If the circuit in the motorcycle runs off a standard 12 V battery, and one of the headlights is switched on, current flowing through the headlight is measured at 3.75 A. What is the power usage of the headlamp? Report the answer to two significant digits

Answer:

20.0 mph

Explanation:

Let x represent the speed in still air, and t represent the time of flight;

Speed of wind = 2 mph

When flying with the wind, it's Speed is;

x + 2

When flying against wind, it's speed is;

x - 2

distance = speed × time;

When flying with the wind, distance is;

(x+2)t = 88 .....1

When flying against wind, distance is;

(x-2)t = 72 .......2

Solving the simultaneous equation;

Divide equation 1 by 2

(x+2)t/(x-2)t = 88/72

(x+2)/(x-2) = 1.222

x+2 = 1.222(x-2)

x+2 = 1.222x - 2.444

x(1.222-1) = 2+2.444

x(0.222) = 4.444

x = 4.444/0.222

x = 20.0 mph

the rate of the bird in still​ air is 20.0 mph

Answer:

B: The electromagnetic waves reach earth, while the mechanical waves do not.

correct Answer:

The electromagnetic waves reach Earth, while the mechanical waves do not

Answer:

1- Short circuit

2- ammeter

Answer:

276 N and 225 N

Explanation:

Draw a free body diagram.  There are three forces on the hoist:

Weight force 350 N pulling down,

Tension force T₁ pulling up and left 50° from the horizontal,

Tension force T₂ pulling up and right 38° from the horizontal.

Sum of forces in the x direction:

∑F = ma

T₂ cos 38° − T₁ cos 50° = 0

T₂ cos 38° = T₁ cos 50°

T₂ = T₁ cos 50° / cos 38°

Sum of forces in the y direction:

∑F = ma

T₂ sin 38° + T₁ sin 50° − 350 = 0

T₂ sin 38° + T₁ sin 50° = 350

Substitute:

(T₁ cos 50° / cos 38°) sin 38° + T₁ sin 50° = 350

T₁ cos 50° tan 38° + T₁ sin 50° = 350

T₁ (cos 50° tan 38° + sin 50°) = 350

T₁ = 350 / (cos 50° tan 38° + sin 50°)

T₁ = 276 N

T₂ = T₁ cos 50° / cos 38°

T₂ = 225 N

Given that,

Length of bar = 600 mm

Diameter of bar = 40 mm

Diameter of hole = 30 mm

Length of hole = 100 mm

Modulus of elasticity = 85 GN/m²

Load = 180 kN

We need to calculate the area of cross section without hole

Using formula of area

[tex]A=\dfrac{\pi\times d^2}{4}[/tex]

Put the value into the formula

[tex]A=\dfrac{\pi\times40^2}{4}[/tex]

[tex]A=1256.6\ mm^2[/tex]

We need to calculate the area of cross section with hole

Using formula of area

[tex]A=\pi\times\dfrac{(d_{b}^2-d_{h}^{2})}{4}[/tex]

Put the value into the formula

[tex]A=\pi\times\dfrac{(40^2-30^2)}{4}[/tex]

[tex]A=549.77\ mm^2[/tex]

We need to calculate the total contraction on the bar

Using formula of total contraction

Total contraction = contraction in bar without hole part + contraction in bar with hole part

[tex]Total\ contraction = \dfrac{F\times L_{1}}{A_{1}\times E}+\dfrac{F\times L_{2}}{A_{2}\times E}[/tex]

Where, F = load

L = length

A = area of cross section

E = modulus of elasticity

Put the value into the formula

[tex]Total\ contraction=\dfrac{180\times10^3}{85\times10^{3}}(\dfrac{500}{1256.6}+\dfrac{100}{549.77})[/tex]

[tex]Total\ contraction = 1.227\ mm^2[/tex]

Hence, The total contraction on the bar is 1.227 mm²

Answer:

A) Arrangement A

Explanation:

The rate of heat conduction is given by Fourier's Law of Heat Conduction. It is given as follows:

Q = KAΔT/L

where,

Q = Rate of Heat Transfer or Conduction

K = Thermal Conductivity

A = Cross-Sectional Area

ΔT = Difference in Temperature

L = Thickness

So, it is clear from the formula that for a constant temperature difference and value of thermal conductivity, the rate of heat transfer is directly proportional to the cross-sectional area and it is inversely proportional to the thickness.

