A section of a rollercoaster is shown below. If the 17,500 kg rollercoaster momentarily comes to rest at point A, where it is 19 m in the air, how fast is it going at point D?

Answer:

[tex]\boxed {\boxed {\sf 2 \ m/s^2}}[/tex]

Explanation:

Acceleration is the rate of change in velocity with respect to time. It is calculated by dividing the change in velocity by the change in time. The formula is:

[tex]a= \frac{ \Delta v}{\Delta t}[/tex]         or          [tex]a= \frac{v_f-v_i}{\Delta t}[/tex]

The change in velocity is the difference between the initial velocity and the final velocity. The motorcycle starts at rest, or 0 meters per second and reaches 6 meters per second. The change in time is 3 seconds.

[tex]\bullet \ v_f= 6 \ m/s\\\bullet \ v_i= 0 \ m/s \\\bullet \ \Delta t = 3 \ s[/tex]

Substitute the values into the formula

[tex]a= \frac { 6 \ m/s - 0 m/s}{3 \ s}[/tex]

Solve the numerator.

[tex]a= \frac{6 \ m/s}{3 \ s}[/tex]

[tex]a= 2 \ m/s^2[/tex]

The motorcyclist's acceleration is 2 meters per second squared.

Answer:

875 Watts

Explanation:

P = W/t = mgh/t = 700(10)/8 = 875 Watts

Complete Question

Complete Question is attached below

Answer:

a)  [tex]D=0.7[/tex]

b)  [tex]W=1787.5N[/tex]

c)  [tex]\rho'=998.19kg/m^3[/tex]  

Explanation:

From the question we are told that:

Hot air:

Temperature [tex]T_a=360K[/tex]

Pressure [tex]P_a=100kPa[/tex]

Distance [tex]d=12m[/tex]

Weight [tex]W=1400N[/tex]

Water:

Temperature [tex]T_w=300K[/tex]

Pressure [tex]P_w=100kPa[/tex]

Since we have The same Reynolds number

a)

Generally the equation for equal Reynolds number is mathematically given by

Re_{air}=Re_{water}

Therefore

[tex]\frac{\rho V D}{\mu_{air}}=\frac{p_{water}*V*D}{\mu_{water]}}[/tex]

[tex]\frac{100*12}{300*0.28*1.81*10^{-3}}}=\frac{998*D}{0.000890}[/tex]

[tex]D=0.7[/tex]

b)

Generally the equation for Weight of scale is mathematically given by

[tex]W=\rho*V*g[/tex]

[tex]W=998*\frac{4}{3}*\pi*0.35^3*9.81[/tex]

[tex]W=1787.5N[/tex]

c)

Generally the equation for Density of buoyant material is mathematically given by

[tex]\rho'=\frac{w}{g*V}[/tex]

[tex]\rho'=\frac{1781.5}{\frac{4}{3}*\pi*0.35^3*9.81}[/tex]

[tex]\rho'=998.19kg/m^3[/tex]

Answer:

θ' =  14.44 × [tex]10^{6}[/tex]

Explanation:

given data

total distance is d = 4820

radius = 1.4 cm

solution

we get here total angle by which the wheel rotates traveling is express as

⇒  [tex]\theta=2.40\times10^6\ \rm{rev}=2.40\times 2\pi\times10^6\ \rm{rad}[/tex]     ................1

and

total angle (θ)  and the total distance (d) express as

⇒ d = r × θ       ...............2

here r is radius

and here rotated through some other angle θ' so put value in given equation and find revolutions    

⇒  d = (r+r)θ'      ........3

here  r = d/θ

so

⇒ [tex]d = ( \frac{d}{\theta}+r) \theta'[/tex]    

so put value and get θ'

⇒  θ' = 2.40 × 2π × [tex]10^{6}[/tex] × [tex]\frac{4820 \times 10^3}{4820 \times 10^3 +0.014 \times 2.40 \times 2 \times \pi \times 10^6}[/tex]  

Answer:

the masses of the two stars are; m₁ = m₂ = 4.92 × 10³⁰ kg

Explanation:

