A bicycle rider pushes a 13kg bicycle up a steep hill. the incline is 24 degree and the road is 275m long. the rider pushes the bike parallel to the road with a force of 25N.
A. How much work does the rider do on the bike?
B. How much work is done by the force of gravity on the bike?

Answer:

3

Explanation:

Answer:

whether a net force acts on the object

The time between when the bat emits the sound and when it hears the echo is 0.05 s

From the question given above, the following data were obtained:

Velocity of sound (v) = 343 m/s

Distance (x) = 8.42 m

We can obtain obtained the time as illustrated below:

v = 2x / t

343 = 2 × 8.42 / t

343 = 16.84 / t

Cross multiply

343 ×  t = 16.84

Divide both side by 343

t = 16.84/343

Thus, the time between  when the bat emits the sound and when it hears the echo is 0.05 s

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Answer:

0.0491

Explanation:

Answer:

1.33

Explanation:

According to snell's law,

n = sin(a)/sin(b)

Given

a = 25.0°

b = 18.5°

Substitute

n = sin25/sin18.5

n = 0.4226/0.3173

n = 1.33

Hence the index of refraction nb for this type of glass is 1.33

Answer:

greater speed will be obtained for the elastic collision,

Explanation:

To answer this exercise we must find the speed that the sail acquires after each impact.

Let's start by hitting a ball of clay.

The system is formed by the candle and the clay balls, therefore the forces during the collision are internal and the moment is conserved.

initial instant. before the crash

         p₀ = m v₀

where m is the mass of the ball and vo its initial velocity, we are assuming that the candle is at rest

final instant. After the crash

the mass of the candle is M

         p_f = (m + M) v

the moment is preserved

          p₀ = p_f

          m v₀ = (m + M) v

          v = [tex]\frac{m}{m+M} \ v_o[/tex]

for when n balls have collided

          v = [tex]\frac{m}{n \ m + M}[/tex]  v₀

Now let's analyze the case of the bouncing ball (elastic)

initial instant

        p₀ = m v₀

final moment

        p_f = m v_{1f} + M v_{2f}

        p₀ = p_f

        m v₀ = m v_{1f} + M v_{2f}

       m (v₀ - v_{1f}) = M v_{2f}

this case corresponds to an elastic collision whereby the kinetic energy is conserved

        K₀ = K_f

        ½ m v₀² = ½ m v_{1f}² + ½ M v_{2f}²

        v₁ = v_{1f}            v₂ = v_{2f}

        m (v₀² - v₁²) = M v₂²

let's use the identity

         (a² - b²) = (a + b) (a-b)

we write our equations

         m (v₀ - v₁) = M v₂                       (1)

         m (v₀ - v₁) (v₀ + v₁) = M v₂²

let's divide these equations

         v₀ + v₁ = v₂

Let's look for the final speeds

we substitute in equation 1

          m (v₀ - v₁) = M (v₀ + v₁)

          v₀ (m -M) = (m + M) v₁

          v₁ = [tex]\frac{m-M}{m + M}[/tex]   v₀

we substitute in equation 1 to find v₂

            [tex]\frac{M}{m}[/tex]  v₂ = v₀ -  [tex]\frac{m-M}{m+M}[/tex]   v₀

            v₂ = [tex]\frac{m}{M} ( 1 - \frac{m-M}{m+M} ) \ v_o[/tex]

            v₂ = [tex]\frac{m}{M} ( \frac{2M}{m+M} ) \ \ v_o[/tex]

            v₂ = [tex]\frac{2m}{m +M} \ v_o[/tex]  

Let's analyze the results for inelastic collision with each ball that collides with the sail, the total mass becomes larger so the speed increase is smaller and smaller.

In the case of elastic collision, the increase in speed is constant with each ball since the total mass remains invariant.

Consequently, greater speed will be obtained for the elastic collision, that is, the ball will bounce.

Answer:

There is no graph i can aswer it soww

Explanation:

Answer:

Recessive, dominant

Explanation:

umm thats the answer

Answer:

put it in a volcano

Explanation:

Answer: i may not be 100% accurate but i believe it is iron, there is a fire work that uses iron to shine a certain way and is found on stars.

