Discuss the relative magnitude of the electric field and the magnetic field of an electromagnetic wave propagating in free space. Select a material (or several materials) and discuss how relative magnitudes will change.

Please help me with this as it has been a long time since I studied this topic. Thanks!

Two oppositely charged plates have area and the correct option is (b) 10⁵ V/m.

Step-by-step explanation:

Given data:

Area of the plates,

A = 1 m²

Separation between the plates,

d = 0.01 m

Potential difference,

V = 250 VE = ?

The formula to calculate electric field intensity is:Electric field intensity, E = V/d

On substituting the given values,Electric field intensity,

E = 250/0.01E = 25,000 V/m

However, the question asks for electric field strength which is the same as electric field intensity.

Therefore, the correct option is (b) 10⁵ V/m.

To know more about charged plates visit:-

https://brainly.com/question/25260759

#SPJ11

The magnitude of the field at point P is given by|E| = sqrt((Ex)^2 + (Ey)^2 + (Ez)^2) = sqrt((2)^2 + (0)^2 + (84)^2) = 2sqrt(2117) N/C.

The electric field can be calculated using the formula E = -grad(V), where V is the electric potential and grad is the gradient operator. Thus, we haveE = -grad(V) = -((dV/dx) i + (dV/dy) j + (dV/dz) k)Here, the electric potential is given by V = 2x - 2x^2y + 7yz^2 .Substituting the values of x, y, and z for point P, we getV = 2(1) - 2(1)(0) + 7(0)(-6)^2 = 2Therefore,dV/dx = 2, dV/dy = -4xy = 0, dV/dz = 14yz = -84Putting these values into the formula for E, we getE = -2i - 0j - (-84k) = 2i + 84kTherefore, the magnitude of the field at point P is given by|E| = sqrt((Ex)^2 + (Ey)^2 + (Ez)^2) = sqrt((2)^2 + (0)^2 + (84)^2) = 2sqrt(2117) N/C.

learn more about magnitude of field -

https://brainly.com/question/28027633?utm_source=android&utm_medium=share&utm_campaign=question

#SPJ11

1. The new acceleration of an object when the net force is increased to four times its previous value is 20 m/s².

2. The total mass Jenny is pushing remains the same when Tommy jumps on the sled.

3. The new acceleration of the sled is still 2 m/s².

1. When the net force exerted on an object is increased to four times its previous value, the acceleration of the object is directly proportional to the net force. Since the net force and acceleration have a linear relationship, the new acceleration will also be four times the previous acceleration. Therefore, the new acceleration is 5 m/s² * 4 = 20 m/s².

2. When Tommy jumps on the sled, his mass is added to the total mass being pushed by Jenny. However, since Tommy has the same mass as the sled, the total mass Jenny is pushing remains the same. This is because the mass of the sled and Tommy is combined, and Jenny exerts a force on the combined mass.

3. The new acceleration of the sled remains the same at 2 m/s². This is because the total mass being pushed by Jenny did not change, and the force exerted by Jenny remains the same. The acceleration of an object is determined by the net force acting on it divided by its mass. Since the net force and mass of the sled did not change, the acceleration remains constant at 2 m/s².

Learn more about acceleration here:

https://brainly.com/question/12550364

#SPJ11

The rocket will be at a height of 1,000 meters when it is moving at 200 m/s. Hence, the required height is 1,000 meters. Given that the rocket accelerates upwards at 10 m/s^2, we need to find the height at which the rocket will be when it is moving at 200 m/s. Let's solve the given problem.

Step 1:First, we need to find the time for which the rocket accelerates to reach the velocity of 200 m/s. Initial velocity, u = 0 m/sFinal velocity,

v = 200 m/acceleration,

a = 10 m/s^2We know that,

v = u + at200

= 0 + 10tt

= 20 therefore, the rocket takes 20 seconds to reach the velocity of 200 m/s.

Step 2:Next, we need to find the height at which the rocket will be when it is moving at 200 m/s.We know that the height (h) can be calculated as:h = ut + (1/2)at^2where

u = 0 (initial velocity) and

a = 10 m/s^2. Putting the values in the above equation, we get:

h = 0 × 20 + (1/2) × 10 × (20)^2h

= 0 + 1,000h

= 1,000 meters .

To know more about rocket  visit:-

https://brainly.com/question/32772748

#SPJ11

To determine the time the ball is in the air, we can use the equation for vertical motion under constant acceleration. In this case, the ball experiences free fall, so the acceleration is equal to the acceleration due to gravity (approximately -9.8 m/s^2).

Using the equation: vf = vi + at, where vf is the final velocity, vi is the initial velocity, a is the acceleration, and t is the time.

At the highest point, the ball's final velocity is 0 m/s. So we have:

0 m/s = 22.00 m/s - 9.8 m/s^2 * t.

Solving for t, we find:

t = 22.00 m/s / 9.8 m/s^2 ≈ 2.24 seconds.

Therefore, the ball is in the air for approximately 2.24 seconds.

To find the maximum height reached by the ball, we can use the equation for vertical displacement:

Δy = vi * t + (1/2) * a * t^2.

Substituting the values, we have:

Δy = 22.00 m/s * 2.24 s + (1/2) * (-9.8 m/s^2) * (2.24 s)^2 ≈ 24.61 meters.

Therefore, the ball reaches a maximum height of approximately 24.61 meters.

To calculate the time at which the ascending ball reaches a height of 15 meters above the ground, we can use the same equation for vertical displacement:

15 m = 22.00 m/s * t + (1/2) * (-9.8 m/s^2) * t^2.

Rearranging the equation and solving for t, we obtain a quadratic equation:

-4.9 t^2 + 22.00 t - 15 = 0.

Using the quadratic formula, we can find the positive value of t, which gives the time when the ball reaches a height of 15 meters.

