onsider a 20-cm-thick granite wall with a thermal conductivity of 2.79 W/m·K. The temperature of the left surface is held constant at 50°C, whereas the right face is exposed to a flow of 22°C air with a convection heat transfer coefficient of 15 W/m2·K. Neglecting heat transfer by radiation, find the right wall surface temperature and the heat flux through the wall.

The height of the cliff is 16.97 m, the maximum height reached by the arrow is 24.47 m, and the impact speed of the arrow just before hitting the cliff is 16.47 m/s.

In this scenario, we can apply the equations of motion to calculate the height of the cliff, the maximum height reached by the arrow, and its impact speed.

Firstly, the height of the cliff can be calculated using the equation Y = Yo + Vy*t - 0.5*g*t^2, where g is the gravity, t is the time, Vy is the initial vertical speed, and Yo is the initial height. Given Yo = 1.7m, Vy = 26sin(60°), t = 3.4s, and g = 9.8 m/s^2, the height H of the cliff is 16.97 m.

Secondly, the maximum height reached by the arrow can be calculated by the equation Hmax = Yo + Vy*t - 0.5*g*(t)^2, where t is the time it takes to reach the maximum height, which can be Ve/g. Ve is the initial vertical velocity whose value is Vy = 26sin(60°). Hence the maximum height Hmax is 24.47 m.

Finally, the arrow’s impact speed can be calculated by using Pythagoras' theorem. The impact speed V = sqrt((Vx)^2 + (Vy)^2), where Vx is the horizontal velocity and Vy is the final vertical velocity. Given Vx = 26cos(60°) and Vy = Ve - g*t, with Ve = 26sin(60°) and t = 3.4s, the impact speed V of the arrow just before hitting the cliff is 16.47 m/s.

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The height of the cliff is 21.6 m, and the arrow’s impact speed just before hitting the cliff is approximately 16.8 m/s. These calculations use projectile motion equations for both components.

A. To determine the height of the cliff, we can use the vertical motion equation:

[tex]y = y_0 + v_{0y}t - 0.5gt^2[/tex]

Where:

Substituting these values into the equation:

y = 1.7 + 22.5(3.4) - 0.5(9.8)(3.4)2

y = 1.7 + 76.5 - 56.6 = 21.6 m

Therefore, the height of the cliff is 21.6 m.

B. To find the impact speed, we need to calculate both the final vertical and horizontal components of velocity:

The total impact speed is found using the Pythagorean theorem:

[tex]v_f = \sqrt{(v_x^2 + v_{fy}^2)} = \sqrt{(132 + (-10.8)2)} \approx 16.8 m/s[/tex]

Thus, the arrow's impact speed is approximately 16.8 m/s.

Answer:

at t=46/22, x=24 699/1210 ≈ 24.56m

Explanation:

The general equation for location is:

x(t) = x₀ + v₀·t + 1/2 a·t²

Where:

x(t) is the location at time t. Let's say this is the height above the base of the cliff.

x₀ is the starting position. At the base of the cliff we'll take x₀=0 and at the top x₀=46.0

v₀ is the initial velocity. For the ball it is 0, for the stone it is 22.0.

a is the standard gravity. In this example it is pointed downwards at -9.8 m/s².

Now that we have this formula, we have to write it two times, once for the ball and once for the stone, and then figure out for which t they are equal, which is the point of collision.

Ball: x(t) = 46.0 + 0 - 1/2*9.8 t²

Stone: x(t) = 0 + 22·t - 1/2*9.8 t²

Since both objects are subject to the same gravity, the 1/2 a·t² term cancels out on both side, and what we're left with is actually quite a simple equation:

46 = 22·t

so t = 46/22 ≈ 2.09

Put this t back into either original (i.e., with the quadratic term) equation and get:

x(46/22) = 46 - 1/2 * 9.806 * (46/22)² ≈ 24.56 m

The velocity of the cart relative to the ground is approximately 89.44 km/h in a direction about 63.43 degrees north of west.

To find the velocity of the cart relative to the ground when a plane is traveling north at 100.0 km/h with a 40.0 km/h crosswind blowing west and the cart is rolling at 20.0 km/h towards the tail of the plane, we can use vector addition to determine the resultant velocity.

1. First, break down the velocities into their horizontal (west-east) and vertical (north-south) components:

- Plane's velocity (north): 100.0 km/h

- Crosswind velocity (west): 40.0 km/h

- Cart's velocity (towards tail): 20.0 km/h

2. The horizontal component of the cart's velocity is the crosswind velocity (40.0 km/h), and the vertical component is its velocity towards the tail of the plane (20.0 km/h).

3. To find the resultant velocity, we can use vector addition by adding the horizontal and vertical components of the velocities separately:

Horizontal component: 40.0 km/h (west) - 0 km/h (east) = 40.0 km/h (west)

Vertical component: 100.0 km/h (north) - 20.0 km/h (south) = 80.0 km/h (north)

4. Now, we can use the Pythagorean theorem to find the magnitude of the resultant velocity:

Resultant velocity = √(40.0^2 + 80.0^2)

Resultant velocity = √(1600 + 6400)

Resultant velocity = √8000

Resultant velocity ≈ 89.44 km/h

5. To find the direction of the resultant velocity, we can use trigonometry:

Direction = arctan(vertical component / horizontal component)

Direction = arctan(80.0 / 40.0)

Direction = arctan(2)

Direction ≈ 63.43 degrees north of west

6. Therefore, the velocity of the cart relative to the ground is approximately 89.44 km/h in a direction about 63.43 degrees north of west.

