In a thundercloud there may be an electric charge of 24 C near the top of the cloud and −24 C near the bottom of the cloud. If these charges are separated by about 2 km, what is the magnitude of the electric force between these two sets of charges? The value of the electric force constant is 8.98755 × 109 N · m2 /C 2 .

Answer:

(a) 1000 N/C

Explanation:

Kinetic energy of electron, K = 1.6 x 10^-17 J

distance, d = 10 cm = 0.1 m

Let the potential difference is V and the electric field is E.

(a) The relation between the kinetic energy and the potential difference is

K = e V

V = K / e

Where, e be the electronic charge = 1.6 x 10^-19 C

V = [tex]\frac{1.6\times 10^{-17}}{1.6\times 10^{-19}}[/tex]

V = 100 V

The relation between the electric field and the potential difference is given by

V = E x d

100 = E x 0.1

E = 1000 N/C

(b) The force acting on the electron, F = q E

where q be the charge on electron

So, F = -e x E

It means the direction of electric field and the force are both opposite to each other.

The direction of electric field and the force on electron is shown in the diagram.

Answer:

Doubles

Explanation:

Before starting to fall, the object will have energy in the form of potential energy. This will be the energy that will be transformed into kinetic energy, once the object starts its fall. And the potential energy that the object had before starting to fall, will be equal to the kinetic energy acquired during the fall. The potential energy is given by:

[tex]E_p = m*g*h[/tex]

Where m is the mass of the object, g is the gravity, and h is the distance to the floor. As you can see, The potential energy is proportional to the height of the fall, so if this height doubles, the potential energy and therefore, the kinetic energy acquired will doubles.

Answer:

The constant acceleration, [tex]a_{c} = 2.044\times 10^15 m/s^2[/tex]  

Solution:

As per the question:

[tex]v_{1} = 3 x 10^5 m/s[/tex]

[tex]v_{2} = 6.40 x 10^6 m/s[/tex]  

length of the re region, l = 1 cm = 0.01 m

Now,

Using the third equation of motion:

[tex]v_{2}^2 = v_{1}^2 + 2a_{c}l[/tex]  

[tex]a_{c} = \frac{v_{2}^2 - v_{1}^2}{2l}[/tex]  

[tex]a_{c} = \frac{(6.40 x 10^6)^2 - (3 x 10^5)^2}{2\times 0.01}[/tex]  

[tex]a_{c} = 2.044\times 10^15 m/s^2[/tex]  

Answer:

Given, average lifetime of muons at rest is To = 2.21μs

Average lifetime of moving muons, T = 7.04μs

mass of muon = 207 mass of electron

a)

rest mass energy of an electron is 0.511 MeV.

[tex] T =\frac{T_o}{\sqrt{1-\beta^2}}[/tex]

Substitute the values:

[tex]7.04 =\frac{2.21}{\sqrt{1-\beta^2}}\\ \beta^2 = 0.90146\\ \beta = 0.95[/tex]

b) K = (γ-1) m₀c²

[tex]\gamma = \frac{1}{\sqrt{1-\beta^2}} = 3.185[/tex]

K = (3.185-1) (207 ×9.1×10⁻³¹) (3×10⁸)² = 3.7×10⁻¹¹J

c) p = γmv = (3.185)((207 ×9.1×10⁻³¹)(0.95×3×10⁸)= 1.71×10⁻¹⁹ N/s

Answer:

Explanation:

As we know that the ball is projected upwards so that it will reach to maximum height of 16 m

so we have

[tex]v_f^2 - v_i^2 = 2 a d[/tex]

here we know that

[tex]v_f = 0[/tex]

also we have

[tex]a = -9.81 m/s^2[/tex]

so we have

[tex]0 - v_i^2 = 2(-9.81)(16)[/tex]

[tex]v_i = 17.72 m/s[/tex]

Now we need to find the height where its speed becomes half of initial value

so we have

[tex]v_f = 0.5 v_i[/tex]

now we have

[tex]v_f^2 - v_i^2 = 2 a d[/tex]

[tex](0.5v_i)^2 - v_i^2 = 2(-9.81)h[/tex]

[tex]-0.75v_i^2 = -19.62 h[/tex]

[tex]0.75(17.72)^2 = 19.62 h[/tex]

[tex]h = 12 m[/tex]

Answer:

option C

Explanation:

the correct answer is option C.

