A cosmic-ray proton in interstellar space has an energy of 10.0 MeV and executes a circular orbit having a radius equal to that of Mercury’s orbit around the Sun (5.80 × 1010m). What is the magnetic field in that region of space?

Answer:

3.948m/s

Explanation:

To solve this we need to apply Doppler effect theory

So

To find the frequency received by insect will be gotten when the Source and observer both are moving in same direction which is given by

f1 = f0 x (V - Vo)/(V - Vs)

f0 = 30.0 kHz

V = 344 m/s

Vs will now be the speed of the bat and

Vo will be the speed of the object which is = 0 m/s

So substituting we have

f1 = 30 x 10^3 x (344- 0)/(344- Vs)

Next to find the frequency reflected by wall we use

f2 = f1 x (V + Vs)/(V + Vo)

So substituting the value of f1 calculated above we have

f2 = 30 x 10^3 x (344 + Vs) x (344 - 0)/[(344 - Vs) x (344 + 0)]

f2 = 30 x 10^3 x (344 + Vs)/(344- Vs)

But the beat frequency detected by bat is 700 Hz,

So we say

f2 - f0 = 700 Hz

30 x 10^3 x (344+ Vs)/(344 - Vs) - 30x 10^3 = 700

(344 + Vs)/(344 - Vs) = 1 + 700/30000 = 1.023

344 + Vs = 344 x 1.023 - Vs x 1.0233

Vs = 344 x ( 1.023 - 1)/(1 + 1.023)

So finally

Vs = Speed of source that is the bat is = 3.949m/s

Answer: Yes.

Explanation: Because of the crystalline structure.

Answer:

Answer:

B .There are multiple paths the electrons can take through the circuit, and it is possible for the electron to pass through one circuit component but not another.

Explanation:

Because in a parallel circuit, all components are connected across each other, forming exactly two sets of electrically common points.

This question is incomplete, the complete question is;

A weightlifter holds a 1,300 N barbell 1 meter above the ground. One end of a 2-meter-long chain hangs from the center of the barbell. The chain has a total weight of 400 N. How much work (in J) is required to lift the barbell to a height of 2 m?

What is the average force (average with respect to height of the barbell from the ground) exerted by the weightlifter in the process?

Answer: Average force exerted by the weightlifter in the process = 1600N

Explanation:

To find Work done to lift a barbell and half of the hanging chain we say;

W₁ = ( 1300N + (1/2 × 400N)) × 1m

W₁ = (1300 + 200) Nm

W₁ = 1500J

now work done to lift the upper half of the chain we say:

W₂ = (1/2 × 400N) ×  (1/2 × 1m)

W₂ = 200N × 0.5m

W₂ = 100J

So total work done will be

W = W₁ + W₂

W = 1500J + 100J

W = 1600J

To find the average force exerted by the weight lifter, we say;

F = W/D

F = (1600 / 1m) N

F = 1600N

∴Average force = 1600N

1.085m

Explanation:

Using

a= lambda/sinစ

Sinစ= (587.5*10^-9) x 0.75*10^-3

= 0.000783

Sinစ=0.875*10^-3/d

0.000783= 0.875/d

d= 1.085m

Answer:

Explanation:

A ball thrown horizontally is a type of projectile which moves in a trajectory path. It moves with respect to its momentum and gravitational pull of the earth.

The motion of the thrown ball can be described in two components, horizontal and vertical. In horizontal component of motion of the thrown ball, gravitational force is zero. While in the case of vertical component, the gravitational pull causes the path followed.

The force of gravity pulls the ball towards the surface of the earth gradually, thus forming a trajectory path.

Complete Question

The complete question is shown on the first uploaded image  

Answer:

The theoretical angular magnification lies within the angular magnification range

Explanation:

From the question we are told that

   The  focal length of  B  is  [tex]f_{objective } = 43.0 \ cm[/tex]

    The focal length of  A  is   [tex]f_{eye} = 10.4 \ cm[/tex]

The  theoretical angular  magnification is mathematically represented as

           [tex]m = \frac{f_{objective }}{f_{eye}} = \frac{43.0}{10.4}[/tex]

            [tex]m = \frac{f_{objective }}{f_{eye}} = 4.175[/tex]

Form the question the measured angular magnification ranges from 4 -5

So from the value calculated and the value given we can deduce that the theoretical angular  magnification lies within the angular magnification range

Thus the response above is correct.

