How much excess charge must be placed on a copper sphere 25.0 cm in diameter so that the potential of its center, relative to infinity, is 1.50 kV?

Answer:

223 degree

Explanation:

We are given that

Magnitude of resultant vector= 8 units

Resultant vector makes an angle with positive -x in counter clockwise direction

[tex]\theta=43^{\circ}[/tex]

We have to find the magnitude and angle of the equilibrium vector.

We know that equilibrium vector is equal in magnitude  and in opposite direction  to the given vector.

Therefore, magnitude of equilibrium vector=8 units

x-component of a  vector=[tex]v_x=vcos\theta[/tex]

Where v=Magnitude of vector

Using the formula

x-component of resultant  vector=[tex]v_x=8cos43=5.85[/tex]

y-component of resultant vector=[tex]v_y=vsin\theta=8sin43=5.46[/tex]

x-component of equilibrium vector=[tex]v_x=-5.85[/tex]

y-component of equilibrium vector=[tex]-v_y=-5.46[/tex]

Because equilibrium vector lies in III quadrant

[tex]\theta=tan^{-1}(\frac{v_x}{v_y})=tan^{-1}(\frac{-5.46}{-5.85})=43^{\circ}[/tex]

The angle [tex]\theta'[/tex]lies in III quadrant

In III quadrant ,angle =[tex]\theta'+180^{\circ}[/tex]

Angle of equilibrium vector measured from positive x in counter clock wise direction=180+43=223 degree

Answer:

Distance from starting point= 242.249871 miles

Explanation:

Distance covered on bearing of 40 degree=a= [tex]1.4*110\ mph[/tex]

Distance covered on bearing of 40 degree=a=154 miles

Distance covered on bearing of 130 degree=b=[tex]1.7*110\ mph[/tex]

Distance covered on bearing of 40 degree=b=187 miles

Angle between bearing=[tex]130-40[/tex]

Angle between bearing=90 degree

With the angle of 90 degree, Distance from starting point can be calculated from Pythagoras theorem.

Distance from starting point=[tex]\sqrt{a^2+b^2}[/tex]

Distance from starting point=[tex]\sqrt{154^2+187^2}[/tex]

Distance from starting point= 242.249871 miles

Answer:

Distance from starting point= 242.249871 miles

Distance covered on bearing of 40 degree=a=

Distance covered on bearing of 40 degree=a=154 miles

Distance covered on bearing of 130 degree=b=

Distance covered on bearing of 40 degree=b=187 miles

Angle between bearing=

Angle between bearing=90 degree

With the angle of 90 degree, Distance from starting point can be calculated from Pythagoras theorem.

Distance from starting point=

Distance from starting point=

Distance from starting point= 242.249871 miles

I got this from the other person so mark them brainliest

In space, spatial coordinates can be roughly divided into measures of Right ascension and declination. The declination is measured in degrees while the ascent is measured in hours, minutes, seconds. When you have objects in space such as those of the characteristics presented we will have to they are not necessarily close together in the sky because we can find two stars on the same right ascension but on different declination lines (Which means they can be very far apart from each other)

Answer:

a. 6.69m/s

b. y=4.48m

c. t=1.43secs

Explanation:

Data given, acceleration,a=35m/s^2

distance covered,d=64cm=0.64m,

a. to determine the speed, we use the equation of motion

initial velocity,u=0m/s

if we substitute values we arrive at

[tex]v^{2}=u^{2}+2as\\v^{2}=0+2*35*0.64\\v^{2}=44.8m/s\\v=\sqrt{44.8}\\ v=6.69m/s\\[/tex]

b. After taking the shot,the acceleration value is due to gravity i.e a=9.81m/s^2

and the distance becomes (y-2.2) above the ground. When it reaches the maximum height, the final velocity becomes zero and the initial velocity becomes 6.69m/s.

Hence we can write the equation above again

[tex]v^{2}=u^{2}-2a(y-2.2)\\[/tex]

if we substitute values we have

[tex]v^{2}=u^{2}-2a(y-2.2)\\0=6.69^{2}-2*9.81(y-2.2)\\y-2.2=\frac{44.76}{19.62} \\y=2.28+2.2\\y=4.48m[/tex]

c. the time it takes to arrive at 1.83m is obtain by using the equation below

[tex]1.83-2.2=6.69t-\frac{1}{2} *9.81t^{2}\\4.9t^{2}-6.69t-0.37\\using \\t= \frac{-b±\sqrt{b^{2}-4ac} }{2a}\\ where \\a=4.9, b=-6.69, c=-0.37[/tex]

if we insert the values, we solve for t , hence t=1.43secs

(a) The speed of the shot when Sam releases it 6.75 m/s.

(b) The height risen by the shot above the ground is 4.52 m.

(c) The time taken for the shot to return to 1.8 m above the ground is 1.44 s.

The given parameters;

The speed of the shot when Sam releases it is calculated as;

[tex]v^2 = u^2 + 2as\\\\v^2 = 0 + 2a(\Delta h)\\\\v = \sqrt{2a(\Delta h)} \\\\v = \sqrt{2\times 35(0.65)} \\\\v = 6.75 \ m/s[/tex]

The height risen by the shot is calculated as follows;

[tex]v^2 = u^2 + 2gh\\\\at \ maximum \ height , v = 0\\\\0 = (6.75)^2 + 2(-9.8)h\\\\19.6h = 45.56 \\\\h = \frac{45.56}{19.6} \\\\h =2.32 \ m[/tex]

The total height above the ground = 2.20 m + 2.32 m = 4.52 m.

