At a certain elevation, the pilot of a balloon has a mass of 125 lb and a weight of 119 lbf. What is the local acceleration of gravity, in ft/s2, at that elevation? If the balloon drifts to another elevation where g = 32.05 ft/s2, what is her weight, in lbf, and mass, in lb?

Answer:

The distance is 2 cm

Solution:

According to the question:

Magnetic field of Earth, B_{E} = [tex]5.0\times 10^{- 5} T[/tex]

Current, I = 5.0 A

We know that the formula of magnetic field is given by:

[tex]B = \farc{\mu_{o}I}{2\pi d}[/tex]

where

d = distance from current carrying wire

Now,

[tex]d = \frac{\mu_{o}I}{2\pi B}[/tex]

[tex]d = \frac{4\pi\times 10^{- 7}\times 5.0}{2\pi\times 5.0\times 10^{- 5}}[/tex]

d = 0.02 m 2 cm

The distance from the wire where the magnitude of the magnetic field equals the Earth's magnetic field is approximately 0.4 meters.

To find the distance from the wire where the magnitude of the magnetic field is equal to the strength of the Earth's magnetic field, we can use the equation:

B = μ0 * I / (2π * r)

Where B is the magnetic field, μ0 is the permeability of free space (4π x [tex]10^-7[/tex]A), I is the current, and r is the distance from the wire.

Plugging in the given values, we have:

B_wire = μ0 * 5.0A / (2π * r) and B_earth = 5.0 x [tex]10^-5 T[/tex]

Setting B_wire equal to B_earth and solving for r:

5.0 x[tex]10^-5[/tex]* 5.0A / (2π * r)

Solving for r, we find that the distance from the wire where the magnitude of the magnetic field is equal to the Earth's is approximately 0.4 meters.

Answer:

acceleration, a = [tex]20 m/s^{2}[/tex]

distance, d = 80 m

Given:

Initial velocity of object, v = 20 m/s

Final velocity of object, v' = 60 m/s

Time interval, [tex]\Delta t = 2.0 s[/tex]

Solution:

(a) Acceleration is the rate at which velocity changes and constant acceleration is when the velocity changes by equal amount in equal intervals of time.

Thus

acceleration, a = [tex]\frac{v' - v}{\Delta t}[/tex]

a = [tex]\frac{60 - 20}{2.0} = 20 m/s^{2}[/tex]

(b) Now, the distance covered is  given by:

[tex]d = vt + \frac{1}{2}at^{2}[/tex]

[tex]d = 20\times 2.0 + \frac{1}{2}20\times 2^{2} = 80 m[/tex]

Answer:

A) The maximum height of the blue ball is 29.7 m above the ground.

B) The height of the blue ball after 1.8 s of throwing the red ball is 22.3 m

C) The balls are at the same height 1.41 s after the red ball is thrown

Explanation:

A) At maximum height, the velocity of the blue ball is 0 because for that instant, the ball does not go up nor down.

The equation for velocity for an accelerated object moving in a straight line is:

v = v0 + a*t

where

v = velocity.

v0 = initial velocity.

a = acceleration, in this case, it is the acceleration of gravity, g, 9.81 m/s².

t = time.

Then:

0 = v0 + g * t  (if the origin of the reference system is the ground, then g is negative)

0 = 23.8 m/s - 9.81 m/s² * t

-23.8 m/s / -9.81 m/s² = t

t = 2.43 s

With this time, we can calculate the position of the blue ball. The equation for position is:

y = y0 + v0 * t + 1/2 * g * t²

y = 0.8 m + 23.8 m/s * 2.43 s - 1/2 * 9.81 m/s² * (2.43 s)²

y = 29.7 m

the maximum height of the blue ball is 29.7 m above the ground.

B) 1.8 s after throwing the red ball, the blue ball was in the air for (1.8 s - 0.6) 1.2 s. Then, using the equation for the position of the blue ball:

y = 0.8 m + 23.8 m/s * 1.2 s - 1/2 * 9.81 m/s² * (1.2 s)² = 22.3 m

The height of the blue ball after 1.8 s of throwing the red ball is 22.3 m

C) Now, we have to find the time at which both positions are equal. Notice that the time of the blue ball is not the same as the time for the red ball. The time for the blue ball is the time of the red ball minus 0.6 s:

t blue = t red - 0.6 s

Then:

position red ball = position blue ball

 y0 + v0 * t + 1/2 * g * t² = y0 + v0 * (t- 0.6) + 1/2 * g * (t-0.6s)²

25 m + 1.2 m/s * t -1/2 * 9.81 m/s² * t² = 0.8 m + 23.8 m/s * (t-0.6 s) - 1/2 * 9.81 m/s² * (t-0.6 s)²

24.2 m + 1.2 m/s * t -4.91 m/s² * t² = 23.8 m/s * t - 14.28 m - 4.91 m/s² * (t-0.6 s)²

38.5 m - 22.6 m/s * t - 4.91 m/s² * t² = -4.91 m/s² (t² - 1.2 s * t + 0.36 s²)

38.5 m - 22.6 m/s * t - 4.91 m/s² * t² = -4.91 m/s² * t² + 5.89 m/s * t - 1.77 m

40.3 m - 28.5 m/s * t = 0

t = -40.3 m / -28.5 m/s

t = 1.41 s

The balls are at the same height 1.41 s after the red ball is thrown and 0.81 s after the blue ball is thrown.

Answer:

The speed of the train is 7.75 m/s towards station.

The speed of the train is 8.12 m/s away from the station.

