An ideal, monotomic gas initially at a temperature of 450K, a pressure of 4.00 atm and a volume of 10.0L, undergoes an adiabatic compression to 1/3 its original volume. Find the final temperature of the gas. A. 72 K B. 150 K C. 216 K D. 936 K E. 1350 K

Answer:

The individual positive plate capacity is 85 Ah.

(D) is correct option.

Explanation:

Given that,

Number of plates = 15

Capacity = 595 Ah

We need to calculate the individual positive plate capacity in motive power cell

We have,

15 plates means 7 will make pair of positive and negative.

So, there are 7 positive cells individually.

The capacity will be

[tex]capacity =\dfrac{power}{number\ of\ cells}[/tex]

Put the value into the formula

[tex]capacity =\dfrac{595}{7}[/tex]

[tex]capacity =85\ Ah[/tex]

Hence, The individual positive plate capacity is 85 Ah.

Answer:

SDFGHJKL

Explanation:

Answer:

Acceleration, [tex]a=-2.48\ m/s^2[/tex]

Explanation:

Initial speed of the skater, u = 8.4 m/s

Final speed of the skater, v = 6.5 m/s

It hits a 5.7 m wide patch of rough ice, s = 5.7 m

We need to find the acceleration on the rough ice. The third equation of motion gives the relationship between the speed and the distance covered. Mathematically, it is given by :

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

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

[tex]a=\dfrac{(6.5)^2-(8.4)^2}{2\times 5.7}[/tex]

[tex]a=-2.48\ m/s^2[/tex]

So, the acceleration on the rough ice [tex]-2.48\ m/s^2[/tex] and negative sign shows deceleration.

Answer:

(a) Magnitude: 14.4 N

(b) Away from the +6 µC charge

Explanation:

As the test charge has the same sign, the force that the other charges exert on it will be a repulsive force. The magnitude of each of the forces will be:

[tex]F_e = K\frac{qq_{test}}{r^2}[/tex]

K is the Coulomb constant equal to 9*10^9 N*m^2/C^2, q and qtest is the charge of the particles, and r is the distance between the particles.

Let's say that a force that goes toward the +6 µC charge is positive, then:

[tex]F_e_1 = K\frac{q_1q_{test}}{r^2}=-9*10^9 \frac{Nm^2}{C^2} \frac{6*10^{-6}C*4*10^{-6}C}{(0.1m)^2} =-21.6 N[/tex]

[tex]F_e_2 = K\frac{q_2q_{test}}{r^2}=9*10^9 \frac{Nm^2}{C^2} \frac{2*10^{-6}C*4*10^{-6}C}{(0.1m)^2} =7.2 N[/tex]

The magnitude will be:

[tex]F_e = -21.6 + 7.2 = -14.4 N[/tex], away from the +6 µC charge

Answer:

21.28 m

Explanation:

height, h = 71 m

velocity of raft, v = 5.6 m/s

let the time taken by the stone to reach to raft is t.

use second equation of motion for stone

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

u = 0 m/s, h = 71 m, g = 9.8 m/s^2

71 = 0 + 0.5 x 9.8 x t^2

t = 3.8 s

Horizontal distance traveled by the raft in time t

d = v x t = 5.6 x 3.8 = 21.28 m

Answer:

d=9.462×10^15 meters

Explanation:

Relation between distance, temps and velocity:

d=v*t

t=1year*(365days/1year)*/(24hours/1day)*(3600s/1h)=31536000s

So:

1 light year=d=3*10^8m/s*3.154*10^7s=9.462×10^15 meters

Answer:

(a) 27.1 m/s

(b) 3.9 second

Explanation:

Let the speed is u.

Maximum horizontal range, R = 75 m

The range is maximum when the angle of projection is 45°.

(a) Use the formula for the maximum horizontal range

[tex]R=\frac{u^{2}}{g}[/tex]

[tex]75=\frac{u^{2}}{9.8}[/tex]

u = 27.1 m/s

(b) Let the time of flight is T.

