If you can read the bottom row of your doctor's eye chart, your eye has a resolving power of one arcminute, equal to 1.67×10^−2 degrees . If this resolving power is diffraction-limited, to what effective diameter of your eye's optical system does this correspond? Use Rayleigh's criterion and assume that the wavelength of the light is 540 nm . Express your answer in millimeters to three significant figures.

Answer:

Explained

Explanation:

a) the work done will be positive since the chicken is scratching the ground. Here displacement is along the direction of force.

b) A person studying does no work in the language of physics because there is no displacement.

C) the work is done on the bucket by the crane and work is positive and here the displacement is in the direction of force.

d) Gravitational force act on the bucket in downward direction, here the work done will be negative as the force displacement are opposite to each other.

E)Negative work will be done as the force applied  by the muscle is in opposite to the displacement.

To estimate the total buoyant force on the 600 balloons filled with helium, we use Archimedes' principle. By calculating the volume of each balloon and using the formula for buoyant force, we can determine the force exerted by each balloon. Multiplying this force by the number of balloons gives us the total buoyant force.

To estimate the total buoyant force on the 600 balloons, we need to calculate the buoyant force on each balloon and then multiply it by the number of balloons. The buoyant force on a balloon can be calculated using Archimedes' principle, which states that the buoyant force is equal to the weight of the fluid displaced by the object. In this case, the fluid is the surrounding air.

The volume of each balloon can be calculated using the formula for the volume of a sphere: V = (4/3)πr^3, where r is the radius of the balloon. In this case, r = 0.54 m. Using this formula, we can calculate the volume of each balloon to be approximately 0.653 m^3.

The buoyant force on each balloon can be calculated using the formula F = ρVg, where F is the buoyant force, ρ is the density of the fluid, V is the volume of the fluid displaced, and g is the acceleration due to gravity. Since the density of air is approximately 1.225 kg/m^3 and g is approximately 9.8 m/s^2, we can calculate the buoyant force on each balloon to be approximately 7.93 N.

Finally, to calculate the total buoyant force on the 600 balloons, we can multiply the buoyant force on each balloon by the number of balloons: F_total = F_per_balloon * number_of_balloons. Plugging in the values, we get F_total ≈ 7.93 N * 600 = 4,758 N.

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The total buoyant force on the 600 balloons that Ian Ashpole used can be calculated using Archimedes' Principle and the principle of Buoyancy. The resulting force is the difference between the weight of the air displaced by the balloons and the weight of the balloons themselves, multiplied by the number of balloons.

This is a classic problem of Archimedes' Principle and Buoyancy, principles in Physics. In simple terms, the buoyant force on a body submerged in a fluid is equal to the weight of the fluid displaced by the body. In the case of a balloon, the buoyant force can be calculated as the difference between the weight of the air displaced by the balloon and the weight of the balloon itself. For a single balloon, this would be:

FB = (weight of the air displaced) - (weight of balloon). But we have 600 balloons, so, we multiply this force by 600 to get the total buoyant force on all the balloons. Given Ian Ashpole used 600 balloons, each with an estimated mass of 0.27 kg and a radius of about 0.54 m, we can calculate the total buoyant force on these balloons.

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Answer:

mean = 10.68 m/s

standard deviation 0.3059

[/tex]\sigma_m = 0.14[/tex]  

Explanation:

1) [tex]Mean = \frac{ 10.2+11+10.7+11+10.5}{5}[/tex]

  mean = 10.68 m/s

2 ) standard deviation is given as

[tex]\sigma = \sqrt{ \frac{1}{N} \sum( x_i -\mu)^2}[/tex]

N = 5

   [tex]\sigma =\sqrt{ \frac{1}{5} \sum{( 10.2-10.68)^2+(11-10.68)^2 + (10.7- 10.68)^2+ (11- 10.68)^2++ (10.5- 10.68)^2[/tex]

SOLVING ABOVE RELATION TO GET STANDARD DEVIATION VALUE

\sigma  = 0.3059

3) ERROR ON STANDARD DEVIATION

[tex]\sigma_m = \frac{ \sigma}{\sqrt{N}}[/tex]

               [tex]= \frac{0.31}{\sqrt{5}}[/tex]

[tex]\sigma_m = 0.14[/tex]  

Answer:

Mean =  = 10.68 m/s

Standard deviation = σ = 0.342 m/s

Error =  0.153 .  