The time taken for the shot to return to 1.8 m above the ground is calculated as follows;

the time taken to reach the maximum height is calculated as;

[tex]h = vt - \frac{1}{2} gt^2\\\\2.32 = 6.75t - (0.5\times 9.8)t^2\\\\2.32 = 6.75t -4.9t^2\\\\4.9t^2 -6.75t + 2.32=0\\\\a = 4.9, \ b = -6.75, \ c = 2.32\\\\t = \frac{-b \ \ + /- \ \ \sqrt{b^2-4ac} }{2a} \\\\t = \frac{-(-6.75) \ +/- \ \ \sqrt{(-6.75)^2 - 4(4.9\times 2.32)} }{2(4.9)} \\\\t = 0.72 \ s\ \ or \ \ 0.66 \ s[/tex]

[tex]t \approx 0.7 \ s[/tex]

height traveled downwards from the maximum height reached = 4.52 m - 1.8 m = 2.72 m

[tex]h = vt + \frac{1}{2} gt^2\\\\h = 0 + \frac{1}{2} gt^2\\\\t = \sqrt{\frac{2h}{g} } \\\\t = \sqrt{\frac{2(2.72)}{9.8} } \\\\t = 0.74 \ s[/tex]

The total time spent in air;

[tex]t = 0.7 \ s \ + \ 0.74 \ s\\\\t = 1.44 \ s[/tex]

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The magnitude of the acceleration of the box is [tex]a=6.41\ m/s^2.[/tex]

First, we need to calculate the force between the two charges using Coulomb's law:

[tex]\[ F = \frac{qQ}{4\pi\epsilon_0r^2} \][/tex]

Given that [tex]\( q = +2.50 \, \mu C = 2.50 \times 10^{-6} \, C \), \( Q = +75.0 \, \mu C = 75.0 \times 10^{-6} \, C \)[/tex], and [tex]\( r = 61.0 \, cm = 0.610 \, m \)[/tex], we can plug these values into the equation:

[tex]\[ F = \frac{(2.50 \times 10^{-6} \, C)(75.0 \times 10^{-6} \, C)}{4\pi(8.85 \times 10^{-12} \, F/m)(0.610 \, m)^2} \][/tex]

Solving this, we get the force [tex]\( F \)[/tex] in newtons.

Next, we need to find the component of this force parallel to the incline, which is given by:

[tex]\[ F_{\parallel} = F \sin(\theta) \][/tex]

However, since the box is on an incline, we use [tex]\( \tan(\theta) \)[/tex] instead of [tex]\( \sin(\theta) \)[/tex] to find the acceleration:

[tex]\[ F_{\parallel} = F \tan(\theta) \][/tex]

Given [tex]\( \theta = 35.0^{\circ} \)[/tex], we can calculate [tex]\( F_{\parallel} \)[/tex].

Finally, the acceleration \( a \) of the box is given by Newton's second law:

[tex]\[ a = \frac{F_{\parallel}}{m} \][/tex]

where[tex]\( m = 495 \ , g = 0.495 \, kg \)[/tex].

Combining all the steps, we have:

[tex]\[ a = \frac{qQ}{4\pi\epsilon_0mr^2} \tan(\theta) \][/tex]

Plugging in the values, we get:

[tex]\[ a = \frac{(2.50 \times 10^{-6} \, C)(75.0 \times 10^{-6} \, C)}{4\pi(8.85 \times 10^{-12} \, F/m)(0.495 \, kg)(0.610 \, m)^2} \tan(35.0^{\circ}) \][/tex]

Calculating the value of [tex]\( a \)[/tex], we find the magnitude of the acceleration of the box.

[tex]a=6.41\ m/s^2[/tex]

Answer:

Both hit at the same time

Explanation:

If air resistant is ignored, then gravitational acceleration g is the only thing that affect the bullets vertical motion, no matter what their horizontal motion is. Since both bullets are starting from rest, vertically speaking their speeds are 0 initially, they are both subjected to the same acceleration g, then they travel at the same rate and would reach the ground at the same time.

Answer:

Explanation:

Given

initial Pressure [tex]P_1=90\ psia[/tex]

elevation [tex]z_1=16\ ft[/tex]

Final Pressure [tex]P_2=30\ psia[/tex]

elevation [tex]z_2=450\ ft[/tex]

Pressure after Pumping(pump inlet pressure ) is given by

[tex]P_{after\ pump}=P_{top}+\Delta P[/tex]

[tex]\Delta P=\rho gh[/tex]

where [tex]\rho [/tex]=density of water[tex](62.2\ lbm\ft^3)[/tex]

g=acceleration due to gravity[tex](32.2\ ft/s^2)[/tex]

h=elevation

[tex]\Delta P=62.2\times 32.2\times (16+450)\times \frac{1\ lbf}{32.174\ lbm\ ft}[/tex]

[tex]\Delta P=28,985\ lbf/ft^2[/tex]

[tex]\Delta P=201.3\ lbf/in.^2[/tex]

[tex]P_{after\ pump}=30+201.3=231.3\ lbf/in.^2[/tex]

Pressure required to be applied

[tex]\Delta P_{pump}=P_{after\ pump}-P_{bottom}[/tex]

[tex]\Delta P_{pump}=231.3-90=141.3\ psi[/tex]                  

Answer:

In New York is at ocean level while Denver is at an altitude of 1 mile from ocean level. In this way, the breaking point of water is lower in Denver than in New York, that is, the water will boil at lower temperatures in Denver than in New York. If the breaking point of the water decreases, at this point it will put aside more effort to cook an egg. Now the time required to cook the egg is higher in Denver than in New York.