Given the data in the question;

Time period = 6 months = 1.577 × 10⁷ s

orbital speed v = 64000 m/s

since its a circular orbit,

v = 2πr / T

we solve for r

r = vT/ 2π

r = ( 64000 × 1.577 × 10⁷ ) / 2π

r = 1.6063 × 10¹¹ m = ( (1.6063 × 10¹¹) / (1.496 × 10¹¹) )AU = 1.0737 AU

Now, from Kepler's law

T² = r³ / ( m₁ + m₂ )

T = 6 months = 0.5 years

we substitute

(0.5)² = (1.0737)³ / ( m₁ + m₂ )

0.25 = 1.2378 / ( m₁ + m₂ )

( m₁ + m₂ ) = 1.2378 / 0.25

( m₁ + m₂ ) = 4.9512

m₁ = m₂  = 4.9512 / 2 = 2.4756 solar mass

we know that solar mass = 1.989 × 10³⁰ kg

so

m₁ = m₂ = 2.4756 × 1.989 × 10³⁰ kg

m₁ = m₂ = 4.92 × 10³⁰ kg

Therefore, the masses of the two stars are; m₁ = m₂ = 4.92 × 10³⁰ kg

Answer: They are equal in magnitude.

- The signs of the two changes in potential energy are opposite

Explanation:

When the proton and electron both move through the same displacement in an electric field, the change in potential energy that is associated with the proton is equal in magnitude.

Also, it should be noted that the signs of the two changes in potential energy are opposite.

(B) 1.00 m

Explanation:

Since the meter stick is traveling with Jill, it will have the same speed as she does so relative to Jill, the meter stick is stationary so its length remains 1.00 m as measured by her.

When astronaut Jill leaves Earth in a spaceship and is now traveling at a speed of 0.280c and measure the meter stick to be 1 meter. Hence, option B is correct.

Length contraction is defined as the phenomenon of the moving object being shorter than its appropriate length, measured in the object's rest frame.

When the object travels with the speed of light, the length of the object gets more contracted than its original length, relative to the observer. It is also known as the Lorentz-Fitgerald contraction.

Length contraction, L = L₀√(1-v²/c²), where L is the original length, L₀ is the contracted length. c² is known as the velocity of light. v² is the velocity of the speed of the object.

From the given,

speed of the spaceship = 0.280c      (c is the speed of the light)

Length contraction, L = L₀ √(1-v²/c²)

The stick also travels in the spaceship. Hence, the length of the meter stick does not change. It remains at its original length of one meter.  Thus, the ideal solution is option B.

To learn more about length contraction:

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a) The mass flow rate through the nozzle is 0.27 kg/s.

b) The exit area of the nozzle is 23.6 cm².

a) The mass flow rate through the nozzle can be calculated with the following equation:

[tex] \dot{m_{i}} = \rho_{i} v_{i}A_{i} [/tex]

Where:

[tex]v_{i}[/tex]: is the initial velocity = 20 m/s

[tex]A_{i}[/tex]: is the inlet area of the nozzle = 60 cm²  

[tex]\rho_{i}[/tex]: is the density of entrance = 2.21 kg/m³

[tex] \dot{m} = \rho_{i} v_{i}A_{i} = 2.21 \frac{kg}{m^{3}}*20 \frac{m}{s}*60 cm^{2}*\frac{1 m^{2}}{(100 cm)^{2}} = 0.27 kg/s [/tex]  

Hence, the mass flow rate through the nozzle is 0.27 kg/s.

b) The exit area of the nozzle can be found with the Continuity equation:

[tex] \rho_{i} v_{i}A_{i} = \rho_{f} v_{f}A_{f} [/tex]

[tex] 0.27 kg/s = 0.762 kg/m^{3}*150 m/s*A_{f} [/tex]

[tex] A_{f} = \frac{0.27 kg/s}{0.762 kg/m^{3}*150 m/s} = 0.00236 m^{2}*\frac{(100 cm)^{2}}{1 m^{2}} = 23.6 cm^{2} [/tex]

Therefore, the exit area of the nozzle is 23.6 cm².