Graph between velocity time which shows the constant velocity is straight line  with slope zero.

constant velocity means velocity is not changing with respect to time hence this condition will be only when graph is straight line with slope zero.

area under velocity time graph shows the displacement of the object.

And where as slope of velocity time graph show acceleration of the object.

Final answer:

The correct option is (d) A straight line sloping upward. A graph of velocity vs time that shows an object moving toward a motion detector at a constant speed is represented as a straight line sloping upward, indicating a constant positive velocity.

Explanation:

When considering an object moving toward a motion detector at a constant speed, the velocity vs. time graph should depict a constant velocity. This is visualized as a straight line on such a graph, since the speed of the object does not change over time.

The correct choice to represent an object moving at a constant speed towards a motion detector is (d) A straight line sloping upward.

The key concept here is that when velocity is constant, the slope of the velocity vs. time graph remains constant. If the object is moving toward the motion detector, the velocity is positive, therefore, the graph slopes upwards, indicating a positive velocity that does not change.

This scenario clearly shows the relationship between velocity and time when an object moves at a constant speed.

Answer:[tex]-18.05 m/s^2[/tex]

Explanation:

Given

Initial velocity of rocket A ([tex]v_A[/tex]) 5300m/s

Initial velocity  of rocket B([tex]v_B[/tex]) 8700 m/s

after time t they have displacement=0

[tex]s_A=5300\times t+\frac{1}{2}\times \left ( -11\right )t^2[/tex]

[tex]0=5300\times t+\frac{1}{2}\times \left ( -11\right )t^2[/tex]

[tex]t=\frac{5300\times 2}{11}=963.63 s[/tex]

for Rocket B

[tex]s_B=8700\times t+\frac{1}{2}\times \left ( a_B\right )t^2[/tex]

time is same for A & B

[tex]0=8700\times 963.63+\frac{1}{2}\times \left ( a_B\right )963.63^2[/tex]

[tex]a_B=\frac{2\times 8700}{963.63}=-18.05 m/s^2[/tex]

Answer: 84.63 m

Explanation:

The speed of sound in the air, at [tex]0 \°C[/tex] is [tex]331.5 m/s[/tex], and for each degree Celsius the temperature rises, the speed of the sound increases by [tex]0.6 m/s[/tex]. So, if we estimate the speed of sound at [tex]30 \°C[/tex] it will be [tex]349 m/s[/tex].

On the other hand, the speed of sound [tex]V[/tex] is defined as the distance traveled [tex]d[/tex] in a especific time [tex]t[/tex]:  

[tex]V=\frac{d}{t}[/tex]  

Where:  

[tex]V=349 m/s[/tex] is the speed of sound  

[tex]t=\frac{0.485 s}{2}=0.2425 s[/tex] is half the time the sound wave travels since you say "hello", the sound wave hits the building and then returns to you

[tex]d[/tex] is the distance between you and the building

[tex]349 m/s=\frac{d}{0.2425 s}[/tex]  

Finding [tex]d[/tex]:  

[tex]d=(349 m/s)(0.2425 s)[/tex]  

Finally:  

[tex]d=84.63 m[/tex]

To find the distance to the building, calculate the speed of sound at 30.6°C, multiply it by the echo time of 0.485 seconds, and divide by 2 to account for the round trip of the sound. Then, convert the distance from meters to feet to get approximately 280.95 feet.