An egg cooks slower in Denver because water boils at a lower temperature due to the city's higher altitude and lower atmospheric pressure. This lower boiling point decreases the rate of the cooking chemical reactions.

An egg cooks more slowly in boiling water in Denver than in New York City due to differences in atmospheric pressure and the effect this has on boiling point temperatures. Denver, known as the Mile-High City, is approximately one mile above sea level. This altitude results in a lower atmospheric pressure than New York City which is virtually at sea level. Because of this, water boils at a slightly lower temperature in Denver (approximately 202 degrees Fahrenheit) than it does in New York City (approximately 212 degrees Fahrenheit).

Since cooking is essentially a series of chemical reactions, and these reactions occur faster at higher temperatures, an egg will cook more slowly in Denver’s boiling water than in New York's. This is because the boiling water in Denver is at a lower temperature due to the effect of pressure on boiling point.

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Answer:

[tex]r=44m[/tex]

Explanation:

β is calculated as:

[tex]\beta =(10dB)log_{10}(I/I_{o} )\\ I=I_{o}10^{\frac{\beta }{10dB} }\\ I=(1.0*10^{-12}W/m^{2} )10^{\frac{\(94dB }{10dB} }\\I=2.51mW/m^{2}[/tex]

The distance r is defined as the radius of spherical wave.solve for r

We have

[tex]I=\frac{P_{source} }{4\pi r^{2} }\\ r=\sqrt{\frac{P_{source}}{4\pi I} }\\ r=\sqrt{\frac{60W}{4\pi (2.51mW/m^{2} )} }\\r=44m[/tex]

Answer:

d. the same within the uncertainty of each measurement method

Explanation:

The density of an object and in general any physical property, has the same value regardless of the method used to measure it, either directly or indirectly. Since two completely different valid methods are used, the results must be the same, taking into account the level of precision of each of the methods.

The height and distance for 30 degrees are 14.4 m and 100 m, for 45 degrees are 28.77 m and 115.11 m, and for 60 degrees are 43.17 m and 100 m, respectively.

In projectile motion, consider the motion along vertical and horizontal separately.

Given:

Initial velocity, [tex]u =33.6\ m/s\\[/tex]

For angle [tex]30^o[/tex], the initial and final velocities are calculated as:

[tex]u_x =33.6\times cos(30.0)\\=29.14m/s\\u_y =33.6\times sin30.0\\ =16.80m/s[/tex]

The height and distance traveled are computed as:

[tex]h=\frac{u_y^2}{2g}\\=\frac{16.80^2}{2\times9.8}\\=14.4\ m\\R= \frac{u^2sin^22\theta}{g}\\= \frac{2\times16.80\times29.14}{9.8}\\=100\ m[/tex]

For angle [tex]45^o[/tex], the initial and final velocities are calculated as:

[tex]u_x =33.6\times cos(45.0)\\=23.75m/s\\u_y =33.6\times sin45\\ =23.75m/s[/tex]

The height and distance traveled are computed as:

[tex]h=\frac{u_y^2}{2g}\\=\frac{23.75^2}{2\times9.8}\\=28.77\ m\\R= \frac{u^2sin^22\theta}{g}\\= \frac{2\times23.75\times23.75}{9.8}\\=115.11\ m[/tex]

For angle [tex]60^o[/tex], the initial and final velocities are calculated as:

[tex]u_x =33.6\times cos(60)\\=16.8m/s\\u_y =33.6\times sin60\\ =29.09m/s[/tex]

The height and distance traveled are computed as:

[tex]h=\frac{u_y^2}{2g}\\=\frac{29.09^2}{2\times9.8}\\=43.17\ m\\R= \frac{u^2sin^22\theta}{g}\\= \frac{2\times16.80\times29.09}{9.8}\\=100\ m[/tex]

Therefore, the distance and height for 30 degrees are 14.4 m and 100 m for 45 degrees are 28.77 m and 115.11 m, and for 60 degrees are 43.17 m and 100 m, respectively.

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To calculate the height and distance traveled by a baseball hit at different angles, we can use the equations of projectile motion. The height can be calculated using the formula h = (v²sin²θ) / (2g), where v is the initial velocity, θ is the launch angle, and g is the acceleration due to gravity. The distance traveled can be calculated using the formula d = v²sin(2θ) / g.

To calculate the height and distance traveled by a baseball hit at different angles, we can use the equations of projectile motion. The height can be calculated using the formula h = (v²sin²θ) / (2g), where v is the initial velocity, θ is the launch angle, and g is the acceleration due to gravity (approximately 9.8 m/s²). Once we have the height, we can calculate the distance traveled using the formula d = v²sin(2θ) / g.