Use the formula for the time of flight

[tex]T=\frac{2uSin\theta}{g}[/tex]

[tex]T=\frac{2\times 27.1 \times Sin45}{9.8}[/tex]

T = 3.9 second

Answer:

A and B

Explanation:

To find the speed at which the shoe hits the ground, we use the equation v = √(u²+2as). To determine the time it takes for the shoe to hit the ground, we use the formula t = √(2h/g). Substituting the given values into these equations, we can find the speed at which the shoe hits the ground and the time it takes to do so.

The question is both a physics problem related to gravity and the free fall of an object. First, we need to find the final velocity of the shoe when it hits the ground. To calculate this, we can use the following equation from physics: v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time. Since the shoe starts from the initial velocity of the parachutist (15.1 m/s), the acceleration a is the acceleration due to gravity (9.81 m/s²), and it falls a distance of 41.2m, we will need to use the equation v = √(u²+2as) to solve for v.

Next, to determine when the shoe hits the ground, we use the equation for time in free fall: t = √(2h/g), where h is the height (41.2 m), and g is acceleration due to gravity (9.81 m/s²). Substituting the given values, we can find the times when the shoe will hit the ground.

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Answer:

44.1 m

Explanation:

Given:

Assumptions:

We know that the distance traveled by the sound in a particular medium is equal to the product of the speed of sound in that medium and the time taken.

For traveling sound through the rod, we have

[tex]l=V_r t\\\Rightarrow t = \dfrac{l}{V_r}[/tex]..........eqn(1)

For traveling sound through the air to the women ear for traveling the same distance, we have

[tex]l=V_aT\\\Rightarrow l=V_a(t+0.12)\\\Rightarrow l=V_a(\dfrac{l}{V_r}+0.12)\,\,\,\,\,\,(\textrm{From eqn (1)})\\\Rightarrow l=V_a(\dfrac{l}{15V_a}+0.12)\\\Rightarrow l=\dfrac{l}{15}+0.12V_a\\\Rightarrow l-\dfrac{l}{15}=0.12V_a\\\Rightarrow \dfrac{14l}{15}=0.12V_a\\\Rightarrow l = \dfrac{15}{14}\times 0.12V_a\\\Rightarrow l = \dfrac{15}{14}\times 0.12\times 343\\\Rightarrow l = \dfrac{15}{14}\times 0.12\times 343\\\Rightarrow l = 44.1\ m[/tex]

Hence, the length of the rod is 44.1 m.

The length of the rod can be calculated using the difference in hearing times and the speed of sound in the rod and air. Using the formula for distance (speed x time), and given that the speed of sound in the rod is 15 times the speed of sound in air, the length of the rod is found to be approximately 41.16 meters.

In this problem, we know that the speed of sound in the rod is 15 times the speed of sound in the air, and that the woman hears the sound of the strike twice with a 0.12 second gap. The first sound is transmitted through the rod and the second, through the air. Therefore, we can use this information to conclude that the difference in time is the amount of time it takes for the sound to travel the length of the rod in air after it already traveled through the rod.

The speed of sound in the rod is 15 times the speed of sound in air, which is given as 343 m/s. So, the speed of sound in the rod is 15 * 343 = 5145 m/s.

We are looking for the distance travelled, which is the length of the rod. We can find the distance by using the formula distance = speed x time. In this case we are calculating distance as time taken for sound to travel through air minus the time taken to travel through the rod. Therefore, the length of the rod can be calculated to be 343 m/s * 0.12 s = 41.16 meters.

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Answer:

Option C

Explanation:

given,

velocity of airplane = 80 m/s

angle with the horizontal = 15°

speed of the ground= ?

when the plane is taking off the horizontal component of the velocity is v cosθ

so,      

        ground speed of the airplane is = [tex]v\times cos\theta[/tex]

                                                              = [tex]80 \times cos 15^0[/tex]

                                                           v  =  77.27 m/s

horizontal velocity of the air plane comes out to be 77.27 m/s ≅ 77 m/s

so, the correct option is Option C

Answer:

0.68 m/s²

Explanation:

Given:

Mass of the dog,  m = 25 kg

Tension in the rope = 10 N

Force applied by the girl on the dog = 7 N

Now,

since the boy is pulling the dog and the girl is pushing the dog

Thus,

The net force on the dog = 10 N + 7 N = 17 N

also,

Net force on the dog = Mass × Acceleration

thus,

25 kg × Acceleration = 17 N

or

Acceleration = [tex]\frac{\textup{17 N}}{\textup{25 kg}}[/tex]

or

Acceleration = 0.68 m/s²

Answer:

a) 3.56m/s

b) 1.73m

Explanation:

We have to treat this as a parabolic motion problem:

we will use the next formulas:

[tex]V=Vo+a*t\\Vy^2=Vyo^2+2*a*Y\\X=Vox*t[/tex]

we first have to calculate the initial velocity of the basketball player:

[tex]Vy^2=Vyo^2+2*a*Y\\0^2=Vyo^2+2*(-9.8)*(0.650)\\Vyo=\sqrt{2*9.8*0.650} \\Vyo=3.56 m/s[/tex]

the final velocity is zero when he reaches the maximun height.

To answer the second part we need to obtain the time to reach the maximun height, so:

[tex]V=Vo+a*t\\\\0=3.56+(-9.8)*t\\t=0.36 seconds[/tex]

Now having that time, let's find the distance on the X axis, the X axis behaves as constant velocity movement, so:

[tex]X=Vox*t\\X=4.80*0.36\\X=1.73m[/tex]

Answer:

(a) 34.47 cm

(b) [tex]24.09^\circ[/tex] south of west

Explanation:

Let us draw a figure representing the individual displacement vectors in the four jumps as shown in the figure attached with this solution.

Now, let us try to write the four displacement vectors in in terms of unit vectors along the horizontal and the vertical axis.

[tex]\vec{d}_1= 31\ cm\ west = -31\ cm\ \hat{i}\\\vec{d}_2= 26\ cm\ south\ of\ west = -26\cos 44^\circ\ \hat{i} -26 \sin 44^\circ\ \hat{j}=(-18.72\ \hat{i}-18.06\ \hat{i})\ cm\\\vec{d}_3= 22\ cm\ south\ of\ east = 22\cos 56^\circ\ \hat{i} -22 \sin 56^\circ\ \hat{j}=(12.30\ \hat{i}-18.23\ \hat{i})\ cm\\\vec{d}_4= 23\ cm\ north\ of\ east = 23\cos 75^\circ\ \hat{i} +23\sin \sin 75^\circ\ \hat{j}=(5.95\ \hat{i}+22.22\ \hat{i})\ cm\\[/tex]

Now, the vector sum of all these vector will give the resultant displacement vector.

[tex]\vec{D} = \vec{d}_1+\vec{d}_2+\vec{d}_3+\vec{d}_4\\\Rightarrow \vec{D} = -31\ cm\ \hat{i}+(-18.72\ \hat{i}-18.06\ \hat{i})\ cm+(12.30\ \hat{i}-18.23\ \hat{i})\ cm+(5.95\ \hat{i}+22.22\ \hat{i})\ cm\\\Rightarrow \vec{D} =(-31.47\ \hat{i}-14.07\ \hat{i})\ cm[/tex]

Part (a):

The magnitude of the resultant displacement vector is given by:

[tex]D=\sqrt{(-31.47)^2+(-14.07)^2}\ m = 34.47\ m[/tex]

Part (b):

Since the resultant displacement vector indicates that the final position of the vector lies in the third quadrant, the vector will make some positive angle in the direction south of west given by:

[tex]\theta = \tan^{-1}(\dfrac{14.07}{31.47})= 24.09^\circ[/tex]

To find the resultant displacement of the grasshopper, we can break down the vectors into their x and y components, and then sum up the components separately. After performing the calculations, we find that the magnitude of the resultant displacement is approximately 39.4 cm and the direction is approximately 38.3° south of west.

To find the resultant displacement of the grasshopper, we need to add the individual displacement vectors. We can do this by breaking down each vector into its x and y components.

For vector (1) with a magnitude of 31.0 cm due west, the x component is -31.0 cm and the y component is 0.

Similarly, for the other vectors, the x and y components are:

Now, we can sum up the x components and y components separately to find the resultant displacement.

