Two resistors of 5.0 and 9.0 ohms are connected inparallel. A
4.0 Ohm resistor is then connected in series withthe parallel
combination. A 6.0V battery is then connected tothe series-parallel
combination. What is the current throughthe 9.0 ohm resistor?

Answer:

16 m

Explanation:

given,

power of hydraulic plant = 770 kW

volume of water pass through the turbine = 300 m³

density of water = 1000 kg/m³

m  =ρ × V

mass of water pass each minute  = 300 × 1000 = 3 × 10⁵

assume height of the fall be h

potential head of the water = mgh

[tex]\dfrac{mgh}{60}= 770 \times 10^3[/tex]

[tex]3\times 10^5 \times 9.81\times h= 770 \times 10^3\times 60[/tex]

[tex]h = \dfrac{770 \times 10^3\times 60}{3\times 10^5 \times 9.81}[/tex]

h = 15.69 m ≈ 16 m

the distance of the water fall is equal to 16 m.

Answer:B

Explanation:

Given

First motor has three times as much as power as another

Let the power of smaller motor be P

Therefore bigger motor has power 3 P

and we know that energy [tex]E=power\times time[/tex]

Larger motor [tex]E=3P\times t=3Pt[/tex]

Smaller motor [tex]E=P\times 3t=3Pt[/tex]

Therefore smaller motor will do same work if it takes three times the time taken by Larger motor

Answer:

Force, F = 550000 N

Explanation:

Given that,

Pressure at the bottom of the Marianas Trench, [tex]P=1100\ kPa=11\times 10^5\ Pa[/tex]

Surface area, [tex]A=0.5\ m^2[/tex]

We need to find the force with which the water pressure at the bottom of the Marianas Trench push on a fish. Mathematically, the pressure is given by :

[tex]P=\dfrac{F}{A}[/tex]

[tex]F=P\times A[/tex]

[tex]F=11\times 10^5\ Pa\times 0.5\ m^2[/tex]

F = 550000 N

So, the force with which the water pressure at the bottom of the Marianas Trench push on a fish is 550000 N. Hence, this is the required solution.

The water pressure at the bottom of the Marianas Trench would exert a force of 550,000 Newtons on a fish with a surface area of 0.50 m^2.

The water pressure exerted on a fish at the bottom of the Marianas Trench can be determined by using the formula for pressure (P), which is force (F) divided by area (A). Given the pressure at the bottom of the trench is approximately 1,100 kPa and the surface area of the fish is 0.50 m2, the force can be calculated using the equation F = P x A. By inserting the given values, we get F = 1,100,000 Pa x 0.50 m2 = 550,000 N. Therefore, the water pressure would push on the fish with a force of 550,000 Newton.

Answer:

magnitude of the average acceleration is 4.16 m/s²

Explanation:

given data

initial velocity u = 5 m/s

distance s = 3 m

final velocity v = 0

to find out

magnitude of the average acceleration

solution

we will apply here linear motion equation that is

v²-u² = 2×a×s   ..............1

put here all these value

v is final velocity and u is initial velocity and s is distance

0²-5² = 2×a×3

a = -4.16

so magnitude of the average acceleration is 4.16 m/s²

Answer: The magnitude of the average acceleration is 4.16m/s^2

Explanation:

Given that,

Number density [tex]n= 1\ atom/cm^{3} =10^{6}\ atom/m^3[/tex]

Temperature = 2.7 K

(a). We need to calculate the pressure in interstellar space

Using ideal gas equation

[tex]PV=nRT[/tex]

[tex]P=\dfrac{nRT}{V}[/tex]

[tex]P=\dfrac{10^{6}\times8.314\times2.7}{6.023\times10^{23}}[/tex]

[tex]P=3.727\times10^{-17}\ Pa[/tex]

[tex]P=36.78\times10^{-23}\ atm[/tex]

The pressure in interstellar space is [tex]36.78\times10^{-23}\ atm[/tex]

(b). We need to calculate the root-mean square speed of the atom

Using formula of rms

[tex]v_{rms}=\sqrt{\dfrac{3RT}{Nm}}[/tex]

Put the value into the formula

[tex]v_{rms}=\sqrt{\dfrac{3\times8.314\times2.7}{1.007\times10^{-3}}}[/tex]

[tex]v_{rms}=258.6\ m/s[/tex]

The root-mean square speed of the atom is 258.6 m/s.

(c). We need to calculate the  kinetic energy

Average kinetic energy of atom

[tex]E=\dfrac{3}{2}kT[/tex]

Where, k = Boltzmann constant

Put the value into the formula

[tex]E=\dfrac{3}{2}\times1.38\times10^{-23}\times2.7[/tex]

[tex]E=5.58\times10^{-23}\ J[/tex]

The kinetic energy stored in 1 km³ of space is [tex]5.58\times10^{-23}\ J[/tex].

Hence, This is the required solution.