Answer:

44.755 m

Explanation:

Given:

Height of the movie star, H = 1.75 m = 1750 mm

Height of the image, h = - 8.25 mm

Focal length of the camera = 210 mm

Let the distance of the object i.e the distance between camera and the movie star be 'u'

and

distance between the camera focus and image be 'v'

thus,

magnification, m = [tex]\frac{\textup{h}}{\textup{H}}[/tex]

also,

m = [tex]\frac{\textup{-v}}{\textup{u}}[/tex]

thus,

[tex]\frac{\textup{-v}}{\textup{u}}=\frac{\textup{h}}{\textup{H}}[/tex]

or

[tex]\frac{\textup{-v}}{\textup{u}}=\frac{\textup{-8.25}}{\textup{1750}}[/tex]

or

[tex]\frac{\textup{1}}{\textup{v}}=-\frac{\textup{1750}}{\textup{-8.25}}\times\frac{1}{\textup{u}}[/tex]  ....................(1)

now, from the lens formula

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

on substituting value from (1)

[tex]\frac{\textup{1}}{\textup{210}}=\frac{\textup{1}}{\textup{u}}+-\frac{\textup{1750}}{\textup{-8.25}}\times\frac{1}{\textup{u}}[/tex]

or

[tex]\frac{\textup{1}}{\textup{210}}=\frac{\textup{1}}{\textup{u}}(1 -\frac{\textup{1750}}{\textup{-8.25}})[/tex]

or

u = 210 × ( 1 + 212.12 )

or

u = 44755.45 mm

or

u = 44.755 m

Answer:

a )  4.5 N.s

b) V =5 m/s    

Explanation:

given,

mass of rifle(M)  = 0.9 kg

mass of bullet(m)  = 6 g = 0.006 kg

velocity of the bullet(v)  = 750 m/s

a) momentum of bullet = m × v

                                  = 750 × 0.006

                                  = 4.5 N.s

b) recoil velocity                                                      

m × u + M × U = m × v + M × V

0  + 0  = 0.006 × 750 -  0.9 × V

V = [tex]\dfrac{4.5}{0.9}[/tex]

V =5 m/s                    

Final answer:

The momentum of the bullet after being fired is 4.5 kg*m/s. The rifle's recoil velocity, while ignoring external forces, is -5 m/s, indicating direction opposite to that of the bullet's motion.

Explanation:

The question asks about the momentum of a bullet after being fired from a rifle and the subsequent recoil velocity of the rifle. To solve this problem, we use the principle of conservation of momentum.

Part A: Bullet Momentum

The momentum of the bullet (pbullet) can be calculated using the formula p = m * v, where m is the mass and v is the velocity. For the bullet:

Mass of the bullet (mbullet): 0.006 kg

Muzzle velocity of the bullet (vbullet): 750 m/s

Therefore, the momentum of the bullet is:

pbullet = mbullet * vbullet = 0.006 kg * 750 m/s = 4.5 kg*m/s.

Part B: Rifle Recoil Velocity

By conservation of momentum, the total momentum before the bullet is fired is equal to the total momentum after. Since the rifle was at rest initially, its initial momentum is zero, and the total momentum after must also be zero. This means the momentum of the rifle (prifle) should be equal and opposite to that of the bullet:

Mass of the rifle (mrifle): 0.9 kg

Let the recoil velocity of the rifle be vrifle. The equation is:

0 = mrifle * vrifle + mbullet * vbullet
Solving for vrifle gives us:

vrifle = - (mbullet * vbullet)/mrifle = - (0.006 kg * 750 m/s) / 0.9 kg = -5 m/s.

The negative sign indicates that the rifle's velocity is in the opposite direction to the bullet's velocity, which is expected in the recoil motion.

Answer:

Because of height and lower atmospheric pressure.

Explanation:

Atmospheric pressure affects aerodynamic drag, lower pressure means less drag. At the altitude of Denver the air has lower pressure, this allows baseball players to hit balls further away.

Another aerodynamic effect is the Magnus effect. This effect causes spinning objects to curve their flightpath, which is what curveball pitchers do. A lower atmospheric pressure decreases the curving of the ball's trajectory.

Denver's high altitude results in lower air pressure which benefits homerun hitters as the baseball can travel further. However, this is disadvantageous for curveball pitchers as the lesser air pressure makes it harder to produce a good curve.

Baseball home run hitters and curveball pitchers react differently to playing in Denver. Denver is located at a high altitude, which means the air pressure is lower than in many other cities. A lower air pressure means there’s less air resistance. For hitters, less air resistance means that the baseball can travel further when hit, increasing the likelihood of hitting a home run.

However, for pitchers who throw curveballs, the low air pressure is not beneficial. This is because the curve of a curveball is produced by the difference in air pressure on either side of the ball. Notably, the spin that the pitcher puts on the ball makes the air pressure higher on one side of the ball and lower on the other. However, the reduced air density in Denver reduces the overall air pressure difference, making it harder to get good curves on their pitches. Thus, hitters like to play in Denver while pitchers prefer places with denser air.

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Answer:

b. the air

Explanation:

David Scott's experiment was performed on the moon, where there is gravity but there is no air. This experiment consisted of letting a hammer and a feather fall at the same time. The result was that the two objects touch the ground simultaneously.

Since these two objects obviously have a different mass, the experiment shows that in a vacuum, objects fall with the same acceleration regardless of their mass.

Answer:

d = 3.44 *10^{-7} m  

Explanation:

given data:

length of metal plates = 16.50 cm

capacitor charge = 18.5 nC

potential difference = 37.8 V

capacitance of parallel plate capacitor

[tex]C = \frac{A\epsilon _{0}}{d}[/tex]

area of the individual plate

A=[tex] a^2 = (16.5*10^{-2})^2 = 272.25 *10^{-4}[/tex] m2

capacitance

[tex]C = QV = 18.5 *10^{-9} *37.8 = 699.3 * 10^{-9} C[/tex]

separation between plates d  is given as[tex] = \frac{A\epsilon _{0}}{C }[/tex]

[tex]d =  \frac{272.25 *10^{-4} *8.85*10^{-12}}{699.3 *10^-9}[/tex]

d = 3.44 *10^{-7} m  

Answer:

[tex]v=0[/tex]

Explanation:

Knowing that the formula for average velocity is:

[tex]v=\frac{x_{2}-x_{1}}{t_{2}-t_{1}}[/tex]

Being said that, we know that the person's displacement is zero because it returns to its starting point

[tex]x_{2}=x_{1}[/tex]

That means [tex]x_{2}-x_{1}=0[/tex]

[tex]v=\frac{0}{t_{2}-t_{1}}=0[/tex]

Answer:

The answer is [tex] 1.956 \times 10^7\ m/s[/tex]

Explanation:

The amount of energy is not enough to apply the relativistic formula of energy [tex]E = mc^2[/tex], so the definition of energy in this case is

[tex]E = \frac{1}{2}m v^2[/tex].