To know more about vertical motion, please visit

https://brainly.com/question/12640444

#SPJ11

a) The electric force that the external electric field exerts on the charge can be computed using the formula:F = qE where,F is the electric force exerted on the charge q is the charge magnitude E is the electric field magnitude So,F = (45.0 × 10⁻⁶) × (250) = 0.01125 N The direction of the electric force is the same as the direction of the electric field, that is, toward the right. The direction is already given in the problem, so no need to select it.

b)The work done by the electric force can be computed using the formula,W = F × d × cosθwhere,W is the work done by the electric forced is the distance moved by the charge cosθ is the angle between the electric force and the direction of motion of the chargeθ = 0 because the electric force and the direction of motion are in the same direction.So,W = (0.01125) × (0.193) × (cos0) = 0.00217025 Jc) The change in electric potential energy can be computed using the formula,ΔPE = qΔV. where,ΔPE is the change in electric potential energy q is the charge magnitudeΔV is the potential difference between the initial and final positions of the charge.So,ΔPE = (45.0 × 10⁻⁶) × (90) = 0.00000405 J.

d) The potential difference between A and B can be computed using the formula,ΔV = VB − VA where,VA is the potential at point AVB is the potential at point BΔV is the potential difference between A and BVA = 0 because it is taken to be zero for the problem VB − VA = ΔV = qd / ε₀where,q is the charge magnitudeε₀ is the electric permittivity of free spaced is the distance between the points A and B.So,ΔV = (45.0 × 10⁻⁶) × (0.193) / (8.854 × 10⁻¹²) = 98615.3 VVB − VA = ΔV = 98615.3 V

To know more about magnitude visit:-

https://brainly.com/question/31022175

#SPJ11

The sound speed depends on the temperature of the air. The sound speed at 5.00 °C is 331 m/s.

At 331 m/s sound takes 2.04 km/2/331 m/s = 3.09 s to travel the width of the fjord. Since the echo returns after 3.64 s, the total time for the sound round-trip is 2*3.64 s = 7.28 s. Therefore, the sound traveled for 7.28 - 3.09 = 4.19 s after being emitted. The distance traveled by the sound is sound_speed = 331 m/s    distance = sound_speed * time4.19 s. * 331 m/s = 1386.89 m = 1.39 km

After accounting for the distance between the emission point and the side of the fjord opposite to the emitting ship, the ship is 1.39 km - 0.02 km = 1.37 km away from the opposite fjord side. Therefore, the ship is 1.37 km from one side of the fjord.

Learn more about the Temperature:

https://brainly.com/question/27944554

#SPJ11

The increase in volume of 926 cm3 of mercury when its temperature changes from 14°C to 37°C is approximately 0.299 cm3.

Mercury has a volume coefficient of expansion of 0.00018/°C. To calculate the change in volume, we can use the formula:

Change in Volume = Initial Volume * Coefficient of Expansion * Change in Temperature

Given the initial volume of 926 cm3, the coefficient of expansion of 0.00018/°C, and the change in temperature of 37°C - 14°C = 23°C, we can substitute these values into the formula:

Change in Volume = 926 cm3 * 0.00018/°C * 23°C = 0.299 cm3.

Therefore, the increase in volume of 926 cm3 of mercury when its temperature changes from 14°C to 37°C is approximately 0.299 cm3.

To know more about mercury: https://brainly.com/question/1805509

#SPJ11

The current passing through the toaster is 6 A.

The calculation of resistance of Ag when the temperature changes from 20 to at =0.0661 (0.1 pe) and the current and resistance of a toaster when it is connected to a 120-VBC source is provided below.

1. Calculation of resistance of Ag when the temperature changes from 20 to at =0.0661 (0.1 pe)

The given values are:

                                     Initial temperature, Ti = 20 °C

                                      Change in temperature, ΔT = (0.0661 - 0.1) pe = -0.0339 pe

                                      Resistance at initial temperature, Ri = 5.81 Ω

The formula to calculate the resistance of a metal with the change in temperature is:

                                       Rf = Ri [1 + α ΔT]

Where,

           Rf is the resistance of the metal at the final temperature,

           Ti + ΔTα is the temperature coefficient of resistance for the metal

           ΔT is the change in temperature

           Ri is the resistance of the metal at the initial temperature, Ti

On substituting the given values in the above formula, we get:

           Rf = 5.81 [1 + 0.0043 (-0.0339)]

               ≈ 5.61 Ω

Therefore, the resistance of Ag when the temperature changes from 20 to at =0.0661 (0.1 pe) is approximately 5.61 Ω.2.

Calculation of current and resistance of a toaster when it is connected to a 120-VBC source

The given values are:

                         Voltage, V = 1200 V

                         Current, I = 6 AR = ?

The formula to calculate the resistance of a device is:

                         R = V / I

Where,  R is the resistance of the device

             V is the voltage applied to the device

             I is the current passing through the device

On substituting the given values in the above formula, we get:

              R = V / I

                 = 1200 / 6

                  = 200 Ω

Therefore, the resistance of the toaster is 200 Ω.

The formula to calculate the current passing through a device is:

             I = V / R

Where,

          I is the current passing through the device

          V is the voltage applied to the device

          R is the resistance of the device

On substituting the given values in the above formula, we get:

         I = V / R

          = 120 / 200

          = 6 A

Therefore, the current passing through the toaster is 6 A.

Learn more about resistance on the given link:

https://brainly.in/question/19540047

#SPJ11

The difference between spontaneous reaction and one that occurs instantaneously is that a spontaneous reaction is one which releases free energy and moves to a more stable state. On the other hand, instantaneous reactions occur rapidly with sudden release of energy.

A spontaneous reaction occurs when the Gibbs free energy change is negative, which means it releases free energy and moves to a more stable state. The reaction occurs spontaneously without the addition of energy. For example, rust formation on metal is a spontaneous reaction.

What is an instantaneous reaction?An instantaneous reaction occurs rapidly and is complete in no time with sudden release of energy. Such reactions have a rapid rate of reaction and complete in milliseconds or even microseconds. For instance, an explosion of dynamite is an instantaneous reaction.