Answer:

3.83 s

Explanation:

Initially for 50 m A falls with acceleration equal to g ie 9.8 m /s².In this journey his initial velocity u = 0 , a = 9.8 , v = ? , t = ?

h = ut + 1/ 2 g t²

50 = .5 x 9.8x t²

t = 3.19 s

v = u +gt

= o + 9.8 x 3.19

= 31.26 m /s

For motion under deceleration

initial speed  u = 31.26 m/s

Final speed v = 14 m/s

deceleration a = - 27 m/s²

v = u - at

14 = 31.26 - 27 t

t = 0.64 s

So total time in the air

= 3.19 + .64 = 3.83 s

Answer:

0.6 m

Explanation:

When a spring is compressed it stores potential energy. This energy is:

Ep = 1/2 * k * x^2

Being x the distance it compressed/stretched.

When the spring bounces the ice cube back it will transfer that energy to the cube, it will raise up the slope, reaching a high point where it will have a speed of zero and a potential energy equal to what the spring gave it.

The potential energy of the ice cube is:

Ep = m * g * h

This is vertical height and is related to the distance up the slope by:

sin(a) = h/d

h = sin(a) * d

Replacing:

Ep = m * g * sin(a) * d

Equating both potential energies:

1/2 * k * x^2 = m * g * sin(a) * d

d = (1/2 * k * x^2) / (m * g * sin(a))

d= (1/2 * 25 * 0.1^2) / (0.05 * 9.81 * sin(25)) = 0.6 m

Using the conservation of mechanical energy principle, we relate the initial potential energy in the compressed spring to the final potential energy due to the height gained by the ice cube sliding up a slope. In doing so, we calculate how far the ice cube travels up the slope before reversing direction to be approximately 0.607 meters.

The question pertains to the concept of energy conservation involving the energies of spring compression (potential energy) and gravity (also potential energy). Here the energy initially stored in the compressed spring is used to propel the ice cube up the slope until its potential energy (due to height gained) fully consumes the initial energy from the spring.

The conservation of mechanical energy principle states the initial total energy is equal to the final total energy. Initially we have spring potential energy and finally it's all converted to gravitational potential energy. So, 1/2*k*x^2=m*g*h, where k is the spring constant, x is the spring compression, m is mass of the ice block, g is acceleration due to gravity, and h is height gained by the ice cube.

We solved for h to get h = (1/2*25*(0.100)^2) / (50*9.8) = 0.255 m. This is the vertical height gained, not the distance along the slope. To get the slope distance traveled by the cube (d), we divide the height by sin(25 degrees) to get: d = 0.255/sin(25) = 0.607 m.

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Answer:

Part a)

[tex]L_o = 6.3181 cm[/tex]

Part b)

[tex]T = 131.3 ^oC[/tex]

Explanation:

Let the length of the rod at 0 degree Celsius is given as Lo

now we have

[tex]L = L_o( 1 + \alpha \Delta T)[/tex]

now we know that

[tex]L_o[/tex] = Length of rod at zero degree C

Part a)

[tex]6.3243 = L_o(1 + \alpha (16 - 0))[/tex]

[tex]6.3568 = L_o(1 + \alpha (100 - 0))[/tex]

now we have

[tex]\frac{6.3568}{6.3243} = \frac{1 + 100 \alpha}{1 + 16 \alpha}[/tex]

[tex]1.005(1 + 16 \alpha) = 1 + 100 \alpha[/tex]

[tex]83.918\alpha = 5.138\times 10^{-3}[/tex]

[tex]\alpha = 6.12 \times 10^{-5}[/tex]

now we have

[tex]L_o = 6.3181 cm[/tex]

Part b)

length of the rod is 6.3689 cm

now we have

[tex]L = L_o(1 + \alpha\Delta T)[/tex]

[tex]6.3689 = 6.3181(1 + \alpha \Delta T)[/tex]

[tex]6.3689 = 6.3181(1 + (6.12 \times 10^{-5})(T - 0))[/tex]

[tex]1.008 = 1 + (6.12 \times 10^{-5})T[/tex]

[tex]T = 131.3 ^oC[/tex]

Final answer:

The speed of the box as it leaves the spring can be calculated using conservation of energy, accounting for the initial potential energy in the spring and work done by friction. The maximum speed of the box is achieved when all the spring's potential energy is converted into kinetic energy, before friction does any work.

Explanation:

To solve for the speed of the box at the instant it leaves the spring and its maximum speed during the motion, we can apply the principle of conservation of energy and the work-energy theorem.

Part (a): Speed of Box When It Leaves the Spring

The initial potential energy stored in the compressed spring is equal to the kinetic energy of the box plus the work done by friction as the box slides the distance where the spring is compressed:

PE_spring = KE_box + Work_friction

Using the formula for potential energy PE_spring = (1/2)[tex]kx^2[/tex], where k is the spring constant and x is the compression distance, and the work done by friction Work_friction = μ_kmgx, where μ_k is the coefficient of kinetic friction, m is the mass, and g is the acceleration due to gravity, we can find the kinetic energy which will then be used to calculate the speed using KE = (1/2)[tex]mv^2[/tex].

To find the actual speed, we rearrange to v = √(2(PE_spring - Work_friction)/m).

Part (b): Maximum Speed of the Box

The maximum speed of the box occurs at the point where all the potential energy in the spring has been converted to kinetic energy and before any work has been done by friction. Therefore, v_max = √(2PE_spring/m).