Learn more about behavioral approach here:

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Answer:

(A.) (- 4.33, 6.33 , 0); (B.) (- 3.66, 7.5, 0); (C.) average at (A) (- 4.33, 6.33 , 0) ; (D.) (- 0.2165, 0.3165, 0)

Explanation:

Given the following :

Time - - - - - - - 6.6s - - - - - - - - - 6.9s - - - - - 7.2s

Position - (1.8,5.0,0) - (0.5,6.9,0) - - (−0.4,9.5,0)

(a) Between 6.6 s and 6.9 s, what was the bee's average velocity?

Vavg = Distance / time

[(0.5,6.9,0) - (1.8,5.0,0)] / 6.9 - 6.6

Vavg = [(0.5 - 1.8), (6.9 - 5.0), (0 - 0)] / 0.3

Vavg = - 1.3 / 0.3, 1.9/0.3, 0/3

Vavg = (- 4.33, 6.33 , 0)

b) Between 6.6 s and 7.2 s, what was the bee's average velocity?

Vavg = [(−0.4,9.5,0) - (1.8,5.0,0)] / 7.2 - 6.6

Vavg = - 2. 2/0.6, 4.5/0.6, 0/0.6

Vavg = (- 3.66, 7.5, 0)

c.) Of the two averages (- 4.3, 6.3 , 0) is closer to the instantaneous Velocity at 6.6s

D.) (d) Using the best information available, what was the displacement of the bee during the time interval from 6.6 s to 6.65 s?

Displacement = Velocity * time

Vavg between 6.6 to 6.9 ; time = (6.65 - 6.6) = 0.05 s

= (- 4.33, 6.33 , 0) * 0.05

= (- 0.2165, 0.3165, 0)

Explanation:

Given:

v₀ = 30 m/s

v = 0 m/s

a = -7.0 m/s²

Find: t and Δx

v = at + v₀

0 m/s = (-7.0 m/s²) t + 30 m/s

t = 4.3 seconds

v² = v₀² + 2aΔx

(0 m/s)² = (30 m/s)² + 2 (-7.0 m/s²) Δx

Δx = 64 meters

Answer:

C. Newton's 3rd law

Explanation:

Swimmers must stroke downward in the water to stay afloat and propel forward. This movement is equal and opposite to the force the water exerts against the swimmer to stop them from moving.

Answer:

c. Newton's Third Law

Explanation:

I got it correct on the test

Answer: -5 m/s^2

Explanation: a = v - u/t

                         = 32 - 47/3

                         = -15/3

                         = -5 m/s^2

Answer:  Kb = 1.89 × 10⁻⁶

Explanation:

R-NH₃+ (aq) <----------> R-NH₂ (aq) + H^+ (aq)

Ecell = - ( 0.0592 / n ) log Kₐ

Where, Ecell = 0.731 - 0.241 = 0.490 V

Therefore, 0.490 = - (0.0592 / 1 ) log Kₐ

Therefore, Kₐ = 5.30 × 10⁻⁹

Thus, Kb = Kw / Ka = ( 1.0 × 10⁻¹⁴ / 5.30 × 10⁹ ) = 1.89 × 10⁻⁶

Answer:

A. We are given t= 40s, x=1.25t²

So The horizontal distance it is from the airport will be = 1.25(40)²= 2000meters

B. y= 0.03t³

So y= 0.03(40)³= 1920m altitude

C. To find speed

We say

dx/dt=Vx= 2.5t= 100

dy/dy = 0.09t= 144

So speed= √100²+144²= 175.31m/s

Answer:

The Gravitational force.

Answer:

First to find deceleration of the train we use

v²= u²+ 2as

56²= 73²+ 2(0.5)a

a= -2193mi/hr²

Then we find time in which the train does the intersection

Using

S= ut+ 1/2 at²

1= 73t-1/2(1293)t²

t =68.5s

But since the train is to intersect in 3.9s the time will be the difference which is

65.68s

So finding acceleration

S= ut + 1/2at²

1.3mi= 45/3600mi/s(65.58s)+ 1/2a(65.5)²

So a= 1.179ft/s²

To find velocity we use

V= u + at

= 45/3600mi/s + (-2.33E-4mis²)(65.58s)

V= 0.0271mi/s

= 97.6ft/s

Answer: 0.757m; 0.881m; 432.70Hz; 371.89Hz

Explanation:

Give the following :