To calculate the distance of the building when an echo is heard 0.485 seconds after shouting, we need to consider the speed of sound at the given air temperature of 30.6°C. Since the speed of sound varies with temperature, we use the formula v = v_0 + 0.6t, where v_0 is the speed of sound at 0°C (approximately 331.5 m/s), t is the temperature in Celsius, and v is the speed of sound at t degrees Celsius. At 30.6°C, the speed of sound is v = 331.5 m/s + 0.6(30.6°C), which calculates to roughly 349.86 m/s. Since the echo travels the distance to the building and back, the actual distance to the building is half the total distance the sound wave traveled.

Therefore, the one-way distance to the building is 0.485 s × 349.86 m/s ÷ 2. Converting meters to feet (1 meter = 3.28084 feet), we find that the building is approximately 280.95 feet away.

Answer:

distance in east is 1273.78 m

Explanation:

given data

average velocity = 1.32 m/s west =

hike = 5.21 km = 5.21 × 10³ m

average velocity = 3.49 m/s west

average velocity = 0.687 m/s east

to find out

distance in east

solution

we consider here distance in east  is = x

so distance from starting point = 5.21 × 10³ - x        ...................1

and we can say time required to reach end

time required = distance / speed

time required = [tex]\frac{5.21 *10^3 - x}{1.32}[/tex]    ................2

and

time required for 6.44 km west

time required = [tex]\frac{5.21 *10^3 - x}{3.49}[/tex]    ................3

and time required for distance x

time required = [tex]\frac{x}{0.687}[/tex]    ................4

so from equation 2 , 3 and 4

[tex]\frac{5.21 *10^3 - x}{1.32}[/tex] = [tex]\frac{5.21 *10^3 - x}{3.49}[/tex]  + [tex]\frac{x}{0.687}[/tex]

x = 1273.78 m

so distance in east is 1273.78 m

Answer:

a) 2.24°

b) 11.99 km

c) 0.9992 miles

Explanation:

We can think of the road as a triangle with a 12 km hypotenuse and a 0.5 km side. Then:

l = 12

h = 0.5

and

sin(a) = h / l

a = arcsin(h / l)

a = arcsin(0.5 / 12)

a = 2.24°

That is the slope.

The map distance travelled would be the other side of the triange.

cos(a) = d / l

d = l* cos(a)

d = 12 * cos(2.24) = 11.99 km

And for miles

d = 1 mile * cos(2.24) = 0.9992 miles

(a) V = [tex]\frac{8.3}{P}[/tex]

(b) (i) the value of [tex]\frac{dV}{dP}[/tex] when P = 50kPa is - 0.00332 [tex]\frac{m^{3} }{kPa}[/tex]

(ii) the meaning of the derivative [tex]\frac{dV}{dP}[/tex] is the rate of change of volume with pressure.

(iii) and the units are [tex]\frac{m^{3} }{kPa}[/tex]

Boyle's law states that at constant temperature;

P ∝ 1 / V

=> P = k / V

=> PV = k   -------------------------(i)

Where;

P = pressure

V = volume

k = constant of proportionality

According to the question;

When;

V = 0.106m³, P = 50kPa

Substitute these values into equation (i) as follows;

50 x 0.106 = k

Solve for k;

k = 5.3 kPa m³

(a) To write V as a function of P, substitute the value of k into equation (i) as follows;

PV = k

PV = 8.3

Make V subject of the formula in the above equation as follows;

V = 8.3/P

=> V = [tex]\frac{8.3}{P}[/tex]        -------------------(ii)

(b) Find the derivative of equation (ii)  with respect to V to get dV/dP as follows;

V = [tex]\frac{8.3}{P}[/tex]

V = 8.3P⁻¹

[tex]\frac{dV}{dP}[/tex] = -8.3P⁻²

[tex]\frac{dV}{dP}[/tex] = [tex]\frac{-8.3}{P^{2} }[/tex]

Now substitute P = 50kPa into the equation as follows;

[tex]\frac{dV}{dP}[/tex] = [tex]\frac{-8.3}{50^{2} }[/tex]   [[tex]\frac{kPam^{3} }{(kPa)^{2} }[/tex]]               -----  Write and evaluate the units alongside