Answer:

a) [tex]y=y_{0}+v_{oy} t+\frac{1}{2} gt^{2}  =640 ft+(65ft/s)(2.018s)+(\frac{1}{2})(-32.2ft/s^{2} )(2.018s)^{2}[/tex]

[tex]y=705.6ft[/tex]

b) [tex]t=8.63 s[/tex]

Explanation:

We start the exercise knowing that a ball is thrown up with an initial velocity of 65 ft/s with an initial height of 680 ft.

To calculate the maximum heigh, we know that at the top of the motion the ball stop going up and start going down because of the gravity

First of all, we need to calculate the time that takes the ball to reach the maximum point.

a) [tex]v_{y}=v_{oy}+gt[/tex]

[tex]t=\frac{-v_{oy} }{g}=\frac{-65ft/s}{-32.2ft/s^{2} } =2.018s[/tex]

Knowing that time, we can calculate the height to which the ball rises:

[tex]y=y_{0}+v_{oy} t+\frac{1}{2} gt^{2}  =640 ft+(65ft/s)(2.018s)+(\frac{1}{2})(-32.2ft/s^{2} )(2.018s)^{2}[/tex]

[tex]y=705.6ft[/tex]

b) Now, to know the time that the ball reach the bottom of the cliff, we know that the final height is y=0ft

[tex]y=y_{0}+v_{oy} t+\frac{1}{2} gt^{2}[/tex]

[tex]0=640ft+(65ft/s)t-\frac{1}{2}(32.2ft/s^{2})t^{2}[/tex]

This is a classic quadratic equation, that can be solve using the quadratic formula

[tex]t=\frac{-b±\sqrt{b^{2}-4ac } }{2a}[/tex]

a=-16.1

b=65

c=640

Solving for t, we have that

[tex]t=-4.6015 s[/tex] or [tex]t=8.6387s[/tex]

Since the time can not be negative:

[tex]t=8.6387s[/tex]

Answer:

DMM should be placed in the series combination with the circuit.

Explanation:

DMM is the digital multi meter. It can measure the voltage, current and resistance at a time.

Answer:

a) [tex]F=-3.1465 N[/tex]

b) Greater attractive force

Explanation:

Assuming they are point charges, we can use coulomb's law to know the magnitude of the force between these objects:

[tex]F=\frac{kq_1q_2}{d^2}[/tex]

Here, K is the Coulomb constant, [tex]q_1[/tex] and [tex]q_2[/tex] are the point charges and d is the distance bewteen the charges.

a) So, we have:

[tex]F=\frac{(8.99*10^9\frac{Nm^2}{C^2})(7*10^{-6}C)(-6*10^{-6}C)}{12*10^{-2}m}\\F=-3.1465 N[/tex]

b) There will be a greater amount of charge on the side closest to the object with opposite charge, therefore this will increase the charge at that point, increasing the attractive force.

The attractive force between a 0.700 µC charged glass rod and a -0.600 µC charged silk cloth 12.0 cm apart is 0.2625 N, as calculated by Coulomb's law. If charges are distributed over an area instead of acting as point charges, the force calculation might differ, depending on the geometry and distribution of the charges.

To calculate the attractive force between a glass rod with a 0.700 µC charge and a silk cloth with a -0.600 µC charge that are 12.0 cm apart, we can use Coulomb's law. Coulomb's law is given by the formula F = k * |q1*q2| / r^2, where F is the force between the charges, q1 and q2 are the values of the charges, r is the distance between the charges, and k is Coulomb's constant (approximately 8.988 × 10^9 Nm^2/C^2). Plugging in the values, we get F = (8.988 × 10^9 Nm^2/C^2) * |0.700 × 10^-6 C * -0.600 × 10^-6 C| / (0.12 m)^2, which results in a force of approximately 0.2625 N.

Discussing how the answer to this problem might be affected if the charges are distributed over some area rather than acting like point charges: The resulting force might differ from the point charge approximation. This is because the distribution of charge affects the electric field produced by the charges, potentially reducing the force if the charges are spread out over a larger area compared to being concentrated at a point. The effect typically depends on the actual distribution and geometry of the charges.

Answer:

a. x= 83.03 m at t= 2 s

b. v=346.7 m/s  at t= 4s

c. t=10.5s : maximum displacement occurs

d.  t= 7s  : maximum velocity

Explanation:

Definitions

acceleration :a(t) = dv/dt :Derived from velocity with respect to time

Velocity : V(t)=dx/dt : Derived from Displacement: with respect to time

Displacement: X(t)

Developing of problem

we have a(t) =10 (-t + 2 ) (t-5)+ 100

a(t) =(-10 t + 20) (t-5)+ 100  = -10 t²+50t+ 20t-100+ 100=--10 t²+70t

a(t)=--10 t²+70t  Equation (1)

a(t) = dv/dt

-10 t²+70t=dv/dt

dv=(-10 t²+70t)dt

We apply integrals to both sides of the equation

∫dv=∫(-10 t²+70t)dt

[tex]v=\frac{-10t^{3} }{3}   +\frac{70t^{2} }{2} +C_{1}[/tex]

