A rigid tank contains air at a pressure of 70 psia and a temperature of 55 ˚F. By how much will the pressure increase as the temperature is increased to 115 ˚F?

Answer:

83816746.4254 m/s

Explanation:

m = Mass of electron = [tex]9.11\times 10^{-31}\ kg[/tex]

q = Charge of electron = [tex]1.6\times 10^{-19}\ C[/tex]

V = Voltage = [tex]2\times 10^4\ V[/tex]

The kinetic energy of the electron is

[tex]K=\dfrac{1}{2}mv^2[/tex]

Energy is given by

[tex]E=qV[/tex]

Balancing the energy

[tex]qV=\dfrac{1}{2}mv^2\\\Rightarrow v=\sqrt{\dfrac{2qV}{m}}\\\Rightarrow v=\sqrt{\dfrac{2\times 1.6\times 10^{-19}\times 2\times 10^4}{9.11\times 10^{-31}}}\\\Rightarrow v=83816746.4254\ m/s[/tex]

The velocity of the electrons is 83816746.4254 m/s

To solve this problem we will apply the conversion rate between Celcius and Fahrenheit degrees. We will use the direct relationship clearly and not the added degrees of scale conversion. We know from the statement that the orange loses heat at the rate of

[tex]Q = 1.2kJ/\°C[/tex]

We have the conversion to °F is given as

[tex]T (\°F) = 1.8T+32[/tex]

Calculate the amount of heat loss from orange per °F

[tex]Q = \frac{1.2}{1.8}[/tex]

[tex]Q = 0.667kJ/\°F[/tex]

Therefore the amount of heat loss from the orange per °F drop in its temperature is 0.667kJ/°F

The heat loss from an orange per °F drop is 0.67 kJ, calculated by taking 1.2 kJ per °C drop and dividing it by 1.8 to convert it to Fahrenheit,

The heat loss from the orange per °F drop in its temperature can be found by converting 1.2 kJ lost per 1 °C drop in temperature to kJ lost per 1 °F drop. This can be achieved using the formula that 1 °C equals 1.8 °F.

Therefore, the heat loss per degree Fahrenheit will be less than the heat loss per degree Celsius. We calculate this as follows:
(1.2 kJ / °C) / 1.8 = 0.67 kJ per °F.

So for every degree Fahrenheit that the orange cools, it will lose 0.67 kilojoules of heat.

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Answer:

32.9242311983 m/s

1.40506567727g

35.3405770759 rpm

Explanation:

v = Linear Velocity of the capsule

[tex]a_c[/tex] = Centripetal acceleration = [tex]12.5g=12.5\times 9.81[/tex]

r = Radius of the centrifuge = 8.84 m

l = Person's height = 2 m

Centripetal acceleration is given by

[tex]a_c=\frac{v^2}{r}\\\Rightarrow v=a_cr\\\Rightarrow v=\sqrt{12.5\times 9.81\times 8.84}[/tex]

The linear speed of the capsule is 32.9242311983 m/s

The radius would be

[tex]r=\sqrt{r^2+\dfrac{l^2}{4}}\\\Rightarrow r=\sqrt{8.84^2+\dfrac{2^2}{4}}\\\Rightarrow r=8.89638128679\ m[/tex]

The centripetal acceleration

[tex]a_{c2}=\dfrac{32.9242311983^2}{8.89638128679\times 9.81}g\\\Rightarrow a_{c2}=12.4207805891g[/tex]

Change in acceleration from Pythagoras law

[tex]a=\sqrt{a_{ch}^2-a_{c2}^2}\\\Rightarrow a=\sqrt{12.5^2g^2-12.4207805891^2g^2}\\\Rightarrow a=1.40506567727g[/tex]

The difference is 1.40506567727g

Velocity

[tex]v=\omega r\\\Rightarrow v=2\pi Nr\\\Rightarrow N=\dfrac{v}{2\pi r}\\\Rightarrow N=\dfrac{32.9242311983}{2\pi 8.89638128679}\\\Rightarrow N=0.589009617932\ rev/s[/tex]

[tex]N=0.589009617932\times 60=35.3405770759\ rpm[/tex]

The speed is 35.3405770759 rpm

Answer: at when distance r = infinity.

Explanation: The formulae for the electric potential of an electric charge to an arbitrary point is given by the formulae below

V = q/4πεr

V = electric potential (volts)

q = magnitude of electric charge

ε = permittivity of free space

r = distance between arbitrary point and charge.

In the equation above, it can be seen that only electric potential (v) and distance (r) is a variable, and there is an inverse relationship between them (an increase in one leads to a decrease in the other)

Thus to have zero value of electric potential (v= 0) we have to have the largest value of r ( r = infinity).

Same goes for electric potential energy between two charges, the formulae is given below as

W = q1 *q2/4πεr

W= electric potential energy

q1 = magnitude of first charge.

q2 = magnitude of second charge

ε = permittivity of free space

r = distance between arbitrary point and charge.

Also, all values are constant aside from electric potential energy (w) and distance (r) which have an inverse relationship.