Explanation:

The data has 5 readings.

Let each of the readings be Y

Take average and find the mean X = (10.2+11+10.7+11+10.5)/5 = 53.4/5 = 10.68 m/s.

Take the difference between the data values and the mean and square them individually.

(10.2 - 10.68)² =(-0.48)² = 0.23

(11 - 10.68)² = 0.32² = 0.102

(10.7 - 10.68)² = (-0.02)² = 0.0004

(11-10.68)² =0.32² = 0.102

(10.5-10.68)² = (-0.18)² = 0.0324

Standard deviation = [tex]\sigma = \sqrt{\frac{\sum(Y-X)^2 }{n-1}}[/tex]

                                = [tex]\sqrt{(0.23+0.102+0.0004+0.102+0.0324)/(5-1)}[/tex]

                                 = [tex]\sqrt{0.1167}[/tex] = 0.342 m/s

Error = Standard deviation / [tex]\sqrt{n}[/tex] = 0.342/5 = 0.153 .

Answer:

3.324 km

Explanation:

d1 = 2 km south

d2 = 1.5 km at 37° east of south

Write the displacements in vector form

[tex]\overrightarrow{d_{1}}=-2\widehat{j}[/tex]

[tex]\overrightarrow{d_{2}}=1.5\left (Sin37\widehat{i}-Cos37\widehat{j}  \right )=0.9\widehat{i}-1.2\widehat{j}[/tex]

The resultant displacement is given by

[tex]\overrightarrow{d} = \overrightarrow{d_{1}}+ \overrightarrow{d_{2}}[/tex]

[tex]\overrightarrow{d} = \left ( 0.9 \right )\widehat{i}+\left ( -2-1.2 \right )\widehat{j}[/tex]

[tex]\overrightarrow{d} = \left ( 0.9 \right )\widehat{i}+\left ( -3.2\right )\widehat{j}[/tex]

The magnitude of displacement is given by

[tex]d=\sqrt{0.9^{2}+\left ( -3.2 \right )^{2}}=3.324 km[/tex]

Thus, the bird has to travel 3.324 km in a straight line to return to its original place.

Answer:

x=22.33m

Explanation:

Kinematics equation for constant deceleration:

[tex]x =v_{o}*t - 1/2*at^{2}=8.9*6.3-1/2*1.70*6.3^{2}=22.33m[/tex]

In a uniform electric field, the electric potential or voltage increases or decreases linearly with distance based on the direction. Given initial voltage of 80V and field strength of 50 N/C, the voltage 1.0 m west increases by 50V to 130V.

For the given scenario, we are dealing with a uniform electric field and we need to determine the voltage at a certain distance in the field.

The key relationship between the electric field (E) and voltage (V) in a uniform electric field is that E = ΔV / Δd, where, ΔV is the potential difference and Δd is distance.

Given the electric field strength E = 50 N/C, and the initial voltage V1 = 80 V, we want to find the potential V2 a distance d = 1.0 m to the west, against the direction of the field. Since the electric field is uniform and points towards decreasing potential, the potential a distance d against the direction of field would increase. Therefore, ΔV = E * d = 50 N/C * 1.0 m = 50 V. Thus, the voltage at the point 1.0 m west would be V2 = V1 + ΔV = 80V + 50V = 130V.

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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:

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:[tex]v_b=71.86 mph[/tex]

Explanation:

Given

Velocity of soft ball is 56.9 mph

Distance between Pitching rubber to home plate is 47.9 ft

In major league distance is 60.5 ft

Let velocity of baseball is [tex]v_b[/tex]

Let t be the time for ball to reach to batter and its reaction time

since t is same for both case

[tex]\frac{47.9}{56.9}=\frac{60.5}{v_b}[/tex]

[tex]v_b=56.9\times \frac{60.5}{47.9}[/tex]

[tex]v_b=71.86 mph[/tex]

To determine the speed at which the baseball must be thrown to allow for equal reaction times between the softball and baseball batters, we can set up a ratio using the distances from the pitching rubber to home plate in both sports and solve for the desired speed.