Velocity of train (Vt) = 26m/s

Frequency of sound (Fs) = 420Hz

Speed of sound (Vs) = 344m/s

1) wavelength = (Vs - Vt) / Fs

Wavelength = (344 - 26) / 420 = 318/420 = 0.757m

11) Wavelength = (Vs + Vt) / Fs

Wavelength = (344 + 26) / 420 = 370/420 = 0.881m

111) According to the doppler effect :

Fl = [(V + Vl) / (V + Vs)] * fs

Fl = frequency of listener ; fs = frequency of sound source ; V = speed of sound ; Vl = Velocity of listener ; Vs = speed of sound source

Vs = - ve (train moving towards listener)

Fl = [(V + Vl) / (V + Vs)] * fs

Fl = [(344 + 0) / (344 - 26)] * 400

Fl = (344 / 318) * 400 = 432.70Hz

1V) Vs = + ve (train moving away listener)

Fl = [(V + Vl) / (V + Vs)] * fs

Fl = [(344 + 0) / (344 + 26)] * 400

Fl = (344 / 370) * 400 = 371.89Hz

Answer:

A. We know that amplitude at x is

Asin (kx)

But k= 2πf/v

k= 2*3.132*235/193= 7.65

So A = 0.35*sin( 7.65x 0.18)= 0.00841m

C

Vmax = Amplitude x angular velocity

= 0.0084 x 2πf

= 0.0084* 2*3.142* 235= 12.4m/s

D. Maximum acceleration = omega² x Amplitude

= (2πf)²* 0.00841= 183.40m/s²

Answer:

we measure sound intensity in Decibels.

Answer:

Explanation:

We measure  sound power or sound pressure in decibels.

They  were named in honour of Alexander Graham Bell,( the inventor of both the telephone and the audiometer).

Answer:

1) A) Lens converging, B) lens 1 converging

2) A and B object is at a greater distance than its focal

3)

Explanation:

For this exercise we will give some more important characteristics of the lenses.

we use the equation of the constructor

         1 / f = 1 / p + 1 / q

where f is the focal length and p and qlas the distance to the object and image, respectively.

We must also use how the lenses are combined

       1 / = 1 / f₁ + 1 / f₂

where feq is the equivalent focal length and it is assumed that the two lenses are in contact.

By applying the equation of the constructor you get some characteristics of the image

Converging lens

* Object farther than focal length real and inverted image

* Oject after focal virtual image and right

Diverging lens

* for all virtual image distance and right

with this information let's analyze the different configurations

1) let's do an analysis for each scenario

A) since lens 1 and lens 2 create real image they must be converging lenses

B) Lens 1 must be convergent and lens 2 can be of both types

C) Lens 1 can be convergent with the object that is at a shorter distance than at focal point (p <f₁). The same analysis diverges for lens 2, the object that is the image of the other lens is shorter than its focal distance p₂ <f₂

D) the object is at a shorter distance than the focal length p₁ <f₁, the object for this lens is at a greater distance than its focal point p₂> f₂

2) let's analyze for each scenario

A) for each lens the object is at a greater distance than its focal length p₁> f₁ and p₂> f₂

B) for lens 1 p₁> f₁

for lens 2

if it is convergent p₂ <f₂

and if it is divergent any distance is possible

C) lens 1 p₁ <f₁ and for lens 2 p₂ <f₂

D) lens 1 p₁ <f₁ and for lens 2 p₂> f₂

3) see attached

The distance with subscript 1 is measured with respect to line1 and the distance with subscript 2 is measured with respect to the lens with subscript 2

Answer:

Acceleration, [tex]a=45\ m/s^2[/tex]

Explanation:

The velocity of a particle is given by :

[tex]V=5t^2[/tex]

t is time in seconds

The acceleration in terms of velocity is given by :

[tex]a=\dfrac{dv}{dt}\\\\a=\dfrac{d(5t^2)}{dt}\\\\a=5\times \dfrac{t^3}{3}[/tex]

We need to find the acceleration of the particle at t = 3 s. Put t = 3 s in the expression of a.

So,

[tex]a=\dfrac{5\times 3^3}{3}\\\\a=45\ m/s^2[/tex]

So, the acceleration of the particle at t = 3 s is [tex]45\ m/s^2[/tex].

Answer:

Perfectly elastic collision

Explanation:

In a closed system, an elastic collision is a type of collision between two bodies, where the total momentum and kinetic energy are conserved.

We are told from the problem that the ball bounces back with its original velocity. For the ball to bounce back to the thrower in the first place, this is our first hint that the collision is elastic. If the collision was inelastic, the ball would most likely have stuck to the wall.