Height (h) = (33.6²sin²30.0°) / (2 * 9.8) = 19.22 meters

Distance (d) = 33.6²sin(2 * 30.0°) / 9.8 = 152.19 meters

Height (h) = (33.6²sin²45.0°) / (2 * 9.8) = 38.45 meters

Distance (d) = 33.6²sin(2 * 45.0°) / 9.8 = 203.43 meters

Height (h) = (33.6²sin²60.0°) / (2 * 9.8) = 57.67 meters

Distance (d) = 33.6²sin(2 * 60.0°) / 9.8 = 228.05 meters

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Answer:

The initial speed of the ball is 30 m/s.

Explanation:

It can be assumed that the ball is thrown at an angle of 45 degrees to the ground. The ball lands 90 m away. We need to find the initial speed of the ball. We know that the horizontal distance covered by the projectile is called its range. It is given by :

[tex]R=\dfrac{u^2\ sin2\theta}{g}[/tex]

u is the initial speed of the ball.

[tex]v=\sqrt{\dfrac{Rg}{sin2\theta}}[/tex]

[tex]v=\sqrt{\dfrac{90\times 9.8}{sin2(45)}}[/tex]

v = 29.69 m/s

or

v = 30 m/s

So, the initial speed of the ball is 30 m/s. Hence, this is the required solution.                                      

The exact initial speed requires the angle [tex]\( \theta \)[/tex]; assuming [tex]\( \theta = 45^\circ \)[/tex], the initial speed is 29.71 m/s.

To solve this problem, we need to use the kinematic equations for projectile motion. The horizontal distance (range) \( R \) that the ball travels can be found using the following equation:

[tex]\[ R = \frac{{v_0^2 \sin(2\theta)}}{g} \][/tex]

Given that the ball lands 90 m away and assuming that the angle of projection [tex]\( \theta \)[/tex] is known (but not provided in the question), we can rearrange the equation to solve for [tex]\( v_0 \)[/tex]:

[tex]\[ v_0^2 = \frac{R \cdot g}{\sin(2\theta)} \][/tex]

[tex]\[ v_0 = \sqrt{\frac{R \cdot g}{\sin(2\theta)}} \][/tex]

However, since the angle [tex]\( \theta \)[/tex] is not given, we cannot calculate the exact initial speed [tex]\( v_0 \)[/tex] without additional information. The problem as stated is incomplete because it requires the value of [tex]\( \theta \)[/tex] to find [tex]\( v_0 \)[/tex].

If we assume that the angle [tex]\( \theta \)[/tex] is such that [tex]\( \sin(2\theta) \)[/tex] is maximized (which occurs at [tex]\( \theta = 45^\circ \)[/tex], then [tex]\( \sin(2 \cdot 45^\circ) = \sin(90^\circ) = 1 \)[/tex], and the equation simplifies to:

[tex]\[ v_0 = \sqrt{R \cdot g} \][/tex]

[tex]\[ v_0 = \sqrt{90 \, \text{m} \cdot 9.81 \, \text{m/s}^2} \][/tex]

[tex]\[ v_0 = \sqrt{882.9 \, \text{m}^2/\text{s}^2} \][/tex]

[tex]\[ v_0 =29.71 \, \text{m/s} \][/tex]

 Therefore, under the assumption that [tex]\( \theta = 45^\circ \)[/tex], the initial speed of the ball would be [tex]\( 29.71 \, \text{m/s} \)[/tex].

In conclusion, without the value of [tex]\( \theta \)[/tex], we cannot determine the exact initial speed of the ball. However, if we assume the most favorable condition for maximum range, where [tex]\( \theta = 45^\circ \)[/tex], the initial speed [tex]\( v_0 \)[/tex] would be [tex]\( 29.71 \, \text{m/s} \)[/tex].

Answer:

[tex]Q=4704000J\\Q=4.7MJ[/tex]

Explanation:

Given data

Heat capacity c=4.2 J/gC

Water weigh m=1000g

Temperature Risen T=20°C to 90°C

To find

Heat Q

Solution

The formula to find Heat is:

Q=mcΔT

Where

m is mass

c is heat capacity

ΔT is change in temperature

So

[tex]Q=4.2(16*1000)*(90-70)\\Q=4704000J\\Q=4.7MJ[/tex]

The automobile cooling system would absorb approximately 4,699,840 Joules of heat when its temperature rises from 20°C to 90°C, according to the formula for heat absorption Q = mcΔT.

To answer this question, we apply the formula for heat absorption: Q = mcΔT.

 Here, 'Q' is the total heat absorbed, 'm' is the mass of the substance (water in this case), 'c' is the specific heat capacity of the substance, and 'ΔT' is the change in temperature. The mass can be defined as the volume of water multiplied by the density of water (1 g/cm^3). So for 16L, m = 16000g. The specific heat capacity of water is approximately 4.186 J/g °C, and the change in temperature, ΔT, is 70°C (90°C - 20°C).

Plugging in these values, we get Q = 16000g× 4.186 J/g °C × 70°C = 4,699,840 Joules. Therefore, it's estimated that the automobile cooling system would absorb around 4,699,840 Joules of heat.