To determine the speed at which the baseball must be thrown, we can set up a ratio using the distances from the pitching rubber to home plate in softball and baseball. Since the times for the batters to react should be equal, the distance ratio is equal to the speed ratio. Therefore, we can write the proportion:

(56.9 mph)/(47.9 ft) = x/(60.5 ft)

Where x represents the speed of the baseball. To solve for x, we can cross-multiply and solve for x:

x = (56.9 mph * 60.5 ft) / 47.9 ft

Calculating the right-hand side of the equation gives us the speed of the baseball:

x ≈ 71.92 mph

Therefore, the baseball must be thrown at approximately 71.92 mph to allow for equal reaction times between the softball and baseball batters.

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Final answer:

The corresponding blackbody temperature of the radiation with a wavelength of 970 um is approximately 2.725 K (Kelvin). The blackbody radiation observed in interstellar space is known as the cosmic microwave background (CMB). This temperature represents the average temperature of the universe at the time the CMB was emitted.

Explanation:

The corresponding blackbody temperature of the radiation with a wavelength of 970 um is approximately 2.725 K (Kelvin).

The blackbody radiation observed in interstellar space is known as the cosmic microwave background (CMB). It is the afterglow of the Big Bang and fills all of space. The blackbody spectrum of the CMB has a temperature of 2.725 K, as determined from observations.

This temperature represents the average temperature of the universe at the time the CMB was emitted. It provides valuable insights into the early universe and supports the idea of the expanding universe.

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:

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 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:

[tex]F = \frac{4kqQb}{(b^2 + \frac{d^2}{2})^{1.5}}[/tex]

Explanation:

Since all the four charges are equidistant from the position of Q

so here we can assume this charge distribution to be uniform same as that of a ring

so here electric field due to ring on its axis is given as

[tex]E = \frac{k(4q)x}{(x^2 + R^2)^{1.5}}[/tex]

here we have

x = b

and the radius of equivalent ring is given as the distance of each corner to the center of square

[tex]R = \frac{d}{\sqrt2}[/tex]

now we have

[tex]E = \frac{4kq b}{(b^2 + \frac{d^2}{2})^{1.5}}[/tex]

so the force on the charge is given as

[tex]F = QE[/tex]

[tex]F = \frac{4kqQb}{(b^2 + \frac{d^2}{2})^{1.5}}[/tex]

The magnitude of the force exerted by object W on object Z is F.

The magnitude of the force exerted by object W on object Z can be determined by analyzing the geometry and the charges involved. Since objects X and Z are at the midpoints of the sides of the square, they are equidistant from all four charges. As a result, the magnitude of the force exerted by each charge on object Z will be the same as the magnitude of the force exerted by each charge on object X. Therefore, the magnitude of the force exerted by object W on object Z is also F.

Answer:

[tex]V=\dfrac{PT}{m}\ ln\dfrac{M+m}{M}-gT[/tex]

Explanation:

Given that

Constant rate of leak =R

Mass at time T ,m=RT

At any time t

The mass = Rt

So the total mass in downward direction=(M+Rt)

Now force equation

(M+Rt) a =P- (M+Rt) g

[tex]a=\dfrac{P}{M+Rt}-g[/tex]

We know that

[tex]a=\dfrac{dV}{dt}[/tex]

[tex]\dfrac{dV}{dt}=\dfrac{P}{M+Rt}-g[/tex]

[tex]\int_{0}^{V}V=\int_0^T \left(\dfrac{P}{M+Rt}-g\right)dt[/tex]

[tex]V=\dfrac{P}{R}\ ln\dfrac{M+RT}{M}-gT[/tex]

[tex]V=\dfrac{PT}{m}\ ln\dfrac{M+m}{M}-gT[/tex]

This is the velocity of bucket at the instance when it become empty.

The velocity of the bucket when it becomes empty can be found by setting up an equation using Newton's second law and the equation for final velocity. When the bucket becomes empty, its final velocity is equal to the force exerted by the rope multiplied by the time taken for the bucket to become empty, divided by the initial mass of the bucket.

When the bucket becomes empty, the force exerted by the rope pulling the bucket up will all be used to accelerate the bucket. We can set up an equation using Newton's second law, which states that the net force on an object is equal to its mass multiplied by its acceleration: F_net = M * a. Since there are no other forces acting on the bucket except for the force exerted by the rope, we can equate the force to the force exerted by the rope: P = M * a.