The magnitude of the resultant displacement can be found using the formula:

resultant magnitude = sqrt((sum of x components)^2 + (sum of y components)^2)

The direction of the resultant displacement can be found using the formula:

resultant direction = atan2((sum of y components), (sum of x components))

Plugging in the values and performing the calculations, we find that the magnitude of the resultant displacement is approximately 39.4 cm and the direction of the resultant displacement is approximately 38.3° south of west.

Answer:

[tex]\frac{F}{W} = 9.37 \times 10^{-4}[/tex]

Explanation:

Radius of the pollen is given as

[tex]r = 12.0 \mu m[/tex]

Volume of the pollen is given as

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

[tex]V = \frac{4}{3}\pi (12\mu m)^3[/tex]

[tex]V = 7.24 \times 10^{-15} m^3[/tex]

mass of the pollen is given as

[tex]m = \rho V[/tex]

[tex]m = 7.24 \times 10^{-12}[/tex]

so weight of the pollen is given as

[tex]W = mg[/tex]

[tex]W = (7.24 \times 10^{-12})(9.81)[/tex]

[tex]W = 7.1 \times 10^{-11}[/tex]

Now electric force on the pollen is given

[tex]F = qE[/tex]

[tex]F = (-0.700\times 10^{-15})(95)[/tex]

[tex]F = 6.65 \times 10^{-14} N[/tex]

now ratio of electric force and weight is given as

[tex]\frac{F}{W} = \frac{6.65 \times 10^{-14}}{7.1 \times 10^{-11}}[/tex]

[tex]\frac{F}{W} = 9.37 \times 10^{-4}[/tex]

Answer:

1) We will have excess of electrons

2) The number of electrons transferred equals [tex]1.343\times 10^{10}[/tex]

Explanation:

Part a)

Since we know that the charge transfer occurs by the transfer of electrons only as it is given that the carpet has acquired a positive charge it means that it has lost some of the electron's since electrons are negatively charged and if a neutral body looses negative charge it will become positively charged. The electron's that are lost by the carpet will be acquired by the feet of the human thus making us negatively charged.Hence we will gain electrons making us excess in electrons.

Part b)

From charge quantinization principle we have

[tex]Q=ne[/tex]

where

Q = charge of body

n = no of electrons

e = fundamental charge

Applying values in the above equation we get

[tex]2.15\times 10^{-9}C=n\times 1.6\times 10^{-19}C\\\\\therefore n=\frac{2.15\times 10^{-9}C}{1.6\times 10^{-19}C}=1.343\times 10^{10}[/tex]

Answer:

Explanation:

Electric field due to a charge Q at a point d distance away is given by the expression

E = k Q / d , k is a constant equal to 9 x 10⁹

Field due to charge = 3 X 10⁻⁹ C

E = E = [tex]\frac{9\times 10^9\times3\times10^{-9}}{d^2}[/tex]

Field due to charge = 4 X 10⁻⁹ C

[tex]E = [tex]\frac{9\times 10^9\times4\times10^{-9}}{(2-d)^2}[/tex]

These two fields will be equal and opposite to make net field zero

[tex]\frac{9\times 10^9\times3\times10^{-9}}{d^2}[/tex] = [tex][tex]\frac{9\times 10^9\times4\times10^{-9}}{(2-d)^2}[/tex][/tex]

[tex]\frac{3}{d^2} =\frac{4}{(2-d)^2}[/tex]

[tex]\frac{2-d}{d} =\frac{2}{1.732}[/tex]

d = 0.928

The bridge was approximately 11.01 meters high. The swimmers were moving at a speed of around 14.72 m/s when they hit the water. If the bridge were twice as high, the drop time would be about 2.12 seconds.