[tex]\frac{dV}{dP}[/tex] = [tex]\frac{-8.3}{2500 }[/tex]    [[tex]\frac{kPam^{3} }{k^{2} Pa^{2} }[/tex]]

[tex]\frac{dV}{dP}[/tex] = - 0.00332 [[tex]\frac{m^{3} }{kPa}[/tex]]

Therefore,

(i) the value of [tex]\frac{dV}{dP}[/tex] when P = 50kPa is - 0.00332 [tex]\frac{m^{3} }{kPa}[/tex]

(ii) the meaning of the derivative [tex]\frac{dV}{dP}[/tex] is the rate of change of volume with pressure.

(iii) and the units are [tex]\frac{m^{3} }{kPa}[/tex]

Boyle’s Law describes the inverse relationship between the pressure and volume of a gas at constant temperature. To express volume as a function of pressure, use the formula V = k/P. The derivative of volume with respect to pressure indicates the rate at which volume changes as pressure changes, measured in cubic meters per kilopascal (m3/kPa).

Boyle’s Law states that for a given amount of gas at a constant temperature, the volume V is inversely proportional to the pressure P. This relationship can be mathematically represented as PV = k, where k is a constant.

(a) Given a sample of air at 25°C with a volume of 0.106 m3 and a pressure of 50 kPa, we can write the volume as a function of pressure by rearranging the formula to V = k/P. Using the initial conditions, we find that k = P × V = 50 kPa × 0.106 m3 = 5.3 kPa · m3, so V(P) = 5.3 kPa · m3 / P.

(b) To calculate dV/dP when P = 50 kPa, we differentiate the function V(P) with respect to P, resulting in dV/dP = -5.3 kPa · m3 / P2. At P = 50 kPa, this becomes dV/dP = -5.3 / (502) = -0.00212 m3/kPa. The derivative represents the rate of change of volume with respect to pressure, which in this case means that for each increase of 1 kPa in pressure, the volume decreases by 0.00212 m3. The units of the derivative are m3/kPa.

Answer:

The gravity at Equator is 9.780 m/s2 and the gravity at poles is 9.832 m/s2. The gravity at poles are bigger than at equator, principally because the Earth is not totally round. The gravity is inversely proportional to the square of the radius, that is the reason for the difference of gravity (The radius at Poles are smaller than at Equator).

If Earths would have a net charge Q. The Electric field of Earth would be inversely proportional to the square of the radius of Earth (Electric field definition for a charge), the same case as for gravity. So there would be a difference between the electric field at poles and equator, too.

Answer:

t = 0.018 s  

Explanation:

given,

nerve impulse in human body travel at a speed of  = 100 m/s

height of the man = 1.80 m

time taken by the impulse to travel from foot to brain = ?

distance = speed × time                      

1.80 = 100 × t                                              

t =[tex]\dfrac{1.80}{100}[/tex]                                    

t = 0.018 s                            

hence, the time taken by the nerve to reach brain from toe is 0.018 s

The pebble's initial speed is approximately 22 m/s. It takes approximately 1.45 seconds for the pebble to first reach a height of 18.5 m.

To determine the pebble's initial speed, we can use the equation for projectile motion:

h = v0yt - (1/2)gt2

Where h is the height, v0y is the initial vertical speed, t is the time, and g is the acceleration due to gravity.

Since the pebble is launched straight up, the final height is equal to the initial height. Plugging in the given values, we have:

37 m = v0y(2.3 s) - (1/2)(9.8 m/s2)(2.3 s)2

Simplifying this equation gives us the value of v0y, the initial vertical speed. To find the pebble's initial speed, we can use the Pythagorean theorem:

v0 = √(v0x2 + v0y2)

Where v0 is the initial speed and v0x is the initial horizontal speed. Since the pebble is launched straight up, v0x = 0. Plugging in the calculated value of v0y, we can solve for v0.

(a) The pebble's initial speed is approximately 22 m/s.