[tex]v=\frac{-10t}{3} +35t^{2} +C_{1}[/tex]

at time t=0, v=0, then, C₁=0

[tex]v=\frac{-10t^{3}}{3} +35t^{2}}[/tex] Equation (2)

[tex]v=\frac{dx}{dt}[/tex]

dx= vdt

We apply integrals to both sides of the equation and we replace v(t) of the equation (2)

[tex]\int\limits {\, dx =\int\limits{\frac{(-10t^{3} }{3}+35t^{2} ) } \, dt[/tex]

[tex]x=\frac{-5t^{4} }{6} +\frac{35t^{3} }{3} +C_{2}[/tex]

at time t=0, x=3,  then,C₂=3

[tex]x=\frac{-5t^{4} }{6} +\frac{35t^{3} }{3} +3}[/tex] Equation (3)

a)Displacement at t=2s

We replace t=2s in the  Equation (3)

[tex]x=\frac{-5(2)^{4} }{6} +\frac{35(2)^{3} }{3} +3}[/tex]

x= 83.03 m at t= 2 s

b) Velocity at  t=4s

We replace t=4s in the  Equation (2)

[tex]v=\frac{-10(4)^{3}}{3} +35(4)^{2}}[/tex]

v=346.7 m/s  at t= 4s

c)Time at which maximum displacement occurs

at maximum displacement :v= [tex]\frac{dx}{dt} =0[/tex]  

in the Equation (2) :

[tex]v=\frac{-10t^{3}}{3} +35t^{2}}[/tex] =0

[tex]35t^{2} } =\frac{10t}{3}[/tex]  ,we divide by t on both sides of the equation

[tex]t=\frac{35*3}{10}[/tex]

t=10.5s

d.  maximum velocity over the interval

at maximum velocity :a= [tex]\frac{dv}{dt} =0[/tex]  

in the Equation (1)

-10 t²+70t = 0   we multiply the equation by -1 and factor

10t ( t-7) =0

t= 7 s

Answer:

The runners are 0.159 mi to the west of the flagpole.

Explanation:

Let´s place the origin of the frame of reference at the point where the flagpole is located.

The initial position of runner A is 3.91 mi and his velocity is -6.04 mi/h

The initial position of runner B is -2.95 mi and his velocity is  5.00 mi/h

When both runners meet, their position is the same. The equation of the position of each runner is:

x = x0 + v · t

Where

x = position at time t

x0 = initial position

v = velocity

t = time

Then, at the meeting point:

x runner A = x runner B

3.91 mi - 6.04 mi/h · t = -2.95 mi + 5.00m/h · t

Solving for t:

3.91 mi + 2.95 mi =  5.00m/h · t + 6.04 mi/h · t

6.86 mi = 11.04 mi/h · t

t = 0.621 h

Now, we use this time to find the meeting point. We can use the equation of any runner. Let´s use the position of runner A:

x = 3.91 mi - 6.04 mi/h · 0.621 h = 0.159 mi

Since the position is positive, the runners met 0.159 mi to the west of the flagpole.

Answer:

Explanation:

If t be the time of fall of drops from a height of 181 s ,

181 = 1/2 g t²

t = 6 s approx

There are  4 drops in vertical line , one touching the floor, one  at 181 m height on the verge of falling down , and two drops in mid air.

The time interval between dripping of two consecutive drops

= 6 / 3

2 s

time duration of fall of second drop = 2 x 2 = 4 s

Position of drop below nozzle

= 1/2 g x 4²

= 78.4 m

b )

Time duration of fall of third drop

= 2 s

Position of drop below nozzle

= 1/2 9.8 x 2²

=  19.6 m

Answer:

440 N

Explanation:

Force applied, F = 440 N

mass of crate, m = 170 kg

μs = 0.5, μk = 0.2

g = 9.81 m/s^2

The normal reaction acting on the crate, N = m g = 170 x 9.81 = 1667.7 N

The maximum value of static friction force acting on the crate

[tex]f_{s}=\mu _{s}N=0.5 \times 1667.7 = 833.85 N[/tex]

The maximum value of static friction force is more than the applied force so the crate does not move and teh applied force becomes friction force.

thus, the friction force acting on the crate is 440 N.

The frictional force exerted by the carpet on the crate is equal to the force of static friction.

To find the frictional force exerted by the carpet on the crate, we need to calculate the force of static friction first. The maximum force of static friction is given by fs(max) = μsN, where μs is the coefficient of static friction and N is the normal force. The normal force is equal to the weight of the crate, which is given by N = mg. In this case, the normal force N = (170 kg)(9.81 m/s^2) = 1666.7 N.

Therefore, the maximum force of static friction is fs(max) = (0.5)(1666.7 N) = 833.35 N. Since the worker is pushing with a force of 440 N, which is less than the maximum force of static friction, the crate remains at rest and the frictional force exerted by the carpet on the crate is equal to the force of static friction, which is 833.35 N.