The subject in question involves principles of physics, specifically gravitational acceleration. When an object is dropped, it falls under the influence of gravity. This is usually around 9.81 m/s2 on Earth. Using the formula for motion d=1/2gt^2, where d is the distance or height, g is the acceleration due to gravity, and t is the time it takes for the object to fall.

a.) For calculation of the bridge's height, plug in the values of time (t=1.5s) and gravity (g=9.81m/s^2) into the equation, we get d=1/2*(9.81)*(1.5)^2 = 11.01 meters. The height of the bridge from which they jumped is approximately 11 meters.

b.) To calculate how fast the swimmers were moving when they hit the water, we can use the equation v=gt, where v is the velocity or speed. Plugging in the values, we get v=(9.81)*(1.5) = 14.72 m/s. So, the swimmers were moving at about 14.7 meters per second when they hit the water.

c.) If the bridge were twice as high, the time of drop would be found using the equation t= sqrt(2d/g), where d is now twice the original distance = 2*11.01 = 22.02 meters. Substituting the given values, we get t= sqrt(2(22.02)/9.81) = 2.12 seconds. The swimmer's drop time would be approximately 2.12 seconds.

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Answer:

The time taken by missile's clock is [tex]4.6\times 10^{6} s[/tex]

Solution:

As per the question:

Speed of the missile, [tex]v_{m = 6.5\times 10^{3}} m/s[/tex]

Now,

If 'T' be the time of the frame at rest then the dilated time as per the question is given as:

T' = T + 1

Now, using the time dilation eqn:

[tex]T' = \frac{T}{\sqrt{1 + (\frac{v_{m}}{c})^{2}}}[/tex]

[tex]\frac{T'}{T} = \frac{1}{\sqrt{1 + (\frac{v_{m}}{c})^{2}}}[/tex]

[tex]\frac{T + 1}{T} = \frac{1}{\sqrt{1 + (\frac{v_{m}}{c})^{2}}}[/tex]

[tex]1 + \frac{1}{T} = \frac{1}{\sqrt{1 + (\frac{v_{m}}{c})^{2}}}[/tex]

[tex]1 + \frac{1}{T} = (1 + (\frac{v_{m}}{c})^{2})^{- \frac{1}{2}}[/tex]         (1)

Using binomial theorem in the above eqn:

We know that:

[tex](1 + x)^{a} = 1 + ax[/tex]

Thus eqn (1) becomes:

[tex]1 + \frac{1}{T} = 1 - \frac{- 1}{2}.\frac{v_{m}^{2}}{c^{2}}[/tex]

[tex]T = \frac{2c^{2}}{v_{m}^{2}}[/tex]

Now, putting appropriate values in the above eqn:

[tex]T = \frac{2(3\times 10^{8})^{2}}{(6.5\times 10^{3})^{2}}[/tex]

[tex]T = 4.6\times 10^{6} s[/tex]

Answer:

25

Explanation:

Answer:Increase

Explanation:

Given

You are holding 2 kg mass in each outstreched hand

If the masses are dropped then Moment of inertia will decease by [tex]2mr^2[/tex]

Where m=2 kg

r=length of stretched arm

Since angular momentum is conserved therefore decrease in Moment of inertia will result in increase of angular velocity

as I[tex]\omega [/tex]=constant

I=Moment of inertia

[tex]\omega [/tex]=angular velocity

Answer: 1 microgram is equal to 0.001 miligrams

Explanation: The factor micro is equal 10^-3 while the factor mili is equal to 10^-3 so to converte the micro to mile we have to multiply by 0.001.

Answer:

it move with velocity 1571 m/s

Explanation:

given data

wavelength λ = 4.63 × [tex]10^{-7}[/tex] m

to find out

how fast is it moving

solution

we will use here de Broglie wavelength equation

that is

wavelength λ = [tex]\frac{h}{mv}[/tex]    ..........1

here h is planck constant = 6.626068 × [tex]10^{-34}[/tex]

and m is mass of electron i.e = 9.10938188 × [tex]10^{-31}[/tex]

and v is velocity

put all value we find velocity  in equation 1

wavelength λ = [tex]\frac{h}{mv}[/tex]  

v = [tex]\frac{6.626068*10^{-34}}{9.10938188*10^{-31}*4.63*10^{-7}}[/tex]

v =  1571.035464

so it move with velocity 1571 m/s

Answer:

[tex]v=1.57*10^{3}\frac{m}{s}[/tex]

Explanation:

As DeBroglie equation proved by Davisson-Germer experiment says, the wavelength of an electron is related with its velocity with the equation:

λ = [tex]\frac{h}{mv}[/tex]

where m is the mass of the electron [tex]m=9.11*10^{-31}kg[/tex], h is the Planck´s constant [tex]h=6.626*10^{-34}J.s[/tex] and v its velocity.