Explanation:

Given that,

Frequency of the whistles f= 175 Hz

Beat frequency [tex]\Delta f= 4.05 Hz[/tex]

Speed of observer = 0

We need to calculate the frequency

Using formula of beat frequency

[tex]\Delta f=f'-f[/tex]

[tex]f'=\Delta f+f[/tex]

[tex]f'=4.05+175[/tex]

[tex]f'=179.05\ Hz[/tex]

When the train moving towards station, then the frequency heard is more than the actual

Using Doppler effect

[tex]f'=f(\dfrac{v-v_{o}}{v-v_{s}})[/tex]

[tex]v=v-\dfrac{vf}{f'}[/tex]

Put the value into the formula

[tex]v=343-\dfrac{343\times175}{179.05}[/tex]

[tex]v=7.75\ m/s[/tex]

The speed of the train is 7.75 m/s towards station.

When the train moving away form the station

Again beat frequency

[tex]\Delta f=f-f'[/tex]

[tex]f'=f-\Delta [/tex]

[tex]f'=175-4.05[/tex]

[tex]f'=170.95\ Hz[/tex]

We need to calculate the speed

Using Doppler effect

[tex]f'=f(\dfrac{v-v_{o}}{v+v_{s}})[/tex]

[tex]v=\dfrac{vf}{f'}-v[/tex]

Put the value into the formula

[tex]v=\dfrac{343\times175}{170.95}-343[/tex]

[tex]v=8.12\ m/s[/tex]

The speed of the train is 8.12 m/s away from the station.

Hence, This is the required solution.

Answer:

Navier Stokes equation

( 1 )  it is a partial differential equation that is describe the flow of incompressible fluids

Reynolds equation

(1) it is partial differential equation that governs the pressure distribution of thin viscous fluid in lubrication

Explanation:

Navier Stokes equation

( 1 )  it is a partial differential equation that is describe the flow of incompressible fluids

(2) Navier Stokes equation is used to model weather and ocean current and water flow in the pipe and air flow around wing

( 3) equation is

[tex]\nabla .\overrightarrow{v} = 0[/tex]   momentum equation

[tex]\rho \frac{d\overrightarrow{v}}{dt} = \nabla p + \rho \overrightarrow{g} + \mu \nabla ^2 v^2[/tex]

here [tex]\nabla p[/tex] is pressure gradient and [tex]\rho \overrightarrow{g}[/tex] is body force and [tex]\mu \nabla ^2 v^2[/tex] is diffusion term

and

Reynolds equation

(1) it is partial differential equation that governs the pressure distribution of thin viscous fluid in lubrication

(2) it is drive in 1886 from Navier Stokes law

(3) equation is attach

here

Answer:

carrier power is 7.8 kW

Explanation:

given data

power = 10 kW

modulation percentage = 75 %

to find out

carrier power

solution

we will use here power transmitted equation that is

power = [tex]carrier power * ( 1+  \frac{modulation}{2})[/tex]   .................1

put here value in equation 1  we get carrier power

10 = [tex]carrier power *(1+ \frac{0.75}{2})[/tex]

carrier power = 7.8

so carrier power is 7.8 kW

The speed of the ball when it leaves Sarah's hand is 8.2 m/s.

Given that;

The ball leaves Sarah's hand at a distance of 1.5 meters above the ground, reaches a maximum height of 8 m above the ground, and takes 1.619 s to get directly over Julie's head.

For the speed of the ball when it leaves Sarah's hand, use the equations of motion and consider the vertical motion of the ball.

Since the ball is thrown vertically upward and then comes back down, the time taken to reach the maximum height is half of the total time of flight.

Therefore, the time to reach the maximum height is,

t/2 = 1.619 s / 2.

So, the time to reach the maximum height is 0.8095 s.

Now, let's find the initial vertical velocity of the ball using the equation:

v = u + at

where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.

At the maximum height, the final vertical velocity is 0 m/s because the ball momentarily stops.

The acceleration due to gravity, a, is -9.8 m/s² (negative because it acts downward).

Using the equation, we have:

0 m/s = u + (-9.8 m/s²) × 0.8095 s

Simplifying the equation:

u = 7.93 m/s

So, the initial vertical velocity of the ball is 7.93 m/s.

Since the ball travels in a parabolic path, the time taken to reach Julie's horizontal position is the same as the time taken to reach the maximum height, which is 0.8095 s.

Now, let's calculate the initial horizontal velocity of the ball, using the equation:

s = ut

where s is the horizontal distance travelled, u is the initial horizontal velocity, and t is the time.

The horizontal distance travelled is equal to the horizontal distance between Sarah and Julie, which we don't have.

However, since we are only interested in the initial horizontal velocity, we can assume that the horizontal distance travelled is equal to the distance between Sarah and Julie.

Therefore, s = 1.5 m.

Using the equation, we have:

1.5 m = u × 0.8095 s

u = 1.5 m / 0.8095 s

Calculating u, we find:

u ≈ 1.853 m/s

So, the initial horizontal velocity of the ball is 1.853 m/s.

Finally, the speed of the ball when it leaves Sarah's hand by combining the horizontal and vertical components of velocity using the Pythagorean theorem:

speed = √(horizontal velocity² + vertical velocity²)

speed = √(1.853 m/s)² + (7.951 m/s)²

Calculating the speed, we find:

speed ≈ 8.2 m/s

Therefore, the speed of the ball when it leaves Sarah's hand is 8.2 m/s.

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Final answer:

Using the kinematic equations for vertical motion, the initial speed of the ball Sarah threw is calculated to be approximately 7.94 m/s, based on the given maximum height and time to reach Julie's horizontal position.

Explanation:

To find the initial speed of the ball when it leaves Sarah's hand, we need to use the information about the ball's motion under gravity. We know it reaches a maximum height of 8 m and takes 1.619 s to get directly over Julie's head, starting from a height of 1.5 m.