From the last equation,

[tex]v= \sqrt{2E/m}[/tex]

where

[tex]E = 2 MeV = 3.204 \times 10^{-13} J[/tex]

and the mass of the neutron is

[tex]m = 1.675\times 10^{-27}\ Kg[/tex].

Then

[tex]v = 1.956 \times 10^7\ m/s[/tex]

the equivalent of [tex]0.065[/tex] the speed of light.

Answer:

(a-1) d₂=4.89 m: The object slides 4.89 m along the rough surface

(a-2) Work (Wf) done by the friction force while the mass is sliding down the in- clined plane:

Wf=  -20.4 J    is negative

(b) Work (Wg) done by the gravitational force while the mass is sliding down the inclined plane:

Wg= 58.8 J is positive

Explanation:

Nomenclature

vf: final velocity

v₀ :initial velocity

a: acceleleration

d: distance

Ff: Friction force

W: weight

m:mass

g: acceleration due to gravity

Graphic attached

The attached graph describes the variables related to the kinetics of the object (forces and accelerations)

Calculation de of the components of W in the inclined plane

W=m*g

Wx₁ = m*g*sin30°

Wy₁=  m*g*cos30°

Object kinematics on the inclined plane

vf₁²=v₀₁²+2*a₁*d₁

v₀₁=0

vf₁²=2*a₁*d₁

[tex]v_{f1} = \sqrt{2*a_{1}*d_{1}  }[/tex]  Equation (1)

Object kinetics on the inclined plane (μ= 0.2)

∑Fx₁=ma₁  :Newton's second law

-Ff₁+Wx₁ = ma₁   , Ff₁=μN₁

-μ₁N₁+Wx₁ = ma₁      Equation (2)

∑Fy₁=0   : Newton's first law

N₁-Wy₁= 0

N₁- m*g*cos30°=0

N₁  =  m*g*cos30°

We replace   N₁  =  m*g*cos30 and  Wx₁ = m*g*sin30° in the equation (2)

-μ₁m*g*cos30₁+m*g*sin30° = ma₁   :  We divide by m

-μ₁*g*cos30°+g*sin30° = a₁  

g*(-μ₁*cos30°+sin30°) = a₁  

a₁ =9.8(-0.2*cos30°+sin30°)=3.2 m/s²

We replace a₁ =3.2 m/s² and d₁= 3m in the equation (1)

[tex]v_{f1} = \sqrt{2*3.2*3}  }[/tex]

[tex]v_{f1} =\sqrt{2*3.2*3}[/tex]

[tex]v_{f1} = 4.38 m/s[/tex]

Rough surface  kinematics

vf₂²=v₀₂²+2*a₂*d₂   v₀₂=vf₁=4.38 m/s

0   =4.38²+2*a₂*d₂  Equation (3)

Rough surface  kinetics (μ= 0.3)

∑Fx₂=ma₂  :Newton's second law

-Ff₂=ma₂

--μ₂*N₂ = ma₂   Equation (4)

∑Fy₂= 0  :Newton's first law

N₂-W=0

N₂=W=m*g

We replace N₂=m*g inthe equation (4)

--μ₂*m*g = ma₂   We divide by m

--μ₂*g = a₂

a₂ =-0.2*9.8= -1.96m/s²

We replace a₂ = -1.96m/s² in the equation (3)

0   =4.38²+2*-1.96*d₂

3.92*d₂ = 4.38²

d₂=4.38²/3.92

d₂=4.38²/3.92

(a-1) d₂=4.89 m: The object slides 4.89 m along the rough surface

(a-2) Work (Wf) done by the friction force while the mass is sliding down the in- clined plane:

Wf = - Ff₁*d₁

Ff₁= μ₁N₁= μ₁*m*g*cos30°= -0.2*4*9.8*cos30° = 6,79 N

Wf= -  6.79*3 = 20.4 N*m

Wf=  -20.4 J    is negative

(b) Work (Wg) done by the gravitational force while the mass is sliding down the inclined plane

Wg=W₁x*d= m*g*sin30*3=4*9.8*0.5*3= 58.8 N*m

Wg= 58.8 J is positive

The work done by the friction force is negative, and the work done by the gravitational force is positive as the object slides down the inclined plane.

The work done by the friction force while the object slides down the inclined plane is negative. The work done by the gravitational force while the object slides down the inclined plane is positive.

When the object slides down the inclined plane, the friction force acts in the opposite direction to its motion. Since friction always opposes the motion, the work done by friction is negative.

The gravitational force, on the other hand, acts in the same direction as the object's motion. Therefore, the work done by the gravitational force is positive.

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Answer:

The fraction of time for turn on is 0.3852

Solution:

As per the question:

Temperature at which oil bath is maintained, [tex]T_{o} = 50.5^{\circ}[/tex]

Heat loss at rate, q = 4.68 kJ/min

Resistance, R = [tex]60\Omega[/tex]

Operating Voltage, [tex]V_{o} = 110 V[/tex]

Now,

Power that the resistor releases, [tex]P_{R} = \frac{V_{o}^{2}}{R}[/tex]

[tex]P_{R} = \frac{110^{2}}{60} = 201.67 W = 12.148 J/min[/tex]

The fraction of time for the current to be turned on:

[tex]P_{R} = \frac{q}{t}[/tex]

[tex]12.148 = \frac{4.68}{t}[/tex]

t = 0.3852

Answer:

c. 15 m

Explanation:

We apply Newton's second law in the x direction:

∑Fₓ = m*a

120*cos(60°) = 50*a

[tex]a = \frac{120*cos(60^o)}{50}  = 1.2 \frac{m}{s^2}[/tex]

Block kinematics

The block moves with uniformly accelerated movement, so we apply the following formula to calculate the distance

[tex]d = V_o*t + \frac{1}{2}*a*t^2[/tex]

[tex]d = 0 + \frac{1}{2}*1.2*5^2[/tex]

d = 15m

Answer:

The force exerted by three charges on the fourth is [tex]F_{resultant}=2.74\times10^{-5}\ \rm N[/tex]

Explanation:

Given:

According to coulombs Law the force F between any two charge particles is given by

[tex]F=\dfrac{kQq}{r^2}[/tex]

where r is the radial distance between them.