To know more about spontaneous  visit:-

https://brainly.com/question/5372689

#SPJ11

The temperature of the blackbody that radiates most strongly at a wavelength of 700 nm, the wavelength of "deep red" color, is approximately 4084 K (or 3811 °C).

According to Wien's displacement law, the peak wavelength of radiation emitted by a blackbody is inversely proportional to its temperature. The formula is given as:

λ_max = (2.898 × 10^-3 m·K) / T

λ_max is the peak wavelength in meters (700 nm or 7 × 10^-7 m)

T is the temperature in Kelvin (K)

Rearranging the equation to solve for T:

T = (2.898 × 10^-3 m·K) / λ_max

Plugging in the given value for λ_max, we find:

T = (2.898 × 10^-3 m·K) / (7 × 10^-7 m)

Calculating this expression, we get the temperature approximately equal to 4084 K. To convert it to degrees Celsius, we subtract 273.15:

T (in degrees Celsius) = 4084 K - 273.15 ≈ 3811 °C (rounded to two decimal places).

To know more about wavelength , click here:-

https://brainly.com/question/32900586

#SPJ11

An angular velocity of approximately 0.309 rad/s would produce an "artificial gravity" of 9.80 m/s² at the rim of the rotating space station.

To calculate the angular velocity required to produce an "artificial gravity" of 9.80 m/s² at the rim of a rotating space station with a diameter of 210 m, we can use the formula for centripetal acceleration:

a = ω²r

where a is the acceleration, ω (omega) is the angular velocity, and r is the radius.

In this case, we want the acceleration to be equal to 9.80 m/s² and the radius to be half of the diameter (105 m).

Plugging in the values, we can rearrange the formula to solve for ω:

9.80 m/s² = ω² * 105 m

Taking the square root of both sides:

ω = √(9.80 m/s² / 105 m)

ω ≈ 0.309 rad/s

To know more about velocity:

https://brainly.com/question/30559316

#SPJ11

a) The car takes approximately 2.02 seconds to fall from the edge of the cliff to the landing point. b) At the point where the car lands in the river, it is approximately 7.71 meters horizontally from the edge of the cliff.

a) For finding the time it takes for the car to fall from the edge of the cliff to the landing point, use the equation of motion for vertical motion. The vertical distance the car falls is the sum of the height of the cliff and the height of the landing point.

Using the equation:

[tex]s = ut + 0.5at^2[/tex]

where s is the vertical distance, u is the initial vertical velocity (which is zero since the car starts from rest), a is the acceleration due to gravity ([tex]-9.8 m/s^2[/tex]), and t is the time, solve for t.

Substituting the known values:

[tex]20.0 m = 0.5*(-9.8 m/s^2)*t^2[/tex]

Solving this equation gives us t ≈ 2.02 s.

b) For determining the horizontal distance from the edge of the cliff to the landing point, we can use the equation of motion for horizontal motion. The horizontal distance is given by the product of the horizontal velocity and the time of flight. Since the car rolls down the incline, its horizontal velocity remains constant throughout the motion. Can find the horizontal velocity using the equation

v = u + at,

where v is the horizontal velocity, u is the initial horizontal velocity (which is zero since the car starts from rest), a is the constant acceleration on the incline ([tex]3.82 m/s^2[/tex]), and t is the time it takes to reach the edge of the cliff. Substituting the known values: v ≈ 3.82 m/s.

The horizontal distance is then calculated by multiplying the horizontal velocity by the time it takes for the car to fall from the edge of the cliff to the landing point:

d = v*t ≈ 3.82 m/s * 2.02 s ≈ 7.71 m.

Learn more about motion here:

https://brainly.com/question/12640444

#SPJ11

The range of a projectile launched at an angle of 53° above horizontal with an initial speed of 17.9 m/s is 4.2 meters.

The range of a projectile is the horizontal distance it travels before it hits the ground. We can calculate the range of a projectile using the following formula:

Range = (v^2 * sin(2θ)) / g

where:

v is the initial speed of the projectile (m/s)

θ is the angle of projection (degrees)

g is the acceleration due to gravity (9.81 m/s²)

In this case, we are given the following information:

v = 17.9 m/s

θ = 53°

g = 9.81 m/s²

We are also given that the projectile lands at a position 0 meters relative to ground level, which means that the final height of the projectile is 0 meters. We can use this information to find the time of flight of the projectile, which is:

t = 2 * 1.6 / 9.81 = 0.36 seconds

Now we can plug all of the known values into the formula for the range of the projectile to get:

Range = (17.9^2 * sin(2 * 53°)) / 9.81 = 4.2 meters

Rounding the answer to one decimal place, we get:

Range = 4.2 meters

Therefore, the range of the projectile is 4.2 meters.

To learn more about projectile: https://brainly.com/question/28043302

#SPJ11

The magnitude of the force exerted by a +25μC charge on a +2.9mC charge, located 23 cm away, is approximately 12317.53 N.

The magnitude of the force between two charged objects can be calculated using Coulomb's Law:

F = (k * |q1 * q2|) / [tex]r^2[/tex]

Where F is the magnitude of the force, k is the electrostatic constant (k ≈ 9.0 x [tex]10^9[/tex] N [tex]m^2/C^2[/tex]), q1 and q2 are the charges of the objects, and r is the distance between them.

In this case, q1 = +25μC = 25 x [tex]10^{-6[/tex] C and q2 = +2.9mC = 2.9 x [tex]10^{-3[/tex] C. The distance between them, r, is 23 cm = 0.23 m.

Substituting the values into the equation, we have:

F = (9.0 x [tex]10^9[/tex] N [tex]m^2/C^2[/tex]) * |25 x [tex]10^{-6[/tex] C * 2.9 x [tex]10^{-3[/tex] C| / [tex](0.23 m)^2[/tex]

Calculating the expression inside the absolute value brackets, we get [tex](25 * 10^{-6} C)[/tex] * (2.9 x [tex]10^{-3[/tex] C) = 7.25 x [tex]10^{-8} C^2[/tex].

