If the velocity of a pitched ball has a magnitude of 47.0 m/s and the batted ball's velocity is 56.5 m/s in the opposite direction, find the magnitude of the change in momentum of the ball and of the impulse applied to it by the bat.

Answer:

The correct answer is the number 4. A wave on a pond is a mechanical wave which requires a medium to travel.

Explanation:

Mechanical waves are those that need a material medium to propagate. The waves of the sea and the waves that we produce on a guitar string, the sound, are examples of mechanical waves. Electromagnetic waves are energetic pulses capable of propagating in a vacuum. This way, a wave on a pond is a mechanical wave which requires a medium to travel.

Final answer:

A wave on a pond is a mechanical wave requiring water to propagate, while electromagnetic waves, such as light, don't require a medium and can travel through a vacuum. Therefore, statement 4 is correct. This reflects fundamental differences in how mechanical and electromagnetic waves propagate and their need for a medium.

Explanation:

To compare waves on a pond with electromagnetic waves, we should understand that a wave on a pond is a mechanical wave which requires a medium to travel (option 4). This is because mechanical waves are disturbances that propagate through a material medium and the pond water serves as this medium. On the other hand, electromagnetic waves do not need a medium; they can travel through a vacuum because they consist of oscillating electric and magnetic fields.

So, the correct statement would be that a wave on a pond is a mechanical wave which requires a medium to travel. Therefore, the waves on a pond are similar to sound waves since both are mechanical and need a medium. In contrast, light waves are an example of electromagnetic waves, which means they can travel through the vacuum of space without any medium.

Furthermore, all electromagnetic waves move at the same speed in a vacuum, which is known as the speed of light. This is a fundamental difference between mechanical and electromagnetic waves since the speed of mechanical waves depends on the medium in which they are traveling. The discovery that electromagnetic waves do not require a medium led to the abandonment of the aether theory, which was invented to provide a medium for light and other electromagnetic wave propagation.

Answer:

210,600πft-lb

Explanation:

Force is a function F(x) of position x then in moving from

x = a to x= b

Work done = [tex]\int\limits^b_a {Fx} \, dx[/tex]

Consider a water tank conical in shape

we will make small horizontal section of the water at depth h and thickness dh and also assume radius at depth h is w

we will have ,

[tex]\frac{w}{12} = \frac{(18-h)}{18} \\w = \frac{2}{3} (18-h)a[/tex]

weight of slice under construction

weight = volume × density × gravitational constant

[tex]weight = \pi \times w^2 \times dh \times 62.4\\= (62.4\pi w^2dh)lb[/tex]

Now we can find work done

[tex]W = \int\limits \, dw\\[/tex]

[tex]\int\limits^{18}_{3} {(62.4\pi } \, dx w^2dh)h\\= 62\pi \frac{4}{9} \int\limits^{18}_{3} {(18-h)^2hdh} \,[/tex]

= [tex]62.4\pi \times\frac{4}{9} (\frac{324}{2} h^2-\frac{36}{3} + \frac{h^4}{4})^1^8_3[/tex]

= 210,600πft-lb

To solve this problem we will apply the concepts related to energy conservation. We know that potential energy is transformed into kinetic and vice versa energy. Since the energy accumulated in the upper part is conserved as potential energy, when the object is thrown all that energy will be converted into kinetic energy. Therefore we will have the following relation,

[tex]KE = PE[/tex]

[tex]KE = mgh[/tex]

Here,

m = mass

g = Gravitational acceleration

h = Height

Replacing,

[tex]KE = (6)(9.8)(5)[/tex]

[tex]KE = 294J[/tex]

Therefore the kinetic energy at the bottom is 294J

Answer:

Normal Conversation: i=106i0

i(dB)=60

Power saw a 3 feet: i=1011i0

i(dB)=110

Jet engine at 100 feet: i=1018i0

i(dB)=180

Explanation:

if these are the same as edge, then these are the answers! :)

Normal Conversation: i=106i0

The intensity in decibels is 60,110,180, respectively.

⇒ normal conversation: i =60(dB)

⇒power saw 3 feet: i =110(dB)

⇒jet engine at 100 feet: i = 180(dB)

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Answer:

Explanation:

Acceleration without friction on an inclined plane = g sinθ

Acceleration with friction on inclined plane = g sinθ - μ g cosθ

s = 1/2 a t²

For motion on friction-less surface

s = 1/2 g sinθ t₁²

For motion on frictional surface

s = 1/2( g sinθ - μ g cosθ)  t₂²

t₂ = 2t₁

(  sinθ - μ  cosθ)  t₂² =  sinθ t₁²

(  sinθ - μ  cosθ)  4t₁² =  sinθ t₁²

(  sinθ - μ  cosθ)  4 =  sinθ

4sinθ - sinθ = 4μ  cosθ

3sinθ = 4μ  cosθ

3 / 4 tanθ =μ  

μ  = .75  tanθ

The minimum coefficient of static friction between the pavement and the tires is 0.156.