Since the force acting on the charge particle will be in different directions so according to triangle law of vector addition

[tex]F_{resultant}=\sqrt ((\dfrac{kQq}{L^2})^2+(\dfrac{kQq}{L^2} })^2)+\dfrac{kQq}{(\sqrt{2}L)^2}\\F_{resultant}=\dfrac{kQq}{L^2}(\sqrt{2}-\dfrac{1}{2})\\F_{resultant}=\dfrac{9\times10^9\times10\times10^{-10}\times3\times10^{-9}}{0.03^2}(\sqrt{2}-\dfrac{1}{2})\\F_{resultant}=2.74\times 10^{-5}\ \rm N[/tex]

The force on the fourth charge is calculated by first determining the individual forces exerted by each of the three other charges separately using Coulomb's Law and then adding these forces as vectors. This involves resolving each force into its x and y components, combining them separately, and then determining the resultant force's magnitude and direction.

The problem here involves Coulomb's Law and the superposition principle in physics. Coulomb's Law defines the force between two point charges as directly proportional to the product of their charges, and inversely proportional to the square of the distance between them.

First, you need to calculate the forces exerted on the fourth charge by each of the three other charges separately. This involves calculating the distance from each existing charge to the fourth charge, then subbing these distances, along with the relevant charge values, into the Coulomb's Law formula. Remember that if the charge is positive (like in the case of charge +q), the force vector points directly from the charge, while if the charge is negative, the force vector points towards the charge.

After calculating the force vectors resulting from each charge, you add these vectors together to get the resultant force vector which is the force exerted on the fourth charge. This problem also involves trigonometry as when you add the force vectors, you have to take into account the direction which each force vector is pointing.

Force due to the positive charge at the lower left: F1 is in the first quadrant
Force due to the positive charge at the lower right: F2 is in the fourth quadrant
Force due to the negative charge at the upper left: F3 is in the third quadrant

In each case, you'll need to resolve each force into its x and y components, and then add up all the x and y components separately to get the x and y components of the total force. Finally, calculate the magnitude of the total force using the Pythagorean theorem.

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Answer:

The vector has a magnitude of 33.86 units and a direction of 121.64°.

Explanation:

To find the magnitude, you use the Pitagorean theorem:

[tex]||V|| = \sqrt{x^2 + y^2}= \sqrt{(-26.5)^2 + (43)^2} = 33.86 units[/tex]

In order to find the direction, you can use trigonometry. You have to keep in mind, that as the y component of the vector is positive and the x component is negative, the vector must have an angle between 90 and 180°, or in the second quadrant of the plane.:

[tex]tan(\alpha) = \frac{y}{x} \\\alpha = tan^{-1}(\frac{43}{-26.5})= 180 - tan^{-1}(\frac{43}{26.5}) = 121.64[/tex]° or, 58.36° in the second quadrant

Answer:

Degenerative accelerator:

 The device which is used to study the brain and degenerative diseases like Alzheimers and Parkinson is called degenerative accelerator.

These accelerator have higher specific activity and it is comparable to reactor products.By using these accelerator many radio active nuclides can be produced those can not be produce by neutron reaction.

These generates synchrotron light that can be used for reveal the inorganic and organic structure.

Answer:42 units

Explanation:

Given

Price-production relationship=-0.7D+300

Total cost=Fixed cost+ variable cost

Total cost=10,287+25D

where D is the units produced

Total revenue[tex]=\left ( -0.7D+300\right )D[/tex]

Total revenue[tex]=-0.7D^2+300D[/tex]

For Break even point

Total revenue=Total cost

[tex]10,287+25D=-0.7D^2+300D[/tex]

[tex]7D^2-2750D+102870=0[/tex]

[tex]D=\frac{2750\pm \sqrt{2750^2-4\times 7\times 102870}}{2\times 7}[/tex]

[tex]D=41.869\approx 42[/tex] units

Answer:

It take 1.78033 second get away

Explanation:

We have given that a car takes 30 m to stop when its speed is 25 m/sec

As the car stops its final speed v = 0 m/sec

Initial speed u = 25 m/sec

Distance s = 30 m

From third law of motion [tex]v^2=u^2+2as[/tex]

So [tex]0^2=25^2+2\times a\times 30[/tex]

[tex]a=-10.4166m/sec^2[/tex]

Now in second case distance s = 28 m

So [tex]v^2=25^2+2\times -10.4166\times 28[/tex]

[tex]v^2=41.666[/tex]

v = 6.4549 m/sec

Now from first equation of motion v=u+at

So [tex]6.4549=25-10.4166\times t[/tex]

t = 1.78033 sec

Answer : The new or final volume of gas will be, 11.4 L

Explanation :

Charles' Law : It is defined as the volume of gas is directly proportional to the temperature of the gas at constant pressure and number of moles.

[tex]V\propto T[/tex]

or,

[tex]\frac{V_1}{V_2}=\frac{T_1}{T_2}[/tex]

where,

[tex]V_1[/tex] = initial volume of gas = 12.1 L

[tex]V_2[/tex] = final volume of gas = ?

[tex]T_1[/tex] = initial temperature of gas = [tex]9^oC=273+9=282K[/tex]

[tex]T_2[/tex] = final temperature of gas = [tex]-7^oC=273+(-7)=266K[/tex]

Now put all the given values in the above formula, we get the final volume of the gas.