(b) To find the time it takes for the pebble to first reach a height of 18.5 m, we can use the equation for height:

18.5 m = v0yt - (1/2)gt2

Solving for t gives the time it takes for the pebble to reach the desired height.

(b) It takes approximately 1.45 seconds for the pebble to first reach a height of 18.5 m.

Answer:

9 m/s

Explanation:

mass of cannon, M = 100 kg

mass of cannon ball, m = 10 kg

velocity of cannon ball, v = 90 m/s

Let the recoil velocity of cannon is V.

Us ethe conservation of linear momentum, as no external force is acting on the system, so the linear momentum of the system is conserved.

Momentum before the firing = momentum after the firing

M x 0 + m x 0 = M x V + m x v

0 = 100 x V + 10 x 90

V = - 9 m/s

Thus, the recoil velocity of cannon is 9 m/s.

Answer:

q₂ = 6.8 x 10⁻⁶C

q₁ = 3.2 x 10⁻⁶ C

Explanation:

First charge is q₂ and second charge will be ( 10 x10⁻⁶ - q₂ )

Force between them F = 8 X 10⁻³ N distance between them d = 4.1 m

F = k ( q₁ xq₂ ) / d²

8 x 10⁻³ = [tex]\frac{8.99\times10^9\times q_2\times (10^{-5}-q_2)}{(4.1)^2}[/tex]

14.95 x 10⁻¹² = 10⁻⁵q₂ - q₂²

q₂²-10⁻⁵q₂ + 14.95 x 10⁻¹² =0

This is a quadratic equation .The solution of it gives the value of q₂

q₂ = 6.8 x 10⁻⁶C

q₁ = 3.2 x 10⁻⁶ C

q₁ = 3.2 x 10⁻⁶ C

q₂ = 6.8 x 10⁻⁶C

Charge is a characteristic of a unit of matter that expresses the extent to which it has more or fewer electrons than protons.

It is also known as electric charge, electrical charge, or electrostatic charge and symbolized q.

Given,

First charge is q₂ and second charge will be ( 10 x10⁻⁶ - q₂ )

F = 8 X 10⁻³ N

d = 4.1 m

By using the Formula,

[tex]F =\frac{k ( q_{1} * q_{2} ) }{d^{2} }[/tex]

8 x 10⁻³ = [tex]\frac{8.99*10^{9}*q_{2} *10^{-5} -q_{2} }{(4.1)^{2} }[/tex]

14.95 x 10⁻¹²

= 10⁻⁵q₂ - q₂²

= q₂²-10⁻⁵q₂ + 14.95 x 10⁻¹²=0

Therefore,

The above equation is a quadratic equation , it gives the value of q₂

q₂ = 6.8 x 10⁻⁶C

q₁ = 3.2 x 10⁻⁶ C

Learn more about Charges here:https://brainly.com/question/14467306

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Answer:

40 m

Explanation:

A car travelling at 45 ms has a kinetic energy of

Ec = 1/2 * m * v^2

We do not know the mass, so we can use specific kinetic energy:

ec = 1/2 * v^2

In this case

ec = 1/2 * 45^2 = 1012 J/kg

If it stops in 10 m, the braking force performed a sppecific work of 1012 J/kg in 10 m

L = F * d

F = L / d

F = 1012 / 10 = 101.2 N/kg

If the car is running at 90 m/s, the specific kinetic energy of:

ec2 = 1/2 * 90^2 = 4050 J/kg

With the same braking force the braking distance is:

d2 = L2 / F

d2 = 4050 / 101.2 = 40 m

Answer:

Part A)   x=40m at t= 2.5s

Part B)    x=240m at t=15 s

Explanation:

Data:

x₁=0 , t₁=0

x₂=80m,  t₂=5 s

t₃= 2.5 s

t₄= 15 s

problem development

We set the ratios x / t:

Part A)

[tex]\frac{x_{3} }{t_{3} } =\frac{x_{2} }{t_{2} }[/tex]

[tex]\frac{x_{3} }{2.5} =\frac{80}{5 }[/tex]

x₃=16*2.5

x₃=40m

Part B)

[tex]\frac{x_{4} }{t_{4} } =\frac{80 }{5 }[/tex]

[tex]\frac{x_{4} }{15} =\frac{80}{5 }[/tex]

x₄=16*15

x₄=240m

Final answer:

The car's position at t = 2.5 s will be 40 m as it is half of the interval to reach 80 m. At t = 15 s, three times the interval, the car will be at 240 m assuming constant velocity.