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Answer:

They would be [tex]r=9.7\,km [/tex] apart

Explanation:

Electric force between two charged objects is:

[tex] F_{e}=k\frac{\mid q_{1}q_{2}\mid}{r^{2}}[/tex] (1)

With q1, q2 the charges of the humans, r the distance between them and k the constant [tex] k=9.0\times10^{9}\,\frac{Nm^{2}}{C^{2}}[/tex], so if we want the electric force between them will be equal to their 600 N weight, we should make W=Fe=600 N on (1):

[tex]600=k\frac{\mid q_{1}q_{2}\mid}{r^{2}} [/tex]

solving for r:

[tex]r^{2}=k\frac{\mid q_{1}q_{2}\mid}{600}[/tex]

[tex] r=\sqrt{k\frac{\mid q_{1}q_{2}\mid}{600}}[/tex]

[tex]r=\sqrt{(9.0\times10^{9})\frac{\mid(-2.5)(2.5)\mid}{600}} =9682 m[/tex]

[tex]r=9.7\,km [/tex]

Answer:

Th steps is as shown in the attachment

Explanation:

from the diagram, its indicates that twelve identical charges are distributed evenly on the circumference of the circle. assuming one of gthe charge is shifted to the centre of he circle alomng the x axis, as such the charge is unbalanced and there is need ot balanced all the identical charges for the net force to be equal to zero.

The mathematical interpretation is as shown in the attachment.

Answer:

The distance traveled by the sperm whale is 12.46 meters.

Explanation:

Given that,

Acceleration of a sperm whale, [tex]a=0.12\ m/s^2[/tex]

Initial speed of the whale, u = 1.1 m/s

Final speed of the whale, v = 2.05 m/s

We need to find the distance traveled by the sperm whale. It can be calculated using third equation of motion as :

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

d is the distance covered

[tex]d=\dfrac{v^2-u^2}{2a}[/tex]

[tex]d=\dfrac{(2.05)^2-(1.1)^2}{2\times 0.12}[/tex]

d = 12.46 meters

So, the distance traveled by the sperm whale is 12.46 meters. Hence, this is the required solution.

The ratio of the electric force on the bee to the bee’s weight (Fe/Fg) is; 2.041 * 10⁻⁶

The electric field strength is; 4.9 * 10⁷ N/C

We are given;

Mass of bee; m = 0.12g = 0.00012 kg

Charge acquired by the bee q = +24 pC = 24 * 10⁻¹² C

Electric field; E = 100 N/C

B)  The force F on a charge in electric field E is given by:

F = qE

F = 24 * 10⁻¹² * 100

F = 24 * 10⁻¹⁰ N

Weight is;

F_g = mg

F_g = 0.00012 * 9.8

F_g = 11.76 × 10⁻⁴ N

Ratio of force to weight is;

F_e/F_g = (24 * 10⁻¹⁰)/(11.76 × 10⁻⁴)

F_e/F_g = 2.041 * 10⁻⁶

C) To get the electric field strength, we will use the formula;

E' = mg/q

E' = (0.00012 * 9.8)/(24 * 10⁻¹²)

E' = 4.9 * 10⁷ N/C

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Final answer:

The honeybee gained 15,000 electrons. The ratio of the electric force on the bee to its weight is (100 N/C) x (24 x 10⁻¹²  C) / [(0.12 g) x (9.8 m/s²)]. To hang suspended in the air without effort, the electric field strength should have the same magnitude as the gravitational field strength (100 N/C downward), but in the opposite direction (upward).

Explanation:

(a) How many electrons were added to or removed from the honeybee?

To determine the number of electrons added to or removed from the honeybee, we can use the formula:

q = ne

Where q is the charge in Coulombs, n is the number of electrons, and e is the elementary charge, which is the charge of a single electron.

Given that the charge acquired by the honeybee is +24 pC (pC = picocoulombs = 10⁻¹²  C), we can substitute the values into the formula:

24 x  10⁻¹² C = n x (1.6 x 10⁻¹⁹ C)

Solving for n, we find:

n = 15,000 electrons

Therefore, the honeybee gained 15,000 electrons.

(b) What is the ratio of the electric force on the bee to the bee’s weight (Fe/Fg)?

To calculate the ratio of the electric force on the bee to its weight, we can use the formula:

F = Eq

Where F is the force, E is the electric field strength, and q is the charge.

Given that the electric field near the surface of the Earth is 100 N/C downward, we can substitute the values into the formula:

F = (100 N/C) x (24 x 10⁻¹² C)

Using the weight formula Fg = mg, where m is the mass and g is the acceleration due to gravity, we can find the weight of the bee.

Fg = (0.12 g) x (9.8 m/s²)

Taking the ratio of Fe to Fg, we get:

Fe/Fg = (100 N/C) x (24 x 10⁻¹² C) / [(0.12 g) x (9.8 m/s²)]

(c) What electric field strength and direction would allow the bee to hang suspended in the air without effort?

The electric field needed for the bee to hang suspended in the air without any effort would need to balance the gravitational force acting on the bee. The gravitational force on the bee can be calculated using the formula Fg = mg, where m is the mass of the bee and g is the acceleration due to gravity. We can then set this force equal to the electric force using the formula F = Eq. Rearranging the formula, we get E = F/q.

Since we want the bee to hang suspended without any effort, the electric force needs to be equal in magnitude and opposite in direction to the gravitational force. This means the electric field strength should have the same magnitude as the gravitational field strength (100 N/C downward), but in the opposite direction (upward).