[tex]\frac{12.1L}{V_2}=\frac{282K}{266K}[/tex]

[tex]V_2=11.4L[/tex]

Therefore, the new or final volume of gas will be, 11.4 L

Final answer:

Using Charles's Law, the new volume of the helium gas in the balloon when taken from an inside temperature of 9°C to an outside temperature of -7°C, at constant pressure, is calculated to be 11.4 L.

Explanation:

To calculate the new volume of the helium gas in the balloon when it is taken outside to a colder temperature, we can use Charles's Law, which states that for a given mass of gas at constant pressure, the volume is directly proportional to its temperature in kelvins (V/T = k). We need to convert the temperatures from Celsius to Kelvin (K = °C + 273.15) and then apply Charles's Law (V1/T1 = V2/T2).

First, convert the temperature from Celsius to Kelvin:

Inside temperature: T1 = 9 °C + 273.15 = 282.15 K  

Outside temperature: T2 = -7 °C + 273.15 = 266.15 K

Next, apply Charles's Law to find the new volume (V2):

V1/T1 = V2/T2

Plugging in the known values:

12.1 L / 282.15 K = V2 / 266.15 K

Solving for V2, we get:

V2 = (12.1 L × 266.15 K) / 282.15 K

V2 = 11.4 L (rounded to three significant digits)

Therefore, the new volume of the balloon when taken outside will be 11.4 L.

Answer:

(a) Maximum height = 53.88 meters

(b) Range of the ball = 924.36 meters

Explanation:

The ball has been launched at a speed = 44.4 meters per second

Angle of the ball with the horizontal = 25°

Horizontal component of the speed of the ball = 44.4cos25° = 40.24 meters per second

Vertical component = 44.4sin25° = 18.76 meters per second

We know vertical component of the speed decides the height of the ball so by the law of motion,

v² = u² - 2gh

where v = velocity at the maximum height = 0

u = initial velocity = 18.76 meter per second

g = gravitational force = 9.8 meter per second²

Now we plug in the values in the given equation

0 = (18.76)² - 2(9.8)(h)

19.6h = 352.10

h = [tex]\frac{352.10}{19.6}[/tex]

h = 17.96 meters

By another equation,

[tex]v=ut-\frac{1}{2}gt^{2}[/tex]

Now we plug in the values again

[tex]0=(18.76)t-\frac{1}{2}(9.8)t^{2}[/tex]

18.76t = 4.9t²

18.76 = 4.9t

t = [tex]\frac{18.76}{4.9}=3.83[/tex]seconds

Since time t is the time to cover half of the range.

Therefore, time taken by the ball to cover the complete range = 2×3.83 = 7.66 seconds

  So the range of the ball = Horizontal component of the velocity × time

                                           = 40.24 × 7.66

                                           = 308.12 meters

This we have calculated all for our planet.

Now we take other planet.

(a) Since the golfer drives the ball 3 times as far as he would have on earth then maximum height achieved by the ball = 17.96 × 3 = 53.88 meters

(b) Range of the ball = 3×308.12 = 924.36 meters

Answer:

total kinetic energy is 8 × [tex]10^{-19}[/tex] J

Explanation:

given data

potential difference = 5 V

e = 1.60 × [tex]10^{-19}[/tex] C

to find out

what is kinetic energy

solution

we will apply here conservation of energy that is

change in potential energy is equal to change in kinetic energy

so

change potential energy is e × potential difference

change potential energy =  1.60 × [tex]10^{-19}[/tex] × 5

change potential energy = 8 × [tex]10^{-19}[/tex] J

so change in kinetic energy  = 8 × [tex]10^{-19}[/tex] J

and we know proton start from rest that mean ( kinetic energy is 0 ) so

change in KE is total KE

total kinetic energy is 8 × [tex]10^{-19}[/tex] J

Answer:

Zero

Explanation:

The overall work done by gravitational force on completion of a complete turn is zero.

Since the work done by gravitaional force is conservative and depends only on the initial and end position here height and the path followed does not matter.

Since, in a complete turn the wheel return to its final position as a result of which displacement is zero and as work is the dot product of Force exerted and displacement, the work done is zero.

Also the work done in half cycle by gravity is counter balance by the work which is done against the gravity in the other half cycle.

The net work done by the gravitational force when a person is riding on a Ferris Wheel and it makes one complete turn is zero.

To understand why the net work done by the gravitational force is zero, we need to consider the definition of work and the nature of the motion on a Ferris Wheel. Work done by a force is defined as the product of the force and the displacement in the direction of the force. Mathematically, this is expressed as:

[tex]\[ W = F \cdot d \cdot \cos(\theta) \][/tex]

where ( W ) is the work, ( F ) is the force, ( d ) is the displacement, and [tex]\( \theta \)[/tex] is the angle between the force and the displacement.

In the case of the Ferris Wheel, the gravitational force acts vertically downward towards the center of the Earth, while the displacement of the person on the Ferris Wheel is along the circumference of the wheel, which is horizontal at any given point. Since the force and displacement are perpendicular to each other at every point in the circle, the angle \( \theta \) between them is always 90 degrees. Therefore, the cosine of 90 degrees is zero, which means that the work done by the gravitational force at each point is zero:

[tex]\[ W = F \cdot d \cdot \cos(90^\circ) = F \cdot d \cdot 0 = 0 \][/tex]

Moreover, over one complete turn of the Ferris Wheel, the initial and final positions of the person are the same. This means that the total displacement over one complete cycle is zero. Even if we consider the components of the gravitational force along the direction of displacement during different parts of the cycle, the net effect is zero because the person is raised to a certain height and then lowered back to the starting point. The work done against gravity to raise the person is equal in magnitude and opposite in sign to the work done by gravity as the person descends, resulting in a net work of zero for the entire cycle.

Therefore, the gravitational force does no net work on the person over one complete turn of the Ferris Wheel.

Answer:

q₁= +0.5nC

Explanation:

Theory of electrical forces

Because the particle q3 is close to three other electrically charged particles, it will experience two electrical forces and the solution of the problem is of a vector nature.