Therefore, the arrangement with larger cross-sectional area and smaller thickness will be the one with the greatest heat transfer rate and as a result greatest heat shall be conducted through that arrangement.

It is clear that the parallel arrangement that is arrangement A, has higher cross-sectional area and smaller thickness. Therefore, the correct option is:

A) Arrangement A

A very small source of light that radiates uniformly in all directions produces an electric field with an amplitude of ܧ ௠at a distance R from the source. What is the amplitude of the magnetic field at a point 2R from the source?

If the distance from the source is doubled. The amplitude of the magnetic field is smaller 4 times.

Answer:

[tex]y_n(x) =C_n \sin \sqrt{\frac{\rho}{T} } w_nx=C_n \sin \sqrt{\frac{\rho}{T} } \sqrt{\frac{T}{\rho} } \frac{n \pi}{L} x[/tex]

[tex]y_n(x) = C_n \sin \frac{n \pi x}{L}[/tex]

Explanation:

The given differential equation is

[tex]T\frac{d^2y}{dx^2} + \rho w ^2y=0[/tex] and y(0) = 0, y(L) =0

where T and ρ  are constants

The given rewrite as

[tex]\frac{d^2y}{dx^2} + \frac{\rho w^2}{T} y=0[/tex]

auxiliary equation is

[tex]m^2+ \frac{\rho w^2}{T} =0\\\\m= \pm\sqrt{\frac{\rho}{T} } wi[/tex]

Solution of this de is

[tex]y(x)=C_1 \cos \sqrt{\frac{\rho}{t} } wx + C_2 \sin \sqrt{\frac{\rho}{T} } wx[/tex]

y(0)=0 ⇒ C₁ = 0

[tex]y(x) = C_2 \sin \sqrt{\frac{\rho}{T} } wx[/tex]

y(L) = 0 ⇒

[tex]C_2 \sin \sqrt{\frac{\rho}{T} } wL=0[/tex]

we need non zero solution

⇒ C₂ ≠ 0 and

[tex]\sin \sqrt{\frac{\rho}{T} } wL=0[/tex]

[tex]\sin \sqrt{\frac{\rho}{T} } wL=0 \rightarrow \sqrt{\frac{\rho}{T} } wL=n \pi[/tex]

[tex]w_n = \sqrt{\frac{T}{\rho} } \frac{n \pi}{L}[/tex]

solution corresponding these [tex]w_n[/tex] values

[tex]y_n(x) =C_n \sin \sqrt{\frac{\rho}{T} } w_nx=C_n \sin \sqrt{\frac{\rho}{T} } \sqrt{\frac{T}{\rho} } \frac{n \pi}{L} x[/tex]

[tex]y_n(x) = C_n \sin \frac{n \pi x}{L}[/tex]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The correct option is the second option

Explanation:

 Generally the electric force exerted by the charge Q  on the  charge  (q) is mathematically represented as

           [tex]F_Q = \frac{kqQ}{r^2}[/tex]

 Generally the electric force exerted by the charge 2Q  on the  charge  (q) is mathematically represented as

         [tex]F_{2Q} = \frac{kq2Q}{2r^2}[/tex]    

Now the net force exerted on q is

       [tex]F_{net} = \frac{kqQ}{r^2} - \frac{2k q Q}{4r^2}[/tex]

        [tex]F_{net} = \frac{4kqQ- 2kqQ}{4r^2}[/tex]

        [tex]F_{net} = \frac{kqQ}{2r^2}[/tex]

Looking at the resulting equation we see that [tex]F_{net} > 0[/tex]

This implies that the charge q would move to the right

Answer:

Explanation:

TJVFKBFVN

Answer: The force at t = 1s is 14 N

Explanation:

I guess that the position equation actually is: (by looking at the units of C and D)

S = A - B*t + C*t^2 - D*t^3

As you may know by Newton's third law, if we want to find the force, we first need to find the acceleration, and before that, the velocity.

the velocity can be found by integrating over time, the velocity is:

V = 0 - B + 2*C*T - 3*D*t^2

For the acceleration we integrate again:

A = 0 + 2*C - 6D*T

and we know that F = m*A

then:

Force(t) = 1kg*(2*10m/s^2 - 6*1m/s^3*t)

We want the force at t = 1s, so we replace t by 1s.