Since the bucket is initially at rest, its initial velocity is zero. As the bucket becomes empty, the mass of the bucket decreases, but the force exerted by the rope remains constant. The acceleration of the bucket will therefore increase, resulting in an increasing velocity. When the bucket becomes empty, its final velocity can be determined using the equation v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time taken for the bucket to become empty. Rearranging the equation, we have v = 0 + Pt/M.

Answer:

Explanation:

We know that the equation for time dilation will be:

[tex]\Delta t = \frac{\Delta t'}{\sqrt{1-\frac{v^2}{c^2}}}[/tex]

where Δt its the time difference measured from Earth, and Δt' is the time difference measured by the astronaut.

Lets work a little the equation

[tex] \sqrt{1-\frac{v^2}{c^2}} = \frac{\Delta t'}{\Delta t}[/tex]

[tex] 1-\frac{v^2}{c^2}= (\frac{\Delta t'}{\Delta t})^2[/tex]

[tex] \frac{v^2}{c^2}= 1 - (\frac{\Delta t'}{\Delta t})^2[/tex]

[tex] \frac{v}{c}= \sqrt{ 1 - (\frac{\Delta t'}{\Delta t})^2 }[/tex]

[tex] v = \sqrt{ 1 - (\frac{\Delta t'}{\Delta t})^2 } c [/tex]

So, we got our equation. Knowing that Δt=211 years and Δt'=15 years

then

[tex] v = \sqrt{ 1 - (\frac{15 \ y}{211 \ y})^2 } c [/tex]

[tex] v = 0.99747 c [/tex]

To calculate the necessary speed, we can use the time dilation formula. Given the time experienced by the astronaut and the time experienced on Earth, we can solve for the velocity using the Lorentz factor. Using the approximate trip time of 211 years, we can calculate the necessary speed with the given equation.

To calculate the speed necessary, we can use the time dilation formula:

Δt' = Δt / γ

Where Δt' is the time experienced by the astronaut, Δt is the time experienced on Earth, and γ is the Lorentz factor given by γ = 1 / √(1 - (v² / c²)), where v is the velocity of the astronaut and c is the speed of light.

Given that the astronaut wishes to age only 15 years during the trip, we can approximate the dilated trip time to be 211 years. Substituting these values, we have:

15 = 211 / γ

Simplifying the equation, we find:

γ = 211 / 15

Using this value of γ, we can calculate the velocity of the astronaut:

v = √((1 - (1 / γ²)) * c²)

Substituting the value of γ, we have:

v = √((1 - (1 / (211 / 15)²)) * c²)

Answer:

The charge transferred equals [tex]-4.96\times 10^{-6}Columbs[/tex]

Explanation:

The charge quantizatrion principle states that the charge is always transferred in terms of product of fundamental charges.

Mathematically [tex]Q=ne[/tex]

where

'e' is the fundamental magnitude of charge

Thus charge transferred by [tex]4.4\times 10^{13}[/tex] electrons equals

[tex]Q_{1}=4.4\times 10^{13}\times -1.6\times 10^{-19}=-7.04\times 10^{-6}Columbs[/tex]

Similarly charge transferred by [tex]1.3\times 10^{13}[/tex] protons equals

[tex]Q_{2}=1.3\times 10^{13}\times 1.6\times 10^{-19}=2.08\times 10^{-6}Columbs[/tex]

Thus the net charge transferred equals

[tex]Q_{1}+Q_{2}=-7.04\times 10^{-6}+2.08\times 10^{-6}\\\\=-4.96\times 10^{-6}columbs[/tex]

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.

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

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:

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:

1.43 m/s^2

Explanation:

Each time you see mass and force, you will probably be going to need to use Newton's second Law. This law basically shows the relationship between the force being applied on an object and its mass and acceleration:

[tex]F = m*a[/tex]

Now, the force that the cable exerts on the elevator, not only has to accelarate it, but it also has to counter gravity. The maximum tension of the cable minus the weigth of the elevator would give us the net force being applied on the elevator:

[tex]T_{cable} - W_{elevator} = m_{elevator}*a[/tex]

[tex]21920 N - 1950kg*9.81 m/s^2 = 1950 kg*a\\a = \frac{21920 N - 1950kg*9.81 m/s^2}{1950kg} = 1.43 m/s^2[/tex]