To solve this problem we apply Coulomb's law:

Two point charges (q1, q2) separated by a distance (d) exert a mutual force (F) whose magnitude is determined by the following formula:

o solve this problem we apply Coulomb's law:  

Two point charges (q₁, q₂) separated by a distance (d) exert a mutual force (F) whose magnitude is determined by the following formula:  

F=K*q₁*q₂/d² Formula (1)  

F: Electric force in Newtons (N)

K : Coulomb constant in N*m²/C²

q₁,q₂:Charges in Coulombs (C)  

d: distance between the charges in meters

Data:

Equivalences

1nC= 10⁻⁹ C

1cm= 10⁻² m

Data

q₃=+5.00 nC =+5* 10⁻⁹ C

q₂= -2.00 nC =-2* 10⁻⁹ C

d₂= 5.00 cm= 5*10⁻² m

d₁= 2.50 cm=  2.5*10⁻² m

k = 8.99*10⁹ N*m²/C²

Calculation of magnitude and sign of q1

Fn₃=0 : net force on q3 equals zero

F₂₃:The force F₂₃ that exerts q₂ on q₃ is attractive because the charges have opposite signs,in direction +x.

F₁₃:The force F₂₃ that exerts q₂ on q₃ must go in the -x direction so that Fn₃ is zero, therefore q₁ must be positive and F₂₃ is repulsive.

We propose the algebraic sum of the forces on q₃

F₂₃ - F₁₃=0

[tex]\frac{k*q_{2} *q_{3} }{d_{2}^{2}  } -\frac{k*q_{1} *q_{3} }{d_{1}^{2}  }=0[/tex]

We eliminate k*q₃ of the equation

[tex]\frac{q_{1} }{d_{1}^{2}  } = \frac{q_{2} }{d_{2}^{2}  }[/tex]

[tex]q_{1} =\frac{q_{2} *d_{1} ^{2} }{d_{2}^{2}  }[/tex]

[tex]q_{1} =\frac{2*10^{-9}*2.5^{2}*10^{-4}   }{5^{2}*10^{-4}  }[/tex]

q₁= +0.5*10⁻⁹ C

q₁= +0.5nC

Final answer:

The magnitude and sign of charge q1 can be found by setting the electric force by q1 on q3 equal and opposite to the force exerted by q2 on q3 using Coulomb's Law and solving for q1 given the known distances.

Explanation:

The question is asking to find the magnitude and sign of charge q1 such that the net electrostatic force on charge q3 is zero when placed in a line with charges q2 and q3. To solve this, we can apply Coulomb's law, which states that the electrostatic force between two point charges is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between them.

To have the net force on q3 be zero, the force exerted on q3 by q1 must be equal in magnitude and opposite in direction to the force exerted on q3 by q2. We can set up equations based on Coulomb's law and solve for q1. First, determine the force F31 between q3 and q1, and the force F32 between q3 and q2. Since these forces must be equal and opposite to cancel each other out, we set F31 = F32 and solve for q1. Since the distance between q3 and q1 is 2.50 cm and between q3 and q2 is 5.00 cm, and accounting for the sign of q2, we can calculate the magnitude and sign of q1 that makes the net force on q3 zero.

Answer:

0.02442 × 10⁻⁹

Explanation:

Given:

Diameter of copper ball = 2.00 mm = 0.002 m

Charge on ball = 40 nC = 40 × 10⁻⁹ C

Density of copper = 8900 Kg/m³

Now,

The number of electrons removed, n = [tex]\frac{\textup{Charge on ball}}{\textup{Charge of an electron}}[/tex]

also, charge on electron = 1.6 × 10⁻¹⁹ C

Thus,

n = [tex]\frac{40\times10^{-9}}{1.6\times10^{-19}}[/tex]

or

n = 25 × 10¹⁰ Electrons

Now,

Mass of copper ball = volume × density

Or

Mass of copper ball =  [tex]\frac{4}{3}\pi(\frac{d}{2})^3[/tex]  × 8900

or

Mass of copper ball =  [tex]\frac{4}{3}\pi(\frac{0.002}{2})^3[/tex]  × 8900

or

Mass of copper ball = 0.03726 grams

Also,

molar mass of copper = 63.546 g/mol

Therefore,

Number of mol of copper in  0.03726 grams = [tex]\frac{ 0.03726}{63.546}[/tex]

or

Number of mol of copper in  0.03726 grams = 5.86 × 10⁻⁴ mol

and,

1 mol of a substance contains = 6.022 × 10²³ atoms

Therefore,

5.86 × 10⁻⁴ mol of copper contains = 5.86 × 10⁻⁴ × 6.022 × 10²³ atoms.

or

5.86 × 10⁻⁴ mol of copper contains = 35.88 × 10¹⁹ atoms

Now,

A neutral copper atom has 29 electrons.  

Therefore,

Number of electrons in ball = 29 × 35.88 × 10¹⁹ = 1023.37 × 10¹⁹ electrons.

Hence,

The fraction of electrons removed = [tex]\frac{25\times10^{10}}{1023.37\times10^{19}}[/tex]

or

The fraction of electrons removed = 0.02442 × 10⁻⁹

The fraction of electrons removed from the copper ball can be calculated by comparing its net charge to the charge of a single electron. Using the provided information, we can find that approximately 2.5 x 10^10 electrons have been removed from the ball. To determine the fraction, we need to compare this number to the total number of electrons in the ball, which can be calculated using the mass, density, and atomic mass of copper. By plugging in the values, we can find the fraction of electrons removed.

To determine the fraction of electrons that have been removed from the copper ball, we need to compare the net charge of the ball to the charge of a single electron. The net charge of the ball is 40 nC, which is equivalent to 40 x 10^-9 C. The charge of a single electron is 1.60 x 10^-19 C.