The minimum coefficient of static friction is calculated by applying Newton's second law of motion in determining the net force.

[tex]F_c = Wsin(\theta) + \mu_s W cos(\theta)\\\\\frac{mv^2}{r} = mg sin(\theta) + \mu_s mg cos(\theta)\\\\\mu_s mg cos(\theta)\ = \frac{mv^2}{r} - mg sin(\theta) \\\\\mu _ s = \frac{mv^2 \ - \ mgr sin(\theta)}{mg rcos(\theta)}[/tex]

where;

[tex]\mu _ s = \frac{1200 \times 18.1^2 \ - \ 1200 \times 9.8 \times 100 sin(10.4)}{1200 \times 9.8 \times 100 \times cos(10.4)}\\\\\mu_s = 0.156[/tex]

Thus, the minimum coefficient of static friction between the pavement and the tires is 0.156.

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Answer:

[tex]44,000 Nm^2/C[/tex]

Explanation:

The electric flux through a certain surface is given by (for a uniform field):

[tex]\Phi = EA cos \theta[/tex]

where:

E is the magnitude of the electric field

A is the area of the surface

[tex]\theta[/tex] is the angle between the direction of the field and of the normal to the surface

In this problem, we have:

[tex]E=1.1\cdot 10^4 N/C[/tex] is the electric field

L = 2.0 m is the side of the sheet, so the area is

[tex]A=L^2=(2.0)^2=4.0 m^2[/tex]

[tex]\theta=0^{\circ}[/tex], since the electric field is perpendicular to the surface

Therefore, the electric flux is

[tex]\Phi =(1.1\cdot 10^4)(4.0)(cos 0^{\circ})=44,000 Nm^2/C[/tex]

The electric flux through the sheet will be "44,000 Nm²/C".

An electric field seems to be an area of space that surrounds an electrically charged particle as well as object whereby an electric charge will indeed experience attraction.

According to the question,

Magnitude of electric field, E = 1.1 × 10⁴ N/C

Length, L = 2.0 m

We know,

The area will be:

→ A = L²

By substituting the value,

      = (2.0)²

      = 4.0 m²

hence,

The electric flux will be:

→ [tex]\Phi[/tex] = EA Cosθ

By substituting the values,

      = [tex](1.1.10^4)(4.0)(Cos 0^{\circ})[/tex]

      = 44,000 Nm²/C

Thus the above answer is appropriate.  

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Answer:

  B = -0.215 s⁻¹,  A₀ = 1.537º

Explanation:

To be able to make this graph we must linearize the data, the best procedure is to calculate the logarithm of the values

           A = A₀ [tex]e^{-Bt}[/tex] + C

The constant creates a uniform displacement of the graph if we start the graph at the angle of 1, the constant disappears, this is done by subtracting 1 from each angle, so the equation is

            (A-1) = A₀  e^{-Bt}

We do the logarithm

           Log (A-1) = log Ao –Bt log e

Make this graph because paper commercially comes in logarithm 10, if we use graph paper if we can calculate directly in base logarithm e, let's perform this calculation

              Ln (A-1) = Ln A₀ - Bt

To graph the points we subtract 1 from each angle and calculate the logarithm, the data to be plotted are

              θ’= ln (θ -1)

θ(º)    t(s)        θ'(º)

6.2      2     1.6487

4.7      4     1.3083

3.2      6    0.7885

2.4      8    0.3365

1.8     10   -0.2231

We can use a graph paper and graph on the axis and the primary angle (θ') and on the x-axis time, mark the points and this graph is a straight line, we see that the point for greater time has a linearity deviation , so we will use the first three for the calculations

To find the line described by the equation

              y -y₀ = m (x -x₀)

             m = (y₂ -y₁) / (x₂ -x₁)

Where m is the slope of the graph e (x₀, y₀) is any point, let's start as the first point of the series

            (x₀, y₀) = (2, 1.65)

           (x₂, y₂) = (2, 1.65)

           (x₁, y₁) = (6, 0.789)