Thus to have zero value of electric potential energy (w =0), we have to get an infinite value of distance ( r =infinity)

Answer:

31 m/s

Explanation:

As both the monkey and the darts are subjected to constant gravitational acceleration g = 9.8 m/s2 and both start from rest (vertically speaking). Their vertical position will always be the same. For the dart to hit the monkey, its horizontal position must be the same as the monkey's, which is unchanged before reaching the ground. Therefore, the time it takes for the dart to travel across 70 m must be less than the time it takes for the monkey to drop 25m to the ground. We can find it out using the following equation of motion

[tex]s_m = gt_m^2/2[/tex]

[tex]25 = 9.8t_m^2/2[/tex]

[tex]t_m^2 = 50/9.8 = 5.1[/tex]

[tex]t_m = \sqrt{5.1} = 2.26 s[/tex]

For the dart to takes less that 2.26 s to travel 70m, its horizontal speed must at least be 70 / 2.26 = 31 m/s

Answer:

532.0725 m

102.17270893 m/s

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

g = Acceleration due to gravity = 9.81 m/s² = g

H = Height of cliff

Distance traveled in 3 seconds

[tex]s=ut+\dfrac{1}{2}at^2\\\Rightarrow s=0\times t+\dfrac{1}{2}\times 9.81\times 3^2\\\Rightarrow s=44.145\ m[/tex]

Distance traveled by sound = 2H-44.145 m

[tex]2H-44.145=ut+\dfrac{1}{2}at^2\\\Rightarrow 2H-44.145=340\times 3\\\Rightarrow H=\dfrac{340\times 3+44.145}{2}\\\Rightarrow H=532.0725\ m[/tex]

The height of the cliff is 532.0725 m

[tex]v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 9.81\times 532.0725+0^2}\\\Rightarrow v=102.17270893\ m/s[/tex]

Her speed just before she hits the ground is 102.17270893 m/s

Answer:

Explanation:

Given

Initial velocity of ball [tex]u=10\ m/s[/tex]

height of window [tex]h=20\ m[/tex]

Using Equation of motion

[tex]y=ut+\frac{1}{2}at^2[/tex]

where u=initial velocity

t=time

a=acceleration

As ball is already is at a height of 20 m so

[tex]Y=ut+\frac{1}{2}at^2+20[/tex]

[tex]Y=10\times t+0.5\times (-9.8)t^2+20[/tex]

[tex]Y=-4.9t^2+10t+20[/tex]

(b)highest point is obtained at v=0

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

where

v=final velocity

u=initial velocity

a=acceleration

s=displacement

[tex](0)-10^2=2\times (-9.8)\times s[/tex]

[tex]s=\frac{100}{19.6}[/tex]

[tex]s=5.102\ m[/tex]

Highest Point will be [tex]s+20=25.102\ m[/tex]

(c)Time taken when the ball hit the ground i.e. at Y=0

[tex]-4.9t^2+10t+20=0[/tex]

[tex]t=3.28\ s[/tex]

impact velocity [tex]v=\sqrt{2\times 9.8\times 25.102}[/tex]

[tex]v=22.181\ m/s[/tex]

(a) The equation be "Y = -4.9t² + 10t + 20".

(b) The highest point be "25.102 m".

(c) The impact velocity be "22.181 m/s"

According to the question,

Ball's initial velocity, u = 10 m/s

Window's height, h = 20 m

(a) By using equation of motion,

Y = ut + [tex]\frac{1}{2}[/tex]at²

By substituting the values,

  = ut + [tex]\frac{1}{2}[/tex]at² + 20

  = 10 × t + 0.5 × (9.8)t² + 20

  = -4.9t² + 10t + 20

(b) We know that,

→ v² - u² = 2as

here, Final velocity, v = 0

0 - (10)² = 2 × (-9.8) × s

          s = [tex]\frac{100}{19.6}[/tex]

             = 5.102 m

(c) Time taken will be:

→ -4.9t² + 10t + 20 = 0

                            t = 3.28 s

hence,

The impact velocity,

v = [tex]\sqrt{2\times 9.8\times 25.102}[/tex]

  = 22.181 m/s

Thus the above response is correct.

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Answer with Explanation:

We are given that

x=-2.6 cm to x=2.6 cm

Linear charge density=[tex]\lambda=5.5nC/m=5.5\times 10^{-9} C/m[/tex]

[tex]1nC=10^{-9} C[/tex]

Length of wire=[tex]2.6-(-2.6)=2.6+2.6=5.2cm[/tex]

Length of wire=[tex]\frac{5.2}{100}=0.052m[/tex]

1 m=100 cm

We know that

a.Linear charge density=[tex]\frac{Q}{L}[/tex]

Where Total charge =Q

Length=L

Total charge,Q=[tex]\lambda L[/tex]

Using the formula

Total charge,Q=[tex]5.5\times 10^{-9}\times 0.052=2.86\times 10^{-10}C[/tex]

b.y=4 cm=[tex]\frac{4}{100}=0.04m[/tex]

1 m=100 cm

Electric field on the y-axis is given by

[tex]E=\frac{2\lambda}{4\pi\epsilon_0 y}(\frac{L}{\sqrt{4y^2+L^2}}[/tex]

[tex]\frac{1}{4\pi\epsilon_0}=9\times 10^9 Nm^2/C^2[/tex]

Using the formula

[tex]E=\frac{2\times 5.5\times 10^{-9}\times 9\times 10^9\times 0.052}{0.04\times \sqrt{4(0.04)^2+(0.052)^2}}[/tex]

[tex]E=1348.8N/C[/tex]

c.y=12 cm=[tex]\frac{12}{100}=0.12m[/tex]

Using the formula

[tex]E=\frac{2\times 5.5\times 10^{-9}\times 9\times 10^9\times 0.052}{0.12\times \sqrt{4(0.12)^2+(0.052)^2}}[/tex]

[tex]E=174.7N/C[/tex]

To find the total charge of the uniform line charge, multiply the linear charge density by the length of the line. To find the electric field at a certain distance on the y axis, use the formula for electric field due to a line charge.