F(1s) = 1kg*(20m/s^2 - 6m/s^2) = 14 N

Explanation:

We have,

Mass of a particle is 0.6 kg

Speed at A is 2 m/s

Kinetic energy at B is 7.5 J

(a) The kinetic energy at A is given by :

[tex]K_A=\dfrac{1}{2}mv^2\\\\K_A=\dfrac{1}{2}\times 0.6\times 2^2\\\\K_A=1.2\ J[/tex]

(b) Kinetic energy at B is given by

[tex]K_B=\dfrac{1}{2}mV^2\\\\V=\sqrt{\dfrac{2K_B}{m}} \\\\V=\sqrt{\dfrac{2\times 7.5}{0.6}} \\\\V=5\ m/s[/tex]

(c) The work done on the particle as it moves form A to B is given by work energy theorem as :

[tex]W=\dfrac{1}{2}m(V^2-v^2)\\\\W=\dfrac{1}{2}\times 0.6\times (5^2-2^2)\\\\W=6.3\ J[/tex]

Answer:

Probably just enforcing it little by little. (If you mean a habit. If you mean a certain topic it's different.)

Explanation:

If you are trying to teach your students a habit-

Try to correct them everytime they don't follow it. Expect them to do so and give a reward when they earn it.

If you mean a certain topic-

Make a slideshow or a presentation with an interactive game. Making it not boring and having them more attentive.

I gave a more general answer, so if you would like to go deeper into the subject I would be more happy to.

(If this helped, please let me know. If you have any questions do not be afriad to ask!)

Answer:

E = 0.0130 V/m.

Explanation:

The electric field is related to the potential difference as follows:

[tex] E = \frac{\Delta V}{d} [/tex]

Where:

E: is electric field

ΔV: is the potential difference = 3.95 mV  

d: is the distance of a person's chest = 0.305 m

Then, the electric field is:

[tex]E = \frac{\Delta V}{d} = \frac{3.95 \cdot 10^{-3} V}{0.305 m} = 0.0130 V/m[/tex]

Therefore, the maximum electric field created is 0.0130 V/m.

I hope it helps you!

Answer:

The weight at a distance 2 RE from surface of earth is W/9

Explanation:

For the value of acceleration due to gravity (g), we have a formula, that is:

g = (G)(ME)/(RE)²    ----- equation (1)

where,

G = Gravitational Constant

ME = Mass of Earth

RE = Radius of Earth

g = Acceleration due to gravity on surface of earth = 9.8 ms²

When the person goes 2RE, distance above earth's surface. Then the total distance from center of earth becomes: 2RE + RE = 3RE.

Therefore, equation (1) becomes:

gh = (G)(ME)/(3RE)²

where,

gh = acceleration due to gravity at height

gh = (G)(ME)/(RE)²9

using equation (1), we get:

gh = g/9

Now, he weight is given by formula:

W = mg   ------- equation (2)

At height 2RE

Wh = (m)(gh)

where,

Wh = Weight at height = ?

m = mass of astronaut

Therefore, using vale of gh, we get:

Wh = mg/9

Using equation (2), we get:

Wh = W/9

Answer:

A) 1236 N

B) Nw = µ_s•N

C) d_max = 1.525 m

Explanation:

From the question, "smooth vertical wall" means that there is no friction there and thus the only vertical forces are the weights of the ladder and painter and the normal force at the floor.

a) Mass of ladder = 14 kg

Mass of painter = 8M = 8 * 14 = 112 kg

Thus, magnitude of normal force is;

N = total mass x acceleration due to gravity = (14 + 112)9.8

N = 1236 N

(b) Sum of the moments about the base of the ladder:

ΣH = 0

Nw - µ_s•N = 0

Nw = µ_s•N

c) Since they are the only two horizontal forces in play, we know that

Nw = Ff where Ff is the friction force at the floor.

