The drag force Fd, imposed by the surrounding air on a vehicle moving with velocity V is given by:
Fd =C dA 2​ pV2
where Cd is a constant called the drag coefficient, A is the projected frontal area of the vehicle, and \rhorho is the air density.
Determine the power, in hp, required to overcome aerodynamic drag for an automobile moving at
(a) 25 miles per hour,
(b) 70 miles per hour.
Assume Cd=0.28,
A= 25ft2
and p=0.075Ib/ft2

Answer:

Answer:

Option (a).

Explanation:

Let the angular velocity is w.

The centripetal acceleration is given by

[tex]a = r w^2[/tex]

where, r is the distance between the axle and the spoke.

So, more is the distance more is the centripetal acceleration.

(a) For the points on the spoke closer to the hub than for points closer to the rim is larger distance, so the centripetal force is more.

The statement is true.  

(b) The direction of centripetal acceleration is always towards the center, so the statement is false.

(c) It is false.

(d) It is false.

Option (a) is correct.

Answer: Heat energy transfer by conduction, convection and radiation

Heat energy is a very difficult energy to store as it can transfer in three different ways from warm surroundings to cooler surroundings. The three processes are conduction, convection or radiation.

Understanding energy, how it is transferred and how the amount of energy that is usefully transferred can be used to measure the efficiency is very important to physics and to the world.

Answer:

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Answer:

The wavelength of the wave is 1.419 cm.

Explanation:

wavelength emitted by the one micro wave = 1.41 cm

beat frequency =  170 M Hz

Let the frequency of the wave is f.

[tex]f=\frac{c}{\lambda }\\\\f =\frac{3\times 10^8}{0.0141}\\\\f = 2.13\times 10^{10} Hz[/tex]

Let the frequency of the other wave  is f'.

[tex]f' = 2.13\times 10^{10}-170\times10^6\\\\f' = 21130\times 10^6 Hz[/tex]

The wavelength is given by

[tex]f' = 21130\times 10^6 Hz\\\lambda = \frac{3\times 10^8}{21130\times 10^6}\\\\\lambda = 0.01419 m = 1.419 cm[/tex]

The time required to lift the weight is 5 seconds.

Time is the measure of past or present events or occurrences. The S.I unit of time is seconds (s).

To calculate the time required to lift the weight, we use the formula below.

Formula:

Where:

make t the subject of the equation.

From the question,

Given:

Substitute these values into equation 2

Hence, The time required to lift the weight is 5 seconds.

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Newton's third law states that for every action (force) in nature there is an equal and opposite reaction. If object A exerts a force on object B, object B also exerts an equal and opposite force on object A. In other words, forces result from interactions.

Hope it helps

Answer:

Newton's third law state that every action there is equal but opposite reactions

hope this will help you more

Answer:

W = 354.9 J

Explanation:

Given that,

The mass of a bale of hay, m = 23 kg

The displacement, d = 3.9 m

The horizontal force exerted on the hay, F = 91 N

We need to find the work done. We know that,

We know that,

Work done, W = Fd

So,

W = 91 N × 3.9 m

W = 354.9 J

So, the required work done is 354.9 J.

Answer:

C the same as pressure on top

Answer:

a. 59 ev. helpful answer

Can you please fill in whatever goes in the blanks ?

Without them, the question makes no sense and has no answer.

Two forces (___) and (___) are applied to an object whose mass is 11.8 kg. The larger force is (___) . When both forces point due east, the object's acceleration has a magnitude of 0.408 m/s2. However, when (___) points due east and (___) points due west, the acceleration is 0.227 m/s2, due east. Find (a) the magnitude of (___) and (b) the magnitude of (___) .

Answer:

C. 6.9 watts

Explanation:

Power = work/time

if work = force×distance...

Then... power= (force×distance)/time

Power = (20×12)/35

= 6.9 watts

Answer:

Fr = 1.91 * 9*10⁹*q²/L²

Explanation:

Let´s say that the corners of the square are  A B C and D

We are going to find out the force on the charge placed on B  ( the charge placed in the upper right corner.

As all the charges are positive (the same sign), then all the three forces on the charge in B are of rejection.