You can find another example of mass flow rate here: https://brainly.com/question/13346498?referrer=searchResults

I hope it helps you!                                                                   

a) Mass flow rate through the nozzle: 0.265 kilograms per second, b) Exit area of the nozzle: 23.202 square centimeters.

We determine the Mass Flow Rate through the nozzle and the Exit Area of the nozzle by means of the Principle of Mass Conservation. A nozzle is a device that works at Steady State, so that Mass Balance can be reduced into this form:

[tex]\dot m_{in} = \dot m_{out}[/tex] (1)

Where:

[tex]\dot m_{in}[/tex] - Inlet mass flow, in kilograms per second.

[tex]\dot m_{out}[/tex] - Outlet mass flow, in kilograms per second.

Given that air flows at constant rate, we expand (1) by dimensional analysis:

[tex]\rho_{in} \cdot A_{in}\cdot v_{in} = \rho_{out}\cdot A_{out}\cdot v_{out}[/tex] (2)

Where:

[tex]\rho_{in}, \rho_{out}[/tex] - Air density at inlet and outlet, in kilograms per cubic meter.

[tex]A_{in}, A_{out}[/tex] - Inlet and outlet area, in square meters.

[tex]v_{in}, v_{out}[/tex] - Inlet and outlet velocity, in meters per second.

a) If we know that [tex]\rho_{in} = 2.21\,\frac{kg}{m^{3}}[/tex], [tex]A_{in} = 60\times 10^{-4}\,m^{2}[/tex] and [tex]v_{in} = 20\,\frac{m}{s}[/tex], then the mass flow rate through the nozzle is:

[tex]\dot m = \rho_{in}\cdot A_{in}\cdot v_{in}[/tex]

[tex]\dot m = \left(2.21\,\frac{kg}{m^{3}} \right)\cdot (60\times 10^{-4}\,m^{2})\cdot \left(20\,\frac{m}{s} \right)[/tex]

[tex]\dot m = 0.265\,\frac{kg}{s}[/tex]

The mass flow rate through the nozzle is 0.265 kilograms per second.

b) If we know that [tex]\rho_{in} = 2.21\,\frac{kg}{m^{3}}[/tex], [tex]A_{in} = 60\times 10^{-4}\,m^{2}[/tex], [tex]v_{in} = 20\,\frac{m}{s}[/tex], [tex]\rho_{out} = 0.762\,\frac{kg}{m^{3}}[/tex] and [tex]v_{out} = 150\,\frac{m}{s}[/tex], then the exit area of the nozzle is:

[tex]\rho_{in} \cdot A_{in}\cdot v_{in} = \rho_{out}\cdot A_{out}\cdot v_{out}[/tex]

[tex]A_{out} = \frac{\rho_{in}\cdot A_{in}\cdot v_{in}}{\rho_{out}\cdot v_{out}}[/tex]

[tex]A_{out} = \frac{\left(2.21\,\frac{kg}{m^{3}} \right)\cdot (60\times 10^{-4}\,m^{2})\cdot \left(20\,\frac{m}{s} \right)}{\left(0.762\,\frac{kg}{m^{3}} \right)\cdot \left(150\,\frac{m}{s} \right)}[/tex]

[tex]A_{out} = 2.320\times 10^{-3}\,m^{2}[/tex]

[tex]A_{out} = 23.202\,cm^{2}[/tex]

The exit area of the nozzle is 23.202 square centimeters.

Answer:

The final pressure in bar will be "[tex]\frac{10}{3} \ Bar[/tex]".

Explanation:

As we know,

PV = nRT

[tex]\frac{P_1}{T_1} =\frac{P_2}{T_2} =CONST[/tex]

then,

⇒ [tex]\frac{2 \ bar}{300 \ K} = \frac{P_2}{500 \ K}[/tex]

⇒ [tex]P_2=(\frac{2}{300}\times 500 )Bar[/tex]

        [tex]=\frac{10}{3} \ Bar[/tex]

Thus the above is the correct answer.