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Answer:

Distance from the net when she meets the return: 14,3 m

Explanation:

First we need to know how much time it takes the ball to reach the net, the kinematycs general equation for position:

(1) [tex]x = x_{0}  + V_{s}  t + \frac{1}{2}  a t^{2}[/tex]

Taking the net as the origin (x = 0), [tex]x_{0} = 25m[/tex], velocity will be nagative [tex]V_{s} = - 50m/s[/tex] and assuming there is no friction wit air acceleration would be 0, so:

(2) [tex]x = x_{0}  + V_{s}  t [/tex]

we want to know the time when it reaches the net so when x=0, replacing de values:

(3) [tex]0 = 25m  - 50 m/s t_{net} [/tex]

So:  [tex]t_{net}  = 0,5 s [/tex]

The opponent will return the ball at [tex]V_{ret} = 25m/s[/tex], the equation for the return of the ball will be:

(4) [tex]y = y_{0}  + V_{ret}  t + \frac{1}{2}  a t^{2}[/tex]

Note that here it start from the origin, [tex]y_{0} = 0[/tex], as in the other case acceleration equals 0, and here we have to consider that the time starts when the ball reaches the net ([tex]t_{net} [/tex]) so the time for this equiation will be [tex]t - t_{net} [/tex], this is only valid for [tex]t >= t_{net} [/tex]:

(5) [tex]y = V_{ret}  (t - t_{net}) [/tex]

Coco starts running as soon as he serves so his equiation for position will be:

(6) [tex]z = z_{0}  + V_{c}  t + \frac{1}{2}  a t^{2}[/tex]

As in the first case it starts from 25m, [tex]z_{0} = 25m[/tex], acceleration equals 0 and velocity is negative [tex]V_{c} = - 10m/s[/tex]:

(7) [tex]z = z_{0} + V_{c}  t[/tex]

To get the time when they meet we have that [tex]z = y[/tex], so from equiations (5) and (7):

[tex]V_{ret}  (t - t_{net}) =  z_{0} + V_{c}  t[/tex]

[tex]t = \frac{z_{0} + V_{ret}* t_{net}}{V_{ret}-V_{c}}[/tex]

Replacing the values:

[tex]t = 1,071 s[/tex]

Replacing t in either (5) or (7):

[tex]z = y = 14,3 m[/tex]

This is the distance to the net when she meets the return

Final answer:

To calculate the atmospheric pressure, we can use the formula: Pressure = Density × Acceleration due to gravity × Height. Given the density of mercury, the height difference, and the acceleration due to gravity, we can calculate the atmospheric pressure.

Explanation:

The normal atmospheric pressure is 1.013 × 10^5 Pa. In this case, the height of a mercury barometer drops by 27.1 mm due to the approaching storm. To calculate the atmospheric pressure, we can use the formula:

Pressure = Density × Acceleration due to gravity × Height

Given that the density of mercury is 13.59 g/cm3, the height difference is 27.1 mm, and the acceleration due to gravity is approximately 9.8 m/s2, we can calculate the atmospheric pressure:

Pressure = (13.59 g/cm3)(9.8 m/s2)(27.1 mm)

Answer:

The inside Pressure of the tank is [tex]4499.12 lb/ft^{2}[/tex]

Solution:

As per the question:

Volume of tank, [tex]V = 0.25 ft^{3}[/tex]

The capacity of tank, [tex]V' = 50ft^{3}[/tex]

Temperature, T' = [tex]80^{\circ}F[/tex] = 299.8 K

Temperature, T = [tex]59^{\circ}F[/tex] = 288.2 K

Now, from the eqn:

PV = nRT                      (1)

Volume of the gas in the container is constant.

V = V'

Similarly,

P'V' = n'RT'                       (2)

Also,

The amount of gas is double of the first case in the cylinder then:

n' = 2n

[tex]\]frac{n'}{n} = 2[/tex]

where

n and n' are the no. of moles

Now, from eqn (1) and (2):

[tex]\frac{PV}{P'V'} = \frac{nRT}{n'RT'}[/tex]

[tex]P' = 2P\frac{T'}{T} = 2\times 2116\times \frac{299.8}{288.2} = 4499.12 lb/ft^{2}[/tex]

Answer:

[tex]r_{cm}\ =\ \dfrac{m_2d\ +\ m_1v_0 (t_1\ -\ t_0)}{m_1\ +\ m_2}[/tex]

Explanation:

Given,

Let [tex]v_{cm}[/tex] be the initial velocity of the center of mass of the system of particle at time [tex]t_o[/tex]

[tex]\therefore v_{cm}\ =\ \dfrac{m_1v_1\ +\ m_2v_2}{m_1\ +\ m_2}\\\Rightarrow v_{cm}\ =\ \dfrac{m_1v_0}{m_1\ +\ m_2}[/tex]

Assuming that the first particle is at origin, distance of the second particle from the origin is 'd'

Center of mass of the system of particles

[tex]x_{cm}\ =\ \dfrac{m_1x_1\ +\ m_2x_2}{m_1\ +\ m_2}\\\Rightarrow x_{cm}\ =\ \dfrac{m_2d}{m_1\ +\ m_2}\\[/tex]

Hence, at time [tex]t_0[/tex], the center of mass of the system is at [tex]x_0\ =\ \dfrac{m_2d}{m_1\ +\ m_2}[/tex] at an initial speed of [tex]v_{cm}[/tex]

Both the particles are assumed to be the point masses, therefore at the time [tex]t_1[/tex] the center of mass is at the position of the second particle which should be equal to the total distance traveled by the first particle because the second particle is at rest.