We can calculate the number of electrons that have been removed using the formula:

Number of electrons removed = Net charge of the ball / Charge of a single electron

Number of electrons removed = (40 x 10^-9 C) / (1.60 x 10^-19 C) = 2.5 x 10^10 electrons

To find the fraction of electrons removed, we need to compare the number of electrons removed to the total number of electrons in the ball. The total number of electrons in the ball can be calculated using the formula:

Total number of electrons = Number of copper atoms x Number of electrons per copper atom

The number of copper atoms can be calculated using the formula:

Number of copper atoms = Mass of the ball / Atomic mass of copper

The mass of the ball can be calculated using the formula:

Mass of the ball = Volume of the ball x Density of copper

Given that the diameter of the ball is 2.0 mm, the volume of the ball can be calculated using the formula for the volume of a sphere:

Volume of the ball = (4/3) x pi x (radius)^3

As the ball is a sphere, the radius is half the diameter, so the radius is 1.0 mm or 1 x 10^-3 m.

Using the given density of copper (8900 kg/m^3) and atomic mass of copper (63.5 g/mol), we can now calculate the fraction of electrons removed:

Fraction of electrons removed = Number of electrons removed / Total number of electrons = (2.5 x 10^10 electrons) / (Number of copper atoms x Number of electrons per copper atom)

Answer:

[tex]-1.144\ \mu C[/tex]

Explanation:

Given:

Assume:

We know that the electric field is the electric force applied on a unit positive charge i.e.,

[tex]\vec{E}=\dfrac{\vec{F}}{Q}[/tex]

This means the electric force applied on this additional charge placed in the field is given by:

[tex]\vec{F}=Q\vec{E}\\\Rightarrow \vec{F} =  1.04\times 10^{-8}\ n C\times (148.0\ \hat{i}-110.0\ \hat{j})\ N/C\\\Rightarrow \vec{F} = (1.539\ \hat{i}-1.144\ \hat{j})\ \mu N\\[/tex]

From the above expression of force, we have the following y-component of force on this additional charge.

[tex]F_y = -1.144\ \mu N[/tex]

Hence, the y-component of the electric force on the this charge is [tex]-1.144\ \mu N[/tex].

Answer:

The resultant field will have a magnitude of 241.71 V/m, 30.28° to the left of E1.

Explanation:

To find the resultant electric fields, you simply need to add the vectors representing both electric field E1 and electric field E2. You can do this by using the component method, where you add the x-component and y-component of each vector:

E1 = 99 V/m, 0° from the y-axis

E1x = 0 V/m  

E1y = 99 V/m, up

E2 = 164 V/m, 48° from y-axis

E2x = 164*sin(48°) V/m, to the left

E2y = 164*cos(48°) V/m, up

[tex]Ex: E_{1_{x}} + E_{2_{x}} = 0 V/m - 164 *sin(48) V/m= -121.875 V/m\\Ey: E_{1_{y}} + E_{2_{y}} = 99 V/m + 164 *cos(48) V/m = 208.74 V/m\\[/tex]

To find the magnitude of the resultant vector, we use the pythagorean theorem. To find the direction, we use trigonometry.

[tex]E_r = \sqrt{E_x^2 + E_y^2}= \sqrt{(-121.875V/m)^2 + (208.74V/m)^2} = 241.71 V/m[/tex]

The direction from the y-axis will be:

[tex]\beta = arctan(\frac{-121.875 V/m}{208.74 V/m}) = 30.28[/tex]° to the left of E1.

the resultant electric field has a magnitude of approximately 247.8 V/m and is directed at an angle of approximately 63.5 degrees above the horizontal, not the vertical.

To find the resultant electric field, you should add the horizontal and vertical components of E1 and E2 separately:

Vertical Component:

E1y = 99 V/m (vertical component of E1)

E2y = 164 V/m * sin(48°) ≈ 123.6 V/m (vertical component of E2)

Horizontal Component:

E2x = 164 V/m * cos(48°) ≈ 109.8 V/m (horizontal component of E2)

Add the vertical components:

Ey = E1y + E2y

Ey = 99 V/m + 123.6 V/m

Ey ≈ 222.6 V/m

Add the horizontal components:

Ex = E2x

Ex ≈ 109.8 V/m

Calculate the magnitude of the resultant electric field (E) using the Pythagorean theorem:

E = √(Ex² + Ey²)

E = √((109.8 V/m)² + (222.6 V/m)²)

E ≈ 247.8 V/m

So, the resultant electric field has a magnitude of approximately 247.8 V/m and is directed at an angle of approximately 63.5 degrees above the horizontal, not the vertical, as previously stated.

Learn more about Electric field here:

https://brainly.com/question/8971780

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Answer:

a) [tex]y=16.53m[/tex]

b) [tex]t_{up}=1.83s[/tex]

c)[tex]t_{down}=1.84s[/tex]

d) [tex]v=-18m/s[/tex]

Explanation:

a) To find the highest point of the ball we need to know that at that point the ball stops going up and its velocity become 0

[tex]v^{2} =v^{2} _{o} +2g(y-y_{o})[/tex]

[tex]0=(18)^{2} -2(9.8)(y-0)[/tex]

Solving for y

[tex]y=\frac{(18)^{2} }{2(9.8)}=16.53m[/tex]

b) To find how long does it take to reach that point:

[tex]v=v_{o}+at[/tex]

[tex]0=18-9.8t[/tex]

Solving for t

[tex]t_{up} =\frac{18m/s}{9.8m/s^{2} }= 1.83s[/tex]

c) To find how long does it take to hit the ground after it reaches its highest point we need to find how long does it take to do the whole motion and then subtract the time that takes to go up

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

[tex]0=0+18t-\frac{1}{2}(9.8)t^{2}[/tex]

Solving for t

[tex]t=0 s[/tex] or [tex]t=3.67s[/tex]

Since time can not be negative, we choose the second option

[tex]t_{down}=t-t_{up}=3.67s-1.83s=1.84s[/tex]

d) To find the velocity when it returns to the level from which it started we need to use the following formula:

[tex]v=v_{o}+at[/tex]

[tex]v=18m/s-(9.8m/s^{2} )(3.67s)=-18m/s[/tex]

The sign means the ball is going down

Answer:

The relative strength of the gravitational and solar electromagnetic pressure forces is [tex]7.33\times10^{13}\ N[/tex]

Explanation:

Given that,

Intensity = 1150 W/m²

(a). We need to calculate the magnetic field

Using formula of intensity

[tex]I=\dfrac{E^2}{2\mu_{0}c}[/tex]