Force due to charge placed in A

module   Fₓ =  K* q² / L²   in the direction of x

Force due to charge placed in C

module  Fy = K* q²/L²   in the direction of y

Force due to  the charge placed in D

That force will have the direction of the diagonal of the square, and the distance between charges placed in D and A is the length of the diagonal.

d²  =  L²  +  L²  =  2*L²

d  =  √2 * L

The module of the force due to charge place in D

F₄₅ = K*q²/ 2*L²

To get the force we need to add first  Fₓ  and  Fy  

Fx + Fy  =  F₁

module of  F₁ = √ Fx² + Fy²    the direction will be the same as the diagonal of the square then:

F₁   =   √  ( K* q²/L² )²  +   ( K* q²/L² )²

F₁  =  √ 2  *  K*q²/L²

And now we add forces F₁   and F₄₅   to get the net force Fr on charge in point B.

The direction of Fr is the direction of the diagonal and is of rejection

the module is

Fr  =  F₁  *  F₄₅

Fr  =  √ 2  *  K*q²/L²  +   K*q²/ 2*L²

Fr  = ( √ 2 + 0,5 ) * K*q² /L²

K  =  9*10⁹  Nm²C²

Fr = 1.91 * 9*10⁹*q²/L²

We don´t know units of L and q

Explanation:

Specifically his leg muscles. As the leg muscles expand, they push down on the ground. Newton's 3rd law says that for any action, there's an opposite and equal reaction. That means a downward push into the ground will have the ground push back, more or less, and that's why the kangaroo will jump. The ground (and the earth entirely) being much more massive compared to the animal means that the ground doesn't move while the kangaroo does move. Perhaps on a very microscopic tiny level the ground/earth does move but it's so small that we practically consider it 0.

This experiment can be done with a wall as well. Go up to a wall and lean against it with your hands. Then do a pushup to move further away from the wall, but you don't necessarily need to lose contact with the wall's surface. As you push against the wall, the wall pushes back, and that causes you to move backward. If the wall was something flimsy like cardboard, then you could easily push the wall over and you wouldn't move back very much. It all depends how much mass is in the object you're pushing on.  

Answer:

The answer is "Option D".

Explanation:

Please find the complete question in the attached file.

As plate separation increased to 2d the capacitance get halred but the change remain same

[tex]\therefore V=\frac{Q}{C}[/tex]

The voltage doubles are now electric field remain same because both the distance and voltage get doubled.

[tex]\to E=\frac{v}{d}\ = \frac{2v}{2d}\\\[/tex]

So,

[tex]energy=\frac{1}{2}\ \frac{Q^2}{C}\\\\c'=\frac{C}{2}\\\\E'=2E[/tex]

Explanation:

Much greater. Gravitational force depends on the mass and separation distance. Shrinking the earth means the mass remains the same while the radius gets smaller. Since gravitational force is inversely proportional to the square of the separation distance, as shown by Newton's universal gravitational law

[tex]\:\:\:F_G = G \dfrac{mM_E}{R^2}[/tex]

shrinking the radius even by a factor of 10 will cause your weight, which also happens to be the gravitational force of the earth on you, to be 100 times more.

If the mass of the Earth remains the same and only its volume and radius are decreased, then the average density of the Earth would increase significantly.

Gravity, also known as gravitational force, pulls objects with mass towards each other. We frequently consider the force of gravity from Earth.

If the Earth's mass remains constant while its volume and radius are reduced, the average density of the Earth increases significantly.

This is due to the fact that the mass has been compressed into a much smaller volume.

As a result, standing on the surface, the gravitational force on you would be greater than before the Earth was compressed to the size of a small mountain.

When the Earth is compressed, the distance between you and the center of the Earth decreases while the mass remains constant.

Thus, your gravitational force increases.

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Answer:

The required ratio is 1.99.

Explanation:

We need to find the atio of speeds of a proton and an alpha particle accelerated through the same voltage.

We know that,

[tex]eV=\dfrac{1}{2}mv^2[/tex]

The LHS for both proton and an alpha particle is the same.

So,

[tex]\dfrac{v_p}{v_a}=\sqrt{\dfrac{m_a}{m_p}} \\\\\dfrac{v_p}{v_a}=\sqrt{\dfrac{6.64\times 10^{-27}}{1.67\times 10^{-27}}} \\\\=1.99[/tex]

So, the ratio of the speeds of a proton and an alpha particle is equal to 1.99.