The motion of the ball can be divided into two segments: ascending and descending. During the ascending part, the ball slows down due to gravity until it reaches its maximum height. In the descending part, it accelerates back down. Since the motion is symmetrical, the time to reach the maximum height is half of the total time, which is 1.619 s / 2 = 0.8095 s.

To find the initial velocity (v_i), we can use the kinematic equation for vertical motion:
v_i = v_f - g*t
where v_f is the final velocity (0 m/s at the maximum height), g is the acceleration due to gravity (9.81 m/[tex]s^2[/tex]), and t is the time to reach maximum height.

Plugging in the values, we get:
v_i = 0 m/s - (-9.81 m/[tex]s^2[/tex] * 0.8095 s) = 7.94 m/s
Therefore, the initial speed of the ball when it leaves Sarah's hand is about 7.94 m/s.

Answer:

(a) 0.9 s

(b) 7.78 m/s

Explanation:

height, h = 4 m

Horizontal distance, d = 7 m

Let it takes time t to reach the ground and u be the initial velocity of the jet.

(a) Use second equation of motion in vertical direction

[tex]s = ut + \frac{1}{2}at^{2}[/tex]

In vertical direction, uy = 0 m/s, a = g = - 9.8 m/s^2, h = - 4 m

By substituting the values, we get

[tex]-4 = 0 - \frac{1}{2}\times 9.8\times t^{2}[/tex]

t = 0.9 second

Thus, the time taken by water jet in air is 0.9 second.

(b) Use

Horizontal distance = horizontal velocity x time

d = u t

7 = u x 0.9

u = 7.78 m/s

Thus, the initial velocity of water jet is 7.78 m/s.

The duration the water is in the air is found using the formula for the motion under gravity, which depends on the vertical distance and the acceleration due to gravity. After finding the time, the initial velocity of the water is calculated using the horizontal distance and the fraction of time the water was in motion.

To determine how long the water is in the air (a), and the initial velocity of the water (b) when a water gun is fired horizontally from a hill, we can use the principles of projectile motion. The time a projectile is in the air is solely determined by its vertical motion. Since the water gun is fired horizontally, it has an initial vertical velocity of 0 m/s.

For part (a), the time (t) it takes for the water to reach the ground can be calculated using the formula for the motion under gravity, which is y = 0.5 * g * t2, where y is the vertical distance (4 meters in this case) and g is the acceleration due to gravity (approximated to 9.81 m/s2). Solving for t gives us the time the water is in the air.

For part (b), once we have the time, we can use the horizontal distance (7 meters) to find the initial velocity (v0) using the formula x = v0 * t. This provides the initial horizontal velocity of the water gun's jet. The overall process involves solving for time first and then using that time to find the initial velocity.

Answer:

[tex]K_e_q=22.75878093\frac{N}{m}[/tex]

[tex]f=1.363684118Hz[/tex]

Explanation:

In order to calculate the equivalent spring constant we need to use the next formula:

[tex]\frac{1}{K_e_q} =\frac{1}{K_1} +\frac{1}{K_2} +\frac{1}{K_3} +\frac{1}{K_4}[/tex]

Replacing the data provided:

[tex]\frac{1}{K_e_q} =\frac{1}{113} +\frac{1}{65} +\frac{1}{102} +\frac{1}{101}[/tex]

[tex]K_e_q=22.75878093\frac{N}{m}[/tex]

Finally, to calculate the frequency of oscillation we use this:

[tex]f=\frac{1}{2(pi)} \sqrt{\frac{k}{m} }[/tex]

Replacing m and k:

[tex]f=\frac{1}{2(pi)} \sqrt{\frac{22.75878093}{0.31} } =1.363684118Hz[/tex]

Answer:

a) [tex]x=26m[/tex]

b) [tex]v_{y}=-21.5m/s[/tex]

Explanation:

From the exercise we know initial velocity, initial height

[tex]y_{o}=20m[/tex]

[tex]v_{o} =12m/s[/tex] [tex]\beta =45[/tex]º

a) The range of the stone is defined by how far does it goes. From the theory of free falling objects, we have:

[tex]y=y_{o}+v_{oy}t+\frac{1}{2}gt^{2}[/tex]

The stone strike the water at y=0

[tex]0=20+12sin(45)t-\frac{1}{2}(9.8)t^{2}[/tex]

Solving for t, using the quadratic formula

[tex]t=-1.33s[/tex] or [tex]t=3.06s[/tex]

Since time can't be negative, the answer is t=3.06s

Now, we can calculate the range of the stone

[tex]x=v_{o}t=(12cos(45))m/s(3.06s)=26m[/tex]

b) We can calculate the velocity were the stone strike the water using the following formula  

[tex]v_{y}=v_{oy}+gt=12sin(45)m/s-(9.8m/s^{2})(3.06s)=-21.5m/s[/tex]

The negative sign indicates that the stone is going down

Answer:

(a) x= 26m

(b) vf= 23.15m/s  

Explanation:

Given data

h=20m

Ɵ=45°

to find

(a) range of stone=x=?

(b) velocity=vf=?

Solution

For part (a)

You need to solve for time first using

yf = yi + visinƟt + 1/2gt^2

0 = 20m + 12sin45t + 1/2(-9.8)t^2

and use the quadratic equation to solve for t

t = 3.064 sec

To solve for the distance traveled use

x = xi + vicosƟt + 1/2at^2 there is no acceleration in the x direction so that cancels

x = 12cos(45)(3.064)

x= 26m

For part(b)

For b I'm not sure if you what direction you want the final velocity in the x, y, or the direction its traveling so I'll just give all 3.

