring 15 cm by 25 cm is rotating in a uniform 1.6-T magnetic field with a frequency of 75 Hz. The rotation axis is perpendicular to the direction of the field. If the coil develops a sinusoidal emf of maximum value 56.9 V, what is the value of N?
A) 2
B) 4
C) 6
D) 8
E) 10

Answers

Answer:

A) 2

Explanation:

Given;

magnetic field of the coil, B = 1.6 T

frequency of the coil, f = 75 Hz

maximum emf developed in the coil, E = 56.9 V

area of the coil, A = 0.15 m x 0.25 m = 0.0375 m²

The maximum emf in the coil is given by;

E = NBAω

Where;

N is the number of turns

ω is the angular velocity = 2πf = 2 x 3.142 x 75 = 471.3 rad/s

N = E / BAω

N = 56.9 / (1.6 x 0.0375 x 471.3)

N = 2 turns

Therefore, the value of N is 2

A) 2

The hot glowing surfaces of stars emit energy in the form of electromagnetic radiation. It is a good approximation to assume that the emissivity eee is equal to 1 for these surfaces.

Required:
a. Find the radius RRigel of the star Rigel, the bright blue star in the constellation Orion that radiates energy at a rate of 2.7 x 10^31 W and has a surface temperature of 11,000 K.
b. Find the radius RProcyonB of the star Procyon B, which radiates energy at a rate of 2.1 x 10^23 W and has a surface temperature of 10,000 K. Assume both stars are spherical. Use σ=5.67 x 10−8^ W/m^2*K^4 for the Stefan-Boltzmann constant.

Answers

Given that,

Energy [tex]H=2.7\times10^{31}\ W[/tex]

Surface temperature = 11000 K

Emissivity e =1

(a). We need to calculate the radius of the star

Using formula of energy

[tex]H=Ae\sigma T^4[/tex]

[tex]A=\dfrac{H}{e\sigma T^4}[/tex]

[tex]4\pi R^2=\dfrac{H}{e\sigma T^4}[/tex]

[tex]R^2=\dfrac{H}{e\sigma T^4\times4\pi}[/tex]

Put the value into the formula

[tex]R=\sqrt{\dfrac{2.7\times10^{31}}{1\times5.67\times10^{-8}\times(11000)^4\times 4\pi}}[/tex]

[tex]R=5.0\times10^{10}\ m[/tex]

(b). Given that,

Radiates energy [tex] H=2.1\times10^{23}\ W[/tex]

Temperature T = 10000 K

We need to calculate the radius of the star

Using formula of radius

[tex]R^2=\dfrac{H}{e\sigma T^4\times4\pi}[/tex]

Put the value into the formula

[tex]R=\sqrt{\dfrac{2.1\times10^{23}}{1\times5.67\times10^{-8}\times(10000)^4\times4\pi}}[/tex]

[tex]R=5.42\times10^{6}\ m[/tex]

Hence, (a). The radius of the star is [tex]5.0\times10^{10}\ m[/tex]

(b). The radius of the star is [tex]5.42\times10^{6}\ m[/tex]



48. A patient presents with a thrombosis in
the popliteal vein. This thrombosis most likely
causes reduction of blood flow in which of the
following veins?

Answers

Answer:

the interation blood veins

Explanation:

When light is either reflected or refracted, the quantity that does not change in either process is its

Answers

Answer:

Frequency

Explanation:

When waves travel from one medium to another, it is only the frequency of the wave that remains constant . when a wave is refracted at the boundary between two media, the wave will slow down and its wavelength decreases. The wave usually bends at the interface between the two media. The wavelength and speed of a wave may change at the boundary between two media but its frequency remains the same.

Hence the frequency of light is its only property that remains constant.

An electron has an initial velocity to the south but is observed to curve upward as the result of a magnetic field. This magnetic field must have a component:___________
a) north
b) upwards
c) downwards
d) east
e) west

Answers

Answer:

e) west

Explanation:

According to Lorentz left hand rule, the left hand is used to represent the motion of an electron in a magnetic field. Hold out the left hand with the fingers held out parallel to the palm, and the thumb held at right angle to the other fingers. If the thumb represents the motion of the electron though the field, and the other fingers represent the direction of the field, then the palm will push in the direction of the force on the particle.

In this case, if we point the thumb to the south (towards your body), with the palm facing up, then the fingers will point west.

An electron is accelerated from rest through a potential difference. After acceleration the electron has a de Broglie wavelength of 880 nm. What is the potential difference though which this electron was accelerated

Answers

Answer:

3x10⁴v

Explanation:

Using

Wavelength= h/ √(2m.Ke)

880nm = 6.6E-34/√ 2.9.1E-31 x me

Ke= 6.6E-34/880nm x 18.2E -31.

5.6E-27/18.2E-31

= 3 x 10⁴ Volts

Consult Interactive Solution 27.18 to review a model for solving this problem. A film of oil lies on wet pavement. The refractive index of the oil exceeds that of the water. The film has the minimum nonzero thickness such that it appears dark due to destructive interference when viewed in visible light with wavelength 653 nm in vacuum. Assuming that the visible spectrum extends from 380 to 750 nm, what is the longest visible wavelength (in vacuum) for which the film will appear bright due to constructive interference

Answers

Answer:

Explanation:

In the given case for destructive interference , the condition is,

path difference = (2n+1)λ /2  where n is an integer and λ is wavelength

2 μ d = (2n+1)λ /2

Putting λ = 653 nm

for minimum thickness n = 0

2 μ d = 653 / 2 nm

= 326.5 nm

For constructive interference the condition is

2 μ d = n λ₁

326.5 nm = n λ₁

λ₁ = 326.5 / n  

For n = 1

λ₁ = 326.5 nm ,

or , 326.5nm .

Longest wavelength possible is 326.5

Which notation is better to use? (Choose between 4,000,000,000,000,000 m and 4.0 × 1015 m)

Answers

Answer:

4 x 10¹⁵

Explanation:

A 28.0 kg child plays on a swing having support ropes that are 2.30 m long. A friend pulls her back until the ropes are 45.0 ∘ from the vertical and releases her from rest.
A: What is the potential energy for the child just as she is released, compared with the potential energy at the bottom of the swing?
B: How fast will she be moving at the bottom of the swing?
C: How much work does the tension in the ropes do as the child swings from the initial position to the bottom?

Answers

Answer

A)184.9J

B)=3.63m/s

C) Zero

Explanation:

A)potential energy of the child at the initial position, measured relative the her potential energy at the bottom of the motion, is

U=Mgh

Where m=28kg

g= 9.8m/s

h= difference in height between the initial position and the bottom position

We are told that the rope is L = 2.30 m long and inclined at 45.0° from the vertical

h=L-Lcos(x)= L(1-cosx)=2.30(1-cos45)

=0.674m

Her Potential Energy will now

= 28× 9.8×0.674

=184.9J

B)we can see that at the bottom of the motion, all the initial potential energy of the child has been converted into kinetic energy:

E= 0.5mv^2

where

m = 28.0 kg is the mass of the child

v is the speed of the child at the bottom position

Solving the equation for v, we find

V=√2k/m

V=√(2×184.9/28

=3.63m/s

C)we can find work done by the tension in the rope is given using expresion below

W= Tdcosx

where W= work done

T is the tension

d = displacement of the child

x= angle between the directions of T and d

In this situation, we have that the tension in the rope, T, is always perpendicular to the displacement of the child, d. x= 90∘ and cos90∘=0 hence, the work done is zero.

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