Explanation:

Given that the car moves from x1 = 0 m at t1 = 0 s to x2 = 80 m at t2 = 5.0 s, we can determine its position at other times under the assumption of constant velocity. Because the velocity is constant, the ratios of times to positions are constant as well.

Part A: Car's Position at t = 2.5 s

The time t = 2.5 s is exactly half of the time interval given (5.0 s), so the position should be half of 80 m, which is 40 m.

Part B: Car's Position at t = 15 s

To find the position at t = 15 s, let's first determine the time ratio. The ratio of 15 s to 5.0 s is 3:1. Since the velocity is constant, the positions scale with time. So, the position x at t = 15 s is three times 80 m, giving us 240 m.

Answer:-24,5m/s

Explanation: what we have here is a UALM with these gravity as acceleration (-9.8 m/s^2). The initial position is 25 m and initial speed is 10m/s.

Speed and gravity are increasing in the opposite direction, speed upwards and gravity downwards, while the position is also upwards, depending on your reference system.

The first thing I need to know is the maximum high it will reach.

Hmax=- S(0)^2/2g=

S= speed.

0= initial

G= gravity

Hm= 100/19,6= 5.1 m

So, the ball will go 5,1 m higher than the initial position, and from there it will fall free.

Then, I need to know how long it takes to fall. For that we use UALM equation:

X(t)= X(0) + S(0)*t + (A*t^2)/2.

X: position

S: speed

A: acceleration

T:time

0: initial

0 = 25m +10*t -(9.8 * t^2)/2

Solving the quadratic equation we get

T= 3,5 sec. ( Negative value for time is impossible)

So now we know that the ball to go up and then fall needs 3,5 sec.

Let's see how long it takes to go up:

30,1=25+10*t-4,9*t^2

0=-5,1+10*t-4,9*t^2

T= 1 sec. So it will take 1 sec to the ball to reach the maximum high and 0=speed and then it'll fall during the resting 2,5 sec

Finally, to know the speed just before it touches the ground, we use the following formula:

A= (St-S0)/t

-9.8m/s^2 = (St- 0m/s)/ 2,5s

-24,5 m/s= St

-24,5 m/s is the speed at 3,5 sec, which is the time just before falling

Answer:

r=2.6 cm

Explanation:

Hi!

Lets call steel material 1, and the new alloy material 2. You know their densities:

[tex]\rho_1=8.08*10^3\frac{kq}{m^3}\\\rho_2=3.14*10^3\frac{kq}{m^3}[/tex]

The volume of a sphere with radius r is given by:

[tex]V=\frac{4}{3}\pi r^3[/tex]

Then the masses of the bearings are:

[tex]m_1=\rho_1V_1=\frac{4}{3}\pi r_1^3 \\m_1=\rho_2V_2=\frac{4}{3}\pi r_2^3[/tex]

For the masses to be the same:

[tex]\rho_1 V_1 = \rho_2 V_2\\\frac{V_2}{V_1} =\frac{\rho_1}{\rho_2}\\(\frac{r_2}{r_1})^3 = \frac{8.08}{3.14}=2.57\\r_2=\sqrt[3]{2.57}\;r1 =1.37r_1=1.37*1.9cm=2.6cm[/tex]

Answer:

[tex]\rm 1.374\times 10^6\ N/C.[/tex]

Explanation:

Given:

The surface area of each of the disc is given by

[tex]\rm A=\pi \times Radius^2.\\\\Radius = \dfrac D2=\dfrac{2.4\ cm}{2} = 1.2\ cm = 1.2\times 10^{-2}\ m.\\A = \pi \times (1.2\times 10^{-2} )^2=4.524\times 10^{-4}\ m^2.[/tex]

For the case, when d<<D, the strength of the electric field at a point due to a charged sheet is given by

[tex]\rm E=\dfrac{|\sigma|}{2\epsilon_o}[/tex]

where,

The electric field strength between the discs due to negative disc is given by

[tex]\rm E_1 = \dfrac{|\sigma|}{2\epsilon_o}=\dfrac{|q|}{2A\epsilon_o}.[/tex]

Since, the electric field is directed from positive charge to negative charge, therefore, the direction of this electric field is towards the negative disc.

The electric field strength between the discs due to positive disc is given by

[tex]\rm E_2 = \dfrac{|\sigma|}{2\epsilon_o}=\dfrac{|q|}{2A\epsilon_o}.[/tex]

The direction of this electric field is away from the positive disc, i.e., towards negative disc.

The electric field between the discs due to both the disc is towards the negative disc, therefore the total electric field strength between the discs is given by

[tex]\rm E=E_1+E_2\\=\dfrac{|q|}{2A\epsilon_o}+\dfrac{|q|}{2A\epsilon_o}\\=\dfrac{|q|}{A\epsilon_o}\\=\dfrac{11\times 10^{-9}}{2\times (4.524\times 10^{-4})\times (8.85\times 10^{-12})}\\=1.374\times 10^6\ N/C.[/tex]

Answer:

The depth of well equals 120.47 meters.

Explanation:

The time it takes the sound to reach our ears is the sum of:

1) Time taken by the rock to reach the well floor.

2) Time taken by the sound to reach our ears.

The Time taken by the rock to reach the well floor can be calculated using second equation of kinematics as:

  [tex]h=ut+\frac{1}{2}gt^2[/tex]

where,

'u' is initial velocity

't' is the time to cover a distance 'h' which in our case shall be the depth of well.

'g' is acceleration due to gravity

Since the rock is dropped from rest hence we infer that the initial velocity of the rock =0 m/s.

Thus the time to reach the well base equals

[tex]h=\frac{1}{2}gt_1^2\\\\\therefore t_{1}=\sqrt\frac{2h}{g}[/tex]  

Since the propagation of the sound back to our ears takes place at constant speed hence the time taken in the part 2 is calculated as

[tex]t_{2}=\frac{h}{Speed}=\frac{h}{340}[/tex]

Now since it is given that

[tex]t_{1}+t_{2}=5.50[/tex]

Upon solving we get

[tex]\sqrt{\frac{2h}{g}}+\frac{h}{340}=5.5\\\\\sqrt{\frac{2h}{g}}=5.5-\frac{h}{340}\\\\(\sqrt{\frac{2h}{g}})^{2}=(5.5-\frac{h}{340})^{2}\\\\\frac{2h}{g}=5.5^2+\frac{h^2}{(340^2)}-2\times 5.5\times \frac{h}{340}[/tex]

Solving the quadratic equation for 'h' we get

h = 120.47 meters.

Answer:

37357 sec  

or 622 min

or 10.4 hrs

Explanation:

GIVEN DATA:

Lifting weight 80 kg

1 cal = 4184 J

from information given in question we have

one lb fat consist of 3500 calories = 3500 x 4184 J

= 14.644 x 10^6 J  

Energy burns in 1 lift = m g h

                                  = 80 x 9.8 x 1 = 784 J

lifts required [tex]= \frac{(14.644 x 10^6)}{784}[/tex]

                      = 18679

from the question,

1 lift in 2 sec.

so, total time = 18679 x 2 = 37357 sec  

or 622 min

or 10.4 hrs

Answer:

0.235 nC

Explanation:

Given:

Since the electric field is the electric force applied on a charged body of unit charge.