[tex]E=\sqrt{2\times I\times\mu_{0}c}[/tex]

Put the value into the formula

[tex]E=\sqrt{2\times1150\times4\pi\times10^{-7}\times3\times10^{8}}[/tex]

[tex]E=931.17\ N/C[/tex]

Using relation of magnetic field and electric field

[tex]B=\dfrac{E}{c}[/tex]

Put the value into the formula

[tex]B=\dfrac{931.17}{3\times10^{8}}[/tex]

[tex]B=0.0000031039\ T[/tex]

[tex]B=3.10\times10^{-6}\ T[/tex]

(2). The relative strength of the gravitational and solar electromagnetic pressure forces of the sun on the earth

We need to calculate the gravitational force

Using formula of gravitational

[tex]F_{g}=\dfrac{GmM}{r^2}[/tex]

Where, m = mass of sun

m = mass of earth

r = distance

Put the value into the formula

[tex]F_{g}=\dfrac{6.67\times10^{-11}\times1.98\times10^{30}\times5.97\times10^{24}}{(1.496\times10^{11})^2}[/tex]

[tex]F_{g}=3.52\times10^{22}\ N[/tex]

We need to calculate the radiation force

Using formula of radiation force

[tex]F_{R}=\dfrac{I}{c}\times\pi\timesR_{e}^2[/tex]

[tex]F_{R}=\dfrac{1150}{3\times10^{8}}\times\pi\times(6.371\times10^{6})^2[/tex]

[tex]F_{R}=4.8\times10^{8}\ N[/tex]

We need to calculate the pressure

[tex]\dfrac{F_{g}}{F_{R}}=\dfrac{3.52\times10^{22}}{4.8\times10^{8}}[/tex]

[tex]\dfrac{F_{g}}{F_{R}}=7.33\times10^{13}\ N[/tex]

Hence, The relative strength of the gravitational and solar electromagnetic pressure forces is [tex]7.33\times10^{13}\ N[/tex]

Answer:

The current through [tex]9 \Omega[/tex] is 0.297 A

Solution:

As per the question:

[tex]R_{5} = 5.0 \Omega[/tex]

[tex]R_{9} = 9.0 \Omega[/tex]

[tex]R_{4} = 5.0 \Omega[/tex]

V = 6.0 V

Now, from the given circuit:

[tex]R_{5}[/tex] and [tex]R_{9}[/tex] are in parallel

Thus

[tex]\frac{1}{R_{eq}} = \frac{1}{R_{5}} + \frac{1}{R_{9}}[/tex]

[tex]R_{eq} = \frac{R_{5}R_{9}}{R_{5} + R_{9}}[/tex]

[tex]R_{eq} = \frac{5.0\times 9.0}{5.0 + 9.0} = 3.2143 \Omega[/tex]

Now, the [tex]R_{eq}[/tex] is in series with [tex]R_{4}[/tex]:

[tex]R'_{eq} = R_{eq} + R_{4} = 3.2143 + 4.0 = 7.2413 \Omega[/tex]

Now, to calculate the current through [tex]R_{9}[/tex]:

[tex]V = I\times R'_{eq}[/tex]

[tex]I = {6}{7.2143} = 0.8317 A[/tex]

where

I = circuit current

Now,

Voltage across [tex]R_{eq}[/tex], V':

[tex]V' = I\times R_{eq}[/tex]

[tex]V' = 0.8317\times 3.2143 = 2.6734 V[/tex]

Now, current through [tex]R_{9}[/tex], I' :

[tex]I' = \frac{V'}{R_{9}}[/tex]

[tex]I' = \frac{2.6734}{9.0} = 0.297 A[/tex]

Final answer:

To find the current through the 9.0 ohm resistor in a series-parallel circuit, we can calculate the equivalent resistance of the parallel combination.

Explanation:

To find the current through the 9.0 ohm resistor, we need to first determine the equivalent resistance of the circuit. The two resistors of 5.0 and 9.0 ohms that are connected in parallel have an equivalent resistance given by the formula:

1/Req = 1/R1 + 1/R2

1/Req = 1/5.0 + 1/9.0

1/Req = (9.0 + 5.0)/(5.0 * 9.0)

1/Req = 14.0/45.0

Req = 45.0/14.0 ≈ 3.21 ohms

The equivalent resistance of the parallel combination is approximately 3.21 ohms.

Answer:

[tex]E_x = \frac{2kQ}{\pi R^2}[/tex]

[tex]E_y = \frac{2kQ}{\pi R^2}[/tex]

Explanation:

Electric field due to small part of the circle is given as

[tex]dE = \frac{kdq}{R^2}[/tex]

here we know that

[tex]dq = \frac{Q}{\frac{\pi}{2}R} Rd\theta[/tex]

[tex]dq = \frac{2Q d\theta}{\pi}[/tex]

Now we will have two components of electric field given as

[tex]E_x = \int dE cos\theta[/tex]

[tex]E_x = \int \frac{kdq}{R^2} cos\theta[/tex]

[tex]E_x = \int \frac{k (2Qd\theta) cos\theta}{\pi R^2}[/tex]

[tex]E_x = \frac{2kQ}{\pi R^2} \int_0^{90} cos\theta d\theta[/tex]

[tex]E_x = \frac{2kQ}{\pi R^2} (sin 90 - sin 0)[/tex]

[tex]E_x = \frac{2kQ}{\pi R^2}[/tex]

similarly in Y direction we have

[tex]E_y = \int dE sin\theta[/tex]

[tex]E_y = \int \frac{kdq}{R^2} sin\theta[/tex]

[tex]E_y = \int \frac{k (2Qd\theta) sin\theta}{\pi R^2}[/tex]

[tex]E_y = \frac{2kQ}{\pi R^2} \int_0^{90} sin\theta d\theta[/tex]

[tex]E_y = \frac{2kQ}{\pi R^2} (-cos 90 + cos 0)[/tex]

[tex]E_y = \frac{2kQ}{\pi R^2}[/tex]