We use the slope equation

           m = (1.65 - 0.789) / (2-6)

           m = -0.215  

The equation is

           y - 1.65 = -0.215 (x- 2)

           y = -0.215 x +0.4301

We buy the two equations and see that the slope is the constant B

            B = -0.215 s⁻¹

The independent term is

            b = ln A₀

            A₀ = [tex]e^{b}[/tex]

           A₀ = [tex]e^{0.43}[/tex]

           A₀ = 1.537º

Answer:

correct answer is (b)  Longitudinal lines remain straight

Explanation:

solution

As we know that Deformation of circular shaft in the torsion is associate with twisting of  shaft more than an specify with the yielding limit.

so when any angle of twist is obtain in the torsion and that is beyond the specified safety limit of shaft

than that shaft will be fail.

but it does not regain its original shape and it will cause permanent deformation

so that we can say longitudinal lines which is twist, they will not regain to original back position as straight

but they will remain in curved shape.

so here incorrect statement is b Longitudinal lines remain straight

Though torsion theory assumes that cross sections of a torsionally loaded shaft remain flat, in reality, under heavy loading, they can warp and so this statement is not entirely accurate.

When a circular shaft deforms under torsion, certain assumptions are made about its deformation according to the torsion theory. These assumptions include:

1) cross sections remain flat and perpendicular to the axis of the shaft,

2) longitudinal lines remain straight,

3) circular sections remain circular, and

4) radial lines on the sections remain straight. However, the statement that is not entirely accurate is that cross sections remain flat. In reality, under severe torsional loading, the cross sections might warp and not remain entirely flat.

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Answer:

The centripetal acceleration at highway speed is greater.

Explanation:

We assume the motion of the car is uniformly accelerated. Let the highway speed be v.

By the equation of motion,

[tex]v=u+at[/tex]

[tex]a=\dfrac{v-u}{t}[/tex]

u is the initial velocity, a is acceleration and t is time

Because the car starts from rest, u = 0.

[tex]a_T=\dfrac{v}{t}[/tex]

This is the tangential acceleration of the thread of the tire.

The centripetal acceleration is given by

[tex]a_C=\dfrac{v^2}{r}[/tex]

r is the radius of the tire.

Comparing both accelerations and applying commonly expected values to r and t, the centripetal acceleration is seen to be greater. The radius of a tyre is, on the average, less than 0.4 m. Then the centripetal acceleration is about

[tex]a_C=\dfrac{v^2}{0.3}=2.5v^2[/tex]

The tangential acceleration can only be greater in the near impossible condition that the time to attain the speed is on the order of microseconds.

Answer:

Check attachment for complete questions, the question is not complete

Explanation:

Check attachment for solution

Complete Question

The complete Question is shown on the first and second uploaded image

Answer:

The speed at which they need to push the mass is v = 13.1 m/s

Explanation:

In order to solve this problem we need to consider conservation of energy when the block is at the top of the inclined plane and also when it is on top of the loop

Now Applying the law of conservation of energy

        [tex]mg (2R) + \frac{1}{2} mv^2 = \frac{1}{2} mv_{top}^2 + mg(2R)[/tex]

  where   [tex]mg (2R)[/tex] is potential energy and [tex]\frac{1}{2} mv^2[/tex] is kinetic energy

  and [tex]v_{top}[/tex] is the velocity at the top inclined plane and the top of the loop

        Now considering the formula

                           [tex]\frac{1}{2} mv^2 = \frac{1}{2} mv_{top}^2[/tex]

                            [tex]v^2 = v_{top}^2[/tex]

                            [tex]v = v_{top}[/tex]

Now to obtain [tex]v_{top}[/tex]

   Looking at the question we can say that the centripetal force that made the block move around loop without leaving the track is q=equivalent to the centripetal force  so we have

            [tex]mg = \frac{mv_{top}^2}{R}[/tex]

The m would cancel out each other then cross- multiplying

             [tex]gR = v^2_{top}[/tex]

         [tex]v_{top} = \sqrt{gR}[/tex]

                 [tex]= \sqrt{(9.8 m/s^2)(17.4\ m)}[/tex]

                [tex]= 13.05 m/s[/tex]

                [tex]\approx 13.1 m/s[/tex]

Answer:

The average current that this cell phone draws when turned on is 0.451 A.

Explanation:

Given;

voltage of the phone, V = 3.7 V

electrical energy of the phone battery, E = 3.15 x 10⁴ J

duration of battery energy, t = 5.25 h

The power the cell phone draws when turned on, is the rate of energy consumption, and this is calculated as follows;

[tex]P = \frac{E}{t}[/tex]

where;

P is power in watts

E is energy in Joules

t is time in seconds

[tex]P = \frac{3.15*10^4}{5.25*3600s} = 1.667 \ W[/tex]

The average current that this cell phone draws when turned on:

P = IV

[tex]I = \frac{P}{V} =\frac{1.667}{3.7} = 0.451 \ A[/tex]

Therefore, the average current that this cell phone draws when turned on is 0.451 A.