Answer:

They both move towards South direction.

Explanation:

NB: The car traveling due east must be backing west and the car traveling due south must be backing north.

From their momentum in calculation below:

Momentum of the car traveling east is MU= 1500x90x1000/3600= 37,500kgm/s.

Momentum of car traveling due South =MU = 3000x60x1000/3600= 50,000. So, from here we see the the momentum of the car moving south has the high momemtum so they stick together and move towards the direction of south.

3 trams must be added

Explanation:

In this problem, there are 12 trams along the ring road, spaced at regular intervals.

Calling L the length of the ring road, this means that the space between two consecutive trams is

[tex]d=\frac{L}{12}[/tex] (1)

In this problem, we want to add n trams such that the interval between the trams will decrease by 1/5; therefore the distance will become

[tex]d'=(1-\frac{1}{5})d=\frac{4}{5}d[/tex]

And the number of trams will become

[tex]12+n[/tex]

So eq.(1) will become

[tex]\frac{4}{5}d=\frac{L}{n+12}[/tex] (2)

And substituting eq.(1) into eq.(2), we find:

[tex]\frac{4}{5}(\frac{L}{12})=\frac{L}{n+12}\\\rightarrow n+12=15\\\rightarrow n = 3[/tex]

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Answer: 3

Explanation:

We need to do (1/(5-1))x12

Answer:

[tex]q = 2.17\times 10^{-15}~{\rm C}[/tex]

Explanation:

The relation between the acceleration and the electrical force between the spheres can be found by Newton's Second Law.

[tex]\vec{F} = m\vec{a}\\\vec{F} = (1.4\times 10^{-3})(110) = 0.154~N[/tex]

This is equal to the Coulomb's Force.

[tex]F = \frac{1}{4\pi\epsilon_0}\frac{q_1 q_2}{r^2} = \frac{1}{4\pi\epsilon_0}\frac{2q}{(1.6 \times 10^{-2})^2} = 0.154\\q = 2.17\times 10^{-15}~C[/tex]

Final answer:

To maintain the total charge below 953 µC for capacitors connected in parallel to a 3.00 V battery, a maximum of 45 capacitors, each with a capacitance of 7.00 µF, can be used. This is calculated by dividing the total charge by the charge per capacitor (953 µC / 21 µC per capacitor).

Explanation:

To find out the maximum number of 7.00 µF capacitors that can be connected in parallel while keeping the total charge under 953 µC, we use the formula for charge on a capacitor, Q=CV, where C is the capacitance and V is the voltage. Since the capacitors would be connected in parallel, the voltage across each capacitor would be the same as the voltage of the battery, which is 3.00 V.

The total charge Q for one capacitor is given by:

Q = C × V = 7.00 µF × 3.00 V = 21.00 µC per capacitor.

To find out how many such capacitors we could connect in parallel without exceeding the total charge of 953 µC, we divide the total permissible charge by the charge per capacitor:

Number of capacitors = Total charge / Charge per capacitor = 953 µC / 21.00 µC per capacitor ≈ 45.38.

Since we can't have a fraction of a capacitor, the maximum number you can use is 45 capacitors.

The maximum number of 7.00 µF capacitors that can be connected in parallel with a 3.00 V battery while keeping the total charge stored under 953 μC is 45 capacitors.

To determine the maximum number of 7.00 µF capacitors that can be connected in parallel with a 3.00 V battery while keeping the total charge stored within the capacitor array under 953 μC, follow these steps:

Since the number of capacitors must be an integer, the maximum number of 7.00 µF capacitors that can be connected in parallel is 45 capacitors.

Answer:

Explanation:

Given

Altitude from which Diver steps out [tex]H=4000\ m[/tex]

altitude at which diver opens the Parachutes [tex]h=760\ m[/tex]

total height covered by diver[tex]=H-h=4000-760=3240\ m[/tex]

Initial velocity of diver is zero

using equation of motion

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

where s=displacement

u=initial velocity

a=acceleration

t=time

[tex]3240=0+\frac{1}{2}\times 9.8\times t^2[/tex]

[tex]t^2=661.224[/tex]

[tex]t=25.71\ s[/tex]

Answer:

884 balloons

Explanation:

Assume ideal gas, since temperature is constant, then the product of pressure and volume is constant.

So if pressures reduces from 100 to 1.2, the new volume would be

[tex]V_2 = \frac{P_1V_1}{P_2} = \frac{100*0.15}{1.2} = 12.5 m^2[/tex]

The spherical volume of each of the balloon of 30cm diameter (15 cm or 0.15 m in radius) is

[tex]V_b = \frac{4}{3}\pir^3 = \frac{4}{3}\pi 0.15^3 = 0.014 m^3[/tex]

The number of balloons that 12.5 m3 can fill in is

[tex]V_2/V_b = 12.5 / 0.014 = 884[/tex]

Final answer:

Based on the volume of the tank, temperature, and pressure, we can calculate the number of moles of helium. By dividing this number by the number of moles of helium in one balloon, we can determine how many balloons can be blown up.

Explanation:

To find out how many balloons can be blown up with the given amount of helium, we need to calculate the total volume of helium in the tank and divide it by the volume of each balloon.