Theres no change in the velocity in the x direction so its just vfx = vixcosƟ = 12cos45 = 8.49m/s

For the y direction its vfy^2 = viy^2 + 2g(Δy)

vfy = sqrt((12sin(45))^2 + 2(-9.8)(0-20m)) = 21.54m/s

The velocity the direction the stone is traveling is vf = sqrt(vx^2 + vy^2) = sqrt(8.49^2 + 21.54^2)

vf= 23.15m/s

Answer:

- 3.46 m

Explanation:

initial position, xi = 3.37 m

displacement, Δx = - 6.83 m

Let the final position is xf.

So, displacement = final position - initial position

Δx = xf - xi

- 6.83 = xf - 3.37

xf = 3.37- 6.83

xf = - 3.46 m

Thus, the final position of the squirrel is - 3.46 m.

Answer:

Three linear operators A,B, and C will satisfy the condition [tex][[A, B],C] + [[B,C), A] + [[C, A], B] = 0[/tex].

Explanation:

According to the question we have to prove.

[tex][[A, B],C] + [[B,C), A] + [[C, A], B] = 0[/tex]

Now taking Left hand side of the equation and solve.

[tex][[A, B],C] + [[B,C), A] + [[C, A], B][/tex]

Now use commutator property on it as,

[tex]=[A,B] C-C[A,B]+[B,C]A-A[B,C]+[C,A]B-B[C,A]\\=(AB-BA)C-C(AB-BA)+(BC-CA)A-A(BC-CB)+(CA-AC)B-B(CA-AC)\\=ABC-BAC-CAB+CBA+BCA-CAB-ABC+ACB+CAB-ACB-BCA+BAC\\=0[/tex]

Therefore, it is proved that [tex][[A, B],C] + [[B,C), A] + [[C, A], B] = 0[/tex].

Final answer:

To find the time the football is in the air before hitting the ground, we can analyze the vertical motion using the given initial velocity and launch angle.

Explanation:

To find the time it takes for the football to hit the ground, we need to analyze the vertical motion of the football. We can use the formula:
t = (2 * vy) / g
where t is the time, vy is the vertical component of the initial velocity, and g is the acceleration due to gravity.

Given that the initial velocity is 22.0 m/s and the launch angle is 58.5°, we can find the vertical component of the velocity using the formula:
vy = v * sin(θ)
where v is the initial velocity and θ is the launch angle.

Using this information, we can calculate the time it takes for the football to hit the ground.

Answer:

option (C)

Explanation:

Speed of car = 30 m/s

Let the time taken by the police car to catch the speeding car is t

The distance traveled by the speeding car in t + 1 second is equal to the distance traveled by the police car in time t

Distance traveled by the police car in time t

[tex]s=ut + 0.5 at^{2}[/tex]    .... (1)

Distance traveled by the speeding car in t + 1 second

s = 30 (t + 1) = 300

t + 1 = 10

t = 9 s

Put the value of t in equation (1), we get

300 = 0 + 0.5 x a x 9 x 9

a = 7.41 m/s^2

C. 7.41 meters per square second.

In this question, the car is travelling at constant speed, whereas the police car accelerates uniformly after some delay to catch the car, the respective kinematic formulas are shown below:

Car

[tex]x_{C} = x_{o} + v_{C}\cdot t[/tex] (1)

Police car

[tex]x_{P} = x_{o} + \frac{1}{2}\cdot a_{P}\cdot (t-t')^{2}[/tex] (2)

Where:

If we know that [tex]x_{o} = 0\,m[/tex], [tex]x_{C} = x_{P} = 300\,m[/tex], [tex]t' = 1\,s[/tex] and [tex]v_{C} = 30\,\frac{m}{s}[/tex], then we have the following system of equations:

[tex]300 = 30\cdot t[/tex] (1)

[tex]300 = \frac{1}{2}\cdot a_{P}\cdot (t-1)^{2}[/tex] (2)

By (1):

[tex]t = 10[/tex]

Then we find that acceleration of the police car must be:

[tex]300 = \frac{1}{2}\cdot a_{P}\cdot (10-1)^{2}[/tex]

[tex]a_{P} = 7.407\,\frac{m}{s^{2}}[/tex]

Therefore, the correct choice is C.

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Using Ohm's law and properties of series and parallel circuits, it's possible to find the resistances. In series, resistances are directly added and for parallel, the reciprocal of total resistance is the sum of reciprocals of individual resistances. Applying these principles with given current and voltage, one can solve for resistances.

This question pertains to electricity and specifically the characteristics of resistors when they are connected in series or parallel. Using Ohm's Law, we know that the voltage (V) is the product of the current (I) and the resistance (R). Therefore, when the resistors are connected in series, the combined resistance (Rtotal) is the sum of the individual resistances, while the current remains the same. This gives us Rtotal = V/I = 12.0V / 1.12A.

When in parallel, however, the total resistance can be found differently. In a parallel circuit, the total resistance is given by 1/Rtotal = 1/R+ 1/R. As per the problem, we know that the total current of the circuit connected in parallel is 9.39 A, so we can use the equation Itotal = V/ Rtotal to find the total resistance in the parallel circuit.

Combining the information from both the circuits would allow us to solve two simultaneous equations to get the values of R and R.

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The two resistances connected in series have a total resistance of 10.71 Ω, and connected in parallel have a total resistance of 1.28 Ω. Solving the equations, we find the resistances to be 1.43 Ω and 9.28 Ω.