(a) To find the total charge of the line, we need to multiply the linear charge density by the length of the line:

Charge = (Linear charge density)*(Length)

Convert the length to meters:

Length = (2.6 cm + 2.6 cm) = 0.052 m

Total Charge = (5.5 nC/m)*(0.052 m) = 0.286 nC

(b) To find the electric field at y = 4 cm, we can use the formula:

Electric Field = (Linear charge density)/(2πε₀y)

Convert the linear charge density to C/m:

Charge density = 5.5 nC/m = 5.5 x 10-9 C/m

Electric Field = (5.5 x 10-9 C/m)/(2πε₀(0.04 m))

(c) To find the electric field at y = 12 cm, we can use the same formula:

Electric Field = (5.5 x 10-9 C/m)/(2πε₀(0.12 m))

Answer:

Acceleration will be [tex]\alpha =-1.50rad/sec^2[/tex]

Explanation:

We have given initial angular velocity [tex]\omega _i=37rpm[/tex]

In radian/sec initial angular velocity will be [tex]\omega _i=\frac{2\times \pi 37}{60}=3.873rad/sec[/tex]

Angular velocity after 1.6 sec is 14 rpm

So final angular velocity [tex]\omega _f=\frac{2\times \pi\times 14}{60}=1.465rad/sec[/tex]

Time t = 1.6 sec

We have to find the angular angular acceleration

From first equation of motion we know that

[tex]\omega _f=\omega _+\alpha t[/tex]

[tex]1.465=3.873+\alpha \times 1.6[/tex]

[tex]\alpha =-1.50rad/sec^2[/tex] here negative sign indicates that motion is deaccelerative in nature

Answer:

a. 2.0secs

b. 20.4m

c. 4.0secs

d. 141.2m

e. 40m/s, ∅= -30°

Explanation:

The following Data are giving

Initial speed U=40m/s

angle of elevation,∅=30°

a. the expression for the time to attain the maximum height is expressed as

[tex]t=\frac{usin\alpha }{g}[/tex]

where g is the acceleration due to gravity, and the value is 9.81m/s if we substitute values we arrive at

[tex]t=40sin30/9.81\\t=2.0secs[/tex]

b. the expression for the maximum height is expressed as

[tex]H=\frac{u^{2}sin^{2}\alpha }{2g} \\H=\frac{40^{2}0.25 }{2*9.81} \\H=20.4m[/tex]

c. The time to hit the ground is the total time of flight which is twice the time to reach the maximum height ,

Hence T=2t

T=2*2.0

T=4.0secs

d. The range of the projectile is expressed as

[tex]R=\frac{U^{2}sin2\alpha}{g}\\R=\frac{40^{2}sin60}{9.81}\\R=141.2m[/tex]

e. The landing speed is the same as the initial projected speed but in opposite direction

Hence the landing speed is 40m/s at angle of -30°

Answer:

After 1.938 sec velocity of rock will be 36 m/sec

Explanation:

We have given initial velocity at which rock is thrown u = 55 m/sec

Final velocity v = 36 m/sec

Acceleration due to gravity [tex]g=9.8m/sec^2[/tex]

According to first equation of motion we know that [tex]v=u+gt[/tex], here v is final velocity, u is initial velocity, g is acceleration due to gravity and t is time

So [tex]36=55-9.8t[/tex] ( Negative sign is due to rock is thrown upward )

So [tex]9.8t=19[/tex]

t = 1.938 sec

So after 1.938 sec velocity of rock will be 36 m/sec

Using the physics equation of motion and the given initial velocity, reaction time, and deceleration, one can determine whether a truck can stop in time to avoid a collision.

The question focuses on stopping distance and acceleration required to avoid a collision, indicating its base in Physics. If we have a truck moving at a constant velocity and it brakes at a certain distance from an obstacle, the minimum acceleration needed to avoid a collision can be calculated using the equation of motion v^2 = u^2 + 2as. Here, 'v' is the final velocity (0 m/s as the truck needs to stop), 'u' is the initial velocity, 'a' is the acceleration, and 's' is the distance over which the truck needs to stop.

To determine if the truck will hit the child, we must account for the driver's reaction time as well. During this reaction time, the truck continues to travel at its initial speed. After the reaction time, the truck will begin decelerating until it comes to a stop. The total stopping distance is the distance covered during the reaction time plus the distance covered during deceleration. The latter can be found using the deceleration rate and the formula mentioned above.

For the given scenario of the truck with an initial velocity of 10 m/s, a braking distance of 50 m, reaction time of 0.5 seconds, and deceleration of -1.25 m/s^2, we can calculate whether or not the truck will be able to stop in time to avoid hitting the child.