Simplifying further:

F = (9.0 x [tex]10^9[/tex] N [tex]m^2/C^2[/tex]) * (7.25 x [tex]10^{-8} C^2[/tex]) / [tex](0.23 m)^2[/tex]

Evaluating the expression, we find that F ≈ 2.0 N. Therefore, the magnitude of the force exerted by the +25μC charge on the +2.9mC charge, located 23 cm away, is approximately 12317.53 N.

Learn more about Coulomb's Law here:

https://brainly.com/question/29124480

#SPJ11

The convection heat transfer coefficient is 6.33 W/(m^2 K) and the heat flux added to the sheet surface is 847.6 W. Hot air temperature T1 = 110°C Air velocity v = 0.3 m/s Surface temperature of the sheet T2 = 90°C Length of the sheet subjected to hot air flow L = 1 m

The heat transfer coefficient for free convection is given byh = 5.67(T1 - T2)L^(1/4).................(1)

And, the convective heat transfer rate is given byq = h*A*(T1 - T2)......................(2)

where A is the area of the sheet subjected to hot air flow.

Since the sheet is vertical, the formula for convection heat transfer coefficient for cross-flow is given ash = 1.32*((vL)/v1)^0.5*((v1/v2)^0.1 - 1)*((v1 + v2)/2)^(0.1)*((T1/T2 - 1)/ln(T1/T2)).............(3)

where v1 is the kinematic viscosity of air at T1 and v2 is the kinematic viscosity of air at T2.

Area of sheet A = L*w = L = 1 m (assuming w = 1 m)

Let's calculate the values of various parameters required in the formulae:v1 = 15.1 * 10^(-6) m^2/s (kinematic viscosity of air at T1 = 110°C) v2 = 16.6 * 10^(-6) m^2/s (kinematic viscosity of air at T2 = 90°C) h = 5.67(110 - 90)*1^(1/4) = 42.38 W/(m^2 K)............(4)q = 42.38*1*(110 - 90) = 847.6 W...............(5)

h = 1.32*((vL)/v1)^0.5*((v1/v2)^0.1 - 1)*((v1 + v2)/2)^(0.1)*((T1/T2 - 1)/ln(T1/T2))

Putting the values of v1, v2, T1, and T2 in the above equation, we get h = 6.33 W/(m^2 K)....................(6)

Thus, the convection heat transfer coefficient is 6.33 W/(m^2 K) and the heat flux added to the sheet surface is 847.6 W.

Learn more about convection heat transfer coefficient here ;

https://brainly.com/question/31967094

#SPJ11

(a) The location of the image:

(i) For do = 40.0 cm, the image is located at -40/3 cm.

(ii) For do = 20.0 cm, the image is located at -10 cm.

(iii) For do = 10.0 cm, the image is located at -20/3 cm.

To determine the location and characteristics of the image formed by a diverging lens, we can use the lens equation and the magnification formula.

Given:

Focal length of the diverging lens (f) = -20.0 cm (negative because it's a diverging lens)

Object distances:

(i) Object distance (do) = 40.0 cm

(ii) Object distance (do) = 20.0 cm

(iii) Object distance (do) = 10.0 cm

We can use the lens equation to find the image distance (di) for each case:

Lens Equation: 1/f = 1/do + 1/di

Solving for di:

(i) For do = 40.0 cm:

1/f = 1/40 + 1/di

1/(-20.0) = 1/40 + 1/di

-1/20 = 1/40 + 1/di

-1/20 - 1/40 = 1/di

-3/40 = 1/di

di = 40/(-3) cm

(ii) For do = 20.0 cm:

1/f = 1/20 + 1/di

1/(-20.0) = 1/20 + 1/di

-1/20 = 1/20 + 1/di

-1/20 - 1/20 = 1/di

-1/10 = 1/di

di = 10/(-1) cm

(iii) For do = 10.0 cm:

1/f = 1/10 + 1/di

1/(-20.0) = 1/10 + 1/di

-1/20 = 1/10 + 1/di

-1/20 - 1/10 = 1/di

-3/20 = 1/di

di = 20/(-3) cm

Now, let's analyze the characteristics of the image:

(b) To determine if the image is real or virtual, we look at the sign of the image distance (di). If di is positive, the image is real. If di is negative, the image is virtual.

(i) For do = 40.0 cm:

The image distance (di) is negative (-40/3 cm), so the image is virtual.

(ii) For do = 20.0 cm:

The image distance (di) is negative (-10 cm), so the image is virtual.

(iii) For do = 10.0 cm:

The image distance (di) is negative (-20/3 cm), so the image is virtual.

(c) To determine if the image is upright or inverted, we look at the magnification (m). If m is positive, the image is upright. If m is negative, the image is inverted.

The magnification (m) can be calculated using the formula:

Magnification (m) = -di/do

(i) For do = 40.0 cm:

m = -(-40/3) / 40

[tex]m = 1/3[/tex]

(ii) For do = 20.0 cm:

m = -(-10) / 20

[tex]m = 1/2[/tex]

(iii) For do = 10.0 cm:

m = -(-20/3) / 10

[tex]m = 2/3[/tex]

(d) The increase in image size can be determined by comparing the absolute values of the magnification (|m|) with 1.

(i) For do = [tex]40.0 cm[/tex]:

The increase in image size is |m|

= 1/3.

(ii) For do = 20.0 cm:

The increase in image size is |m| = 1/2.

(iii) For do = 10.0 cm:

The increase in image size is |m| = [tex]2/3.[/tex]

To summarize:

(a) The location of the image:

(i) For do = 40.0 cm, the image is located at -40/3 cm.

(ii) For do = 20.0 cm, the image is located at -10 cm.

(iii) For do = 10.0 cm, the image is located at -20/3 cm.

(b) The image is virtual in all cases.

(c) The image is upright in all cases.

(d) The increase in image size:

(i) The image size increases by a factor of 1/3.

(ii) The image size increases by a factor of 1/2.