Let's solve this step-by-step:

Step 1: Series Connection

When two resistors are connected in series, the total resistance, [tex]R_{total[/tex], is the sum of the resistances:

[tex]R_{total[/tex] = R₁ + R₂

Using Ohm's Law (V = IR), we can find the total resistance in series:

V = 12.0 V;

I = 1.12 A;

[tex]R_{total[/tex] = V/I

[tex]R_{total[/tex] = 12.0 V / 1.12 A = 10.71 Ω

Step 2: Parallel Connection

When the same resistors are connected in parallel, the total resistance, [tex]R_{parallel[/tex], can be found using the formula:

1/[tex]R_{parallel[/tex] = 1/R₁ + 1/R₂

Again, using Ohm's Law, we first find [tex]R_{parallel[/tex]:

V = 12.0 V; [tex]I_{total[/tex] = 9.39 A;

[tex]R_{parallel[/tex] = V/[tex]I_{total[/tex] = 12.0 V / 9.39 A = 1.28 Ω

Step 3: Solving the Equations

We now have two equations:

R₁ + R₂ = 10.71 Ω

(R₁ * R₂) / (R₁ + R₂) = 1.28 Ω (since 1/[tex]R_{parallel[/tex] = 1/R₁ + 1/R₂ )

Let's solve these equations:

Substitute R₂ = 10.71 - R₁ into the parallel equation:

(R₁ * (10.71 - R₁)) / 10.71 = 1.28

R₁ * (10.71 - R₁) = 1.28 * 10.71

10.71R₁ - R₁² = 13.69

R₁² - 10.71R₁ + 13.69 = 0

Solving the quadratic equation using the quadratic formula:

R₁ = [10.71 ± √((10.71)² - 4*1*13.69)] / 2

Solving this, we get R₁ ≈ 1.43 Ω or R₁ ≈ 9.28 Ω

Then, R₂ = 10.71 - 1.43 = 9.28 Ω ,

R₂ = 10.71 - 9.28 = 1.43 Ω

Answer:

The force on X Fx=0 N

The force on Y Fy=-2.18 N

Explanation:

We have an array of charges, we will use the coulomb's formula to solve this:

[tex]F=k*\frac{Q*Q'}{r^2}\\where:\\k=coulomb constant\\r=distance\\Q=charge[/tex]

but we first have to find the distance and the angule of the charge respect the charges on the Y axis:

[tex]r=\sqrt{(7*10^{-2}m)^2+ (3*10^{-2}m)^2} \\r=7.62cm=0.0762m[/tex]

we can notice that it is the same distance from both charges on Y axis.

we can find the angle with:

[tex]\alpha = arctg(\frac{7cm}{3cm})=66.80^o[/tex]

for the charge of 5µC [tex]\alpha =-66.80^o[/tex]

for the charge of -5µC [tex]\alpha =66.80^o[/tex]

the net force on the X axis will be:

[tex]F_{x5u}=9*10^9*\frac{5*10^{-6}*2*10^{-6}}{0.0762^2}*cos(-66.80)\\F_{x5u}=0.465N[/tex]

and

[tex]F_{x(-5u)}=9*10^9*\frac{-5*10^{-6}*2*10^{-6}}{0.0762^2}*cos(66.80)\\F_{x(-5u)}=-0.465N[/tex]

So the net force on X will be Zero.

for the force on Y we have:

[tex]F_{y5u}=9*10^9*\frac{5*10^{-6}*2*10^{-6}}{0.0762^2}*sin(-66.80)\\F_{y5u}=-1.09N[/tex]

and

[tex]F_{y(-5u)}=9*10^9*\frac{-5*10^{-6}*2*10^{-6}}{0.0762^2}*sin(66.80)\\F_{y(-5u)}=-1.09N[/tex]

Fy=[tex]F_{y5u}+F_{y(-5u)}[/tex]

So the net force on Y is Fy=-2.18N

To find the force on a charge of 2 µC on the x axis at x = 3 cm, we can use Coulomb's Law. The force is calculated to be 5.12 N.

To find the force on a charge of 2 µC on the x axis at x = 3 cm, we can use Coulomb's Law. Coulomb's Law states that the force between two charges is directly proportional to the product of their magnitudes and inversely proportional to the square of the distance between them.

The formula for Coulomb's Law is:

F = k * |q1 * q2| / r^2

Where F is the force, k is the electrostatic constant (k ≈ 8.99 * 10^9 N * m² / C²), q1 and q2 are the magnitudes of the charges, and r is the distance between them.

In this case, one charge is 5 µC and the other charge is -5 µC, with a distance of 14 cm between them (7 cm on the positive y-axis and 7 cm on the negative y-axis). The charge we're interested in is 2 µC at x = 3 cm. Plugging these values into the formula:

F = (8.99 * 10^9 N * m² / C²) * |(5 µC) * (2 µC)| / (0.14 m)²

F = 5.12 N

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Answer:1.26 m/s

Explanation:

Given

translation speed of ball =3.5 m/s

Moment of inertia of ball about com [tex]I=\frac{2}{5}mr^2[/tex]

Initial Energy

[tex]E_i=\frac{1}{2}mu^2+\frac{1}{2}I\omega _i^2(\omega =\frac{u}{r})[/tex]

Final Energy

[tex]E_f=\frac{1}{2}mv^2+\frac{1}{2}I\omega _f^2+mgh[/tex]

Equating energy as no energy loss take place

[tex]E_i=E_f[/tex]

[tex]\frac{1}{2}mu^2+\frac{1}{2}I\omega _i^2=\frac{1}{2}mv^2+\frac{1}{2}I\omega _f^2+mgh[/tex]

[tex]\frac{1}{2}mu^2+\frac{1}{2}\times \frac{2}{5}mr^2\times \left ( \frac{u}{r}\right )^2=\frac{1}{2}mv^2+\frac{1}{2}\times \frac{2}{5}mr^2\times \left ( \frac{v}{r}\right )^2+mgh[/tex]

m term get cancel

[tex]\left ( \frac{u^2}{2}\right )+\left ( \frac{2u^2}{10}\right )=\left ( \frac{v^2}{2}\right )+\left ( \frac{2v^2}{10}\right )+gh[/tex]

