A straight nonconducting plastic wire 9.5 cm long carries a charge density of 100 nc/m distributed uniformly along its length. It is lying on a horizontal tabletop. A. Find the magnitude and direction of the electric field this wire produces at a point 4.50 cm directly above its midpoint. B. If the wire is now bent into a circle lying flat on the table, find the magnitude and direction of the electric field it produces at a point 4.50 cm directly above its center.

Answer:

V=22.4m/s;T=2.29s

Explanation:

We will use two formulas in order to solve this problem. To determine the velocity at the bottom we can use potential and kinetic energy to solve for the velocity and use the uniformly accelerated displacement formula:

[tex]mgh=\frac{1}{2}mv^{2}\\\\X= V_{0}t-\frac{gt^{2}}{2}[/tex]

Solving for velocity using equation 1:

[tex]mgh=\frac{1}{2}mv^{2} \\\\gh=\frac{v^{2}}{2}\\\\\sqrt{2gh}=v\\\\v=\sqrt{2*9.8\frac{m}{s^2}*25.6m}=22.4\frac{m}{s}[/tex]

Solving for time in equation 2:

[tex]-25.6m = 0\frac{m}{s}t-\frac{9.8\frac{m}{s^{2}}t^{2}}{2}\\\\-51.2m=-9.8\frac{m}{s^{2}}t^{2}\\\\t=\sqrt{\frac{51.2m}{9.8\frac{m}{s^{2}}}}=2.29s[/tex]

Answer:

Acceleration will be [tex]a=3.185m/sec^2[/tex]

Explanation:

We have given final velocity v = 21.5 m/sec

Time t = 6.75 sec

As cheetah starts from rest so initial velocity u = 0 m/sec

From first equation of motion we know that v = u+at, here v is final velocity, u is initial velocity, a is acceleration and t is time

So [tex]21.5=0+a\times 6.75[/tex]

[tex]a=3.185m/sec^2[/tex]

Answer:

[tex]a=3.185\frac{m}{s^2}[/tex]

Explanation:

Acceleration is the change in velocity for a given period of time, we can express this in the next formula:

[tex]a = \frac{\Delta v}{\Delta t} =\frac{v_{1}-v_{0}}{t_{1}-t_{0}}[/tex]

In this case the values are:

[tex]v_{0}=0\\v_{1}= 21.5 m/s\\t_{0}=0\\t_{1}= 6.75 s\\[/tex]

Inserting known values, the acceleration is:

[tex]a= \frac{21.5 m/s}{6.75 s} \\a=3.185\frac{m}{s^2}[/tex]

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]

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:

Time, t = 4200.23 seconds

Explanation:

Given that,

Speed of the airplane, v = 1000 km/h = 277.77 m/s

Distance covered, d = 1166700 m

Let t is the time taken by the airplane. The formula to find t is given by :

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

[tex]t=\dfrac{1166700\ m}{277.77\ m/s}[/tex]

t = 4200.23 seconds

So, the airplane will take 4200.23 seconds to covered 1166700 meters. Hence, this is the required solution.

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.

https://brainly.com/question/14243397

#SPJ12

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 maximum height is 162.67 m.

Explanation:

Suppose the total height is h.

And at the height of 2/3h the speed of an object is,

[tex]u=32.6m/s[/tex]

And the remaining height will be,

[tex]h'=h-\frac{2}{3}h\\ h'=\frac{1}{3}h[/tex]

So, according to question the initial speed is,

[tex]u=32.6m/s[/tex]

Acceleration in the upward direction is negative,

[tex]a=-9.8m/s^{2}[/tex]

And the final speed will be v m/s which is 0 m/s.

Now according to third equation of motion.

[tex]v^{2} =u^{2} -2as[/tex]

Here, v is the final velocity, u is the initial velocity, a is the acceleration, s is the displacement.

[tex]0^{2} =32.6^{2} +2(-9.8)\dfrac{h}{3} \\h=\dfrac{3\times 1062.76}{2\times 9.8}\\h=162.67 m[/tex]

Therefore, the maximum height is 162.67 m.