Given:

The volume of the tank is 0.150 m³

The temperature of the helium in the tank is 27.0°C

The pressure of the helium in the tank is 100 atm

The volume of each balloon is 30.0 cm in diameter, which is equivalent to a radius of 15.0 cm or 0.15 m

The temperature of the filled balloon is 27.0°C

The absolute pressure of the filled balloon is 1.20 atm

First, we need to convert the volume of the tank to liters:

0.150 m³ * 1000 L/m³ = 150 L

Next, we need to calculate the number of moles of helium in the tank using the ideal gas law:

P * V = n * R * T

n = (P * V) / (R * T)

where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin.

Using the given values:

Pressure (P) = 100 atm

Volume (V) = 150 L

Ideal gas constant (R) = 0.0821 L·atm/K·mol

Temperature (T) = 27.0°C + 273.15 = 300.15 K

Calculating the number of moles:

n = (100 atm * 150 L) / (0.0821 L·atm/K·mol * 300.15 K) = 6.19 moles

Now we can calculate the number of balloons that can be blown up:

Number of balloons = (Number of moles of helium) / (Number of moles of helium in one balloon)

The number of moles of helium in one balloon can be found using the ideal gas law:

P * V = n * R * T

n = (P * V) / (R * T)

Using the given values:

Pressure (P) = 1.20 atm

Volume (V) = (4/3) * π * (0.15 m)³ = 0.141 m³

Ideal gas constant (R) = 0.0821 L·atm/K·mol

Temperature (T) = 27.0°C + 273.15 = 300.15 K

Calculating the number of moles:

n = (1.20 atm * 0.141 m3) / (0.0821 L·atm/K·mol * 300.15 K) = 0.006 moles

Finally, calculating the number of balloons:

Number of balloons = 6.19 moles / 0.006 moles = 1031

Therefore, 1031 balloons can be blown up with the given amount of helium.

Answer:

[tex]v = 21.77\ m/s[/tex]

Explanation:

Given,

Air resistance, = 0.45 m/s² for each 1 m/s

Air resistance for velocity v = 0.45 v.

Terminal velocity = ?

acceleration due to gravity= g = 9.8 m/s²

 now,

a = 0.45 v

we know,

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

[tex]\dfrac{dv}{dt}=0.45 v[/tex]

[tex]\dfrac{dv}{v} = 0.45 dt[/tex]

integrating both side

[tex]\int \dfrac{dv}{v} = 0.45\int dt[/tex]

[tex] ln(v) = 0.45 t[/tex]

[tex]t = e^{0.45 t}[/tex]

When a body is moving at terminal velocity, Force due to gravity is equal to  force due to air resistance.

 m a = m g

    0.45 v = 9.8

     [tex]v = \dfrac{9.8}{0.45}[/tex]

    [tex]v = 21.77\ m/s[/tex]

Hence, the terminal velocity of the skydiver is equal to 21.77 m/s.

Answer:

V= 12.56 m/s

Explanation:

Given that

R= 1 m

The angular speed ,ω = 2 rev/s

We know that 1 rev = 2 π rad

Therefore

ω = 4 π rev/s

The linear speed V is given as

V= ω x R

V = 4 π x 1 m/s

V= 4 π m/s

Therefore the speed of the bicycle will be 4 π m/s

We know that , π = 3.14

V= 12.56 m/s

The linear speed will be 12.56 m/s

Answer:

[tex]heigth=2.86m[/tex]

Explanation:

Given data

time=0.19 s

distance=1.6 m

To find

height

Solution

First we need to find average velocity

[tex]V_{avg}=\frac{distance}{time}\\V_{avg}=\frac{1.6m}{0.19s}\\V_{avg}=8.42m/s[/tex]

Also we know that average velocity

[tex]V_{avg}=(V_{i}+V_{f})/2\\[/tex]

Where

Vi is top of window speed

Vf is bottom of window speed

Also we now that

[tex]V_{f}=V_{i}+gt\\V_{f}=V_{i}+(9.8)(0.19)\\V_{f}=V_{i}+1.862[/tex]

Substitute value of Vf in average velocity

So

[tex]V_{avg}=(V_{i}+V_{f})/2\\where\\V_{f}=V_{i}+1.862\\and\\V_{avg}=8.42m/s\\So\\8.42m/s=(V_{i}+V_{i}+1.862)/2\\2V_{i}+1.862=16.84\\V_{i}=(16.84-1.862)/2\\V_{i}=7.489m/s\\[/tex]

Vi is speed of balloon at top of the window

Now we need to find time

So

[tex]V_{i}=gt\\t=V_{i}/g\\t=7.489/9.8\\t=0.764s[/tex]

So the distance can be found as

[tex]distance=(1/2)gt^{2}\\ distance=(1/2)(9.8)(0.764)^{2}\\ distance=2.86m[/tex]

To determine the height from which the balloon was dropped, we can use the equation for vertical motion: h = 0.5*g*t^2, where h is the height, g is the acceleration due to gravity, and t is the time of flight. Given that the balloon takes 0.19 s to cross the 1.6 m high window, we can solve for the initial height to be approximately 0.01485 m or 14.85 cm above the top of the window.

To determine the height from which the balloon was dropped, we can use the equation for vertical motion: h = 0.5*g*t^2, where h is the height, g is the acceleration due to gravity, and t is the time of flight.