[tex]\frac{7}{10}u^2=\frac{7}{10}v^2+gh[/tex]

[tex]v^2=3.5^2-\frac{10}{7}\times 9.81\times 0.76[/tex]

[tex]v=\sqrt{1.6}=1.26 m/s[/tex]

Answer:

a)[tex]v=6.19m/s[/tex]

b)[tex]v=5.51m/s[/tex]

c)[tex]a=3.3*10^{3}m/s^{2}[/tex]

d)[tex]x=5.78*10^{-3}m[/tex]

Explanation:

h1=195m

h2=1.55

a) Velocity just before the ball strikes the floor:

Conservation of the energy law

[tex]E_{o}=E_{f}[/tex]

[tex]E_{o}=mgh_{1}[/tex]

[tex]E_{f}=1/2*mv^{2}[/tex]

so:

[tex]v=\sqrt{2gh_{1}}=\sqrt{2*9.81*1.95}=6.19m/s[/tex]

b) Velocity just after the ball leaves the floor:

[tex]E_{o}=E_{f}[/tex]

[tex]E_{o}=1/2*mv^{2}[/tex]

[tex]E_{f}=mgh_{2}[/tex]

so:

[tex]v=\sqrt{2gh_{2}}=\sqrt{2*9.81*1.55}=5.51m/s[/tex]

c) Relation between Impulse, I, and momentum, p:

[tex]I=\Delta p\\ F*t=m(v_{f}-v{o})\\ (ma)*t=m(v_{f}-v{o})\\\\ a=\frac{ v_{f}-v{o}}{t}=\frac{ 5.51- (-6.19)}{3.5*10^{-3}}=3.3*10^{3}m/s^{2}[/tex]

d) The compression of the ball:

The time elapsed between the ball touching the ground and it is fully compressed, is half the time the ball is in contact with the ground.

[tex]t_{2}=t/2=3.5/2=1.75ms[/tex]

Kinematics equation:

[tex]x(t)=v_{o}t+1/2*a*t_{2}^{2}[/tex]

Vo is the velocity when the ball strike the floor, we found it at a) 6.19m/s.

a, is the acceleration found at c) but we should to use it with a negative sense, because its direction is negative a Vo, a=-3.3*10^3

So:

[tex]x=6.19*1.75*10^{-3}-1/2*3.3*10^{3}*(1.75*10^{-3})^2=5.78*10^{-3}m[/tex]

Answer:

(a) d = 20 m

(b) d' = 80 m

(c) x = 28 m

(d) x' = 96 m

Solution:

As per the question:

Initial velocity of the object, v = 0

Constant acceleration of the object, [tex]a_{c} = 10 m/s^{2}[/tex]

(a) Distance traveled, d in t = 2.0 s is given by the second eqn of motion:

[tex]d = vt + \frac{1}{2}a_{c}t^{2}[/tex]

[tex]d = 0.t + \frac{1}{2}\times 10\times 2^{2} = 20 m[/tex]

(b) Distance traveled, d' in t = 4.0 s is given by the second eqn of motion:

[tex]d' = vt + \frac{1}{2}a_{c}t^{2}[/tex]

[tex]d' = 0.t + \frac{1}{2}\times 10\times 4^{2} = 80 m[/tex]

Now, when initial velocity, v = 4 m/s, then

(c) Distance traveled, x in t = 2.0 s is given by the second eqn of motion:

[tex]x = vt + \frac{1}{2}a_{c}t^{2}[/tex]

[tex]x = 4\times 2.0 + \frac{1}{2}\times 10\times 2^{2} = 28 m[/tex]

(d) Distance traveled, x' in t = 4.0 s is given by the second eqn of motion:

[tex]x = vt + \frac{1}{2}a_{c}t^{2}[/tex]

[tex]x = 4\times 4.0 + \frac{1}{2}\times 10\times 4^{2} = 96 m[/tex]

Answer:

force is 96 N

Explanation:

given data

mass = 40 kg

acceleration = 2.4 m/s²

to find out

force

solution

we know force is mass time acceleration so

we will apply here force formula that is express as

force = m × a   ..............1

here m is mass and a is acceleration so

put here value in equation 1 we get force

force = 40 × 2.4

force = 96

so force will be 96 N

Answer:

mechanical energy per unit mass is 887.4 J/kg

power generated is 443.7 MW

Explanation:

given data

average velocity = 3 m/s

rate = 500 m³/s

height h = 90 m

to find out

total mechanical energy and power generation potential

solution

we know that mechanical energy is sum of potential energy and kinetic energy

so

E = [tex]\frac{1}{2}[/tex]×m×v² + m×g×h    .............1

and energy per mass unit is

E/m =  [tex]\frac{1}{2}[/tex]×v² + g×h

put here value

E/m =  [tex]\frac{1}{2}[/tex]×3² + 9.81×90

E/m = 887.4 J/kg

so mechanical energy per unit mass is 887.4 J/kg

and

power generated is express as

power generated = energy per unit mass ×rate×density

power generated = 887.4× 500× 1000

power generated = 443700000

so power generated is 443.7 MW

The total mechanical energy per unit mass of the river water is the sum of the potential and kinetic energy. The potential energy is calculated using the height and the gravitational constant, while kinetic energy is calculated using velocity. The power generation potential of the river is the total mechanical energy multiplied by the volume flow rate.

The total mechanical energy per unit mass of the river water at this location can be determined using the principle of mechanical energy conservation and the knowledge of potential and kinetic energy. The total mechanical energy (E) equals the potential energy (PE) plus the kinetic energy (KE) per unit mass.