Let [tex]r_{cm}[/tex] be the distance traveled by the center of mass of the system of particles in the time interval [tex](t_1\ +\ t_0)[/tex]

From the kinematics,

[tex]s\ =\ x_0\ +\ vt\\\Rightarrow r_{cm}\ =\ x_{cm}\ +\ v_{cm}{t_1\ -\ t_0}\\\Rightarrow r_{cm}\ =\ \dfrac{m_2d}{m_1\ +\ m_2}\ +\ \left ( \dfrac{m_1v_0}{m_1\ +\ m_2}\ \right )\times (t_1\ -\ t_0)\\\Rightarow r_{cm}\ =\ \dfrac{m_2d\ +\ m_1v_0 (t_1\ -\ t_0)}{m_1\ +\ m_2}[/tex]

Hence, this is the required distance traveled by the first mass to collide with the second mass which is at rest.

Answer:

0.3067 m

Explanation:

Resolving power of a lens is given by the expression

R.P = 1.22λ / D

where λ is wavelength of light used, D is diameter of the lens

Substituting the data given

R.P in radian =

[tex]\frac{1.22\times550\times10^{-9}}{35\times10^{-2}}[/tex]

R P = 19.17 X 10⁻⁷ radian

If d be the minimum spacing between two objects on the ground that is resolvable by lens

[tex]\frac{d}{160\times10^3} =19.17\times10^{-7}[/tex]

d = 0.3067 m

Answer:

(a): [tex]\rm 1.133\ eV\ \ \ or\ \ \ 1.8128\times 10^{-19}\ J.[/tex]

(b): [tex]\rm 1.298\ eV \ \ \ or \ \ \ 2.077\times 10^{-19}\ J.[/tex]

Explanation:

The energy of the photon that absorbed by a hydrogen atom causes a transition is equal to the difference in energy levels of the hydrogen atom corresponding to that transition.

According to Rydberg's formula, the energy corresponding to [tex]\rm n^{th}[/tex] level in hydrogen atom is given by

[tex]\rm E_n = -\dfrac{E_o}{n^2}.[/tex]

where,

[tex]\rm E_o=13.6\ eV.[/tex]

Part (a): For the electronic transition from the n = 3 state to the n = 6 state.

The energy of the photon which cause this transition is given by

[tex]\rm \Delta E=E_6-E_3.\\\\where,\\E_6=-\dfrac{E_o}{6^2}=-\dfrac{13.6}{36}=-0.378\ eV.\\E_3=-\dfrac{E_o}{3^2}=-\dfrac{13.6}{9}=-1.511\ eV.\\\\\therefore \Delta E = (-0.378)-(-1.511)=1.133\ eV\\or\ \ \ \Delta E = 1.133\times 1.6\times 10^{-19}\ J=1.8128\times 10^{-19}\ J.[/tex]

Part (b): For the electronic transition from the n = 3 state to the n = 8 state.

The energy of the photon which cause this transition is given by

[tex]\rm \Delta E=E_8-E_3.\\\\where,\\E_8=-\dfrac{E_o}{8^2}=-\dfrac{13.6}{64}=-0.2125\ eV.\\E_3=-\dfrac{E_o}{3^2}=-\dfrac{13.6}{9}=-1.511\ eV.\\\\\therefore \Delta E = (-02125)-(-1.511)=1.298\ eV\\or\ \ \ \Delta E = 1.298\times 1.6\times 10^{-19}\ J=2.077\times 10^{-19}\ J.[/tex]

Final answer:

The energy of the photon that can cause an electronic transition in a hydrogen atom from n = 3 to n = 6 or n = 8 is -3.04 x 10^-19 J. The frequency of the photon is -4.60 x 10^14 Hz.

Explanation:

To calculate the energy of a photon that can cause an electronic transition in a hydrogen atom, we can use the equation:

E = Ef - Ei

where Ef is the energy of the final state and Ei is the energy of the initial state. The energy of a photon is given by:

E = hf

where h is Planck's constant (6.626 x 10-34 J·s) and f is the frequency of the photon. By substituting these equations, we can determine the frequency and energy of the photon.

(a) For an electronic transition from the n = 3 state to the n = 6 state, we have:

E = E6 - E3

E = (-3.4 eV) - (-1.5 eV)

E = -1.9 eV

Using the conversion factor 1 eV = 1.6 x 10-19 J, we can convert the energy to joules:

E = -1.9 eV x (1.6 x 10-19 J/eV)

E = -3.04 x 10-19 J

Now, we can find the frequency of the photon by rearranging the equation:

f = E/h

f = (-3.04 x 10-19 J) / (6.626 x 10-34 J·s)

f = -4.60 x 1014 Hz

(b) For an electronic transition from the n = 3 state to the n = 8 state, we follow the same steps:

E = E8 - E3

E = (-3.4 eV) - (-1.5 eV)

E = -1.9 eV

E = -3.04 x 10-19 J

f = E/h

f = (-3.04 x 10-19 J) / (6.626 x 10-34 J·s)

f = -4.60 x 1014 Hz

Therefore, the energy of the photon that can cause the electronic transition from the n = 3 state to the n = 6 state or the n = 8 state is -3.04 x 10-19 J and the frequency is -4.60 x 1014 Hz.

