When a honeybee flies through the air, it develops a charge of +20 pC . Part A How many electrons did it lose in the process of acquiring this charge? Express your answer as a number of electrons.

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:

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:

Option (a) is correct.

Explanation:

The angle of banking of curved path is given as

tan θ[tex]=\frac{v^{2} }{rg}[/tex]

Here, v is linear velocity, r is radius of curved path, θ is bank angle and g is acceleration due to gravity.

We have,  θ= 20.0°, r = 100 m and take [tex]g=9.8 m/s^{2}[/tex]

Substituting these values in above formula, we get

[tex]tan 20 =\frac{v^{2} }{100 m*9.8m/s^{2} }[/tex]

[tex]v^{2} =356.69[/tex]

[tex]v=\sqrt{356.69} =18.9m/s[/tex]

Thus, the ideal speed is 18.9 m/s.

The ideal speed to take a 100 m radius curve banked at a 20.0° angle is 18.9 m/s. The correct Option is a.

The speed of any moving object is the ratio of the distance covered and the time taken to cover that distance.

Speed s = distance d / time t

Given a 100 m radius curve banked at a 20.0° angle.

The angle of banking of curved path is given as

tan θ= V² /2g

Here, V is linear velocity, r is radius of curved path, θ is bank angle and g is acceleration due to gravity.

Substituting the values in above formula, we get

tan 20° = V² /2x9.81

V = 18.9 m/s

Thus, the correct option is a.

Learn more about speed.

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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:

the magnitude of each spaceships velocity is 0.495c

Explanation:

This is a relative velocity problem, that means that the velocity is relative to the observer, in this case, there is an observer on earth, from this reference point the velocity of each spaceship are the same, but for the observer on a spaceship would be different, we can see it as a velocity difference :

[tex]V=Va-Vb\\[/tex]

because the spaceship B is going on the opposite direction from A, the velocity is negative, the observer on earth says the velocity is the same for both o them, so:

[tex]0.99c=va-(-va)\\0.99c=2va\\va=0.495c[/tex]

vb=-va

Answer:

A. structure of the classroom

Explanation:

According to the Creative Curriculum model, the A. structure of the classroom  sets the context for teaching and learning.

Creative curriculum model stands to enhance the skill of the students through a structured class room. Creative curriculum goes beyond the rote learning and focuses on big ideas and individual passion of each students.

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: 8*10^-15 N

Explanation: In order to calculate the force applied on an electron in the middle of the two planes at 500 V we know that,  F=q*E

The electric field between  the plates is given by:

E = ΔV/d = 500 V/0.01 m=5*10^3 N/C

the force applied to the electron is: F=e*E=8*10^-15 N

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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Answer:

Option B

Explanation:

The fundamental quantities are the basic quantities from which other quantities are derived.

There are a total of 7 fundamental quantities, namely Length, Mass, Time, Luminous Intensity, Amount of substance, Electric Current, Temperature.

From these 7 quantities, the other quantities are derived.

Weight is the product of its mass and the gravitational acceleration 'g' which acts on that mass.

Where mass is the fundamental quantity and 'g' is also derived and hence weight is a derived quantity.

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:

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:

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:

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)

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:

[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:

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:

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 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:

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]

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:

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.

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:

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:

Radius of aluminium sphere which has same mass as of sphere of iron with radius 14 m is 40.745 meters.

Explanation:

Let the radius of aluminium sphere be [tex]r_{a}[/tex]

From the relation between density, mass and volume we know that

[tex]Mass=Density\times volume...............(i)[/tex]

Applying equation 'i' separately to iron and aluminium sphere we get

[tex]M_{a}=\rho _{a}\times V_{a}[/tex]

[tex]M_{ir}=\rho _{ir}\times V_{ir}[/tex]

Equating the masses of iron and aluminium spheres we get

[tex]M_{a}=M_{ir}[/tex]

[tex]\rho _{a}\times \frac{4\pi r_a^3}{3}=\rho _{ir}\times \frac{4\pi r_{ir}^3}{3}[/tex]

[tex]\rho _{a}\times r_{a}^3=\rho _{ir}\times r_{ir}^{3}\\\\\therefore r_{a}=(\frac{\rho _{ir}}{\rho _{a}})^{1/3}\cdot r_{ir}\\\\r_{a}=\frac{7.8610}{2.701}\times 14\\\\\therefore r_a=40.745m[/tex]

Answer:

a) ω1 = 18rpm    ω2 = -18rpm

b) ω1 = 102rpm     ω2 = 138rpm

c) ω1 = ω2 = 3.18rpm

Explanation:

For the first case, we know that each wheel will spin in a different direction but with the same magnitude, so:

ωr = 6rpm   This is the angular velocity of the robot

[tex]\omega = \frac{\omega r * D/2}{r_{wheel}}[/tex]  where D is 30cm and rwheel is 5cm

[tex]\omega = \frac{6 * 30/2}{5}=18rpm[/tex]  One velocity will be positive and the other will be negative:

ω1 = 18rpm    ω2 = -18rpm

For part b, the formula is the same but distances change. Rcircle=100cm:

[tex]\omega 1 = \frac{\omega r * (R_{circle} - D/2)}{r_{wheel}}[/tex]

[tex]\omega 2 = \frac{\omega r * (R_{circle} + D/2)}{r_{wheel}}[/tex]

Replacing values, we get:

[tex]\omega 1 = \frac{6 * (100 - 30/2)}{5}=102rpm[/tex]

[tex]\omega 2 = \frac{\omega r * (100 + 30/2)}{5}=138rpm[/tex]

For part c, both wheels must have the same velocity:

[tex]\omega = \frac{V_{robot}}{r_{wheel}}=20rad/min[/tex]

[tex]\omega = 20rad/min * \frac{1rev}{2*\pi rad}=3.18rpm[/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

Answer:

The diameter of the piston of the players equals 55.136 cm.

Explanation:

from the principle of transmission of pressure in a hydraulic lift  we have

[tex]\frac{F_{1}}{A_{1}}=\frac{F_{2}}{A_{1}}[/tex]

Since the force in the question is the weight of the individuals thus upon putting the values in the above equation we get

[tex]\frac{57.0\times 9.81}{\frac{\pi \times (19.0)^{2}}{4}}=\frac{4\times 120\times 9.81}{\frac{\pi \times D_{2}^{2}}{4}}[/tex]

Solving for [tex]D_{2}[/tex] we get

[tex]D_{2}^{2}=\frac{4\times 120}{57}\times 19^{2}\\\\\therefore D_{2}=\sqrt{\frac{4\times 120}{57}}\times 19\\\\D_{2}=55.136cm[/tex]

The diameter of the football players' piston is  [tex]\({55.4 \text{ cm}} \).[/tex]

To solve for the diameter of the football players' piston in the hydraulic lift, we start by using the principle of hydraulic systems, which states that pressure is constant throughout the fluid.

Given data:

- Mass of the cheerleader, [tex]\( m_1 = 57.0 \)[/tex] kg

- Mass of four football players, [tex]\( m_2 = 4 \times 120 \)[/tex] kg

- Height difference, [tex]\( h = 1.10 \)[/tex] m

- Diameter of the cheerleader's piston, [tex]\( D_1 = 19.0 \)[/tex] cm

First, convert the diameter of the cheerleader's piston into meters:

[tex]\[ D_1 = 19.0 \text{ cm} = 0.19 \text{ m} \][/tex]

The force exerted by the cheerleader's piston is:

[tex]\[ F_1 = m_1 g \][/tex]

The force exerted by the football players' piston is:

[tex]\[ F_2 = m_2 g \][/tex]

According to Pascal's principle:

[tex]\[ \frac{F_1}{A_1} = \frac{F_2}{A_2} \][/tex]

where [tex]\( A_1 \)[/tex] and [tex]\( A_2 \)[/tex] are the cross-sectional areas of the cheerleader's and football players' pistons, respectively.

The area [tex]\( A \)[/tex] of a piston is given by:

[tex]\[ A = \frac{\pi D^2}{4} \][/tex]

Now, calculate [tex]\( A_1 \)[/tex]:

[tex]\[ A_1 = \frac{\pi (0.19)^2}{4} \][/tex]

[tex]\[ A_1 = \frac{\pi \times 0.0361}{4} \][/tex]

[tex]\[ A_1 \approx 0.0284 \text{ m}^2 \][/tex]

Next, find [tex]\( F_1 \)[/tex]:

[tex]\[ F_1 = m_1 g = 57.0 \times 9.8 \][/tex]

[tex]\[ F_1 \approx 558.6 \text{ N} \][/tex]

Now, find [tex]\( A_2 \)[/tex]:

[tex]\[ \frac{F_1}{A_1} = \frac{F_2}{A_2} \][/tex]

[tex]\[ A_2 = \frac{F_2 \times A_1}{F_1} \][/tex]

Calculate [tex]\( F_2 \)[/tex]:

[tex]\[ F_2 = m_2 g = 480 \times 9.8 \][/tex]

[tex]\[ F_2 = 4704 \text{ N} \][/tex]

Now, find [tex]\( A_2 \)[/tex]:

[tex]\[ A_2 = \frac{4704 \times 0.0284}{558.6} \][/tex]

[tex]\[ A_2 \approx 0.2397 \text{ m}^2 \][/tex]

Finally, solve for [tex]\( D_2 \)[/tex]:

[tex]\[ A_2 = \frac{\pi (D_2)^2}{4} \][/tex]

[tex]\[ (D_2)^2 = \frac{4 \times 0.2397}{\pi} \][/tex]

[tex]\[ D_2 \approx \sqrt{\frac{0.9588}{\pi}} \][/tex]

[tex]\[ D_2 \approx 0.554 \text{ m} \][/tex]

Convert [tex]\( D_2 \)[/tex] from meters to centimeters:

[tex]\[ D_2 \approx 55.4 \text{ cm} \][/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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