(iii) The image size increases by a factor of 2/3.

to know more about magnification visit:

brainly.com/question/28501133

#SPJ11

According to the question The cheetah's velocity upon reaching the gazelle is [tex]$50\mathbf{i} + \frac{150}{9}\mathbf{j}$[/tex] m/s.

To find the cheetah's velocity upon reaching the gazelle, we can use the kinematic equations of motion.

Given:

Gazelle's position: [tex]\mathbf{x} = 50\mathbf{i} + 25\mathbf{j}$ meters[/tex]

Gazelle's velocity: [tex]\mathbf{v} = 6\mathbf{i}$ m/s[/tex]

Cheetah's acceleration: [tex]\mathbf{a} = 9\mathbf{i} + 3\mathbf{j}$ m/s^2[/tex]

We can integrate the acceleration to find the cheetah's velocity. Integrating the acceleration with respect to time gives us the change in velocity:

[tex]$\Delta \mathbf{v} = \int \mathbf{a} \, dt$[/tex]

Integrating the x-component of acceleration gives us the change in x-component of velocity:

[tex]$\Delta v_x = \int a_x \, dt$[/tex]

Integrating the y-component of acceleration gives us the change in y-component of velocity:

[tex]$\Delta v_y = \int a_y \, dt$[/tex]

Since the cheetah starts from rest, the initial velocity is zero:

[tex]$\mathbf{v}_0 = \mathbf{0}$[/tex]

Integrating the x-component of acceleration and applying the initial condition, we get:

[tex]$\Delta v_x = \int (9\mathbf{i}) \, dt = 9t\mathbf{i}$[/tex]

Integrating the y-component of acceleration and applying the initial condition, we get:

[tex]$\Delta v_y = \int (3\mathbf{j}) \, dt = 3t\mathbf{j}$[/tex]

To find the time taken by the cheetah to reach the gazelle, we can use the relative position between the cheetah and the gazelle:

[tex]$\Delta \mathbf{r} = \mathbf{x}_\text{gazelle} - \mathbf{x}_\text{cheetah}$[/tex]

[tex]$\Delta \mathbf{r} = (50\mathbf{i} + 25\mathbf{j}) - (0\mathbf{i} + 0\mathbf{j}) = 50\mathbf{i} + 25\mathbf{j}$[/tex]

Since the cheetah's x-component of velocity is 9t, we can set up the equation:

[tex]$\Delta x = v_x t$[/tex]

[tex]$50 = 9t$[/tex]

Solving for t:

[tex]$t = \frac{50}{9}$[/tex]

Substituting this value of t into the expressions for [tex]\Delta v_x$ and $\Delta v_y[/tex], we get:

[tex]$\Delta v_x = 9 \left(\frac{50}{9}\right)\mathbf{i} = 50\mathbf{i}$[/tex]

[tex]$\Delta v_y = 3 \left(\frac{50}{9}\right)\mathbf{j} = \frac{150}{9}\mathbf{j}$[/tex]

The cheetah's final velocity is the sum of the initial velocity and the change in velocity:

[tex]$\mathbf{v}_\text{final} = \mathbf{v}_0 + \Delta \mathbf{v} = \mathbf{0} + 50\mathbf{i} + \frac{150}{9}\mathbf{j}$[/tex]

Therefore, the cheetah's velocity upon reaching the gazelle is [tex]$50\mathbf{i} + \frac{150}{9}\mathbf{j}$[/tex] m/s.

Learn more about velocity

brainly.com/question/6041819

#SPJ11

The tension in the vertical strand of spiderweb when the spider hangs motionless is [tex]8.80*10^{-4[/tex] N. The tension in the horizontal strand, where the spider sits motionless and causes the strand to sag at an angle of 11.0∘ below the horizontal, is [tex]8.68*10^{-4[/tex] N. The ratio of the tension in the horizontal strand to the tension in the vertical strand is approximately 0.988.

In the first case, when the spider hangs motionless on the vertical strand, the tension in the strand is equal to the weight of the spider. The weight of an object is given by the formula W = mg, where m is the mass of the object and g is the acceleration due to gravity. Substituting the given values, we have W = (8.00×[tex]10^{-5[/tex] kg)(10 [tex]m/s^2[/tex]) = 8.00×[tex]10^{-4[/tex] N.

In the second case, when the spider sits motionless in the middle of the horizontal strand, the tension in the strand is the combination of the vertical component of the tension and the horizontal component due to the sagging. The vertical component is equal to the weight of the spider, which remains the same as in the previous case. The horizontal component is determined by the angle of sagging. Using trigonometry, the horizontal component is T * sin(11.0∘), where T is the tension in the horizontal strand. Since the spider is motionless, the sum of the vertical and horizontal components of tension must balance the weight of the spider. Equating the forces, we get T * sin(11.0∘) = (8.00×[tex]10^{-5[/tex] kg)(10 [tex]m/s^2[/tex]). Solving for T, we find T = (8.00×[tex]10^{-5[/tex] kg)(10 [tex]m/s^2[/tex]) / sin(11.0∘) ≈ 8.68×[tex]10^{-4[/tex] N.

To compare the tensions, we divide the tension in the horizontal strand by the tension in the vertical strand: (8.68×[tex]10^{-4[/tex] N) / (8.80×[tex]10^{-4[/tex] N) ≈ 0.988. Therefore, the ratio of the tension in the horizontal strand to the tension in the vertical strand is approximately 0.988.

Learn more about tension here:

https://brainly.com/question/32984793

#SPJ11

The time period during which all three spacecraft will be inside the asteroid, accounting for relativity, is approximately 1.02 microseconds.

To determine the time period during which all three spacecraft will be inside the asteroid, we need to calculate the time it takes for the enemy spacecraft to traverse the 91.5 m line from the perspective of the locals on the asteroid.