The average current that the cell phone draws when turned on is approximately [tex]\(0.450 \, \text{A}\)[/tex].

To find the average current drawn by the cell phone, we need to use the relationship between power, voltage, and current. Let's go step by step.

Step 1: Calculate the power consumed by the cell phone

We already have:

- The total energy provided by the battery: [tex]\( \Delta W = 3.15 \times 10^4 \, \text{J} \)[/tex]

- The total time of operation: [tex]\( \Delta t = 5.25 \, \text{hours} \)[/tex]

First, convert the operation time from hours to seconds:

[tex]\[\Delta t = 5.25 \, \text{hours} \times 3600 \, \text{seconds/hour} = 5.25 \times 3600 = 18900 \, \text{s}\][/tex]

Now, calculate the power P using the energy and time:

[tex]\[P = \frac{\Delta W}{\Delta t} = \frac{3.15 \times 10^4 \, \text{J}}{18900 \, \text{s}} \approx 1.666 \, \text{W}\][/tex]

Step 2: Use the power to find the average current

We know that power P, voltage V, and current I are related by the equation:

[tex]\[P = VI\][/tex]

We can solve for the current I

[tex]\[I = \frac{P}{V}\][/tex]

Given the battery voltage [tex]\(V = 3.70 \, \text{V}\)[/tex] and the power [tex]\(P \approx 1.666 \, \text{W}\)[/tex], we can calculate the current:

[tex]\[I = \frac{1.666 \, \text{W}}{3.70 \, \text{V}} \approx 0.450 \, \text{A}\][/tex]

The complete question is this:

The battery for a certain cell phone is rated at 3.70 V. According to the manufacturer it can produce 3.15 x 104 J of electrical energy, enough for 5.25 h of operation, before needing to be recharged. Find the average current that this cell phone draws when turned on.

Ok, so what I did so far was convert time into seconds and found Power:

t = 18900 s

P = ΔW/Δt =student submitted image, transcription available below= 1.6666 W

I think you have to use the problem : P = VabI = I2R = εI - I2R

Answer:

The diagram is in the attachment

Explanation:

An ammeter is use to know the current flowing in a circuit,

A voltmeter is use to know the potential difference across an element.

The ideal voltmeter and the ideal ammeter has zero internal resistance, so as to drop as little voltage as possible as current flows through it.

1. Let analyse the first circuit i.e the ammeter connection

The ammeter is connected rightly and all the current coming from the battery will flow into the bulb and the bulb will glow bright using only the current from the battery and the ammeter work in the circuit is only to measure the current from the battery.

Now let analyse the second circuit, the voltmeter connection.

This is a wrong connection and if this is done it will act has high resistance to the current flow. The connecting of voltmeter in series is equivalent to connecting a very high resistance in series with the circuit. By this only small insignificant amount of current flow through the circuit and nearly results in an open circuit.

Conclusion,

The first connection of ammeter bulb will glow brightly because it uses up all the current from the battery but for the voltmeter connection the current has been reduced due to the high resistance of voltmeter and thus reduces current.

Answer:

Φ = 2.26 10⁶ N m² / C

Explanation:

The electric flow is

           Ф = E .ds = [tex]q_{int}[/tex] / ε₀

Since we have two components for this flow. The uniform electric field and the flux created by the point charge.

The flux of the uniform electric field is zero since the flux entering is equal to the flux leaving the Gaussian surface

The flow created by the point load at the origin is

              Φ = q_{int} /ε₀

Let's calculate

              Φ = 2.00 10⁻⁶ /8.85 10¹²

              Φ = 2.26 10⁶ N m² / C

Answer:

i = f = 0.1 m until the lens moves towards the screen 0.1 m

Explanation:

For this exercise let's use the constructor equation

         1 / f = 1 / o + 1 / i

Where f is the focal distance, or the distance to the object and "i" the distance to the image

It indicates that the focal distance is 100 mm (f = 100 mm), when an object is at infinity the image is formed at its focal length