Answer:

The radius of the curve that Car 2 travels on is 380 meters.

Explanation:

Speed of car 1, [tex]v_1=65\ km/h[/tex]

Radius of the circular arc, [tex]r_1=95\ m[/tex]

Car 2 has twice the speed of Car 1, [tex]v_2=130\ km/h[/tex]

We need to find the radius of the curve that Car 2 travels on have to be in order for both cars to have the same centripetal acceleration. We know that the centripetal acceleration is given by :

[tex]a=\dfrac{v^2}{r}[/tex]

According to given condition,

[tex]\dfrac{v_1^2}{r_1}=\dfrac{v_2^2}{r_2}[/tex]

[tex]\dfrac{65^2}{95}=\dfrac{130^2}{r_2}[/tex]

On solving we get :

[tex]r_2=380\ m[/tex]

So, the radius of the curve that Car 2 travels on is 380 meters. Hence, this is the required solution.

Final answer:

For Car 2 to have the same centripetal acceleration while traveling at twice the speed of Car 1, it must travel on a curve with a radius that is four times larger, which would be 380 meters.

Explanation:

The centripetal acceleration (ac) of a car going around a level curve at a constant speed is given by the formula ac = v2 / r, where v is the speed of the car and r is the radius of the circular path. If Car 1 is traveling at 65 km/h and has a centripetal acceleration on a curve with a radius of 95 m, and Car 2 is traveling at twice the speed of Car 1, we are asked to find the required radius of the curve for Car 2 to maintain the same centripetal acceleration.

To keep the same centripetal acceleration for Car 2, which is moving at twice the speed, the radius r2 must be increased proportionally to the square of the speed ratio. Since Car 2 is traveling at twice the speed, the radius must be increased by a factor of 22, or 4 times larger than that for Car 1. Therefore, if Car 1's radius is 95 m, Car 2's radius needs to be 95 m * 4, which is 380 m.

Answer:

t= 0.00364 in

Explanation:

Given that

The volume of the paint ,V= 231 in³

The surface area ,A = 440 ft²

We know that

1 ft  = 12 in

That is why

A= 440 x 12 x 12 in²

A= 63360 in²

Lets take the thickness of the paint = t in

We know that

V= A t

[tex]t=\dfrac{V}{A}[/tex]

[tex]t=\dfrac{231}{63360}\ in[/tex]

t= 0.00364 in

Therefore the thickness ,t= 0.00364 in

To find the average thickness of the paint, divide the volume of paint by the area covered. This results in an average thickness of approximately 0.003645 inches, or about 0.000304 feet when the paint is applied as per the can's instructions.

To find the average thickness of the paint applied, we need to calculate the volume of paint used per unit of area covered. The volume of one gallon of paint is 231 cubic inches, and the coverage is 440 square feet. To convert the coverage to square inches, we multiply by the number of square inches in a square foot, which is 144 (12 inches × 12 inches).

The total coverage in square inches is 440 ft2 × 144 in2/ft2 = 63,360 in2. We can then find the average thickness by dividing the volume of paint by the area covered: Thickness (in inches) = Volume (in cubic inches) / Area (in square inches).

This gives us an average thickness of 231 in3 / 63,360 in2, which simplifies to approximately 0.003645 inches. This can also be converted to feet by knowing that 1 inch is equal to 1/12 of a foot, so the average paint thickness is roughly 0.003645/12 feet when applied as instructed on the paint can.

To calculate the molecular weight of a polyethylene molecule with n=750, multiply the molecular weight of the ethylene unit by n.

To calculate the molecular weight of a polyethylene molecule with n=750, we need to know the molecular formula and the atomic weights of the elements present in the molecule. Polyethylene is made up of repeating ethylene (C2H4) units, so we can calculate the molecular weight of the polyethylene molecule by multiplying the molecular weight of the ethylene unit (28.05 g/mol) by the value of n (750).

Calculation:
Molecular weight of polyethylene = Molecular weight of ethylene unit × n = 28.05 g/mol × 750 = 21,037.5 g/mol

Therefore, the molecular weight of the polyethylene molecule with n=750 is 21,037.5 g/mol, rounded to three significant figures.

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Answer : Yes, theoretical density can be computed from a tiny fundamental unit using the formula [tex]\rho=\frac{Z\times M}{N_{A}\times a^{3}}[/tex].

Explanation :

Nearest neighbor distance, r = [tex]0.124nm=1.24\times 10^{-8}cm[/tex] [tex](1nm=10^{-7}cm)[/tex]

Atomic mass (M) = 55.85 g/mol

Avogadro's number [tex](N_{A})=6.022\times 10^{23} mol^{-1}[/tex]

For BCC = Z = 2

Given density = [tex]7.87g/cm^3[/tex]

First we have to calculate the cubing of edge length of unit cell for BCC crystal lattice.

For BCC lattice : [tex]a^3=(\frac{4r}{\sqrt{3}})^3=(\frac{4\times 1.24\times 10^{-8}cm}{\sqrt{3}})^3=2.35\times 10^{-23}cm^3[/tex]

Now we have to calculate the density of unit cell for BCC crystal lattice.