Answer:

The scientist finds an equation that predicts how long the rainstrom will last.

Explanation:

I just answered it :)

The wavelength of this radio signal is equal to 3.18 meters.

Given the following data:

Wavelength can be defined as the distance between two (2) successive crests (troughs) of a wave.

Mathematically, the wavelength of a wave is given by this formula:

[tex]\lambda = \frac{V}{F}[/tex]

Where:

Substituting the given parameters into the formula, we have;

[tex]\lambda = \frac{3 \times 10^8}{9.45 \times10^7}[/tex]

Wavelength = 3.18 meters.

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The wavelength of the radio signal travel at speed of light is 3.17m.

Given the data in the question;

Wavelength the spatial period of a periodic wave. That is to say, when the shapes of waves are Wavelength , the distance over which they are repeated is called wavelength. Wavelength  is expressed as;

[tex]\lambda = \frac{v}{f}[/tex]

Where [tex]\lambda[/tex] is wavelength, f is the frequency of the wave and c is the velocity or speed of light ( [tex]c = 3*10^8m/s[/tex] )

We substitute our values into the expression above.

[tex]\lambda = \frac{c}{ f}\\ \\\lambda = \frac{3*10^8m/s}{9.45*10^7s^{-1}} \\\\\lambda = \frac{3*10^8ms/s}{9.45*10^7}\\\\\lambda = 3.17m[/tex]

Therefore, the wavelength of the radio signal travel at speed of light is 3.17m.

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Answer:

the required current is 0.12 A

Option b) 0.12 A is the correct answer

Explanation:

Given the data in the question,

to determine the current needed in the wire, we use the following relation;

B = μ₀n[tex]I[/tex]

[tex]I[/tex] = B / μ₀n

where μ₀ is the magnetic constant   ( 4π × 10⁻⁷ T.m/A )

n is the number of turns ( 1 / 1mm[tex]\frac{10^{-3}m}{1 mm}[/tex]) = 1000 m⁻¹

B is magnitude ( 150μT ( [tex]\frac{10^{-6}m}{1uT}[/tex]) )

so we substitute

[tex]I[/tex] =  [ 150μT ( [tex]\frac{10^{-6}m}{1uT}[/tex]) ] / [ ( 4π × 10⁻⁷ T.m/A ) × 1000 m⁻¹ ]

[tex]I[/tex] =  [ 0.00015 ] / [ 0.00125 ]

[tex]I[/tex] = 0.12 A

Therefore, the required current is 0.12 A

Option b) 0.12 A is the correct answer

Answer:

The mass of the fish is 6.493 kilograms.

Explanation:

The spring-mass system experiments a Simple Harmonic Motion, then the angular frequency ([tex]\omega[/tex]), in radians per second, is determined by the following equation:

[tex]\omega = 2\pi\cdot f[/tex] (1)

Where [tex]f[/tex] is the frequency, in hertz.

In addition, the mass of the fish ([tex]m[/tex]), in kilograms, is:

[tex]m = \frac{k}{\omega^{2}}[/tex] (2)

Where [tex]k[/tex] is the spring constant, in newtons per meter.

And the spring constant is determined by Hooke's Law:

[tex]k = \frac{F}{\Delta x}[/tex] (3)

Where:

[tex]F[/tex] - Elastic force, in newtons.

[tex]\Delta x[/tex] - Spring elongation, in meters.

If we know that [tex]F = 225\,N[/tex], [tex]\Delta x = 0.125\,m[/tex] and [tex]f = 2.65\,hz[/tex], then the mass of the fish is:

[tex]\omega = 2\pi\cdot f[/tex]

[tex]\omega \approx 16.650\,\frac{rad}{s}[/tex]

[tex]k = \frac{F}{\Delta x}[/tex]

[tex]k = 1800\,N[/tex]

[tex]m = \frac{k}{\omega^{2}}[/tex]

[tex]m = 6.493\,kg[/tex]

The mass of the fish is 6.493 kilograms.

This question involves the concepts of Hooke's Law, and the frequency of spring-mass system.