Solving the equation for the velocity of the electron, we have:

v = h/mλ

And replacing the values:

[tex]v=\frac{6.626*10^{-34}J.s}{(9.11*10^{-31}Kg)*(4.63*10^{-7}m)}[/tex]

[tex]v=1570.9\frac{m}{s}[/tex]

[tex]v=1.57*10^{3}\frac{m}{s}[/tex]

Answer:

a) 0°

b) 180°

c) 90°

Explanation:

Hello!

To solve this question let a be the vector whose length is 15 m and b the vector of length 20 m

So:

|a | = 15

|b | = 20

Since we are looking for the angle between the vectors we need to calculate the length of the sum of the two vectors, this is:

[tex]|a+b|^{2} = |a|^{2} + |b|^{2} + 2 |a||b|cos(\theta)[/tex]

Now we replace the value of the lengths:

[tex]|a+b|^{2} = 15^{2} + 20^{2} + 2*15*20*cos(\theta)[/tex]

[tex]|a+b|^{2} = 625 + 600*cos(\theta)[/tex] --- (1)

Now, if:

a) |a+b| = 35

First we can see that 20 + 15 = 35, so the angle must be 0, lets check this:

[tex]35^{2} = 625 + 600*cos(\theta)[/tex]

[tex]1225 = 625 + 600*cos(\theta)[/tex]

[tex]600 =  600*cos(\theta)[/tex]

[tex]1= cos(\theta)[/tex]

and :

[tex]\theta = arccos(1)[/tex]

      θ = 0

b) |a+b|=5

From eq 1 we got:

[tex]\theta = arccos(\frac{|a+b|^{2}-625}{600})[/tex]   --- (2)

[tex]\theta = arccos(\frac{|a+b|^{2}-625}{600})[/tex]

[tex]\theta = arccos(-1)[/tex]

  θ = π     or    θ = 180°

c) |a+b|=25

[tex]\theta = arccos(\frac{|25|^{2}-625}{600})[/tex]

[tex]\theta = arccos(-1)[/tex]

  θ = π/2     or    θ = 90°

Final answer:

In vector addition, an angle of 0° between two vectors gives a resultant of 35 m, an angle of 180° gives a resultant of 5 m, and the angle for a resultant of 25 m can be found using the Law of Cosines.

Explanation:

The question involves the concept of vector addition and the use of trigonometry to determine the resultant displacement when two vectors are combined at various angles. The displacement vectors have magnitudes of 15 m and 20 m, and we are interested in finding the angles that would result in resultant displacements of 35 m, 5 m, and 25 m, respectively.

For (a) a resultant displacement of 35 m, the two vectors must be added in the same direction. This implies that the angle between them is 0°.

For (b) a resultant displacement of 5 m, the two vectors must be in exactly opposite directions. Since the difference in magnitudes is 5 m, this means that the larger vector (20 m) partially cancels out the smaller vector (15 m). Hence, the angle between them is 180°.

For (c) a resultant displacement of 25 m, we can use the Law of Cosines to determine the angle:

c2 = a2 + b2 - 2ab cos(θ)

Where a = 15 m, b = 20 m, and c = 25 m. Solving this equation will give us the value of θ.