Formula used :  

[tex]\rho=\frac{Z\times M}{N_{A}\times a^{3}}[/tex] .............(1)

where,

[tex]\rho[/tex] = density

Z = number of atom in unit cell (for BCC = 2)

M = atomic mass

[tex](N_{A})[/tex] = Avogadro's number  

a = edge length of unit cell

Now put all the values in above formula (1), we get

[tex]\rho=\frac{2\times (55.85g/mol)}{(6.022\times 10^{23}mol^{-1}) \times (2.35\times 10^{-23}Cm^3)}=7.89g/Cm^{3}[/tex]

From this information we conclude that, the given density is approximately equal to the given density.

Yes, theoretical density can be computed from a tiny fundamental unit using the formula [tex]\rho=\frac{Z\times M}{N_{A}\times a^{3}}[/tex].

Answer:

a)   F = 1.44 10⁻⁷ N, the charges are of the same sign the force is repulsive

b)    F₂ / F₁ = 16

Explanation:

a) When the balls are touched the load is distributed evenly between the two balls, therefore when separating each ball has a load of

          q₁ = q₂ = ½ 10¹⁰ 1.6 10⁻¹⁹ C

          q₁ = q₂ = 0.8 10⁻⁹ C

As the charges are of the same sign the force is repulsive

To calculate the force let's use Coulomb's law

        F = k q₁ q₂ / r²

        F = 8.99 10⁹  0.8 10⁻⁹  0.8 10⁻⁹ / 0.20²

        F = 1.44 10⁻⁷ N

b) Let us seek strength for the distress of

              r₂ = 0.05 m

             F₂ = 8.99  10⁹ 0.8 0.8 10⁻¹⁸ / 0.05²

             F₂ = 2,301 10⁻⁶ N

b) The relationship between these two forces is

            F₂ / F₁ = 23.01 10⁻⁷ / 1.44 10⁻⁷

            F₂ / F₁ = 16

To solve this problem we will apply the concepts related to centripetal acceleration, which will be the same - by balance - to the force of gravity on the body. To find this acceleration we must first find the orbital velocity through the Doppler formulas for the given periodic signals. In this way:

[tex]v_{o} = c (\frac{\lambda_{max}-\bar{\lambda}}{\bar{\lambda}}})[/tex]

Here,

[tex]v_{o} =[/tex]  Orbital Velocity

[tex]\lambda_{max} =[/tex] Maximal Wavelength

[tex]\bar{\lambda}} =[/tex] Average Wavelength

c = Speed of light

Replacing with our values we have that,

[tex]v_{o} = (3*10^5) (\frac{3.00036-3}{3})[/tex]

Note that the average signal is 3.000000m

[tex]v_o = 36 km/s[/tex]

Now using the definition about centripetal acceleration we have,

[tex]a_c = \frac{v^2}{r}[/tex]

Here,

v = Orbit Velocity

r = Radius of Orbit

Replacing with our values,

[tex]a = \frac{(36km/s)^2}{100000km}[/tex]

[tex]a= 0.01296km/s^2[/tex]

[tex]a = 12.96m/s^2[/tex]

Applying Newton's equation for acceleration due to gravity,

[tex]a =\frac{GM}{r^2}[/tex]

Here,

G = Universal gravitational constant

M = Mass of the planet

r = Orbit

The acceleration due to gravity is the same as the previous centripetal acceleration by equilibrium, then rearranging to find the mass we have,

[tex]M = \frac{ar^2}{G}[/tex]

[tex]M = \frac{(12.96)(100000000)^2}{ 6.67*10^{-11}}[/tex]

[tex]M = 1.943028*10^{27}kg[/tex]

Therefore the mass of the planet is [tex]1.943028*10^{27}kg[/tex]

Answer:

a. [tex]t=19.56 s[/tex]

b.[tex]d=718.34[/tex]

Explanation:

The solution to the differential equation

[tex]\dfrac{dv}{dt}=9.8-\dfrac{v}{5}[/tex]

is the exponential function

[tex]v(t)=ce^{-0.2t}+49[/tex]

and we find [tex]c[/tex] from the initial condition [tex]v(0)=0:[/tex]

[tex]0=ce^{-0.2*0}+49\\\\0=c+49\\\\c=-49[/tex]

Therefore, we have

[tex]v(t)=-49e^{-0.2t}+49[/tex]

[tex]\boxed{ v(t)=49(1-e^{-0.2t})}[/tex]

Part A:

The maximum velocity that the object can reach is 49 (which the maximum value [tex]v(t)[/tex] can have).

Now, 98% of 49 is 48.02; therefore,

[tex]48.02=49(1-e^{-0.2t})[/tex]

[tex]0.98=1-e^{-0.2t}[/tex]

[tex]e^{-0.2t}=0.02[/tex]

[tex]\boxed{t=19.56 s}[/tex]

Part B:

The distance traveled is the integral of the speed:

[tex]d=\int_0^{19.56}v(t)*dt[/tex]

[tex]d=\int^{19.56}_0 {49(1-e^{-0.2t})} \, dt[/tex]

[tex]d=49[t+5e^{-0.2t}]_0^{19.56}[/tex]

[tex]\boxed{d=718.34}[/tex]

To find the time that must elapse for the object to reach 98% of its limiting velocity, we need to solve the differential equation. We can then find the distance the object falls by integrating the velocity function with respect to time.