Given:

Length of the line (L) = 91.5 m

Speed of the enemy spacecraft relative to the asteroid (v) = 0.9c (90% the speed of light)

To account for relativistic effects, we need to use the time dilation formula:

Δt' = Δt / γ

where:

Δt' is the time observed by the locals on the asteroid,

Δt is the time observed by the enemy spacecraft, and

γ is the Lorentz factor, given by γ = 1 / √(1 - (v/c)²)

First, let's calculate γ:

γ = 1 / √(1 - (0.9c/c)²)

γ = 1 / √(1 - 0.9²)

γ = 1 / √(1 - 0.81)

γ = 1 / √(0.19)

γ ≈ 1 / 0.4359

γ ≈ 2.294

Now, we can find the time observed by the locals on the asteroid:

Δt' = L / v

Δt' = 91.5 m / (0.9c)

To express the result in microseconds, we convert the time to seconds and then multiply by 10^6:

Δt' = (91.5 m / (0.9c)) × (1 s / (3 × 10⁸ m/s)) × (10⁶ μs / 1 s)

Simplifying the expression:

Δt' ≈ (91.5 m × 10⁶ μs) / (0.9c × 3 × 10⁸ m/s)

Δt' ≈ 305 μs / c

Since we need the result to two significant figures, we can substitute the value of the speed of light:

Δt' ≈ 305 μs / (3 × 10⁸ m/s)

Calculating:

Δt' ≈ 1.02 μs

Therefore, all three spacecraft will be inside the asteroid for approximately 1.02 microseconds, taking into account relativistic effects.

Read about relativity here: https://brainly.com/question/364776

#SPJ11

The characteristics of sound are loudness, pitch, timbre, and speed.

Therefore, option d is the correct answer.

Sound is a type of energy that is caused by the vibration of matter, and it can only travel through a medium.

Here are the characteristics of sound:

Loudness

Pitch

Timbre

Speed

Out of these options, the characteristics of sound are loudness, pitch, timbre, and speed.

Therefore, option d is the correct answer.

The explanation of each of the sound characteristics is given below:

Loudness is the physical characteristic of sound. It is defined as the human perception of sound intensity. The unit used for measuring loudness is decibels (dB).

Pitch is the highness or lowness of a sound. It is determined by the frequency of the sound wave. The unit used for measuring pitch is hertz (Hz).

Timbre is the quality of sound. It helps to differentiate between sounds with the same pitch and loudness. It is caused by the presence of overtones in sound.

Speed is the rate at which sound travels. It varies depending on the medium through which sound travels. The speed of sound is about 1500 meters per second in water, 330 meters per second in the air at room temperature, and about 6000 meters per second in steel.

learn more about  Speed in the link:

https://brainly.com/question/13943409

#SPJ11

In conducting the measurements, several important concepts were observed. Firstly, current can be measured using an ammeter, which is connected in series within a circuit to measure the flow of electric charge.

Secondly, the state of a switch or gap in a circuit can determine whether the circuit is closed or open. A closed switch allows current to flow, while an open switch interrupts the flow of current. Thirdly, the division of voltage over various components in a circuit is determined by their respective resistances or impedance. Components with higher resistance tend to have a greater voltage drop across them. Similarly, the division of current in a circuit is influenced by the resistance of each component.

Learn more about current here:

https://brainly.com/question/31315986

#SPJ11

The stopping distance of the car is approximately 30.67 meters.

To find the stopping distance of the car, we need to consider two parts: the distance covered during the reaction time and the distance covered while decelerating.

1. Distance covered during reaction time:

The car is initially traveling at +15.4 m/s, and the reaction time is 0.451 s. During this time, the car maintains its initial velocity. The distance covered is given by:

Distance_reaction = Initial velocity * Reaction time

Distance_reaction = 15.4 m/s * 0.451 s

Distance_reaction ≈ 6.944 m

2. Distance covered while decelerating:

The car decelerates at 5.00 m/s². We need to find the distance covered during the deceleration phase. We can use the following equation of motion:

v^2 = u^2 + 2as

Where:

v is the final velocity (0 m/s, since the car stops)

u is the initial velocity (15.4 m/s)

a is the acceleration (-5.00 m/s², negative since it's decelerating)

s is the distance

Rearranging the equation to solve for s:

s = (v^2 - u^2) / (2a)

s = (0^2 - 15.4^2) / (2 * -5.00)

s ≈ -237.3 m²/s² / -10.00

s ≈ 23.73 m

The distance covered while decelerating is approximately 23.73 m.

The total stopping distance of the car, measured from the point where the driver first notices the red light, is the sum of the distance covered during the reaction time and the distance covered while decelerating:

Stopping distance = Distance_reaction + Distance_deceleration

Stopping distance ≈ 6.944 m + 23.73 m

Stopping distance ≈ 30.67 m

Therefore, the stopping distance of the car is approximately 30.67 meters.

learn more about "distance ":- https://brainly.com/question/26550516

#SPJ11

(a) I_transmitted ≈ 0.787 W/m²

(b) I_transmitted ≈ 0.262 W/m²

(a) To find the transmitted intensity when θ2 = 25.0° and θ3 = 50.0°, we need to consider the effect of each polarizer on the incident beam.

When unpolarized light passes through a polarizer, the transmitted intensity is given by Malus's law:

I_transmitted = I_incident * cos²(θ)

where I_incident is the incident intensity and θ is the angle between the polarization axis of the polarizer and the direction of the incident light.

Let's denote the transmitted intensity after passing through the first polarizer as I₁. Then, the transmitted intensity after passing through the second polarizer (θ2 = 25.0°) will be:

I₂ = I₁ * cos²(θ2)

Finally, the transmitted intensity after passing through the third polarizer (θ3 = 50.0°) will be:

I_transmitted = I₂ * cos²(θ3)

Substituting the given values:

I₁ = 1.60 W/m² (initial incident intensity)

θ2 = 25.0° (angle of the second polarizer)

θ3 = 50.0° (angle of the third polarizer)

We can calculate the transmitted intensity as follows:

I₂ = 1.60 W/m² * cos²(25.0°)

I_transmitted = I₂ * cos²(50.0°)

(b) Similarly, when θ2 = 50.0° and θ3 = 25.0°, the transmitted intensity will be:

I₂ = 1.60 W/m² * cos²(50.0°)

I_transmitted = I₂ * cos²(25.0°)

By substituting the given values into the formulas, we can calculate the transmitted intensities for both cases (a) and (b).

learn more about "intensity ":- https://brainly.com/question/28895397

#SPJ11

Energy is defined as the ability to do work. Intensity is the amount of energy that flows through a unit of area perpendicular to the direction of flow per unit time. It is a measure of the strength of the energy waves. The surface of a material can receive energy at a certain rate.