[tex]\therefore E = \dfrac{F}{q}\\\Rightarrow q =\dfrac{F}{E}\\\Rightarrow q =\dfrac{2.00\times 10^{-6}\ N}{8.50\times 10^{3}\ N/C}\\\Rightarrow q =0.235\times 10^{-9}\ C\\\Rightarrow q =0.235\ nC[/tex]

Hence, the value of q is 0.235 nC.

Answer:

If an object moves twice as fast its kinetic energy quadruples.

Explanation:

The kinetic energy (K₁) of a body of mass (m) that moves with speed (v) is:

K₁= 1/2 * m* v²

If we double the speed of the body, its kinetic energy (K₂) will be:

K₂= 1/2 * m*( 2v)²

K₂= 1/2 * m* 4 *v²

K₂= 4(1/2 * m *v²)

K₂= 4*K₁

Answer:

The charge at the center of the conducting sphere is zero.

Explanation:

A principle of conductor materials is that the electric field inside a conductor in electrostatic state is always zero. The gauss law says that the flux of a electric field in a closed surface is proportional to the charge enclosed by the surface. Then if the Electric field inside of a conductor is zero, imperatively the charge anywhere inside the conductor is zero too, so the charge at the center of the sphere is zero.

Answer:1.624 m

Explanation:

Given

height of 2nd floor =5 m

time recorded is 0.58 sec

Let h be the height of third floor above 2 nd floor

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

here u=0

t=time taken to cover height h

[tex]h=\frac{gt^2}{2}[/tex]   ----------1

Now time taken to complete whole building length

[tex]h+5=\frac{g(t+0.58)^2}{2}[/tex]  ----------2

Subtract 1 form 2

[tex]5=\frac{g(2t+0.58)(0.58)}{2}[/tex]

Taking gravity [tex]g=1 m/s^2[/tex]

1=(2t+0.58)(0.58)

t=0.57 s

Substitute the value of t in 1

h=1.624 m

Answer:

The train should be a 10 meters away before the professor falls

Explanation:

Calculation of the teacher's fall time:

[tex]y(t)=v_{o}t-1/2*g*t^{2}[/tex]

in this case Vo=0 and y=-5:

[tex]-5=-1/2*10*t^{2}[/tex]

[tex]t=1s[/tex]

Calculation how far away is the train when the professor falls:

The distance the train travel while the professor is falling:

[tex]d=v_{train}*t=10m/s*1s=10m[/tex]

So the train should be a 10 meters away before the professor falls

Answer:

Part a)

[tex]\lambda = 4\pi[/tex]

Part b)

[tex]f = 47.7 Hz[/tex]

Part c)

[tex]v = 600 m/s[/tex]

Explanation:

Part a)

As we know that angular wave number is given as

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

[tex]k = 0.500[/tex]

[tex]0.500 = \frac{2\pi}{\lambda}[/tex]

[tex]\lambda = \frac{2\pi}{0.500}[/tex]

[tex]\lambda = 4\pi[/tex]

Part b)

As we know that angular frequency is given as

[tex]\omega = 300 rad/s[/tex]

[tex]\omega = 2\pi f[/tex]

[tex]300 = 2\pi f[/tex]

[tex]f = \frac{300}{2\pi}[/tex]

[tex]f = 47.7 Hz[/tex]

Part c)

Speed of the wave is given as

[tex]v = \lambda \times frequency[/tex]

so we have

[tex]v = 4\pi \times 47.7[/tex]

[tex]v = 600 m/s[/tex]

Answer:

cart displacement is 66 m

Explanation:

given data

velocity = 5 m/s

acceleration = 2 m/s²

time = 6 s

to find out

What is the

magnitude of cart displacement

solution

we will apply here equation of motion to find displacement that is

s = ut + 0.5×at²    .............1

here s id displacement and u is velocity and a is acceleration and time is t here

put all value in equation 1

s = ut + 0.5×at²

s = 5(6) + 0.5×(2)×6²

s = 66

so cart displacement is 66 m