Some runners train with parachutes that trail behind them to provide a large drag force. These parachutes are designed to have a large drag coefficient. One model expands to a square 1.8 mm on a side, with a drag coefficient of 1.4. A runner completes a 240 mm run at 6.0 m/s with this chute trailing behind.

Required:
How much thermal energy is added to the air by the drag force?

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

Complete Question:

Find the resistance of a wire of length 0.65 m, radius 0.25 mm and resistivity 3 * 10^{-6} ohm-metre.

Answer:

Resistance = 9.95 Ohms

Explanation:

Given the following data;

Length = 0.65 m

Radius = 0.25 mm to meters = 0.00025 m

Resistivity = 3 * 10^{-6} ohm-metre.

To find the resistance of the wire;

Mathematically, resistance is given by the formula;

[tex] Resistance = P \frac {L}{A} [/tex]

Where;

First of all, we would find the cross-sectional area of the wire.

Area of circle = πr²

Substituting into the equation, we have;

Area  = 3.142 * (0.00025)²

Area = 3.142 * 6.25 * 10^{-8}

Area = 1.96 * 10^{-7} m²

Now, to find the resistance of the wire;

[tex] Resistance = 3 * 10^{-6} * \frac {0.65}{1.96 * 10^{-7}} [/tex]

[tex] Resistance = 3 * 10^{-6} * 3316326.531 [/tex]

Resistance = 9.95 Ohms

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 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]

Answer:

26280

Explanation:

In current time, good telescope can measure redshift to a galaxy in 10 minutes.  

Thus, in one year that has on an average 365 days, the total time taken to  measure redshifts is = ( 365 *12 *60) minute  

= 262800 minutes .

Hence, the number of redshifts observed in a year = (262800/10) = 26280

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: 0

Explanation: Trust

The magnetic force on the charge is approximately 6.47 x 10^(-4) Newtons.

The magnetic force on a charged particle moving through a magnetic field can be calculated using the formula:

F = q * v * B * sin(θ)

Where:

F is the magnetic force,

q is the charge of the particle (in this case, 4.88 x 10^(-6) C),

v is the velocity of the particle (in this case, 265 m/s),

B is the magnetic field strength (in this case, 0.0579 T),

θ is the angle between the velocity vector and the magnetic field vector (in this case, 90 degrees).

Plugging in the values:

F = (4.88 x 10^(-6) C) * (265 m/s) * (0.0579 T) * sin(90°)

Since sin(90°) is equal to 1, the equation simplifies to:

F = (4.88 x 10^(-6) C) * (265 m/s) * (0.0579 T) * 1

Calculating the value:

F = 6.47 x 10^(-4) N

Therefore, the magnetic force on the charge is approximately 6.47 x 10^(-4) Newtons.

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Answer:

1:Rotation

2:Axis

3:Aphelion

4:orbit

Answer: divide by zero, or square root of a negative

Explanation: If your question is how to get infinity as an answer to a problem, that generally means that the answer is undefined or doesn't exist.

A couple of ways to get that...

Answer:

[tex]\mathbf{f_o =1.87 \times 10^{11} \ Hz}[/tex]

Explanation:

From the given information:

The elastic potential energy can be calculated by using the formula:

[tex]U_o = \dfrac{1}{2}kR_o^2[/tex]

Making K the subject;

[tex]K = \dfrac{2 U_o}{R_o^2}[/tex]

[tex]k = \dfrac{2\times 1.68 \times 10^{-21}}{(3.82\times 10^{-10})^2}[/tex]

k = 2.3 × 10⁻² N/m

Now; the frequency of the small oscillation can be determined by using the formula:

[tex]f_o = \dfrac{1}{2 \pi}\sqrt{\dfrac{k}{m}}[/tex]

where;

m = mass of each atom = 1.66 × 10⁻²⁶ kg

[tex]f_o = \dfrac{1}{2 \pi}\sqrt{\dfrac{2.3 \times 10^{-2} N/m}{1.66 \times 10^{-26} \ kg}}[/tex]

[tex]\mathbf{f_o =1.87 \times 10^{11} \ Hz}[/tex]

Answer:

i believe the answer is D

Explanation:

Hope it works

Answer:

Explanation:

Answer:

The tub turns 37.520 revolutions during the 25-second interval.