The mass of the fish is "6.5 kg".

First, we will calculate the angular speed of the system:

[tex]\omega=2\pi f[/tex]

where,

ω = angular speed = ?

f = frequency = 2.65 Hz

Therefore,

[tex]\omega = 2\pi(2.65\ Hz)\\\omega = 16.65\ rad/s\\[/tex]

Now, we will use the Hooke's Law to find out the spring constant of the spring:

[tex]K =\frac{F}{\Delta x}[/tex]

where,

K = spring constant = ?

F = force applied = maximum weight = 225 N

Δx = extension = 12.5 cm = 0.125 m

Therefore,

[tex]k=\frac{225\ N}{0.125\ m}\\\\k = 1800\ N/m\\[/tex]

Now, we will use the formula for the angular speed of a spring-mass system to find out the mass of the fish:

[tex]\omega = \sqrt{\frac{k}{m}}\\\\m = \frac{k}{\omega^2}\\\\m=\frac{1800\ N/m}{(16.65\ rad/s)^2}[/tex]

m = 6.5 kg

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The attached picture illustrates Hooke's Law.

Answer:

Asch's experiment showed that about 75% of people were "yielders" who conformed and 25% were "independent" who didn't conform. Asch concludes that people ignored reality and gave an incorrect answer in order to follow the rest of the group.

Answer: Energy dissipation

As water passes over a spillway and down the chute, potential energy converts into increasing kinetic energy. Failure to dissipate the water's energy can lead to scouring and erosion at the dam's toe (base).

HOPE THIS HELPS

Answer: Energy dissipation

Explanation: As water passes over a spillway and down the chute, potential energy converts into increasing kinetic energy. Failure to dissipate the water's energy can lead to scouring and erosion at the dam's toe (base). Hope this helps. Can u give me brainiliest

Answer:

R₁ = 0.126 m

Explanation:

Let's use the definition of intensity which is the power per unit area

          I = P / A

the generated power is constant

          P = I A

power is

         P = E / t

if we perform the calculations for a given time, the wave energy is

        E = q V

we substitute

        P = [tex]\frac{q V\ A}{t}[/tex]

we can write this equation for two points, point 1 the antenna and point 2 the receiver

         V₁A₁ = V₂A₂

          A₁ = [tex]\frac{V_2}{V_1} \ A_2[/tex]

          A₁ = 0.1 10⁻³  5 10⁻⁴ /V₁

          A₁ = 5 10⁻⁸ /V₁

In general, the electric field on the antenna is very small on the order of micro volts, suppose V₁ = 1 10⁻⁶ V

let's calculate

          A₁ = 5 10⁻⁸ / 1 10⁻⁶

         A₁ = 5 10⁻² m²

the area of ​​a circle is

          A = π r²

we substitute

         π R1₁²= 5 10⁻²

         R₁ = [tex]\sqrt{ \frac{5 \ 10^{-2} }{\pi } }}[/tex]

         R₁ = 0.126 m

Answer:

(a) 6.283 Wb (b) 69.11 Wb (c) I = 0.628 A

Explanation:

Given that,

The diameter of the loop, d = 40 cm

Radius, r = 20 cm

Initial magnetic field, B = 5 mT

Final magnetic field, B' = 55 mT

Initial magnetic flux,

[tex]\phi_i=BA\\\\=5\times 10^{-3}\times \pi \times 20^2\\\\=6.283\ Wb[/tex]

Final magnetic flux,

[tex]\phi_f=B'A\\\\=55\times 10^{-3}\times \pi \times 20^2\\\\=69.11\ Wb[/tex]

Due to change in magnetic field an emf will be generated in the loop. It is given by :

[tex]\epsilon=-\dfrac{d\phi}{dt}\\\\=\phi_f-\phi_i\\\\=69.11-6.283\\\\=62.827\ V[/tex]

Let I be the current in the loop. We can find it using Ohm's law such that,

[tex]\epsilon=IR\\\\I=\dfrac{\epsilon}{R}\\\\I=\dfrac{62.827}{100}\\\\=0.628\ A[/tex]

Hence, this is the required solution.