Given that the balloon takes 0.19 s to cross the 1.6 m high window, we can plug in these values into the equation to solve for the initial height:

1.6 = 0.5*9.8*(0.19)^2

Simplifying the equation, we find that the balloon was dropped from a height of approximately 0.01485 m or 14.85 cm above the top of the window.

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The resolution of an ADC with a word length of 12 bits and an input range of 100V is approximately 0.0244V. Calculating this gives us a resolution of approximately 0.0244V.

The resolution of an analog-to-digital converter (ADC) is determined by the number of bits used to represent the digital output.

In this case, the ADC has a word length of 12 bits.

The resolution can be calculated using the formula:

Resolution = Full Scale Range / (2^Word Length)

In this case, the Full Scale Range is 100V. Plugging in the values:

Resolution = 100V / (2^12)

Calculating this gives us a resolution of approximately 0.0244V.

Answer:

a.Velocity=[tex]\vec{v}=(6t)\hat{j}+8\hat{k}[/tex]

b.[tex]\vec{a}=6\hat{j}[/tex]

Explanation:

We are given that

A particle moves so that its position (in m) as  function of time is

[tex]\vec{r}=2\hat{i}+(3t^2)\hat{j}+(8t)\hat{k}[/tex]

a.We have to find its velocity

We know that

Velocity,[tex]v=\frac{dr}{dt}[/tex]

Using the formula

Velocity,v=[tex]\frac{d(2i+3t^2j+8tk)}{dt}[/tex]

Velocity=[tex]\vec{v}=(6t)\hat{j}+8\hat{k}[/tex]

b.Acceleration[tex]=\vec{a}=\frac{d\vec{v}}{dt}[/tex]

[tex]\vec{a}=\frac{d((6t)\hat{j}+8\hat{k})}{dt}[/tex]

[tex]\vec{a}=6\hat{j}[/tex]

The velocity and acceleration of a particle can be calculated from its position function using differentiation in respect to time. The first derivative gives the velocity, while the second derivative gives the acceleration. Unit vector notation is used to display the results.

The question involves the calculations of a particle's velocity and acceleration over time, represented in unit vector notation. This scenario falls within the realm of physics, specifically kinematics. Given a particle's motion described by a position function, we can calculate its velocity and acceleration as functions of time.

V(t), the velocity of the particle, is the first derivative of the position function. Hence, to find the velocity, we simply differentiate the position function with respect to time. Now, to calculate A(t), the acceleration of the particle, we take the first derivative of the velocity function, or equivalently, the second derivative of the position function. This gives us the rate of change of velocity, which represents acceleration.

For example, suppose the particle's position function r(t) is given by r(t) = 3t^2 + 2 squares. Differentiating the position function once gives us v(t) = 6t + 2, the velocity function. Differentiating v(t) gives us a(t) = 6, the acceleration function. Both velocity and acceleration are given in unit vector notation.

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Answer:

Number of electrons, [tex]n=9.98\times 10^{19}[/tex]

Explanation:

A strong lightning bolt transfers an electric charge of about 16 C to Earth, q = 16 C

We need to find the number of electrons that transferred. Let there are n electrons transferred. It is given by using quantization of electric charge as :

q = ne

[tex]n=\dfrac{q}{e}[/tex]

e is elemental charge

[tex]n=\dfrac{16}{1.602\times 10^{-19}}[/tex]

[tex]n=9.98\times 10^{19}[/tex]

So, there are [tex]9.98\times 10^{19}[/tex] electrons that gets transferred. Hence, this is the required solution.

A strong lightning bolt that transfers an electric charge of about 16 C to Earth, transfers 1.0 × 10²⁰ electrons (1.7 × 10⁻⁴ moles of electrons).

A strong lightning bolt transfers an electric charge of about 16 C to Earth (or vice versa).

Electric charge is the physical property of matter that causes it to experience a force when placed in an electromagnetic field.

We want to calculate the number of electrons that have a charge of 16 C. We have to consider the following relationships:

16 C × 1 electron/1.602 × 10⁻¹⁹ C = 1.0 × 10²⁰ electron

1.0 × 10²⁰ electron × 1 mol/6.022 × 10²³ electron = 1.7 × 10⁻⁴ mol

A strong lightning bolt that transfers an electric charge of about 16 C to Earth, transfers 1.0 × 10²⁰ electrons (1.7 × 10⁻⁴ moles of electrons).

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Answer:

[tex] f_s = 640 Hz[/tex]

Explanation:

For this case we know that the speed of the sound is given by:

[tex] V_s = 340 m/s[/tex]

And we have the following info provided:

[tex] v_c = 25 m/s [/tex] represent the car leading

[tex] v_s= 25 m/s[/tex] represent the car behind with the source

[tex] f_o = 640 Hz[/tex] is the frequency for the observer

And we can find the frequency of the source [tex] f_s[/tex] with the following formula:

[tex] f_s = \frac{v-v_o}{v-v_s} f_o [/tex]

And replacing we got:

[tex] f_s = \frac{340-25}{340-25} *640 Hz = 640 Hz[/tex]

So then the frequency for the source would be the same since the both objects are travelling at the same speed.

[tex] f_s = 640 Hz[/tex]