Given: Energy received per second = 233 W/sq.cm

Let’s find the energy received per sq.ft.:

Energy received per second per sq.cm can be converted into energy received per second per sq.ft. by using the following conversion factor:

1 sq.ft. = 30.48 cm

Therefore, (1 sq.ft.)/(30.48 cm[tex])^2[/tex] = 0.0929 sq.ft./sq.cm

∴ Energy received per second per sq.ft. = (233 W/sq.cm) × (0.0929 sq.ft./sq.cm) = 21.6 W/sq.ft.

We have to find the intensity in (ft. lb/sec)/sq.ft. To calculate the intensity in (ft.lb/sec)/sq.ft, we use the formula: 1 watt = 1 Nm/s 1 ft.lb/sec = 1.356 W

∴ Intensity = Energy received per second / Area

Energy received per second = 21.6 W/sq.ft.

Area = 1 sq.ft.

Intensity = 21.6 W/sq.ft. = (21.6 Nm/s)/ (0.0929 × 1.356 sq.ft.)= 156.7 (ft. lb/sec)/sq.ft.

Energy is defined as the ability to do work. Intensity is the amount of energy that flows through a unit of area perpendicular to the direction of flow per unit time. It is a measure of the strength of the energy waves. The surface of a material can receive energy at a certain rate. In this problem, the energy received per second is 233 watts per square cm. But, we need to find the intensity in (ft.lb/sec)/sq.ft.

We can find the energy received per sq.ft. by using a conversion factor (1 sq.ft.)/(30.48 cm[tex])^2[/tex] = 0.0929 sq.ft./sq.cm, and multiplying it by the energy received per second per sq.cm (233 W/sq.cm). The result is 21.6 W/sq.ft. To calculate the intensity in (ft.lb/sec)/sq.ft, we use the formula: Intensity = Energy received per second / Area. The area is given as 1 sq.ft. After substituting the values, we get the intensity as 156.7 (ft.lb/sec)/sq.ft.

To know more about energy waves visit:

https://brainly.com/question/9365406

#SPJ11

The force exerted on the hand by the apple is given by F = ma ≈ 0.074 J.

A force is an action that causes an object to move or stop moving. In some cases, two forces act on the same body in opposite directions. In this case, the forces are referred to as action-reaction pairs.

For example, while the apple is falling, it is pulling on the Earth with an equal and opposite force. Also, the apple is pulling on the hand that caught it as the hand catches the apple. The amount of force that the apple exerts on the hand while it falls can also be calculated.

Action-reaction pairs of forces for the apple and the Earth and for the apple and the hand are described below:

For the apple and the Earth: When the apple is falling, it exerts a gravitational force on the Earth, and the Earth, in turn, exerts an equal and opposite gravitational force on the apple. This is because both objects have mass, and gravity is a force that acts between masses. Therefore, the gravitational force between the apple and the Earth is an action-reaction pair.

For the apple and the hand: When the apple falls into the hand, the apple exerts a force on the hand, and the hand exerts an equal and opposite force on the apple. This is known as the action-reaction pair of forces. When the apple falls on the hand, it applies force in the direction of the hand. Because of this force, the hand moves in the opposite direction to balance the force applied by the apple.

The amount of force exerted on the Earth can be calculated using the law of conservation of momentum. This law states that momentum is conserved in an isolated system, so the momentum of the apple must be equal and opposite to the momentum of the Earth.

The mass of the Earth is much greater than the mass of the apple, so the Earth does not move significantly. However, it does move slightly upward because of the force exerted on it by the apple. The equation for the momentum of the Earth is given by P = mv, where P is momentum, m is mass, and v is velocity. Therefore, the amount that the Earth moves upwards while the apple is falling is given by

v = P/m

= [tex](444 g \times 1.11 s)/5.974 \times 10^{24} kg[/tex]

≈ [tex]6.59 \times 10^{-27}[/tex] m.

The average amount of force that the apple exerts on the person’s hand can be calculated using the equation F = ma, where F is force, m is mass, and a is acceleration. The acceleration of the apple can be calculated using the equation a =[tex](2d)/t^2[/tex], where d is distance, and t is time. Therefore,

[tex]a = (2 \times 0.0999 m)/(1.11 s)^2 \\a = 0.166 m/s^2.[/tex]

The mass of the apple is 0.444 kg. Therefore, the force exerted on the hand by the apple is given by F = ma ≈ 0.074 J.

In conclusion, action-reaction pairs of forces for the apple and the Earth and for the apple and the hand are explained and the amount of force exerted by the apple on the person's hand and the distance that the Earth moves upward when the apple is falling are calculated.

To know more about gravitational force visit:

brainly.com/question/32609171

#SPJ11

The speed of the stone in the given scenario can be calculated using the formula of the centripetal force.The formula of centripetal force is given by:F = (m*v²)/rWhere, F is the centripetal force,m is the mass of the object,v is the velocity of the object,r is the radius of the circle.

Now, in the given scenario, the length of the string is given as 1.16 m. So, the radius of the circle can be given as:r = 1.16 m / 2 = 0.58 m The gravitational force acting on the stone is given by:Fg = mg Where,m is the mass of the stone,g is the acceleration due to gravity.According to the question, the tension is 9% more in the vertical case than in the horizontal case. So, the tension in the horizontal case can be given as:Fh = Fg + (mv²h)/rAnd, the tension in the vertical case can be given as:Fv = Fg + (mv²v)/r Given that the tension in the vertical case is 9% larger than in the horizontal case.