Answer:

Do we have to choose out of those three?

Explanation:

Volumes can be transfered to liters only when their measurement units are dm³. a) Which means that: 0.45m³=450dm³ you move the decimal place three places to the right because m³ is larger measurement unit than dm³. 375 cm³ = 0.375 dm³; 75cl = 0.75l because of the prefix centi. 0.6dm³ = 0.6 dm³

So it will be 450l; 0.375l; 0.75l; 0.6l

b) 450l>0.75l>0.6l>0.375l

0.45m³>75cl>0.6dm³>375cm³

I hope i didn't make a mistake, but it would be good if somebody else answered too, so you can compare the answers and see if i made a mistake

The given volumes in liter will be 450 L, 0.375 L, 0.75 L, 0.6 L and the arrangement in descending order will be 450 L > 0.75 L > 0.6 L > 0.37 L .

Every three-dimensional item takes up space in some way. The volume of this area is what is used to describe it. The area covered within an entity's three-dimensional bounds is referred to as its volume. The object's size is another name for it.

As per the given information in the question,

a.

The given volume in a meter cube is 0.45.

Unit conversion:

1 m³ = 1000 L

So, 0.45 m³ = 0.45 × 1000

= 450 L

The given volume in a centimeter cube is 375.

Unit conversion:

1 cm³ = 0.001 L

375 cm³ = 0.001 × 375

= 0.375 L.

The given volume in a centiliter cube is 75.

Unit conversion:

1 cl³ = 0.01 L

75 cl³ = 75 × 100

= 0.75 L

The given volume in the decimeter cube is 0.6.

1 dm³ = 1 L

0.6 dm³  = 0.6 L

b.

So, the volumes in descending order will be,

450 L > 0.75 L > 0.6 L > 0.37 L or,

0.45 m³ > 75 cl³ > 0.6 dm³ > 375 cm³

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Answer:

If we double the length we will have:

[tex]f'=\frac{f}{\sqrt{2}}[/tex]

Explanation:

The equation of the angular frequency of a pendulum is given by:

[tex]\omega=\sqrt{\frac{g}{L}}[/tex]

Where:

By definition the period is:

[tex]T=\frac{2\pi}{\omega}[/tex]

And frequency is 1 over the frequency T:

[tex]f=\frac{1}{T}[/tex]

[tex]f=\frac{1}{2\pi}\sqrt{\frac{g}{L}}[/tex]

Now, if we double the length we will have:

[tex]f'=\frac{1}{2\pi}\sqrt{\frac{g}{2L}}[/tex]

[tex]f'=\frac{1}{2\pi\sqrt{2}}\sqrt{\frac{g}{L}}[/tex]

[tex]f'=\frac{f}{\sqrt{2}}[/tex]

Therefore, the new frequency is the old frequency over √2.

I hope it helps you!

Answer:

Explanation: the major difference between an electromagnet and permanent magnet is that the former can have a magnetic field when electric current flows through it and disappears when the flow of the current stops. ... It will always displays the magnetic behaviour.

Answer:

well

Explanation:

basically it's like

Answer:

In my opinion, ''A" would be the longest wavelength.

Complete question

To reduce the drag coefficient and thus to improve the fuel efficiency, the frontal area of a car is to be reduced. Determine the amount of fuel and money saved per year as a result of reducing the frontal area from 18 to 14 ft2. Assume the car is driven 12,000 mi a year at an average speed of 55 mi/h. Take the density and price of gasoline to be 50 lbm/ft3 and $3.10/gal, respectively; the density of air to be 0.075 lbm/ft3, the heating value of gasoline to be 20,000 Btu/lbm; and the overall efficiency of the engine to be 30 percent. Take the drag coefficient as CD=0.3 for a passenger car.