Answer:

9.81 × 10⁻¹⁰ C

Explanation:

Given:

Distance between the tissue and the tip of the scale, r = 6 cm = 0.06 m

Charge on the ruler, Q = - 12 μC = - 12 × 10⁻⁶ C

Mass of the tissue = 3 g = 0.003 Kg

Now,

The force required to pick the tissue, F = mg

where, g is the acceleration due to gravity

also,

The force between (F) the charges is given as:

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

where,

q is the charge on the tissue

k is the Coulomb's constant = 9 × 10⁹ Nm²/C²

thus,

[tex]mg=\frac{kQq}{r^2}[/tex]

on substituting the respective values, we get

[tex]0.003\times9.81=\frac{9\times10^9\times(-12\times10^{-6})\times q}{0.06^2}[/tex]

or

q = 9.81 × 10⁻¹⁰ C

Minimum charge required to pick the tissue paper is 9.81 × 10⁻¹⁰ C

Answer:

The amount of gasoline is [tex]2.105\times10^{-2}\ gallons[/tex].

Explanation:

Given that,

Energy contained in gasoline [tex]= 1.3\times10^{8}\ J[/tex]

Mass = 2000 kg

Speed = 20 m/s

Energy used propel the car[tex] E=15\%\ of 1.3\times10^{8}\ J[/tex]

[tex]E=\dfrac{15}{100}\times1.3\times10^{8}[/tex]

[tex]E=19500000 = 1.9\times10^{7}\ J[/tex]

[tex]E=1.9\times10^{7}\ J[/tex]

We need to calculate the work done by the frictional force to stop the car

Using formula of work done

[tex]W=\Delta KE[/tex]

[tex]W=\dfrac{1}{2}m(v_{f}^2-v_{0}^2)[/tex]

[tex]W=\dfrac{1}{2}\times2000\times(0-20^2)[/tex]

[tex]W=-4.0\times10^{5}\ J[/tex]

Therefore,

Work done to bring the car back to its original speed

[tex]W=4.0\times10^{5}\ J[/tex]

[tex]Amount\ of\ gasoline\ needed = \dfrac{W}{E}[/tex]

[tex]Amount\ of\ gasoline =\dfrac{4.0\times10^{5}}{1.9\times10^{7}}[/tex]

[tex]Amount\ of\ gasoline =2.105\times10^{-2}\ gallons[/tex]

Hence, The amount of gasoline is [tex]2.105\times10^{-2}\ gallons[/tex].

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.

Answer:

The period of the pendulum is 1.816 sec.

Explanation:

Given that,

Length = 81.9 cm

Mass = 165 g

Angle = 10°

We need to calculate the period of the pendulum

Using formula of period

[tex]T = 2\pi\sqrt{\dfrac{l}{g}}[/tex]

Where, l = length

g = acceleration due to gravity

Put the value into the formula

[tex]T =2\pi\sqrt{\dfrac{81.9\times10^{-2}}{9.80}}[/tex]

[tex]T=1.816\ sec[/tex]

Hence, The period of the pendulum is 1.816 sec.

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:

3073 m/s

Explanation:

The key is that the period of the satellite is 24 hours  because it is synchronized with the rotation of the Earth. You can use Kepler's Third law to find the radius of the orbit:

  [tex]4\pi^2 r^3 = G MT^2[/tex]

[tex]4\pi^2 r^3= 6.67\times10^{-11}\times6\times10^{24}\times86400^2[/tex]

     r =   4.225 x 10^7 meters

So  one complete orbit is a distance of

   [2πr = 2× π × 4.225 x 10^7 = 26.55 x 10^7 meters

So the speed is

  distance / time = 26.55 x 10^7 meters / 86400 seconds =

        =   3073 m/s

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:

5.37 m/s

0.98 seconds

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.81 m/s²

[tex]v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times -9.81\times 10+15^2}\\\Rightarrow v=5.37\ m/s[/tex]

Velocity of the ball when it passes an observer sitting at a window is 5.37 m/s

[tex]v=u+at\\\Rightarrow t=\frac{v-u}{a}\\\Rightarrow t=\frac{5.37-15}{-9.81}\\\Rightarrow t=0.98\ s[/tex]

Time taken by the ball to pass the observer sitting at a window is 0.98 seconds