Answer:

(a). The velocity of the object is -2.496 m/s.

(b).  The total distance of the object travels during the fall is 23.80 m.

Explanation:

Given that,

Time = 1.95 s

Distance = 23.5 m

(a). We need to calculate the velocity

Using equation of motion

[tex]s = ut+\dfrac{1}{2}gt^2[/tex]

Put  the value into the formula

[tex]-23.5=u\times1.95+\dfrac{1}{2}\times(-9.8)\times(1.95)^2[/tex]

[tex]u=\dfrac{-23.5+4.9\times(1.95)^2}{1.95}[/tex]

[tex]u=-2.496\ m/s[/tex]

(b). We need to calculate the total distance the object travels during the fall

Using equation of motion

[tex]v = u+gt[/tex]

Put the value in the equation

[tex]-2.496=0-9.8\times t[/tex]

[tex]t =\dfrac{2.496}{9.8}[/tex]

[tex]t=0.254\ sec[/tex]

The total time is

[tex]t'=t+1.95[/tex]

[tex]t'=0.254+1.95[/tex]

[tex]t'=2.204\ sec[/tex]

We need to calculate the distance

Using equation of motion

[tex]s = ut+\dfrac{1}{2}gt^2[/tex]

Put the value into the formula

[tex]s=0+\dfrac{1}{2}\times9.8\times(2.204)^2[/tex]

[tex]s=23.80\ m[/tex]

Hence, (a). The velocity of the object is -2.496 m/s.

(b).  The total distance of the object travels during the fall is 23.80 m.

Answer:

[tex]\theta = 58.98[/tex]°

Explanation:

given data:

h = 18 ft = 5.48 m

from figure

[tex]h_{max} = 5.48 - 0.50 = 4.98 m[/tex]

[tex]h_{max} = \frac{ v_y^2}{2g}[/tex]

[tex]v_y =\sqrt{2gh_{max}[/tex]

[tex]v_y = \sqrt{2*9.8* 4.98} m/s[/tex]

[tex]v_y = 9.88 m/s[/tex]

[tex]t = \frac{v_y}{g} =\frac{9.88}{9.8} = 1.01 s[/tex]

[tex]v_x = \frac{d}{t} = \frac{6}{1.01}  = 5.49 m/s

[tex]v = \sqrt{v_x^2+v_y^2} = \sqrt{5.95^2+9.88^2}[/tex]

v = 11.53 m/s

[tex]tan\theta = \frac[v_x}{v_y}[/tex]

[tex]\theta = tan^{-1} \frac{9.88}{5.94}[/tex]

[tex]\theta = 58.98[/tex]°

Answer:

a) 0.067 seconds

b) 112.5 m/s²

Explanation:

t = Time taken

u = Initial velocity = 7.5 m/s

v = Final velocity = 0

s = Displacement = 0.25 m

a = Acceleration

Eqaution of motion

[tex]v^2-u^2=2as\\\Rightarrow a=\frac{v^2-u^2}{2s}\\\Rightarrow a=\frac{0^2-7.5^2}{2\times 0.25}\\\Rightarrow a=-112.5\ m/s^2[/tex]

b) The deceleration of the football player is 112.5 m/s²

[tex]v=u+at\\\Rightarrow t=\frac{v-u}{a}\\\Rightarrow t=\frac{0-7.5}{-112.5}\\\Rightarrow t=0.067\ s[/tex]

a) The collision lasts for 0.067 seconds

Answer:

36 rev

Explanation:

See it in the pic

Answer:

The acceleration of the car is [tex]20 m/s^{2}[/tex]

Solution:

According to the question:

Initial velocity of the car, u = 12 m/s

Final velocity of the car, v = 52 m/s

Time interval for the change in velocity, [tex]\DeltaT = 2 s[/tex]

The rate at which the velocity of an object varies is referred to as its acceleration:

[tex]a = \frac{v - u}{\Delta T}[/tex]

[tex]a = \frac{52 - 12}{2} = 20 m/s^{2}[/tex]

Answer:

[tex]8.66\times 10^{-6}\ C[/tex] or [tex]8.66\ \mu C[/tex].

Explanation:

Given:

For the moving particle to circular motion, the electrostatic force between the two must be balanced by the centripetal force on the moving particle.