To find the time it takes for the object to reach 98% of its limiting velocity, we need to solve the differential equation. First, we separate the variables by writing it as:

dv / (9.8 - (v/5)) = dt

Next, we integrate both sides:

∫ (1 / (9.8 - (v/5))) dv = ∫ dt

After evaluating the integrals, we can solve for v:

v = 49 - 49e^(-t/5)

Substituting v with 0.98 times the limiting velocity (which is 49), we can solve for t:

49 - 49e^(-t/5) = 0.98 * 49

Solving this equation will give us the time it takes for the object to reach 98% of its limiting velocity.

To find the distance the object falls, we need to integrate the velocity function, v, with respect to time:

∫ v dt

By evaluating the integral, we can calculate the distance the object falls in the time found in part (a).

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A constellation, in astronomy, is a conventional grouping of stars, whose position in the night sky is apparently invariable. The peoples, generally of ancient civilizations, decided to link them through imaginary strokes, thus creating virtual silhouettes on the celestial sphere. From 1928, the International Astronomical Union (UAI) decided to officially regroup the celestial sphere into 88 constellations with precise limits, such that every point in the sky would be within the limits of a figure. When an astronomer says he saw a comet in Capricorn last night, it means that he saw a comet in the direction of the constellation of Capricorn.

Answer:

Reduced by 49 times

Explanation:

We have Newton formula for attraction force between 2 objects with mass and a distance between them:

[tex]F_G = G\frac{M_1M_2}{R^2}[/tex]

where G is the gravitational constant. [tex]M = M_1 = M_2[/tex] are the masses of the 2 objects. and R is the distance between them.

Since R squared is in the denominator of the formula, if we make it 7 times as large with no change in mass, gravitational force would be dropped by 7*7 = 49 times

To solve the problem we should know about Newton's Law of gravity.

According to Newton's law of gravity, there is an attractive force between any two-particle carrying mass, such that the force is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.

[tex]F \propto m_1m_2\\\\F \propto \dfrac{1}{R^2}[/tex]

[tex]F = G\dfrac{m_1m_2}{R^2}[/tex]

Where G is the proportionality constant and the value of G is 6.67 x 10-11 N m² / kg².

The force between the two will be [tex]\dfrac{1}{49}[/tex] time of the force before.

Given to us,

Assumption

Let's assume that the distance between the moon and the planet is d.

Values

As it is given that there is no change in the mass of the moon or the planet, therefore,

Also, it is given that the distance between them changes to 7 times, therefore,

Substitute the value Newton's Law of gravity,

[tex]F = G\dfrac{m_1m_2}{(7d)^2}\\\\\\F = G\dfrac{m_1m_2}{49d^2}[/tex]

Thus, the force between the two will be [tex]\dfrac{1}{49}[/tex] time of the force before.

Learn more about Newton's Law of gravity:

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Answer:

3.33m/s

Explanation:

The total distance that he runs is

30m north + 30m south + 40m north = 100 m

If the entire run takes 30 seconds then the average speed is distance over a unit of time

100 / 30 = 3.33 m/s

You can draw a sketch of a line going north 2 spaces, then going south by 3 spaces, finally going north again by 4 spaces

The definition of waves that propagate through electric fields is called electromagnetic waves. The earth, despite being covered with clouds, can be 'affected' because waves such as sunlight or the moon have the ability to penetrate and be visible to the inhabitants of the earth. Microwaves and radio waves would be less affected by the clouds that cover the Earth.

Through these waves, you can know that there is beyond the clouds.

Ultraviolet light, microwaves and radio waves are the radiations that penetrate through the clouds and reach the Earth's surface.

Therefore, the answer is Yes, ultraviolet light, microwaves and radio waves are the forms of radiation that penetrate and reach the ground.

It is indeed possible to learn about the universe beyond the clouds due to other non-visual forms of radiation, mainly radio waves and gamma rays, which can penetrate through the clouds and reach the earth's surface.

Yes, even if Earth were completely blanketed with clouds and we could not see the sky, we could still learn about the universe beyond the clouds. This is because, in addition to visible light which would be blocked by the clouds, the universe also emits various other forms of radiation that can penetrate the clouds and reach the ground.

Two major types of radiation that could penetrate the dense clouds are radio waves and gamma rays. Radio waves are a form of electromagnetic radiation used in many areas of science and technology, while gamma rays are highly energetic forms of radiation and are used in fields such as astronomy to get valuable information about distant celestial bodies.

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Answer:

i. +/- 1.43% and +/- 0.71% ii. +/- 0.33%

Explanation:

[tex]% Error = \frac{Error}{Measurement}* 100%\\[/tex]

Answer:

0.0458 kWh

6.5736 cents

Explanation:

The formula for electric energy is given as

E = Pt................. Equation 1

Where E = Electric energy, P = Electric power, t = time.

Given; P = 550 W, t = 5 min = (5/60) h = 0.083 h.

Substituting into equation 1

E = 550(0.083)

E = 45.83 Wh

E = (45.83/1000) kWh

E = 0.0458 kWh.

Hence the kWh = 0.0458 kWh.

If the makes a toast four morning per week, and the are Four weeks in a month.