     1 / f = 1 / inf + 1 / i = 1 / i

      i = f = 100 mm

At this point the screen is placed

For the shortest distance the lens has to move a little so the distance to the image is

        f = 100 mm = 0.1 m

        o = 6 pm = 6 10⁻¹² m

        i = 100 + x

        1 / f = 1 / o + 1 / 100+ x

        1 /( 0.10 + x) = 1 / f - 1 / o

        1 / 0.100+ x = 1/0.100 - 1/6 10-12

        1 / 0.100 + x = 10 - 10¹¹ = -10¹¹

        0.100 + x = -10⁻¹¹

        x = -10⁻¹¹ -10⁻¹

        x = -10⁻¹

        x = - 0.1 m

This negative distance indicates that the lens moves towards the screen 0.1 m

To calculate the resultant magnetic field at different points between the Helmholtz coils, use the formula B = (μ0 * N * I * R^2) / ((R^2 + x^2)^(3/2)). Plug in the given values to find the magnetic field at specific distances from the center of the coil.

To calculate the resultant magnetic field at different points, we can use the formula:

B = (μ0 * N * I * R^2) / ((R^2 + x^2)^(3/2))

Where B is the magnetic field, μ0 is the permeability of free space, N is the number of turns, I is the current, R is the radius of the coil, and x is the distance from the center of the coil.

Using this formula, we can calculate the magnetic field at x1 = 2.8 cm, x2 = 5.5 cm, x3 = 7.3 cm, and x4 = 11.0 cm by plugging in the given values.

Helmholtz coils create a uniform magnetic field using two identical coils with the same current. The formula for calculating the magnetic field at any position along the axis demonstrates this uniformity and the provided distances can be used to find specific magnetic field values.

A Helmholtz coil configuration consists of two identical circular coils, each having N turns, radius R, and carrying the same current I. The coils are separated by a distance equal to R. This setup creates a uniform magnetic field between the coils. Given data are R = 11.0 cm, I = 17.0 A, and N = 300 turns.

Using the formula for the magnetic field along the axis of Helmholtz coils:

This approach allows for the precise calculation of the magnetic field Bx at the given distances from the center, demonstrating the setup’s uniformity and reinforcement of the fields.

Complete Question : Two coils that are separated by a distance equal to their radius and that carry equal currents such that their axial fields add are called Helmholtz coils. A feature of Helmholtz coils is that the resultant magnetic field between the coils is very uniform. Let R = 11.0 cm, I = 17.0 A, and N = 300 turns for each coil. Place one coil in the y-z plane with its center at the origin and the other in a parallel plane at R = 11.0 cm. Calculate the resultant field Bx at x1 = 2.8 cm, x2 = 5.5 cm, x3 = 7.3 cm, and x4= 11.0 cm.

Answer:

    a = 4.8 10⁻⁵ m

Explanation:

The diffraction phenomenon is described by the expression

        a sin θ = m λ

How the pattern is observed on a distant screen

        tan θ = y / L = sin θ / cos θ

Since the angle is very small in these experiments ’we can approximate the tangent function

    tan θ = sin θ = y / L

We substitute

           a y / L = m λ

The first minimum occurs for m = 1

        a = λ L / y

        a = 680 10⁻⁹ 5.5 / 0.078

        a = 4.8 10⁻⁵ m

Answer:

Magnetic field at point having a distance of 2 cm from wire is 6.99 x 10⁻⁶ T

Explanation:

Magnetic field due to finite straight wire at a point perpendicular to the wire is given by the relation :

[tex]B=\frac{\mu_{0}I }{2\pi R }\times\frac{L}{\sqrt{L^{2}+R^{2} } }[/tex]      ......(1)

Here I is current in the wire, L is the length of the wire, R is the distance of the point from the wire and μ₀ is vacuum permeability constant.

In this problem,

Current, I = 0.7 A

Length of wire, L = 0.62 m

Distance of point from wire, R = 2 cm = 2 x 10⁻² m = 0.02 m

Vacuum permeability, μ₀ = 4π x 10⁻⁷ H/m

Substitute these values in equation (1).

[tex]B=\frac{4\pi\times10^{-7}\times 0.7 }{2\pi \times0.02 }\times\frac{0.62}{\sqrt{(0.62)^{2}+(0.02) ^{2} } }[/tex]

B = 6.99 x 10⁻⁶ T

Answer:

[tex]F_{cp}=3947.84N[/tex]

Explanation:

From the formula for centripetal force and acceleration we can deduce that:

[tex]F_{cp}=ma_{cp}=m\frac{v^2}{r}=m\frac{(r\omega)^2}{r}=mr\omega^2=mr\omega^2[/tex]

Since one revolution is [tex]2\pi\ rad[/tex], 0.5 revolutions are [tex]\pi\ rad[/tex], so we have:

[tex]F_{cp}=(4kg)(100m)(\pi\ rad/s)^2=3947.84N[/tex]

Answer: The intensity level of sound in the bedroom is 80dB

Explanation:

Intensity of lawn mower at r=1m is 100dB

Beta1= 10dBlog(I1/Io)

100dB= 10dB log(I1/Io)

10^10= I1/Io

I1= Io(10^10)

10^12)×(10^10)= I1

I1=10^-2w/m^2

Intensity of lawn mower at r=20m

I2/I1=(r1/r2)^2 =(1/20)^2

I2= I1(1/400)

I2=2.5×10^-3W_m^2

Intensity of 4 lown mowers at 20m fro. Window

= 10dBlog(4I2/Io)

= 10^-4/10^-12

=80dB

The angular velocity of the merry-go-round increases as a child walks towards the center, due to the conservation of angular momentum and a decrease in the system's moment of inertia.

When a child walks from the outer edge of a rotating merry-go-round towards the center, the system's angular momentum remains constant due to the conservation of angular momentum. However, the moment of inertia of the system decreases as the child moves closer to the center. Since angular momentum is the product of the moment of inertia and angular velocity (L = I∙ω), and it is conserved, the angular velocity of the merry-go-round must increase to compensate for the decrease in moment of inertia.

This concept can be compared to a figure skater pulling in their arms to spin faster. As the skater's arms come closer to the body, their moment of inertia decreases, causing an increase in their spinning speed, or angular velocity, with no external torques acting on the system.

The same principle applies to the merry-go-round: as the child moves inward, the system's angular velocity increases to conserve angular momentum. This can be observed in playgrounds and is a practical example of conservation laws in rotational motion.

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As the child walks towards the center of the rotating merry-go-round, the system's angular momentum remains constant due to the conservation of angular momentum. The moment of inertia decreases, causing the angular velocity to increase.

A child is standing on the edge of a merry-go-round that is rotating with frequency f.

As the child walks towards the center of the merry-go-round, the angular momentum of the system (child plus merry-go-round) remains constant.

This is due to the conservation of angular momentum, which states that if no external torque acts on the system, the total angular momentum remains unchanged.

Angular momentum (L) is given by the formula:

where I is the moment of inertia and ω is the angular velocity.

complete question:

A child is standing on the edge of a merry-goround that is rotating with frequency f. The child then walks towards the center of the merry-go-round. For the system consisting of the child plus the merry-go-round, what remains constant as the child walks towards the center of the rotating merry-go-round?

The pilot's maximum endurance speed in a circle of 85 m diameter without blacking out can be calculated using the centripetal acceleration formula, taking into account the acceleration limit of 7 g, where 1 g equals 9.80 m/s².

The question requires the application of concepts from physics, specifically the idea of centripetal force and acceleration when a body moves in a circular path.

Given that the maximum acceleration a pilot can endure without blacking out is 7 g, and knowing that 1 g is equivalent to 9.80 m/s², we can calculate the maximum speed under these conditions using the formula for centripetal acceleration, which is `a = v² / r`, where `a` is the acceleration, `v` is the speed, and `r` is the radius of the circle.

The diameter of the circle is given as 85 meters, which means the radius `r` will be half of that, 42.5 meters. The maximum acceleration of 7 g translates to 7 × 9.80 m/s², which is 68.6 m/s². We can rearrange the formula to solve for `v`, leading to `v = √(a × r)`.

Substituting the values provided, we get `v = √(68.6 m/s² × 42.5 m)

v = √2923.175 m²/s² ≈ 54.1 m/s

Therefore, the maximum speed the pilot can tolerate while spinning in the horizontal circle is approximately 54.1 m/s, which is equivalent to about 192 km/h.

Answer:

11.2 Km/Sec

Explanation:

Escape velocity is the velocity that one has to achieve to go out of the gravitational effect of a body. Here they are talking about escaping from the Earth's gravity. To travel to Mars, first we will have to go far enough so that the impact of Earth's gravity is negligible.