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:

after 38.8 years it will double

correct option is D 38.8 years

Explanation:

given data

population grows rate = 1.8%

to find out

how many years will it take to double

solution

we consider here initial population is x

so after 1 year population will be = (100% + 1.8% ) x = 1.018 x

and after n year population will be = [tex]1.018^{n} x[/tex]

so it will double

2x = [tex]1.018^{n} x[/tex]

take log both side

log 2 = n log (1.018)

n = [tex]\frac{log2}{log1.018}[/tex]

n = 38.853

so after 38.8 years it will double

correct option is D 38.8 years

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:

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:

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.

The electric potential midway between two point charges is determined by calculating the potential due to each charge separately and adding them together. Coulomb's constant and the distances to the midpoint are used in this calculation.

The student is asking for the electric potential midway between two point charges. The charges mentioned are +3.40 pC and -6.10 uC, with a separation of 1.20 m. To calculate the potential at the midway point, the contributions of both charges to the potential have to be added algebraically since electric potential is a scalar quantity.

The electric potential due to a single point charge at a distance r is given by the formula V = k * q / r, where V is the electric potential, k is Coulomb's constant (
approximately 8.99 x 109 N*m2/C²), q is the charge, and r is the distance from the charge to the point of interest. Because the point is midway, r will be 0.60 m for both charges.

Calculating the potential for each charge separately, we add the potentials resulting from each charge to find the total electric potential at the midpoint.

Answer:

Distance from start point is 72.5km

Explanation:

The attached Figure shows the plane trajectories from start point (0,0) to (x1,y1) (d1=40km), then going from (x1,y1) to (x2,y2) (d2=56km), then from (x2,y2) to (x3,y3) (d3=100). Taking into account the angles and triangles formed (shown in the Figure), it can be said:

[tex]x1=d1*cos(60), y1=d1*sin(60)\\\\ x2=x1 , y2=y1-d2\\\\ x3=x2-d3*cos(30) , y3=y2+d3*sin(30)[/tex]

Using the Pitagoras theorem, the distance from (x3,y3) to the start point can be calculated as:

[tex]d=\sqrt{x3^{2} +y3^{2} }[/tex]

Replacing the given values in the equations, the distance is calculated.

Final Answer:

The airplane ends up approximately 72.53 km from its starting point.

Explanation:

To determine how far the airplane ends up from its starting point after these displacements, we can use vector addition to find the resultant displacement. Since the movements are given in terms of directions relative to north, we can use a coordinate system where north corresponds to the positive y-axis, and east corresponds to the positive x-axis.

Let's start with the first displacement:
1. The airplane flies 40 km in a direction 30° east of north.
We can resolve this displacement into x and y components:
- The x-component (eastward) is 40 km * sin(30°) because the angle is measured from the north (y-axis).
- The y-component (northward) is 40 km * cos(30°) because the angle is with respect to the vertical (north direction).

Using the fact that sin(30°) = 1/2 and cos(30°) = √3/2:
- x1 = 40 km * 1/2 = 20 km
- y1 = 40 km * √3/2 ≈ 40 km * 0.866 = 34.64 km

Now for the second displacement:
2. The airplane flies 56 km due south.
This movement is along the negative y-axis.
- x2 = 0 km (no movement east or west)
- y2 = -56 km (southward)

For the third displacement:
3. The airplane flies 100 km 30° north of west.
- The x-component (westward) will be -100 km * cos(30°) because we are measuring the angle from the north and going west is negative in our coordinate system.
- The y-component (northward) will be 100 km * sin(30°).

Using the trigonometric values found earlier:
- x3 = -100 km * √3/2 ≈ -100 km * 0.866 = -86.6 km
- y3 = 100 km * 1/2 = 50 km

Having found the components for each displacement, we can now sum them up to find the total displacement.