Ff = µ_s*N = 0.39 × 1236

Ff = 482.04 N

So, to find maximum distance painter can stand without slipping, we'll use the formula ;

Nw(Lsinθ) = (Mgcosθ)(L/2) + (8Mgcosθ * d_max)

Plugging in the relevant values, we have;

482.04(2.7*sin55) = ((14 × 9.8cos55)*(2.7/2)) + (8*14*9.8cos55 * d_max)

1066.133 = 106.238 + 629.5575*d_max

629.5575*d_max = 1066.133 - 106.238

629.5575*d_max = 959.895

d_max = 959.895/629.5575

d_max = 1.525 m

A) The magnitude of the normal force is  1236 N

B) expression for the magnitude of the normal force Nw = µ_s•N

C) The largest distance up the ladder (d) max, d_max = 1.525 m

The friction is defined as when any  object slides on a surface then the force in opposite direction of the external applied force is generated which restrict the motion of the object is called as the friction force.

From the question, "smooth vertical wall" means that there is no friction there and thus the only vertical forces are the weights of the ladder and painter and the normal force at the floor.

a) Mass of ladder = 14 kg

Mass of painter = 8M = 8 * 14 = 112 kg

Thus, the magnitude of normal force is;

N = total mass x acceleration due to gravity = (14 + 112)9.8

N = 1236 N

(b) Sum of the moments about the base of the ladder:

ΣH = 0

Nw - µ_s•N = 0

Nw = µ_s•N

c) Since they are the only two horizontal forces in play, we know that

Nw = Ff where Ff is the friction force at the floor.

Ff = µ_s*N = 0.39 × 1236

Ff = 482.04 N

So, to find maximum distance painter can stand without slipping, we'll use the formula ;

[tex]N_{w}(Lsin\theta) = (Mgcos\theta)(\dfrac{L}{2}) + (8Mgcos\theta \times d_{max)[/tex]

Plugging in the relevant values, we have;

[tex]482.04(2.7\times sin55) = ((14 \times 9.8cos55)\times (\dfrac{2.7}{2})) + (8\times 14\times 9.8cos55 \times d_{max)[/tex]

[tex]1066.133 = 106.238 + 629.5575\times d_{max[/tex]

[tex]629.5575\times d_{max} = 1066.133 - 106.238[/tex]

[tex]d_{max} = \dfrac{959.895}{629.5575}[/tex]

d_max = 1.525 m

Thus, the magnitude of the normal force is  1236 N and an expression for the magnitude of the normal force Nw = µ_s•N . The largest distance up the ladder (d) max, d_max = 1.525 m

To know more about friction, follow

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Answer:

The time taken by the sound is [tex]3.82\times 10^{-4}\ s[/tex]

Explanation:

We have,

A low-frequency sound source is placed to the right of a person, whose ears are approximately 13 cm apart, and the speed of sound generated is 340 m/s.

It is required to find the time taken by the sound when the sound arrives at the right ear and the sound arrives at the left ear. Let t is the time taken. It can be calculated as :

[tex]v=\dfrac{d}{t}\\\\t=\dfrac{d}{v}\\\\t=\dfrac{0.13}{340}\\\\t=3.82\times 10^{-4}\ s[/tex]

So, the time taken by the sound is [tex]3.82\times 10^{-4}\ s[/tex]

Answer:

a) the velocity increases then decreases.

Answer:

A) N = (M + mf + md) g

B) d = md L / (2 mf)

Explanation:

Draw a free body diagram.  There are four forces acting on the board.

Weight force M g pulling down at the center of the board.

Normal force N pushing up at the center of the board.

Weight force mf g pulling down a distance d from the center.

Weight force md g pulling down at the end of the board.