The escape velocity ([tex]v_{e}[/tex]) can be calculate using the given formula:

[tex]v_{e} = \sqrt{\frac{2GM}{r}}[/tex]

where,

G = Gravitational Constant

M = Mass of the object from which we need to escape (Here it is the mass of the Earth)

r = distance from the center of that object (In this case it is equal to the radius of Earth)

For Earth, this comes out to be 11.2 KM/Sec

Explanation:

According to Newton's second law of motion,

            [tex]m_{1}g - T = m_{1}a[/tex] ......... (1)

and,    [tex]T - m_{2}g = m_{2}a[/tex] ......... (2)

When we add both equations, (1) and (2) then the expression obtained for "a" is as follows.

             a = [tex]\frac{m_{1} - m_{2}}{m_{1} + m_{2}} \times g[/tex]

                = [tex]\frac{17.7 - 11.1}{17.7 + 11.1} \times 9.8[/tex]

                = [tex]\frac{6.6}{28.8} \times 9.8[/tex]

                = 2.24 [tex]m/s^{2}[/tex]

Now, putting the value of "a" in equation (1) then we will calculate the tension as follows.

              [tex]m_{1}g - T = m_{1}a[/tex]

              [tex]17.7 \times 9.8 - T = 17.7 \times 2.24[/tex]

                   173.46 - T = 39.648

                       T = 133.812 N

Thus, we can conclude that the magnitude of their acceleration is 2.24 [tex]m/s^{2}[/tex] and the tension T is 133.812 N in the rope.

Answer:

Explanation:

m1 = 17.7 kg

m2 = 11.1 kg

Let a be the acceleration and T be the tension in the string.

use Newton's second law

m1 g - T = m1 x a ....(1)

T - m2 g = m2 x a ..... (2)

Adding both the equations

(m1 - m2) g = ( m1 + m2 ) x a

(17.7 - 11.1 ) x 9.8 = (17.7 + 11.1) x a

64.68 = 28.8 a

a = 2.25 m/s²

Put the value of a in equation (1)

17.7 x 9.8 - T = 17.7 x 2.25

173.46 - T = 39.825

T = 133.64 N

c. The object with the higher initial temperature will have a greater temperature change when brought into thermal contact with another object.

The object with the higher initial temperature will have a greater temperature change when brought into thermal contact with another object. This is because energy transfers from the object with the higher temperature to the object with the lower temperature until they reach thermal equilibrium.

Mass and specific heat do not directly affect the temperature change. The mass only affects the amount of energy transferred, while the specific heat determines how much energy is needed to raise the temperature.

Therefore, the correct answer is c. The one with the higher initial temperature.

Learn more about Thermal contact here:

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Explanation:

The given data is as follows.

           mass (m) = 5 kg

       Height of tower = 15 m

        u = 7 m/s

    air resistance = 610 v

(a)   Now, differential equation for the given mass which is thrown vertically upwards is as follows.

              [tex]m \frac{d^{2}x}{dt^{2}}[/tex] = F

                  -bv = Fr

Here, mg is downwards due to the force of gravity.

              [tex]\frac{md^{2}x}{dt^{2}} = bv - mg[/tex]

       [tex]\frac{md^{2}x}{dt^{2}} + b \frac{dx}{dt} + mg[/tex] = 0

Hence, the differential equation required to solve the problem is as follows.

       [tex]\frac{md^{2}x}{dt^{2}} + b \frac{dx}{dt} + mg[/tex] = 0    

(b)   When final velocity of the object is equal to zero then the object will reach towards its maximum height and it will start to fall downwards.

              F = [tex]\frac{md^{2}x}{dt^{2}}[/tex]

                 = 0

Therefore, the object reach its maximum height at v = 0.

Answer:

a) b) d)

Explanation:

The question is incomplete. The Complete question might be

In an inertial frame of reference, a series of experiments is conducted. In each experiment, two or three forces are applied to an object. The magnitudes of these forces are given. No other forces are acting on the object. In which cases may the object possibly remain at rest? The forces applied are as follows: Check all that apply.

a)2 N; 2 N

b) 200 N; 200 N

c) 200 N; 201 N

d) 2 N; 2 N; 4 N

e) 2 N; 2 N; 2 N

f) 2 N; 2 N; 3 N

g) 2 N; 2 N; 5 N

h ) 200 N; 200 N; 5 N

For th object to remain at rest, sum of all forces must be equal to zero. Use minus sign to show opposing forces

a) 2+(-2)=0 here minus sign is to show the opposing firection of force

b) 200+(-200)=0

c) 200+(-201)[tex]\neq[/tex]0

d) 2+2+(-4)=0

e) 2+2+(-2)[tex]\neq[/tex]0

f) 2+2+(-3) [tex]\neq[/tex]0; 2+(-2)+3[tex]\neq[/tex]0

g) 2+2+(-5)[tex]\neq[/tex]0; 2+(-2)+5[tex]\neq[/tex]0

h)200 + 200 +(-5)[tex]\neq[/tex]0; 200+(-200)+5[tex]\neq[/tex]0

Answer:

3,4,1,2

Explanation:

The filing procedure involves Filing in terms of the first letter.If the first letters are common, file in terms of the second letter then File in terms of surnames.If the surnames are common, file in terms of the initial. If the surname is more than one then file under the first surname etc.