Answer:

244.64m

Explanation:

First, we find the distance traveled with constant velocity. It's simply multiplying velocity time the time that elapsed:

[tex]x = V*t = -8\frac{m}{s} *3s = -24m[/tex]

After this, the ball will start traveling with a constant acceleration motion. Due to the fact that the acceleration is the opposite direction to the initial velocity, this motion will have 2 phases:

1. The velocity will start to decrease untill it reaches 0m/s.

2. Then, the velocity will start to increase at the rate of the acceleration.

The distance that the ball travels in the first phase can be found with the following expression:

[tex]v^2 = v_0^2 + 2a*d[/tex]

Where v is the final velocity (0m/s), v_0 is the initial velocity (-8m/s) and a is the acceleration (+9m/s^2). We solve for d:

[tex]d = \frac{v^2 - v_0^2}{2a} = \frac{(0m/s)^2 - (-8m/s)^2}{2*7m/s^2}= -4.57m[/tex]

Now, before finding the distance traveled in the second phase, we need to find the time that took for the velocity to reach 0:

[tex]t_1 = \frac{v}{a} = \frac{8m/s}{7m/s^2} = 1.143 s[/tex]

Then, the time of the second phase will be:

[tex]t_2 = 9s - t_1 = 9s - 1.143s = 7.857s[/tex]

Using this, we using the equations for constant acceleration motion in order to calculate the distance traveled in the second phase:

[tex]x = \frac{1}{2}a*t^2 + v_0*t + x_0[/tex]

V_0, the initial velocity of the second phase, will be 0 as previously mentioned. X_0, the initial position, will be 0, for simplicity:

[tex]x = \frac{1}{2}*7\frac{m}{s^2}*t^2 + 0m/s*t + 0m = 216.07m[/tex]

So, the total distance covered by this object in meters will be the sum of all the distances we found:

[tex]x_total = 24m + 4.57m + 216.07m = 244.64m[/tex]

Answer:

Part a)

[tex]q_1 = -1.47 \times 10^{-6} C[/tex]

Part b)

[tex]q_2 = 3.75 \times 10^{-6} C[/tex]

Explanation:

Let the charge on two spheres is q1 and q2

now the force between two charges are

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

[tex]0.116 = \frac{(9\times 10^9)(q_1)(q_2)}{0.654^2}[/tex]

[tex]q_1 q_2 = 5.51 \times 10^{-12}[/tex]

now when we connect then with conducting wire then both sphere will equally divide the charge

so we will have

[tex]q = \frac{q_1-q_2}{2}[/tex]

now we have

[tex]0.0273 = \frac{(9\times 10^9)(\frac{q_1- q_2}{2})^2}{0.654^2}[/tex]

[tex]q_1 - q_2 = 2.28\times 10^{-6} C[/tex]

now we will have

Now we can solve above two equations

Part a)

negative charge on the sphere is

[tex]q_1 = -1.47 \times 10^{-6} C[/tex]

Part b)

positive charge on the sphere is

[tex]q_2 = 3.75 \times 10^{-6} C[/tex]

Answer:

distance between the charges is 5.12 × 10⁶ m

Explanation:

charges q₁ = -130.0 C                

              q₂ = 180 C                  

force between the charges = 8 N

force between two charge                              

   [tex]F = \dfrac{k q_1q_2}{r^2}[/tex]

value of  K =8.975 × 10⁹ N.m²/C²

    [tex]8 = \dfrac{8.975 \times 10^{9}\times 130 \times 180}{r^2}[/tex]

    [tex]r^2 = \dfrac{8.975 \times 10^{9}\times 130 \times 180}{8}[/tex]

    [tex]r^2 =2.625 \times 10^{13} [/tex]

    r = 5.12 × 10⁶  m                                          

hence, distance between the charges is 5.12 × 10⁶ m.

Answer:

acceleration = 4.4 m/s²

time is 3.86 s

Explanation:

given data

initial speed = 7 m/s

final speed = 24 m/s

distance = 60 m

to find out

acceleration and time when change speed change

solution

we will apply here equation of motion for acceleration

v²-u² = 2×a×s   .................1

here v is final speed and u is initial speed and s is distance and a is acceleration

put here all these value

24²-7² = 2×a×60

so

a = 4.4

acceleration = 4.4 m/s²

and

now find time by equation of motion

v = u +at

put her value

24 = 7 + 4.4 (t)

t = 3.86

so time is 3.86 s

Answer:

946.92 kJ

Explanation:

This process has 3 parts:

1. The first part, where the temperature of Ethyl alcohol remains constant and it changes from gas to liquid.

2. The second part, where the temperature drops from 78°C to -114°C

3. The third parts, where the temperature remains constant and it changes from liquid to solid.

The energy lost in a phase change is:

Q = m*cl

The energy lost because of the drop in temperature is:

[tex]Q = m c(T_2-T_1)[/tex]

cl is the heat of vaporization or heat of fusion, depending on the type of phase change. c is the specific heat.