Answer:

P = -1 D

Explanation:

For this exercise we must use the equation of the constructor

       / f = 1 / p + 1 / q

where f is the focal length, p and q is the distance to the object and the image, respectively

The far view point is at p =∞  and its image must be at q = -100 cm = 1 m, the negative sign is because the image is on the same side as the image  

        [tex]\frac{1}{f} = \frac{1}{infinity} + \frac{1}{-1}[/tex]

         f = 1 m

         P = 1/f

          P = -1 D

Answer:

The rms voltage of new generator is 3.4 V.

Explanation:

f = 3200 Hz

rms voltage, V = 2 V

rms current, i = 28 mA

Now

f' = 4700 Hz

rms current, i' = 70 mA

let the new rms voltage is V'.

[tex]i = \frac{V}{Xc} = V \times 2\pi fC....(1)\\\\i' = V' \times 2 \pi f' C..... (2)\\\\\frac{i}{i'} =\frac{V f}{V' f'}\\\\\frac{28}{70}=\frac{2\times 3200}{V'\times 4700}\\\\V' = 3.4 V[/tex]

Answer:

the work done on the ball is 0

Explanation:

Given the data in the question;

Katie swings a ball around her head at constant speed in a horizontal circle with circumference 2.1 m.

circle circumference = 2πr = 2.1 m

radius r will be; r = 2.1 m / 2π = 0.33 m

Tension force = 34.4 N

one half revolution means, displacement of the ball is;

d = 2r = 2 × 0.33 = 0.66 m

Now, Work done = force × displacement × cosθ

we know that, the angle between the tension force on string and displacement of object is always 90.

so we substitute

Work done = 34.4 N × 0.66 m × cos(90)

Work done = 34.4 N × 0.66 m × 0

Work done = 0 J

Therefore,  the work done on the ball is 0

The work done by [tex]\vec F[/tex] along the given path C from A to B is given by the line integral,

[tex]\displaystyle \int_C \mathbf F\cdot\mathrm d\mathbf r[/tex]

I assume the path itself is a line segment, which can be parameterized by

[tex]\vec r(t) = (1-t)(-2\,\vec\imath + 5\,\vec\jmath) + t(4\,\vec\imath+7\,\vec\jmath) \\\\ \vec r(t) = (6t-2)\,\vec\imath+(2t+5)\,\vec\jmath \\\\ \vec r(t) = x(t)\,\vec\imath + y(t)\,\vec\jmath[/tex]

with 0 ≤ t ≤ 1. Then the work performed by F along C is

[tex]\displaystyle \int_0^1 \left(6x(t)^3\,\vec\imath-4y(t)\,\vec\jmath\right)\cdot\frac{\mathrm d}{\mathrm dt}\left[x(t)\,\vec\imath + y(t)\,\vec\jmath\right]\,\mathrm dt \\\\ = \int_0^1 (288(3t-1)^3-8(2t+5)) \,\mathrm dt = \boxed{312}[/tex]

Answer:

[tex]u=34cm[/tex]

Explanation:

From the question we are told that:

Far point is [tex]V=34 cm[/tex]

Near point is [tex]u=17 cm[/tex]

Therefore

Focal Length

[tex]f=-34cm[/tex]

Generally the equation for the Lens is mathematically given by

[tex]\frac{1}{u}=\frac{1}{f}-\frac{1}{v}[/tex]

[tex]\frac{1}{u}=\frac{1}{-34}-\frac{1}{-17}[/tex]

[tex]u=34cm[/tex]

Answer:

EMF = 51.01 Volt

Explanation:

A 50-turn coil has a diameter of 15 cm. The coil is placed in a spatially uniform magnetic field of magnitude 0.500~\text{T}0.500 T so that the plane of the coil makes an angle of 30^\circ30 ​∘ ​​ with the magnetic field. Find the magnitude of the emf induced in the coil if the magnetic field is reduced to zero uniformly in 0.100~\text{s}0.100 s

We have,

Number of turn in the coil, N = 50

The diameter of the coil, d = 15 cm

Radius, r = 7.5 cm = 0.075 m

Initial magnetic field, [tex]B_i=0.5\ T[/tex]

The plane of the coil makes an angle of 30° with the magnetic field.