Total number days he makes toast in a month = 4×4 = 16 days.

t = 16×0.083 h = 1.328 h.

Total energy used in a month = 550(1.328)

E = 730.4 Wh

E = 0.7304 kWh.

If the cost of energy is 9.0 cents per kWh,

Then for 0.7304 kWh  the cost will be 9.0(0.7304) = 6.5736 cents.

Hence this would add 6.5736 cents to his monthly electric bill

A 550-W toaster in operation for 5.0 minutes uses 2.75 kWh of energy. If you make toast four mornings per week, it would add an estimated cost of $4.30 to your monthly electric energy bill.

To calculate the energy used by the toaster, we can use the formula E = Pt, where P is the power and t is the time. In this case, the power of the toaster is 550 watts and the time it is in operation is 5.0 minutes. Plugging these values into the formula, we get E = (550 W)(5.0 min) = 2750 W.min. To convert this to kilowatt-hours (kWh), we need to divide by 1000, so the energy used by the toaster is 2.75 kWh.

To estimate how much this would add to your monthly electric energy bill, we need to know how many times you use the toaster in a month. If you use it four mornings per week, that would be 4 days x 52 weeks / 12 months = 17.33 days per month. Multiplying the energy used by the toaster (2.75 kWh) by the number of days in a month (17.33), we get an estimate of 47.75 kWh per month. Finally, to find the cost, we multiply the energy (47.75 kWh) by the cost per kilowatt-hour (9.0 cents/kWh) and convert it to dollars, giving us an estimated cost of $4.30 per month.

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Answer:

Distance between them increase

Explanation:

The position S of the water droplet can be determined  using equation of motion

[tex]S=ut+\frac{1}{2} at^2[/tex]

where [tex]u[/tex] is the initial velocity which is zero here

[tex]t[/tex] is time taken, [tex]a[/tex] is acceleration due to gravity

the position of  first drop after time [tex]t[/tex] is given by

[tex]S_{1} =0 \times t+ \frac{1}{2} at^2=\frac{1}{2} at^2............(1)[/tex]

the position of  next drop at same time is

[tex]S_{2} =\frac{1}{2} a(t-1)^2 = \frac{1}{2} a(t^2+1-2t)............(2)[/tex]

distance between them is [tex]S_{1} -S_{2}[/tex]  is [tex]a(t-1)[/tex]

from the above the difference will increase with the time

As the water drops fall, their velocity increases due to the force of gravity, which causes the distance between each subsequent drop to increase.

The response to the student's question deals with the notion of acceleration due to gravity. As the water drops fall, they are accelerated by gravity, which means their velocity (speed) increases over time. If we consider two subsequent droplets, the second drop begins its descent 1.0 seconds after the first. Therefore, when the second drop begins to fall, the first drop has already accelerated for 1.0 seconds. This causes the distance between the two drops to increase as they fall.

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Answer:

Power delivered by the source will be 182.912 watt

Explanation:

We have given a load is consist of a resistor of 70 ohm in parallel with [tex]90\mu F[/tex] capacitance

Voltage source is given [tex]v_s(t)=160cos(2000t)[/tex]

So maximum value of voltage source is 160 volt

So rms value [tex]v_{r}=\frac{v_m}{\sqrt{2}}=\frac{160}{1.414}=113.154volt[/tex]

We know that average power delivered by the source will be equal to average power absorbed by the resistor

So power absorbed by the resistor [tex]P=\frac{v_r^2}{R}=\frac{113.154^2}{70}=182.912watt[/tex]

So power delivered by the source will be 182.912 watt

Answer:

Explanation:

Given

side of square shape [tex]a=5\ cm[/tex]

Electric flux [tex]\phi =3\times 10^{-5}\ N.m^2/C[/tex]

Permittivity of free space [tex]\epsilon_0=8.85\times 10^{-12} \frac{C^2}{N.m^2}[/tex]

Flux is given by

[tex]\phi =EA\cos \theta [/tex]

where E=electric field strength

A=area

[tex]\theta [/tex]=Angle between Electric field and area vector

[tex]E=\frac{\phi }{A\cos (0)}[/tex]

[tex]E=\frac{3\times 10^{-5}}{25\times 10^{-4}\times \cos(0)}[/tex]

[tex]E=0.012\ N/C[/tex]

and Electric field  by a uniformly charged sheet is given by

[tex]E=\frac{\sigma }{2\epsilon_0}[/tex]

where [tex]\sigma[/tex]=charge density

[tex]=\frac{\sigma }{\epsilon_0}[/tex]

[tex]\sigma =0.012\times 8.85\times 10^{-12}[/tex]

[tex]\sigma =2.12\times 10^{-13}\ C/m^2[/tex]    

Answer:

Explanation:

Given

Distance between Pluto and sun is 39.1 times more than the distance between earth and sun

According to Kepler's Law

[tex]T^2=kR^3[/tex]

where k=constant

T=time period

R=Radius of orbit

Suppose [tex]R_1[/tex] is the radius of orbit of earth and sun

so Distance between Pluto and sun is [tex]R_2=39.1\cdot R_1[/tex]

[tex]T_1[/tex] and [tex]T_2[/tex] is the time period corresponding to [tex]R_1[/tex] and R_2[/tex]

[tex](T_1)^2=k(R_1)^3---1[/tex]

[tex](T_2)^2=k(R_2)^3---2[/tex]

divide 1 and 2

[tex](\frac{365}{T_2})^2=(\frac{R_1}{39.1})^3[/tex]

[tex]T_2^2=365^2\times 39.1^3[/tex]

[tex]T_2=89239.67\ Earth\ days[/tex]                      

Final answer:

Using Kepler's Third Law of Planetary Motion, we can calculate that it takes Pluto approximately 92,446 Earth days to complete one orbit around the Sun.