Explanation:

The total number of revolutions done by the tub of the washer ([tex]\Delta n[/tex]), in revolutions, is the sum of the number of revolutions done in the acceleration ([tex]\Delta n_{1}[/tex]), in revolutions, and deceleration stages ([tex]\Delta n_{2}[/tex]), in revolutions:

[tex]\Delta n = \Delta n_{1} + \Delta n_{2}[/tex] (1)

Then, we expand the previous expression by kinematic equations for uniform accelerated motion:

[tex]\Delta n = \frac{1}{2}\cdot ( \ddot n_{1}\cdot t_{1}^{2} - \ddot n_{2} \cdot t_{2}^{2})[/tex] (1b)

Where:

[tex]\ddot n_{1}, \ddot n_{2}[/tex] - Angular accelerations for acceleration and deceleration stages, in revolutions per square second.

[tex]t_{1}, t_{2}[/tex] - Acceleration and deceleration times, in seconds.

And each acceleration is determined by the following formulas:

Acceleration

[tex]\ddot n_{1} = \frac{\dot n}{t_{1}}[/tex] (2)

Deceleration

[tex]\ddot n_{2} = -\frac{\dot n}{t_{2} }[/tex] (3)

Where [tex]\dot n[/tex] is the maximum angular velocity of the tub of the washer, in revolutions per second.

If we know that [tex]\dot n = 3\,\frac{rev}{s}[/tex], [tex]t_{1} = 13\,s[/tex] and [tex]t_{2} = 12\,s[/tex], then the quantity of revolutions done by the tub is:

[tex]\ddot n_{1} = \frac{3\,\frac{rev}{s} }{13\,s}[/tex]

[tex]\ddot n_{1} = 0.231\,\frac{rev}{s^{2}}[/tex]

[tex]\ddot n_{2} = -\frac{3\,\frac{rev}{s} }{12\,s}[/tex]

[tex]\ddot n_{2} = -0.25\,\frac{rev}{s^{2}}[/tex]

[tex]\Delta n = \frac{1}{2}\cdot ( \ddot n_{1}\cdot t_{1}^{2} + \ddot n_{2} \cdot t_{2}^{2})[/tex]

[tex]\Delta n = \frac{1}{2}\cdot \left[\left(0.231\,\frac{rev}{s^{2}} \right)\cdot (13\,s)^{2}-\left(-0.25\,\frac{rev}{s^{2}} \right)\cdot (12\,s)^{2}\right][/tex]

[tex]\Delta n = 37.520\,rev[/tex]

The tub turns 37.520 revolutions during the 25-second interval.

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:

[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:

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

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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Answer:

yes

Explanation:

it is in the body system

Answer:

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

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

Answer:

right one is more efficient

Explanation:

Answer:

6.28 m/s.

Explanation:

Given that,

The mass of the object, m = 0.2 kg

The radius of the circle, r = 0.5 m

It takes the object 0.500 s to complete one revolution.

We need to find the translational speed of the object. Let it is v. We know that,

[tex]v=\dfrac{2\pi r}{t}\\\\v=\dfrac{2\pi \times 0.5}{0.5}\\\\v=6.28\ m/s[/tex]

So, the transalational speed of the object is 6.28 m/s.

Answer:

deductive reasoning usually follows steps .

Answer:

[tex]F_x = 100N[/tex]

[tex]F_y = 100\sqrt 3 \ N[/tex]

Explanation:

Given

[tex]F = 200N[/tex]

[tex]\theta = 60^o[/tex]

Required

The component of the force in F direction

To do this, we simply calculate the force in the vertical and horizontal direction.

This is calculated as:

[tex]F_x = F * \cos(\theta)[/tex] --- Horizontal

[tex]F_y = F * \cos(\theta)[/tex] ---- Vertical

So, we have:

[tex]F_x = F * \cos(\theta)[/tex] --- Horizontal

[tex]F_x = 200N * \cos(60^o)[/tex]

[tex]F_x = 200N * 0.5[/tex]

[tex]F_x = 100N[/tex]

[tex]F_y = F * \cos(\theta)[/tex] ---- Vertical

[tex]F_y = 200N * \sin(60^o)[/tex]

[tex]F_y = 200N * \frac{\sqrt 3}{2}[/tex]

[tex]F_y = 100\sqrt 3 \ N[/tex]

Explanation:

Total voltage = 9 × 2 = 18v

Resistance = 82 Ω

Ohm's law::

V = IR

18v = 82 Ω × I

18v /82 /Ω = I

18/82 Ampere is the current

Answer:

The explosion is set off by an electrostatic spark. When the mixture ignites, the rapid increase in temperature brings about a huge increase in gas pressure. If the burning vapour were to be confined the resulting rise in pressure could destroy the chamber with a loud explosion.

Explanation:

Answer:

c. physiological reactions; convey emotional states to other members of the species

Explanation:

Darwin believed that emotional expressions began as physiological reactions that came to have evolutionary value because they convey emotional states to other members of the species. These reactions are used by many different types of species to convey various things.