The potential energy is given by: PE = mgh, where m is the mass, g is the acceleration due to gravity (~9.81 m/s^2), and h is the height (90 m above lake surface). But since we are looking for energy per unit mass, m cancels out and we get PE = gh.

The kinetic energy is given by: KE = 0.5mv^2, where v is the velocity (3 m/s). In per unit mass terms, this simplifies to KE = 0.5v^2.

Therefore, E = PE + KE = gh + 0.5v^2.

The power generation potential can be calculated using the equation: P = E * volume flow rate (in our case, 500 m^3/s).

So,​​ P = (gh + 0.5v^2) * 500.

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Answer:

The diameter is 0.378 ft.

Explanation:

Given that,

Mass of shot = 12 lb

Density of fresh water = 62.4 lb/ft

Specific gravity = 6.8

We need to calculate the volume of shot

[tex]V = \dfrac{4}{3}\pi r^3\ ft^3[/tex]

The density of shot is

Using formula of density

[tex]\rho = \dfrac{m}{V}[/tex]

Put the value into the formula

[tex]\rho =\dfrac{12}{ \dfrac{4}{3}\pi r^3}[/tex]

We need to calculate the radius

Using formula of specific gravity

[tex]specific\ gravity =\dfrac{density\ of\ shot}{dnsity\ of\ water}[/tex]

Put the value into the formula

[tex]6.8=\dfrac{\dfrac{12}{\dfrac{4}{3}\pi r^3}}{62.4}[/tex]

[tex]r^3=\dfrac{12}{\dfrac{4}{3}\pi\times6.8\times62.4}[/tex]

[tex]r^3=0.0067514[/tex]

[tex]r =(0.0067514)^{\frac{1}{3}}[/tex]

[tex]r=0.1890\ ft[/tex]

The diameter will be

[tex]d = 2\times r[/tex]

[tex]d =2\times0.1890[/tex]

[tex]d =0.378\ ft[/tex]

Hence, The diameter is 0.378 ft.

Answer:

Explanation:

Let the charge on bead A be q nC  and the charge on bead B be 28nC - qnC

Force F between them

4.8\times10^{-4} = [tex]\frac{9\times10^9\times q\times(28-q)\times10^{-18}}{(5\times10^{-2})^2}[/tex]

=120 x 10⁻⁸ = 9 x q(28 - q ) x 10⁻⁹

133.33 = 28q - q²

q²- 28q +133.33 = 0

It is a quadratic equation , which has two solution

q_A = 21.91 x 10⁻⁹C or q_B = 6.09 x 10⁻⁹ C

The charges [tex]q_A[/tex] = 21.907nC and [tex]q_B[/tex] = 6.093nC are the charges on the beads.

We have to use Coulomb's Law and the principle of charge conservation.

1. Conservation of Charge:

  The total charge before and after they are touched must be the same. Initially, bead A has a charge of 28 nC, and bead B is neutral. After touching, let:

[tex]\[ q_A + q_B = 28 \, \text{nC} \][/tex]

2. Coulomb's Law:

  The force between two charges is given by Coulomb's Law:

[tex]\[ F = k \frac{q_A q_B}{r^2} \][/tex]

Let's set up the equation:

[tex]\[4.8 \times 10^{-4} = 8.99 \times 10^9 \frac{q_A q_B}{(0.05)^2}\][/tex]

Rearrange to solve for [tex]\( q_A q_B \)[/tex]:

[tex]\[q_A q_B = \frac{4.8 \times 10^{-4} \times (0.05)^2}{8.99 \times 10^9}\][/tex]

[tex]\[q_A q_B = \frac{4.8 \times 10^{-4} \times 0.0025}{8.99 \times 10^9}\][/tex]

[tex]\[q_A q_B = \frac{1.2 \times 10^{-6}}{8.99 \times 10^9}\][/tex]

[tex]\[q_A q_B = 1.3348 \times 10^{-16} \, \text{C}^2\][/tex]

Now, we have two equations:

1. [tex]\( q_A + q_B = 28 \times 10^{-9} \, \text{C} \)[/tex]

2. [tex]\( q_A q_B = 1.3348 \times 10^{-16} \, \text{C}^2 \)[/tex]

To solve these, we can substitute [tex]\( q_B = 28 \times 10^{-9} \, \text{C} - q_A \)[/tex] into the second equation:

[tex]\[q_A \left( 28 \times 10^{-9} - q_A \right) = 1.3348 \times 10^{-16}\][/tex]

[tex]\[28 \times 10^{-9} q_A - q_A^2 = 1.3348 \times 10^{-16}\][/tex]

[tex]\[q_A^2 - 28 \times 10^{-9} q_A + 1.3348 \times 10^{-16} = 0\][/tex]

This is a quadratic equation of the form [tex]\( ax^2 + bx + c = 0 \)[/tex], where:

[tex]\[a = 1, \quad b = -28 \times 10^{-9}, \quad c = 1.3348 \times 10^{-16}\][/tex]

Solve using the quadratic formula [tex]\( q_A = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex]:

[tex]\[q_A = \frac{28 \times 10^{-9} \pm \sqrt{(28 \times 10^{-9})^2 - 4 \times 1 \times 1.3348 \times 10^{-16}}}{2}\][/tex]

[tex]\[q_A = \frac{28 \times 10^{-9} \pm \sqrt{7.84 \times 10^{-16} - 5.3392 \times 10^{-16}}}{2}\][/tex]

[tex]\[q_A = \frac{28 \times 10^{-9} \pm \sqrt{2.5008 \times 10^{-16}}}{2}\][/tex]

[tex]\[q_A = \frac{28 \times 10^{-9} \pm 1.5814 \times 10^{-8}}{2}\][/tex]

This gives us two solutions for [tex]\( q_A \)[/tex]:

1. [tex]\( q_A = \frac{28 \times 10^{-9} + 1.5814 \times 10^{-8}}{2} \\q_A = \frac{4.3814 \times 10^{-8}}{2} \\q_A = 2.1907 \times 10^{-8} \, \text{C} \\q_A = 21.907 \, \text{nC} \)[/tex]

2. [tex]\( q_A = \frac{28 \times 10^{-9} - 1.5814 \times 10^{-8}}{2} \\q_A = \frac{1.2186 \times 10^{-8}}{2} \\q_A = 6.093 \times 10^{-9} \, \text{C} \\q_A = 6.093 \, \text{nC} \)[/tex]

Since bead A has the greater charge, we take:

[tex]\[q_A = 21.907 \, \text{nC}\][/tex]

And the charge on bead B:

[tex]\[q_B = 28 \, \text{nC} - 21.907 \, \text{nC} = 6.093 \, \text{nC}\][/tex]

So, the charges on the beads are:

[tex]\[q_A = 21.907 \, \text{nC}, \quad q_B = 6.093 \, \text{nC}\][/tex]

To determine the distance to the epicenter of an earthquake, we can use the time difference between the arrival of P-waves and S-waves. In this case, the P-wave arrives 2.00 minutes before the S-wave. The distance of the earthquake center can be calculated by multiplying the time difference by the speed difference between the two waves.

To determine the distance to the epicenter of an earthquake, we can use the time difference between the arrival of P-waves and S-waves. In this case, the P-wave arrives 2.00 minutes before the S-wave. We know that the speed of the P-wave is 9000 m/s and the speed of the S-wave is 5000 m/s.

We can calculate the distance using the formula: distance = speed × time.

So, the distance of the earthquake center can be calculated as follows:

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Answer:

76 degree  south of west.

Explanation:

We shall represent velocities in vector form , considering east as  x axes and west as Y axes.

V₁ = 4.4 j

V₂ = 3.8, 30 degree  west of south

V₂ = -  3.8 sin 30 i - 3.8 cos 30 j

= - 1.9 i - 3.29 j

Change in velocity

= V₂ - V₁

= - 1.9 i - 3.29 j - 4.4 j

= - 1.9 i - 7.69 j

Acceleration

= change in velocity / time

(- 1.9 i - 7.69 j  ) / 60 ms⁻² .

Direction of acceleration θ

Tan θ = 7.69 / 1.9 =4.047

θ = 76 degree  south of west.

Answer:t=0.54 s

Explanation:

Given

Jacob is traveling 5 m/s in North direction

Jacob throw a ball with a in south direction with a velocity of 5 m/s

Ball is thrown in opposite direction of motion of car therefore it seems as if it is dropped from car as its net horizontal velocity is 5-5=0

Time taken by ball to reach ground

[tex]s=ut+\frac{gt^2}{2}[/tex]

[tex]1.45=0+\frac{9.81\times t^2}{2}[/tex]

[tex]t^2=frac{2\times 1.45}{9.81}[/tex]

t=0.54 s

Motion of ball will be straight line

Answer:

1.93 m

Explanation:

Initial velocity of fish u = 11 m /s

Angle of jump = 34 degree.

Vertical component of its velocity = u sin34 = 11 x.5592 = 6.15 m /s

Considering its motion in vertical direction ,

u = 6.15 m/s

g = - 9.8 m /s

maximum height attained = h

From the formula

v² = u² - 2gh

0 = 6.15x 6.15 - 2 x 9.8 h

h = (6.15 x 6.15 )/ 2 x 9.8

= 1.93 m .

Answer:

Explanation:

Moles of helium ( n ) = 156.7 / 4 = 39.175

Temperature T₁ = 35.73 +273 = 308.73 K

Volume V₁ = V

Pressure P₁ = 3.55 atm

V₂ =1.75 V

a ) For isothermal change

P₁ V₁ = P₂V₂

P₂ = P₁ V₁ / V₂

= 3.55 X V / 1.75 V

= 2.03 atm.

b ) Work done by the gas = nRT ln(V₂/V₁)

= 39.175 X 8.321 X 308.73 X ln 1.75

= 56318.8 J

Work done on the gas = - 56318.8 J

c ) Since there is no change in temperature , internal energy of gas is constant

Q = ΔE + W

ΔE  = 0

Q = W

Work done by gas = heat absorbed

heat absorbed = 56318.8 J

d ) Change in the internal energy of gas is zero because temperature is constant.

Answer:

Explanation:

When saw slices wood by exerting a force on the wood , wood also exerts a reaction force on the saw in opposite direction which is equal to the force of action that is 104 N.

So torque exerted by wood on the blade

= force x perpendicular distance from the axis of rotation

= 104 x .128

=13.312 Nm.

Since this torque opposes the movement of blade , it turns the blade slower.

The speed of the ball when passing its original height on its way down is 20 m/s. The time it takes to reach the ground level is 3.19 seconds. The ball travels a total distance of 70 meters.

The motion of the tennis ball can be analyzed by using the principles of Physics, particularly the laws of motion and the concept of gravity.

(a) The ball will have the same speed when it passes its original height on its way down, which is 20 m/s. This is based on the principle of conservation of energy. Since air resistance is ignored, the speed when it passes the original height on its way down should be the same as its initial speed.

(b) Calculating the total time it takes the ball to reach the ground involves the formula for time in free fall, which is t = sqrt(2h/g), wherein h is the height and g is the gravity. Thus plug in the values: t = sqrt((2*50)/9.8) = 3.19 seconds.

(c) The total distance traveled by the ball includes it going up and falling back down. The ball rises to a height of 20 m/s before falling from that height plus an additional height corresponding to the height of the cliff, total distance is 70 meters.

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