Answer:

Explanation:

From the given information:

a)

Let's have an imaginary view of the rod located at a given distance r from he the mass (m) of the sphere.

Then the equation for the potential energy as related to the small area of the dr of the rod can be computed as:

[tex]dU = -\dfrac{GMm}{L}*\dfrac{dr}{r}[/tex]

where,

G = gravitational constant

[tex]U = - \int^{x+L}_{x}Gm\dfrac{M}{L}*\dfrac{dr}{r}[/tex]

[tex]U = - \dfrac{GMm}{L}\int^{x+L}_{x}\dfrac{dr}{r}[/tex]

By taking the integral within the limit

[tex]U = - \dfrac{GMm}{L} \Big[In \ r\Big]^{x+L}_{x}[/tex]

[tex]\mathbf{\implies - \dfrac{GmM}{L} In \Big(\dfrac{{x+L}}{{x}}\Big)}[/tex]

b)

By using [tex]F= -\dfrac{dU}{dx}[/tex], the magnitude of the gravitational force can be determined as follows:

Here, we have:

[tex]F = -\dfrac{d}{dx} \Big [\dfrac{-GmM}{L}In(\dfrac{x+l}{x}) \Big ] \\ \\ = \dfrac{GmM}{L}\times \dfrac{x}{x+L}\times (0-\dfrac{L}{x^2}) \\ \\ By \ solving \\ \\ \mathbf{ =-\dfrac{GmM}{x(x+L)}}[/tex]

From above, the negative sign indicates an attractive force

Answer:

the charge of the bees must be of the same sign

Explanation:

The electric field is given by the relation

            E = k q / r²

This electric field has outgoing direction if the charge is positive and incoming towards the charge if it is negative.

The force generated by this field on a test charge is

            F = q E

Since the charge is a scalar, the direction of the force is the same as the electric field.

In this case the two flowers are at a certain distance and the two charged bees land on them, so the force on a test charge is the vector sum of the force that each bee creates, so that this force is subtracted from the two bees must have the same charge sign.

The force created by the bee on the left goes to the right and the force created by the bee on the right goes to the left, so the forces are subtracted,

Consequently the charge of the bees must be of the same sign

False

It's not straight up

Answer:

T = 188.5 s, correct is  C

Explanation:

This problem must be worked on using conservation of angular momentum. We define the system as formed by the fan and the paper, as the system is isolated, the moment is conserved

initial instant. Before the crash

        L₀ = r m v₀ + I₀ w₀

the angular speed of the fan is zero w₀ = 0

final instant. After the crash

        L_f = I₀ w + m r v

        L₀ = L_f

        m r v₀ = I₀ w + m r v

angular and linear velocity are related

        v = r w

        w = v / r

        m r v₀ = I₀ v / r + m r v

         m r v₀ = (I₀ / r + mr) v

       v = [tex]\frac{m}{\frac{I_o}{r} +mr} \ r v_o[/tex]

let's calculate

       v = [tex]\frac{0.020}{\frac{1.4}{0.6 } + 0.020 \ 0.6 } \ 0.6 \ 4[/tex]

       v = [tex]\frac{0.020}{2.345} \ 2.4[/tex]

       v = 0.02 m / s

To calculate the time of a complete revolution we can use the kinematics relations of uniform motion

        v = x / T

         T = x / v

the distance of a circle with radius r = 0.6 m

         x = 2π r

we substitute

         T = 2π r / v

let's calculate

         T = 2π 0.6/0.02

         T = 188.5 s

reduce

         t = 188.5 s ( 1 min/60 s) = 3.13 min

correct is  C

Answer:

The number of electrons transferred from one ball to the other is 2.06 x 10¹² electrons

Explanation:

Given;

magnitude of the attractive force, F = 17 mN = 0.017 N

distance between the two objects, r = 24 cm = 0.24 m

The attractive force is given by Coulomb's law;

[tex]F = \frac{1}{4\pi \epsilon _0} \times \frac{Q^2}{r^2} = \frac{kQ^2}{r^2} \\\\Q^2 = \frac{Fr^2}{k} \\\\Q = \sqrt{ \frac{Fr^2}{k}} \\\\Q = \sqrt{ \frac{(0.017)(0.24)^2}{9\times 10^9}} \\\\Q = 3.298 \times 10^{-7} \ C[/tex]