Answer:

22.22%

$57

Explanation:

From the question we are told that

Initial area of frontal area [tex]a_1=18ft^2[/tex]

Final area of frontal area [tex]a_2=14ft^2[/tex]

Distance covered a year [tex]D=12000mile[/tex]

Average speed a year [tex]V_{avg}=55mile/h[/tex]

Density [tex]\rho=50 lbm/ft3[/tex]

Price [tex]P= $3.10/gal[/tex]

Density of air [tex]\rho_{air} 0.075 lbm/ft3[/tex]

Heating value of gasoline [tex]Q=20,000 Btu/lbm[/tex]

Efficiency [tex]\eta=30\%[/tex]

Drag coefficient [tex]CD=0.3[/tex]

[tex]\triangle A=18-14ft^2=4ft^2[/tex]

Generally the equation for drag force [tex]C_D[/tex] is mathematically given as

[tex]F_D=\frac{C_DAPV^2}{2}[/tex]

[tex]F_D=\frac{0.3*18*0.075*(80.685^2)*A}{2}[/tex]

[tex]F_D=73.24Alb[/tex]

where [tex]v=55mil/h*1.467=80.685ft/s[/tex]

Generally the equation for work done W is mathematically given as[tex]W=F_D*L[/tex]

where [tex]L=12000mile*5280[/tex]

[tex]L=63360000[/tex]

[tex]W=73.24A*63360000[/tex]

[tex]W=4.6*10^9A[/tex]

Generally the equation for overall efficiency [tex]\eta[/tex] is mathematically given as

where

[tex]W_{req}=required\ gasoline\ power\ efficiency[/tex]

[tex]\eta=\frac{W}{W_{req}}[/tex]

[tex]W_{req}=\frac{W}{\eta}[/tex]

[tex]W_{req}=\frac{4.6*10^9}{0.3}[/tex]

[tex]W_{req}=1.55*10^{10}A[/tex]

Generally the equation for reduction fee with change in frontal area [tex]\triangle M[/tex] is mathematically given as

[tex]\triangle M =\triangle Vgasoline*cost[/tex]

Where

[tex]\triangle Vgasoline= volume\ reduction\ of\ gasoline[/tex]

[tex]\triangle Vgasoline=\frac{E_{req}}{H*P}[/tex]

[tex]\triangle Vgasoline=\frac{1.55*10^{10}A}{20000*778.169*32.2*50}[/tex]

if

[tex]20000btu/ibm=20000*778.169*32.2(1bm.ft^2/s)[/tex]

[tex]\triangle Vgasoline=\frac{1.55*10^{10}A}{20000*778.169*32.2*50}[/tex]

[tex]\triangle Vgasoline=0.61859A[/tex]

Therefore

[tex]\triangle M =0.61859A*3.10[/tex]

if [tex]1ft^2=7.48gal[/tex]

[tex]\triangle M =0.61859(4)*3.10*7.48[/tex]

[tex]\triangle M = \$ 57.671[/tex]

Generally the equation for reduction of fuel [tex]F_r[/tex]is mathematically given as

[tex]F_r=\frac{\triangle A}{\triangle i}*100[/tex]

where

[tex]\triangle i=18ft^2[/tex]

[tex]F_r=\frac{4}{18}*100[/tex]

Fuel reduction price by reducing front area is 22.22%

Money saved per year is $57

Answer:

It accelerate but will not spin.

Explanation:

If an irregular shaped object is dropped from rest without feeling any form of air resistance it will accelerate without spinning and this is due to the fact that there is no Torque around  the center of gravity

Answer:

(d) 0.15 N

Explanation:

mass of the fish, m₁ = 65 g = 0.065 kg

initial velocity of the fish, u₁ = 0

mass of water expelled by the fish, m₂ = 0.30 g = 0.0003 kg

time during which the water was expelled, t = 5.0 ms = 5.0 x 10⁻³ s

velocity of the water, v = 2.5 m/s

The magnitude of force of the exerted water is equal to the magnitude of force the fish exerted on the water.

The magnitude of force of the exerted water is calculated as follows;

[tex]F = ma = m\frac{v}{t} \\\\F = \frac{mv}{t} \\\\F = \frac{m_2 \ \times \ v}{t} \\\\F = \frac{0.0003 \ \times \ 2.5}{5 \times 10^{-3}} \\\\F = 0.15 \ N[/tex]

The correct option is D.

Therefore, the average force the fish exerts on the drop of water is 0.15 N.