Answer:

225.56V

Explanation:

Q=cv

6.93 µF capacitor was charged to 94 V, therefore Q₁ =6.93 *10^-6 x 94 =651.42µC

1.76 µF capacitor charged to 81 V, therefore Q₂ =1.76 *10^-6 x 81 =142.56µC

When the capacitor are brought together, the charges on them move to maintain equilibrum, so therefore, each capacitor has 0.5(142.56µC + 651.42µC )= 396.99µC

The final potential difference across the 1.76 µF capacitor =  Q/c = 396.99µC/1.76 µF = 225.56V

To find the final potential difference across the 1.76 µF capacitor, use charge conservation by equating the total charge before and after connection. After calculating the charges on each capacitor, rearrange the equation to solve for the final potential difference across the 1.76 µF capacitor.

In order to find the final potential difference across the 1.76 µF capacitor, we can use the concept of charge conservation.

First, let's calculate the total charge on the capacitors before they are connected. The charge on a capacitor is given by Q = CV, where Q is the charge, C is the capacitance, and V is the potential difference across the capacitor.

For the 6.93 µF capacitor, the charge is Q1 = (6.93 µF)(94 V) = 650.42 µC. For the 1.76 µF capacitor, the charge is Q2 = (1.76 µF)(81 V) = 142.56 µC.

After connecting the capacitors, the total charge remains the same. So, we have Q1 + Q2 = Qf, where Qf is the final charge on the capacitors. Rearranging the equation, we can find the final potential difference across the 1.76 µF capacitor:

Vf = (Qf - Q1) / C2 = (142.56 µC - 650.42 µC) / (1.76 µF).

Calculating this expression will give us the final potential difference across the 1.76 µF capacitor.

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Answer:

6104 N/C.

Explanation:

Given:

k = 8.99 × 10^9 Nm2/C^2

Qx = 1.3 × 10^-5 C

rx = 7 m

Qy = 1 × 10−5 C

ry = 4 m

E = F/Q

= kQ/r^2

Ex = (8.99 × 10^9 × 1.3 × 10^−5) ÷ 7^2

= 2385.1 N/C.

Ey = (8.99 × 10^9 × 1.0 × 10^−5) ÷ 4^2

= 5618.75 N/C

Eo = sqrt(Ex^2 + Ey^2)

= sqrt(3.157 × 10^7 + 5.69 × 10^6)

= 6104 N/C.

The magnitude of the electric field at the origin due to the charges fixed at (7 m, 0) and (0, 4 m) is approximately 6107.83 N/C. This was calculated by determining the electric fields from each charge and then using vector addition to find the resultant electric field. The calculation involved using the formula E = k |q| / r² for each charge, followed by determining the resultant using Pythagoras' theorem.

Let's determine the electric field at the origin due to two fixed charges. The first charge is -1.3 × 10⁻⁵ C located at (7 m, 0), and the second charge is 1 × 10⁻⁵C located at (0, 4 m).

The electric field due to a point charge q at a distance r is given by E = k |q| / r², where k = 8.99 × 10⁹ N·m²/C² is the Coulomb constant.

Step-by-Step Calculation

Calculate the distance from each charge to the origin:

Charge 1 at (7 m, 0): r₁ = 7 m

Charge 2 at (0, 4 m): r₂ = 4 m

Compute the electric field due to each charge:

For q₁ = -1.3 × 10⁻⁵ C:
E₁ = k |q₁| / r₁² = (8.99 × 10⁹ N·m²/C²) (1.3 × 10⁻⁵ C) / (7 m)² ≈ 2381.57 N/C along the negative x-direction.

For q₂ = 1 × 10⁻⁵ C:
E₂ = k |q₂| / r₂² = (8.99 × 10⁹ N·m²/C²) (1 × 10⁻⁵ C) / (4 m)² ≈ 5621.88 N/C along the positive y-direction.

Determine the resultant electric field at the origin using vector addition. Since the fields are perpendicular:

Resultant E = √(E₁² + E₂²)
≈ √((2381.57 N/C)² + (5621.88 N/C)²)
≈ √(5662179.6649 + 31643689.7344)
≈ √(37305869.3993)
Resultant E ≈ 6107.83 N/C

The magnitude of the resultant electric field at the origin is therefore approximately 6107.83 N/C.