The electrostatic force on the moving particle due to the charge Q at origin is given by Coulomb's law as:

[tex]\rm F_e = \dfrac{kqQ}{r^2}.[/tex]

where, [tex]\rm k[/tex] is the Coulomb's constant having value [tex]\rm 9\times 10^9\ Nm^2/C^2.[/tex]

The centripetal force on the moving particle due to particle at origin is given as:

[tex]\rm F_c = \dfrac{mv^2}{r}.[/tex]

For the two forces to be balanced,

[tex]\rm F_e = F_c\\\dfrac{kqQ}{r^2}=\dfrac{mv^2}{r}\\\Rightarrow Q = \dfrac{mv^2}{r}\times \dfrac{r^2}{kq}\\=\dfrac{mv^2r}{kq}\\=\dfrac{(0.959\times 10^{-3})\times (47.9)^2\times (0.207)}{(9\times 10^9)\times (5.84\times 10^{-6})}\\=8.66\times 10^{-6}\ C\\=8.66\ \mu C.[/tex]

For the moving particle to execute circular motion, the charge Q must be approximately 6.75 µC and have the same sign as the particle's charge (5.84 µC) to ensure attraction and centripetal motion.

To determine the value of charge Q that will allow the moving particle to execute circular motion, we must consider the forces acting on the charged particle due to the electric field created by Q.

For the charged particle to execute circular motion, the centripetal force required to keep it in circular motion must equal the electric force acting on it due to charge Q.

Centripetal Force:
The centripetal force (
F_c) is given by:
[tex]F_c = \frac{mv^2}{r}[/tex]
where r is the radius of the circular motion. Here, r = 0.207 m (distance from the origin).

Substituting the given values, we get:
[tex]F_c = \frac{0.000959 \text{ kg} \times (47.9 ext{ m/s})^2}{0.207 ext{ m}}[/tex]
[tex]F_c = \frac{0.000959 \times 2299.61}{0.207} \approx 0.008291 \text{ N}[/tex]

Electric Force:
The electric force (F_e) on the particle due to the charge Q at the origin is given by Coulomb's Law:
[tex]F_e = \frac{k |Q| |q|}{r^2}[/tex]
where k is Coulomb's constant [tex]k \approx 8.99 \times 10^9 \text{ N m}^2/ ext{C}^2[/tex].

Setting the electric force equal to the centripetal force for circular motion:
[tex]\frac{k |Q| |q|}{r^2} = F_c[/tex]
Plugging in the numbers:
[tex]\frac{(8.99 \times 10^9) |Q| (5.84 \times 10^{-6})}{(0.207)^2} = 0.008291[/tex]

Solving for Q:
Rearranging for Q:
[tex]|Q| = \frac{0.008291 \cdot (0.207)^2}{8.99 \times 10^9 imes (5.84 \times 10^{-6})}[/tex]
Calculate the right side:
[tex]|Q| = \frac{0.008291 \cdot 0.042849}{8.99 \times 10^9 imes 5.84 \times 10^{-6}}[/tex]
[tex]|Q| \approx \frac{0.000355}{0.05256} \approx 6.75 \times 10^{-6} \text{ C}[/tex] or 6.75 µC.

Answer:

The position of near point is 1.68 m.

Explanation:

Given that,

Power = + 4.4 D

Object distance = 20 cm

We need to calculate the focal length

Using formula of power

[tex]P =\dfrac{1}{f}[/tex]

[tex]f=\dfrac{1}{P}[/tex]

[tex]f=\dfrac{1}{4.4}[/tex]

[tex]f=0.227\ m[/tex]

We need to calculate the image distance

Using formula of lens

[tex]\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}[/tex]

[tex]\dfrac{1}{v}=\dfrac{1}{f}+\dfrac{1}{u}[/tex]

Put the value into the formula

[tex]\dfrac{1}{v}=\dfrac{1}{0.227}-\dfrac{1}{0.20}[/tex]

[tex]\dfrac{1}{v}=-\dfrac{135}{227}[/tex]

[tex]v=-1.68\ m[/tex]

Hence, The position of near point is 1.68 m.

Mary's near point without glasses is 22.73 cm, calculated by using the lens equation and the given lens power of her glasses with a knowledge of optics.

The question involves determining the near point location when the student, Mary, is not using her glasses. Mary has glasses with +4.4D converging lenses, and with the glasses, her near point is at 20 cm. To find the near point without glasses, we must use the lens equation:

1/f = 1/di + 1/do

Where f is the focal length of the lens, di is the image distance, and do is the object distance. For her glasses:  

The lens power P is +4.4D.

The focal length f is the inverse of power, f = 1/P.

Hence, f = 1/4.4 = 0.2273 meters (or 22.73 cm).

Since the near point with glasses is the closest distance at which she can see clearly, that means the image distance di (for her near point with glasses) is at infinity because the lenses correct her vision to normal. So:

1/f = 0 + 1/do

1/22.73 cm = 1/do

Therefore, the near point without glasses, do, is also 22.73 cm, since for the near point with glasses di is infinite - hence the person can see objects clearly at the near point with the glasses when the image is formed at infinity without the glasses.

Answer:

Explanation:

The steam was earlier at 300 ° C. and pressure of 1 MPa. When the gas is allowed to expand against vacuum , work done by the gas is nil because there is no external pressure against which it has to work . Therefore there will not be any change in its internal energy. Since the tank is insulated therefore there is no possibility of external heat to increase its internal energy.

Hence the temperature of gas will remain unaffected.  It will remain stagnant at 300