In addition to that, the velocity of the ball remains fairly unchanged even after the collision. This confirms that the kinetic energy it had before the collision is the same as the kinetic energy it has after the collision.  

As a result of this, the collision is perfectly elastic

Answer:

Its not A..

Explanation:

That was my choice and it was incorrect

The displacement must be towards the spring, to compress it and cause it to force back.

The displacement must be away from the spring, to stretch it and cause it to pull back.

The spring constant is used to define the stiffness of the spring, the greater the value of the spring constant stiffer the spring and it is more difficult to stretch the spring.

The spring's restoring force attempts to return it to its equilibrium position; as a result, the displacement must be away from the spring in order to stretch it and make it a drawback, and it must be toward the spring in order to compress it and force it back.

Thus, the correct answers are option B and option D.

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Answer:

The peak value of the electric field is 489.64 V/m

Explanation:

Given;

power of the laser, P = 1.0 mW = 1 x 10⁻³ W

Radius of the beam, R = 1.0 mm = 1 x 10⁻³ m

Area of the beam = πr² = π(1 x 10⁻³ )² = 3.142 x 10⁻⁶ m²

The average intensity of the light = P / A

The average intensity of the light = ( 1 x 10⁻³) / (3.142 x 10⁻⁶)

The average intensity of the light = 318.27 W/m²

The peak value of the electric field is given by;

[tex]E_o = \sqrt{\frac{2I_{avg}}{c\epsilon_o}}\\\\E_o = \sqrt{\frac{2(318.27)}{(3*10^8)(8.85*10^{-12})}}\\\\E_o = 489.64 \ V/m[/tex]

Therefore, the peak value of the electric field is 489.64 V/m.

Answer:

What forces act on a marble rolling down a ramp?

Answer: Gravity acts vertically downward. A normal force acts from the ruler toward the marble/ball in a direction that is perpendicular to the plane of the ruler. Friction acts in the direction opposite to which the marble/ball is moving. ... Friction slows down the marble/ball.

Answer:

The maximum height to which water could be squirted with the hose is 13.380 meters.

Explanation:

A line of current of a fluid can be explained sufficiently by Bernoulli's Theorem. In this case, the system can be simplified due to neglectance of changes in absolute pressure. Water is squirted with an initial speed and reaches its maximum height, where final speed is zero. That is to say:

[tex]\frac{v_{1}^{2}}{2\cdot g} +z_{1} = \frac{v_{2}^{2}}{2\cdot g} +z_{2}[/tex]

Where:

[tex]z_{1}[/tex], [tex]z_{2}[/tex] - Initial and final height of water, measured in meters.

[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.

[tex]v_{1}[/tex], [tex]v_{2}[/tex] - Initial and final speed of water, measured in meters per second.

If [tex]z_{1} = 0\,m[/tex], [tex]v_{1} = 16.2\,\frac{m}{s}[/tex], [tex]v_{2} = 0\,\frac{m}{s}[/tex] and [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], then:

[tex]z_{2} = \frac{v_{1}^{2}-v_{2}^{2}}{2\cdot g} +z_{1}[/tex]

[tex]z_{2} = \frac{\left(16.2\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}}{2\cdot \left(9.807\,\frac{m}{s^{2}} \right)} +0\,m[/tex]

[tex]z_{2} = 13.380\,m[/tex]

The maximum height to which water could be squirted with the hose is 13.380 meters.

Answer:

B

Explanation:

Answer:

false

Explanation:

Answer:

Explanation:

a )

59° south of west  means (180 + 59 )° counterclockwise from the +x-axis.

so angle θ = 239 ° .

r = 3.5 miles .

polar coordinates ( r ,  θ )

= ( 3.50 , 239° )

As the smallest positive angle in degree is 26 then 26 x 9 = 234

so rounding off 239 to 234 , the angle will be 234

polar coordinates ( 3.50 ,  234 )

b )

Cartesian coordinates (in miles) for the same point

x coordinate = r cosθ = 3.5 cos 234 =   -  2.06 mi

y coordinate  = r sinθ = 3.5 sin 234 = - 2.83  mi

Answer:

This is because as we know the magnitude of a forward voltage drop depends on the material used for constructing the diode and also the magnitude of the band gap, so since LED material are generally gallium arsenide and gallium phosphide which has the higher band gap than silicon and since the higher the band gap the higher the voltage drop so LED has the greater forward voltage drop then the silicon diode