The magnetic field reduced to zero in 0.1 seconds

We need to find the emf induced in the coil. We know that, emf is equal to the rate of change of magnetic flux. So,

[tex]\epsilon=\dfrac{BNA\cos\theta}{t}\\\\\epsilon=\dfrac{0.5\times 50\times \pi \times 0.075\cos(30)}{0.1}\\\\\epsilon=51.01\ V[/tex]

So, the induced emf in the coil is 51.01 V.

Answer:

30ms

Explanation:

you need to multiple the 10ms by 3s which gives you 30ms

Answer:

the correct one is d

Explanation:

The filament of the bulb fulfills the law of ohm

          V = i R

indicate that the filament resistance increases with temperature, as the voltage remains constant if the resistance of the filament increases the current should decrease,

When examining the different statements, the correct one is d

The intensity of radiation is the defined as amount of energy per surface angle which can be used to determine the amount of energy emitting from a source that will hit another surface.

The factors that affect the intensity of radiation are

Type of emission from the source :This  can be alpha, gamma, beta or electromagnetic rays etc

Distance of the detector from the source: The shorter the distance between the source and the detector, the more the effect and vice versa for the longer the distance.

Type of materials between the source and the detector: The type of material between the source and the detector will tell how absorbing and penetrating the radiation is.

Read more on Radiation Intensity here:  https://brainly.com/question/10148635

Answer:

1.5kg

Explanation:

Given data

mass m1= 2.5kg

mass m2=??

velocity of mass one v1= 1.5m/s

velocity of mass two v2= 4.1m/s

common velocity after impact v= 2.5m/s

Let us apply the formula for the conservation of linear momentum for inelastic collision

The expression is given as

m1v1+ m2v2= v(m1+m2)

substitute

2.5*1.5+ m2*4.1= 2.5(2.5+m2)

3.75+4.1m2= 6.25+2.5m2

collect like terms

3.75-6.25= 2.5m2-4.1m2

-2.5= -1.6m2

divide both sides by -1.6

m2= -2.5/-1.6

m2= 1.5 kg

Hence the second mass is 1.5kg

Speed:

Acceleration:

Answer:

[tex]dF=2.5Hz[/tex]

Explanation:

From the question we are told that:

Time [tex]T=0.2sec[/tex]

Generally the Period is given as

 [tex]T= 2 * 0.2 = 0.4[/tex]

Therefore difference in frequency dF

 [tex]dF=\frac{1}{T}[/tex]

 [tex]dF=\frac{1}{0.4}[/tex]

 [tex]dF=2.5Hz[/tex]

Answer:

true

Explanation:

The kinetic energy of a gas is directly proportional to the temperature of the gas.because temperature is the average kinetic energy of a substance

I hope this helps

Answer:

10.78 s

Explanation:

The force on the charge is computed by using the equation:

[tex]F^{\to}= qE^{\to} +q (v^{\to} + B^{\to}) \\ \\ F^{\to} = (9.12 \times 10^{-6}) *278 (-\hat k) +9.12 *10^{-6} *2.1 *0.18 (\hat i * \hat k) \\ \\ F^{\to} = -2.535 *10^{-3} \hat k -3.447*10^{-6} \hat j[/tex]

F = ma

∴

[tex]a ^{\to}= \dfrac{F^{\to}}{m}[/tex]

[tex]a ^{\to}= \dfrac{-1}{3.57\times 10^{-3}}(2.535*10^{-3}\hat k + 3.447*10^{-6} \hat j)[/tex]

[tex]a ^{\to}=-0.710 \hat k -9.656*10^{-4} \hat j[/tex]

At time t(sec; the partiCle velocity becomes [tex]v(t) = 3.78 v_o[/tex]

The velocity of the charge after the time t(sec) is expressed by using the formula:

[tex]v^{\to}= v_{o \ \hat i} + a^{\to }t \\ \\ \implies (2.1)\hat i -0.710 t \hat k -9.656 \times 10^{-4} t \hat j = 3.78 v_o \\ \\ \implies (2.1)^2 +(0.710\ t)^2+ (9.656 *10^{-4}t )^2 = (3.78 *2.1^2 \\ \\ \implies 4.41 +0.5041 t^2 +9.324*10^{-7} t^2 = 63.012 \\ \\ \implies 4.41 +0.5041 t^2 = 63.012\\ \\ 0.5041t^2 = 63.012-4.41 \\ \\ t^2 = \dfrac{58.602}{0.5041} \\ \\ t^2 = 116.25 \\ \\ t = \sqrt{116.25} \\ \\ \mathbf{t = 10.78 \ s}[/tex]

Answer:

X = 10.8387 cm²/s

Explanation:

In this exercise, you're required to convert a value from one unit to another.

Converting 42 ft²/hr to cm²/s;

Conversion:

1 ft² = 929.03 cm²

42 ft² = X cm²

Cross-multiplying, we have;

X = 42 * 929.03

X = 39019.26 cm²

Next, we would divide by time in seconds.

1 hour = 3600 seconds

X = 39019.26/3600

X = 10.8387 cm²/s

This seems to be incomplete, as we do not have any information about the magnetic field surrounding the wire, but we can answer in a general way.

We know that for a wire of length L, with a current I, and in a magnetic field B, the force can be written as:

F = L*(IxB)

if we define the right as the positive x-axis, and knowing that the current flows to the right, we can write:

I = i*(1, 0, 0)

And the field will be some random vector that can't be parallel to the current because in that case, we do not have any force.

To find the direction of the force, which will tell us the direction in which the wire deflects or moves, first, we need to point with our thumb in the direction of the current, in this case, to the right.

Now, with the hand open, using the tip of our other fingers we point in the direction of the magnetic field.

For example, if the magnetic field is in the positive z-axis, we will point upwards.

Now the palm of our hand tells us in which direction the force is applied.

This is the right-hand rule.

For example, in the case that the current goes to the right and the magnetic field is upwards, we could see that the force is to the front.

Answer:

F' = 128 N

Explanation:

The electrostatic force of attraction between two charges is given by Colomb's Law, as follows:

[tex]F = \frac{kq_1q_2}{r^2}\\\\[/tex]

where,

F = Force of attraction = 1 N

G = universal gravitational constant

q₁ = magnitude of the first charge

q₂ = magnitude of the second charge

r = distance between charges

Therefore,

[tex]1\ N = \frac{kq_1q_2}{r^2}[/tex] --------------------- eq(1)

Now, we apply the changes given in the question:

[tex]F' = \frac{k(2q_1)(4q_2)}{(\frac{1}{4}r)^2}\\\\F' = 128(\frac{kq_1q_2}{r^2})[/tex]

using eq (1):

F' = 128(1 N)

F' = 128 N

Answer:

the electric field and the electric potential increase 5 times

Explanation:

The electric field created by a point charge is

         E = k q / r²

in this case the charge changes from q₁ = 1 10⁺⁰ C to q₂ = 5 10⁻⁹ C

with the electric field is proportional to the charge

         E₅ = 5 E₁

the electrical power for a point charge is

         V = k q / r

as the electric power is proportional to the charge

         V₅ = 5 V₁

consequently both the electric field and the electric potential increase 5 times

An electric field is produced by a charged object:

[tex]\to E = \frac{k q}{r^2}\\\\[/tex]

In this situation, the charge shifts:

[tex]\to q_1 = 1 10^{\circ}\ C \\\\ \to q_2 = 5 \times 10^{-9}\ C[/tex]

so with electric field remaining proportional to the charge

[tex]\to E_5 = 5 E_1[/tex]

The electrical power consumed for a point charge:

 [tex]\to V = \frac{k q}{ r}[/tex]

since the electric power is related to the charge

[tex]\to V_5 = 5 V_1[/tex]

As a result, both the electric field as well as the electric potential increase by 5 times.

Learn more:

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Answer:

T = 1/f = 1/50(s)

ω = 2πf = 100π (rad/s)

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