Sum of forces in the y direction:

∑F = ma

N − M g − mf g − md g = 0

N = (M + mf + md) g

Sum of moments about the center of the board:

∑τ = Iα

md (L/2) − mf d = 0

d = md L / (2 mf)

The magnitude of the upward force n exerted by the support on the board will be the same with the magnitude of the weight of the seesaw. Which is

W = 9.8M Newton

The father should sit at L/4 at the other end in order to balance the system at rest

Since the seesaw is of a uniform board, the center of gravity will act at the center. If it is of length L, the support called the fulcrum will be positioned at L/2.

If the daughter is a distance L/2 from the center, Then, she must be positioned at L/4 from the one edge of the board.

A. The magnitude of the upward force n exerted by the support on the board will be the same with the magnitude of the weight of the seesaw. Which is

W = mg

W = 9.8M Newton

The weight of both the father and the daughter will act downward. so, they will not be included.

The father should sit at L/4 at the other end in order to balance the system at rest

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Answer:

The magnification of the image is  [tex]M = -1.5[/tex]

Explanation:

From the question we are told that

      The object distance is  [tex]u = 14 cm[/tex]

      The image distance is  [tex]v = 21 \ cm[/tex]

The magnification of the image is mathematically represented as

        [tex]M = - \frac{v}{u}[/tex]

substituting values

       [tex]M = - \frac{21}{14}[/tex]

        [tex]M = -1.5[/tex]

The negative value means that the image is real , inverted and enlarged

Answer:

Both of Arnell and Adeen are right.

Explanation:

For Adeen, hammering the wire makes it flat, effectively reducing the cross sectional area, and increasing the length of the wire. Recall that the resistivity of a metal conductor increases with length and decreases with cross sectional area. From this, we can see that the resistance of the wire will increase due to the hammering. This means Adeen is correct.

For Arnell, heating a metal causes the atoms along with their electrons to vibrate in a random manner. These random motions is not organised and can be in any direction. For electricity to flow smoothly, the electrons must travel in the same direction in an orderly fashion. The randomly vibrating atoms will continuously bump into the flowing electron, resisting the smooth motion of the electrons. This collisions results in an additional resistance, hence, the resistance is increased. This means Arnell is correct.

Answer:

1) 2.9m/s

2)0.235cm

3)0.509s

Explanation:

1. moments are conserved, when the steak hits the pan:

velocity 'v' of the steak:  

v = √(2gh) = √(2*9.81*0.5) = 3.13 m/s

velocity pan + steak is

v =[tex]m_1v_1/(m_1+m_2)[/tex]= 2.5*3.13/(2.7) = 2.9 m/s

2.initial kinetic energy of pan + steak = spring energy  

1/2 mv² = 1/2 kA²  

where, A = amplitude  

2.7 x 2.9² = 410 x A²

A²=22.707/410

A = 0.235 cm

3. T = 2π√(m/k) = 2π√(2.7/410) = 0.509 s

Therefore, the period of that motion is 0.509s

1) The speed of the pan and steak immediately after the collision is 2.9m/s

2) The  amplitude of the subsequent motion is 0.235cm.

3) The period of that motion is 0.509s.

1.

Since

v = √(2gh) = √(2*9.81*0.5) = 3.13 m/s

Also,

v == 2.5*3.13/(2.7)

= 2.9 m/s

2.

Now

initial kinetic energy of pan + steak = spring energy  

So,

1/2 mv² = 1/2 kA²  

Here,

A = amplitude  

2.7 x 2.9² = 410 x A²

A²=22.707/410

A = 0.235 cm

3.

Now

T = 2π√(m/k)

= 2π√(2.7/410)

= 0.509 s

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Answer:

a. 6666.67 V/m

b. 2. the positively charged plat

Explanation:

a. The computation of the magnitude of the electric field between the plates is shown below:

As we know that

[tex]E = \frac{V}{d}[/tex]

[tex]= \frac{400 V}{0.06 m}[/tex]

= 6666.67 V/m

hence, the magnitude of the electric field is 6666.67 V/m

b. Based on this the higher potential is positively charged plate as the flow of the current goes from positive to negative and it is inverse in case of th electron

We simply applied the above formula  

Answer:

v= 0.94 m/s

Explanation:

In order to calculate the speed of the waves, you use the following formula for the speed of the waves:

[tex]v=\lambda f[/tex]          (1)

v: speed of the wave

λ: wavelength of the wave

f: frequency of the wave

The frequency is calculated by using the information about the time that boat takes to travel from its highest point to its lowest point. This time is a half of a period:

[tex]\frac{T}{2}=3.5s\\\\T=7.0s[/tex]

Then, the frequency is:

[tex]f=\frac{1}{T}=\frac{1}{7s}=0.142Hz[/tex]  

The wavelength of the wave is the distance between crests of the wave

[tex]\lambda=6.6m[/tex]

With the values of the frequency and the wavelength, you can find the speed of the wave by using the equation (1):

[tex]v=(6.6m)(0.142Hz)=0.94\frac{m}{s}[/tex]

The speed of the wave is 0.94m/s

Answer:

481 m

Explanation:

To fall 235 m, the time required is

t = √(2H/g)

t= √(2[tex]\times[/tex]235/9.8)

t=6.92 seconds.

The supplies will travel forward

6.92 [tex]\times[/tex] 69.4 ≈ 481 m

Therefore, the goods must be dropped 481  m in advance of the recipients.

Answer:

Momentum is mainly related to forces

Answer:

The visible light spectrum is the segment of the electromagnetic spectrum that the human eye can view. More simply, this range of wavelengths is called visible light. Typically, the human eye can detect wavelengths from 380 to 700 nanometers.

Explanation:

Answer:

Visible light, which travels at a dizzying 186,282 miles per second through space, is just one part of light's broad spectrum, which encompasses all electromagnetic radiation. We can detect visible light because of cone-shaped cells in our eyes that are sensitive to the wavelengths of some forms of light

Explanation:

Answer:

The axial region of the body consists of the bones of the head, trunk of a vertebrate, skull, vertebral column, and thoracic cage. The human skeleton consists of 80 bones.

Explanation:

The axial region of the body consists of the bones of the head, trunk of a vertebrate, skull, vertebral column, and thoracic cage. The human skeleton consists of 80 bones.

It is composed of the following six parts:

1. Skull (22 bones)

2. Ossicles of the middle ear

3. Hyoid bone

4. Rib cage

5. Sternum

6. Vertebral column

The axial region of the body forms the vertical axis of the body as the axial skeleton supports the head, neck, back, and chest.

Answer:

Explanation:

(a)

  [tex]\dfrac{6.2\cdot 10^2\,\text{rev}}{60\,\text{s}}\times\dfrac{2\pi\,\text{rad}}{\text{rev}}=\dfrac{62\pi\,\text{rad}}{3\,\text{s}}\approx\boxed{64.93\,\text{rad/s}}[/tex]

__

(b)

  [tex]d=r\theta=(3.0\,\text{m})(2.0\cdot 10^2\,\text{s})\left(\dfrac{62\pi\,\text{rad}}{3\,\text{s}}\right)=124\pi\cdot 10^2\,\text{m}=\boxed{38\,956\,\text{m}}[/tex]

Answer:

The force of gravity at the shell will be extremely great on me due to the huge mass collapsed into the small radius.

At the center of the shell, the gravitational forces all around should cancel out, giving me a feeling of weightlessness; which will be a lesser force compared to that felt while standing on the shell.

Explanation:

For the collapsed earth:

mass = 5.972 × 10^24 kg

radius = 1 ft

according to Newton's gravitation law, the force of gravity due to two body with mass is given as

Fg = GMm/[tex]R^{2}[/tex]

Where Fg is the gravitational force between the two bodies.

G is the gravitational constant

M is the mass of the earth

m is my own mass

R is the distance between me and the center of the earths in each case

For the case where I stand on the shell:

radius R will be 1 ft

Fg = GMm/[tex]1^{2}[/tex]

Fg = GMm

For the case where I stand stand inside the shell, lets say I'm positioned at the center of the shell. The force of gravity due to my mass will be balanced out by all other masses around due to the shell of the hollow earth. This cancelling will produce a weightless feeling on me.