Total x-component (x_total) = x1 + x2 + x3 = 20 km + 0 km - 86.6 km = -66.6 km (westward)
Total y-component (y_total) = y1 + y2 + y3 = 34.64 km - 56 km + 50 km = 28.64 km (northward)

Now, we can determine the magnitude of the resultant displacement vector using the Pythagorean theorem:

R = √(x_total^2 + y_total^2)
R = √((-66.6 km)^2 + (28.64 km)^2)
R = √(4440.96 km^2 + 820.5696 km^2)
R = √(5261.5296 km^2)
R ≈ 72.53 km

So, the airplane ends up approximately 72.53 km from its starting point.

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]

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.

https://brainly.com/question/29545516

#SPJ12

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:

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:

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

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

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.

To calculate the common height from which the rocks were released, use the equations of motion. Substitute the given values and solve for the height using the equations h = (1/2)gt^2 and h = v0t + (1/2)gt^2.

To calculate the common height from which the rocks were released, we need to use the equations of motion. Let's assume the common height is h. For Wes, the time taken to reach the ground is given as 1.20 s. Using the equation h = (1/2)gt^2, where g is the acceleration due to gravity, we can substitute the values and solve for h. For Lindsay, the time taken to reach the ground is the same, 1.20 s. Using the equation h = v0t + (1/2)gt^2, where v0 is the initial velocity, we can substitute the values and solve for h. By calculating the common height from these two equations, we can determine the height from which the rocks were released.

Answer:

u₀ = 17.14 m/s

Explanation:

given,

bridge height = 44 m

initial speed of the first stone = 0 m/s

initial speed of the second stone = ?

difference after which the second stone is thrown = 1.72 s

for stone 1

[tex]h = ut + \dfrac{1}{2}gt^2[/tex]

[tex]h =\dfrac{1}{2}gt_1^2[/tex]

for stone 2

[tex]h = u_0 (t_1-t) + \dfrac{1}{2}g (t_1-t) ^2[/tex]

[tex]t_1 =\sqrt{\dfrac{1}{2}gh}[/tex]

[tex]t_1 = \sqrt{\dfrac{1}{2}\times 9.81\times 44}[/tex]

t₁ = 14.69 s

[tex]44 = u_0 \times 1.72 + \dfrac{1}{2}g\times 1.72 ^2[/tex]

u₀ = 17.14 m/s

Answer:

A. The specific learning behaviors you are expecting from the lesson

Explanation:

Lesson plan is systematic way of approaching subject learning in schools and colleges. A Lesson plan has various section among them there is section of Objectives.

Objectives are defined precise and focused goals of the learning from the particular topic that the student must learn. It is a goal oriented method where  aim is already known before its accomplishment at the end of the chapter. Hence option A seem most appropriate answer.

To produce 5000 kg/h of tapioca flour with 5% moisture from cassava granules with 66% moisture, 13970.59 kg of granules must be dried, resulting in 8966.59 kg of water being removed.

The question pertains to the process of drying cassava root to produce tapioca flour, which involves reducing the moisture content from 66 wt % to 5%. To find the weight of cassava granules needed to produce 5000 kg/h of flour, we utilise mass balance concepts.

Let x be the amount (kg) of granules required. These granules initially contain 66% moisture, so there are 0.34x kg of dry solids in them. After drying to 5% moisture, the 5000 kg of flour contains 95% dry solids, or 0.95 x 5000 kg.

Assuming no loss of solid material during drying:

0.34x = 0.95 x 5000

x = (0.95 x 5000) / 0.34

x ≈ 13970.59 kg

The initial weight of water in the granules is the total weight of granules minus the weight of dry solids:

Initial water weight = x - 0.34x

Initial water weight = 0.66x

Initial water weight = 0.66 x 13970.59 kg

Initial water weight ≈ 9216.59 kg

The final weight of water in the 5000 kg of flour at 5% moisture is:

Final water weight = 0.05 x 5000 kg

Final water weight = 250 kg

The amount of water removed during the drying process is the initial water weight minus the final water weight.

Water removed = 9216.59 kg - 250 kg

Water removed ≈ 8966.59 kg

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: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:

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:

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.

https://brainly.com/question/506926

#SPJ3

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.