So, the energy lost in each part is:

1. [tex]Q_1 = 0.651kg*879 kJ/kg =  572.23 kJ[/tex]

2. [tex]Q_2 = 0.651kg*2.43 kJ/kgK(78.0^oC - (-114^oC)) = 303.73 kJ[/tex]

3. [tex]Q_3 = 0.651kg*109kJ/kg = 70.96 kJ[/tex]

Then, the total energy removed should be:

Q = Q1 + Q2 + Q3 = 572.23 kJ + 303.73kJ + 70.96kJ = 946.92 kJ

Explanation:

Given that,

Weight of water = 25 kg

Temperature = 23°C

Weight of mass = 32 kg

Distance = 5 m

(a). We need to calculate the amount of work done on the water

Using formula of work done

[tex]W=mgh[/tex]

[tex]W=32\times9.8\times5[/tex]

[tex]W=1568\ J[/tex]

The amount of work done on the water is 1568 J.

(b). We need to calculate the internal-energy change of the water

Using formula of internal energy

The change in internal energy of the water equal to the amount of the  work done on the water.

[tex]\Delta U=W[/tex]

[tex]\Delta U=1568\ J[/tex]

The  change in internal energy is 1568 J.

(c). We need to calculate the final temperature of the water

Using formula of the change internal energy

[tex]\Delta U=mc_{p}\Delta T[/tex]

[tex]\Delta U=mc_{p}(T_{2}-T_{1})[/tex]

[tex]T_{2}=T_{1}+\dfrac{\Delta U}{mc_{p}}[/tex]

[tex]T_{2}=23+\dfrac{1568}{25\times4.18\times10^{3}}[/tex]

[tex]T_{2}=23.01^{\circ}\ C[/tex]

The final temperature of the water is 23.01°C.

(d). The amount of heat removed from the water to return it to it initial temperature is the change in internal energy.

The amount of heat is 1568 J.

Hence, This is the required solution.

The work done on the water is 1568 Joules, which is also the internal-energy change of the water. The final temperature of the water is 23.015°C and to return the water to its initial temperature, 1568 Joules of heat must be removed.

(a) The amount of work done on the water is calculated using the formula for gravitational potential energy which depends on the weight's height, mass and acceleration due to gravity. Therefore, work done= mass × gravity × height = 32 kg × 9.8 m·s−2 × 5 m = 1568 Joules.

(b) As per the Law of Conservation of Energy, the work done on the water is converted completely into the internal energy of the water, so the internal-energy change of the water is 1568 Joules.

(c) The final temperature of the water can be calculated using the formula q = m × c × Δt, where 'q' is heat-transfer, 'm' is mass, 'c' is specific heat capacity and 'Δt' is change in temperature. Rearranging, we find Δt = q /(m × c). Substituting the known values gives Δt = 1568 J /(25 kg × 4.18 kJ/kgC) = 0.015 °C. Adding this to the initial temperature, we find the final temperature of the water is 23.015°C.

(d) To return the water to its initial temperature, the heat equal to the increase in internal energy must be removed. Hence, the amount of heat to be removed from the water = 1568 Joules.

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Answer:

Momentum of bullet

[tex]P = 2.72 kg m/s[/tex]

momentum of baseball

[tex]P = 8 kg m/s[/tex]

Explanation:

As we know that momentum is defined as the product of mass and velocity

here we know that

mass of the bullet = 8 gram

velocity of bullet = 340 m/s

momentum of the bullet is given as

[tex]P = mv[/tex]

[tex]P = (\frac{8}{1000})(340)[/tex]

[tex]P = 2.72 kg m/s[/tex]

Now we have

mass of baseball = 0.2 kg

velocity of baseball = 40 m/s[/tex]

momentum of baseball is given as

[tex]P = (0.2)(40)[/tex]

[tex]P = 8 kg m/s[/tex]