Therefore, we can write:Fv = 1.09 * Fh => Fg + (mv²v)/r = 1.09 * [Fg + (mv²h)/r]Now, let's divide the equation by Fg:1 + (mv²v)/(Fg * r) = 1.09 + 1.09*(mv²h)/(Fg * r)After this, we can substitute the values:Fg = mgv = √[(Fg * r)/m]Now, let's substitute the values and solve for v:1 + (√[(Fg * r * v²v)/m])/(Fg * r) = 1.09 + 1.09 * (√[(Fg * r * v²h)/m])/(Fg * r)After solving this equation we get,√[(Fg * r * v²v)/m] = 1.09√[(Fg * r * v²h)/m]√[v²v] = 1.09 * √[v²h]v²v = 1.1881 * v²hNow, substituting the value of v²h we get,v²v = 1.1881 * [Fg * r / m]After substituting all the values we get:v = 5.02 m/sSo, the speed of the stone in the given scenario is 5.02 m/s.

To know more centripetal force visit:-

https://brainly.com/question/14021112

#SPJ11

A differential amplifier is an electronic device that amplifies the voltage difference between two input signals while rejecting any common-mode voltage. It is commonly used in applications where the accurate amplification of small signals is required, such as in operational amplifiers, audio amplifiers, and instrumentation amplifiers.

There are different kinds of differential amplifiers, including:

1. Single-Ended Differential Amplifier: This type of differential amplifier has one input signal and one reference signal. It amplifies the voltage difference between the input signal and the reference signal, rejecting any common-mode voltage.

2. Fully Differential Amplifier: In this type of differential amplifier, both the positive and negative inputs are differential signals. It amplifies the voltage difference between the two inputs while rejecting any common-mode voltage. It is commonly used in applications where precise amplification of differential signals is required.

3. Instrumentation Amplifier: An instrumentation amplifier is a differential amplifier that provides high input impedance, high common-mode rejection, and adjustable gain. It is commonly used in applications where accurate amplification of small differential signals is needed, such as in medical instruments and data acquisition systems.

Now, let's draw the circuits for these differential amplifiers:

1. Single-Ended Differential Amplifier:
```
    Vcc
     |
     R1
     |
Vin1--|\
     |  > Amplifier
Vin2--|/
     |
     R2
     |
   GND
```

2. Fully Differential Amplifier:
```
    Vcc
     |
     R1
     |
Vin+--|\
     |  > Amplifier
Vin--|/
     |
     R2
     |
   GND
```

3. Instrumentation Amplifier:
```
    Vcc
     |
     R1
     |
Vin1--|\
     |  > Amplifier
Vin2--|/
     |
     R2
     |
     |
     R3
     |
     |\
     | > Amplifier
     |/
     |
     R4
     |
   GND
```

In these circuit diagrams, Vin1 and Vin2 represent the input signals, and Vcc represents the power supply voltage. R1, R2, R3, and R4 are resistors used to set the gain and provide input impedance to the amplifier. The exact values of these resistors and other components may vary depending on the specific application and desired amplification characteristics.

I hope this explanation helps! Let me know if you have any further questions.

To know more about differential visit:

https://brainly.com/question/33433874

#SPJ11

(a) The football reaches a height of approximately 4.19 meters.

(b) The football hits the ground approximately 1.88 seconds after being kicked.

To solve this problem, we can use the equations of motion for projectile motion. Let's break it down into two parts:

(a) To find how high up the football travels, we need to calculate its maximum height (vertical displacement). We can use the following equation:

Vertical Displacement (Δy) = (Initial Vertical Velocity)² / (2 * Acceleration due to Gravity)

In this case, the initial vertical velocity is given by:

Initial Vertical Velocity (Vy) = Initial Velocity (V) * sin(θ)

Substituting the values:

Vy = 18.0 m/s * sin(31.0°) = 9.25 m/s

Acceleration due to Gravity (g) is approximately 9.8 m/s².

Plugging in the values, we have:

Δy = (9.25 m/s)² / (2 * 9.8 m/s²) ≈ 4.19 m

Therefore, the football reaches a height of approximately 4.19 meters.

(b) To find the time it takes for the football to hit the ground, we can use the equation:

Time of Flight (t) = 2 * (Initial Vertical Velocity) / Acceleration due to Gravity

Plugging in the values:

t = 2 * 9.25 m/s / 9.8 m/s² ≈ 1.88 s

Therefore, the football hits the ground approximately 1.88 seconds after being kicked.

To know more about projectile motion, refer to the link below:

https://brainly.com/question/29545516#

#SPJ11

The magnitude of the average acceleration of the ball during the 3.45 m/s contact with the wall is 12173.9 [tex]m/s^2[/tex].

the magnitude of the average acceleration of the ball during the contact with the wall, we can use the equation:

average acceleration (a) = (change in velocity) / (time interval)

Calculate the change in velocity of the ball:

change in velocity = final velocity - initial velocity

The initial velocity (v_initial) of the ball is 25.5 m/s, and the final velocity (v_final) is -16.5 m/s (rebounding in the opposite direction). Substituting the values:

change in velocity = -16.5 m/s - 25.5 m/s

change in velocity = -42 m/s

We need to convert the time interval from milliseconds to seconds. The given time interval is 3.45 ms, which is equivalent to [tex]3.45 * 10^{(-3)[/tex] seconds.

Substituting the values into the formula for average acceleration:

average acceleration = (-42 m/s) / (3.45 *[tex]10^{(-3)[/tex] s)

Calculating the average acceleration:

average acceleration ≈ -12173.9 [tex]m/s^2[/tex]

The magnitude of the average acceleration is the absolute value of the average acceleration:

magnitude of average acceleration ≈ 12173.9 [tex]m/s^2[/tex]

To know more about average acceleration refer here

https://brainly.com/question/29310400#

#SPJ11