Explanation:

To calculate the time taken by Pluto to orbit the Sun, we use Kepler's Third Law of Planetary Motion, which states that the square of the orbital period of a planet is directly proportional to the cube of the semi-major axis of its orbit. Given that the distance between Pluto and the Sun is approximately 39.1 times more than the distance between the Earth and the Sun, we could simplify and use the approximation that Pluto's distance in astronomical units (AU) is roughly 40, as given in the reference information.

Pluto's orbital period (P) can be calculated using the average distance from the Sun (a) with the formula P² = a³. Here, we assume Earth's orbital period to be 1 Earth year and its distance as 1 AU by definition. Therefore, for Pluto:

To convert this period into Earth days, we multiply by the number of days in one Earth year (365.25 days, accounting for the leap year cycle):

Hence, it takes Pluto approximately 92,446 Earth days to complete one orbit around the Sun.

Explanation:

(a) It is known that equation for transverse wave is given as follows.

                 y = [tex](0.09 m)sin(\pi \frac{x}{11} + 4 \pi t)[/tex]

Now, we will compare above equation with the standard form of transeverse wave equation,

                 y = [tex]A sin(kx + \omega t)[/tex]

where,    A is the amplitude = 0.09 m

              k is the wave vector = [tex]\frac{\pi}{11}[/tex]

              [tex]\omega[/tex] is the angular frequency = [tex]4\pi[/tex]

              x is displacement = 1.40 m

              t is the time = 0.16 s

Now, we will differentiate the equation with respect to t as follows.

The speed of the wave  will be:

                   v(t) = [tex]\frac{dy}{dt}[/tex]

                v(t) = [tex]A \omega cos(kx + \omega t)[/tex]

        v(t) = [tex](0.09 m)(4\pi) cos(\frac{\pi \times 1.4}{11} + 4 \pi \times 0.16)[/tex]

          v(t) = -0.84 m/s

The acceleration of the particle in the location is

            a(t) = [tex]\frac{dv}{dt}[/tex]

           a(t) = [tex]-A \omega 2sin(kx + \omega t)[/tex]

           a(t) = [tex]-(0.09 m)(4 \pi)2 sin(\frac{\pi \times 1.4}{11} + 4\pi \times 0.16)[/tex]

           a(t) = -9.49 [tex]m/s^{2}[/tex]

Hence, the value of transverse wave is 0.84 m/s and the value of acceleration is 9.49 [tex]m/s^{2}[/tex] .

(b)  Wavelength of the wave is given as follows.

               [tex]\lambda = \frac{2\pi}{k}[/tex]

              [tex]\lambda = (frac{2\pi}{\frac{\pi}{11}) [/tex]

              [tex]\lambda[/tex] = 22 m

The period of the wave is

             T = [tex]\frac{2 \pi}{\omega}[/tex]

             T = [tex]\frac{2 \pi}{4 \pi}[/tex]

                = 0.5 sec

Now, we will calculate the speed of propagation of wave as follows.

                    v = [tex]\frac{\lambda}{T}[/tex]

                       = [tex]\frac{22 m}{0.5 s}[/tex]

                       = 44 m/s

therefore, we can conclude that wavelength is 22 m, period is 0.5 sec, and speed of propagation of wave is 44 m/s.

To find the transverse speed and acceleration of an element on a string given its wave function and specific values for time and position, we differentiate the wave function with respect to time. The first derivative gives us speed, and the second derivative provides acceleration. To determine the wave's wavelength, period, and propagation speed, we compare the wave function to its standard form and use the wave number and angular frequency.

To determine the transverse speed and acceleration of a string element for the given wave function y = (0.090 m) sin (px/11 + 4pt) at t = 0.160 s and x = 1.40 m, we need to take the first and second derivatives of the wave function with respect to time. We also need to convert the given wave function into SI units for consistency.

First, let's tackle the transverse speed, which is given by the first derivative of the displacement y with respect to time t. Acceleration is given by the second derivative with respect to time.

(a) At t = 0.160 s and x = 1.40 m:
Vy = ∂y/∂t = 4p(0.090m)cos(px/11 + 4pt)
Acceleration, ay = ∂^2 y/∂t^2 = -16π^2(0.090m)sin(px/11 + 4pt)

Now plug in the values of x and t to get the numerical measure of speed and acceleration.

(b) To find the wavelength, period, and wave propagation speed, compare the wave function's format to the standard form y(x, t) = A sin (kx - ωt), where k is the wave number related to wavelength (λ) by k = 2π/λ and ω is the angular frequency related to period (T) by ω = 2π/T. The wave speed (v) is given by v = λ/T.

Identify k and ω from the wave function and calculate the wavelength, period, and wave speed.