The charge of 1 electron = 1.602 x 10⁻¹⁹ C

n(1.602 x 10⁻¹⁹ C) = 3.298 x 10⁻⁷

[tex]n = \frac{3.298 \times 10^{-7}}{1.602 \times 10^{-19}} = 2.06 \times 10^{12} \ electrons[/tex]

Therefore, the number of electrons transferred from one ball to the other is 2.06 x 10¹² electrons

Answer:

[tex]a=0.2*10^{-5}g[/tex]

Explanation:

From the question we are told that:

Mass [tex]M=180=>0.18kg[/tex]

Charge [tex]Q=18mC=18*10^-^3C[/tex]

Velocity [tex]v=2.2m/s[/tex]

Length of Wire [tex]L=8.6cm=>0.086[/tex]

Current [tex]I=30A[/tex]

Generally the equation for Magnetic Field of Wire B is mathematically given by

 [tex]B=\frac{\mu_0*I}{2\pi*l}[/tex]

 [tex]B=\frac{4*3.14*10^-^7*I}{2*3.14*8.6}[/tex]

 [tex]B=6.978*10^{-5}T[/tex]

Generally the equation for Force on the plane F is mathematically given by

 [tex]F=qvB[/tex]

Therefore

 [tex]ma=qvB[/tex]

 [tex]a=\frac{qvB}{m}[/tex]

 [tex]a=\frac{18*10^{-5}83.4*6.978*10^{-5}}{0.18kg}[/tex]

 [tex]a=2.37*10^{-5}[/tex]

Therefore in Terms of g's

 [tex]a=\frac{2.37*10^{-5}}{9.8}[/tex]

 [tex]a=0.2*10^{-5}g[/tex]

Answer:

Solid is the state in which Matter maintains a fixed volume

Answer:

The state of matter that has a fixed volume is Solid.

Explanation:

Solid substances will maintain a fixed volume and shape.

Answer:

yes

Explanation:

it is in the body system

Answer:

it would show clearly because it is a metal piece in the body.

Answer:

1:Rotation

2:Axis

3:Aphelion

4:orbit

Answer:

The required wavelength is 1.19 μm

Explanation:

In the double-slit study, the formula below determines the position of light fringes [tex]y_m[/tex] on-screen.

[tex]y_m = \dfrac{m \lambda D}{d}[/tex]

where;

m = fringe order

d = slit separation

λ = wavelength

D = distance between screen to the source

For the first bright fringe, m = 1, and we make (d) the subject, we have:

[tex]d = \dfrac{(1) \lambda D}{y_1}[/tex]

[tex]d = \dfrac{ \lambda D}{y_1}[/tex]

replacing the value from the given question, we get:

[tex]d = \dfrac{ (597 \ nm )\times (3.00 \ m)}{4.84 \ mm} \\ \\ d = \dfrac{ (597 \ nm \times (\dfrac{1 \ m}{10^9\ nm}) )\times (3.00 \ m)}{4.84 \ mm(\dfrac{1 \ m}{1000 \ mm })} \\ \\ d = 3.7 \times 10^{-4} \ m[/tex]

In the double-slit study, the formula which illustrates the position of dark fringes [tex]y_m[/tex] on-screen can be illustrated as:

[tex]y_m = (m+\dfrac{1}{2}) \dfrac{\lambda D}{d}[/tex]

The value of m in the dark fringe first order = 0

∴

[tex]y_0 = (0+\dfrac{1}{2}) \dfrac{\lambda D}{d}[/tex]

[tex]y_0 = (\dfrac{1}{2}) \dfrac{\lambda D}{d}[/tex]

making λ the subject of the formula, we have:

[tex]\lambda = \dfrac{2y_o d}{D} \\ \\ \lambda = \dfrac{2(4.84 \ mm) \times \dfrac{1 \ m}{1000 \ mm} (3.7 \times 10^{-4} \ m) }{3.00 \ m}[/tex]

[tex]\lambda = 1.19 \times 10^{-6} \ m ( \dfrac{10^6 \mu m }{1\ m}) \\ \\ \lambda = 1.19 \mu m[/tex]