Which, if any, of the following statements concerning the work done by a conservative force is NOT true? All of these statements are true. It can always be expressed as the difference between the initial and final values of a potential energy function. None of these statements are true. It is independent of the path of the body and depends only on the starting and ending points. When the starting and ending points are the same, the total work is zero.

Answer:

0.06 Nm

Explanation:

mass of object, m = 3 kg

radius of gyration, k = 0.2 m

angular acceleration, α = 0.5 rad/s^2

Moment of inertia of the object

[tex]I = mK^{2}[/tex]

I = 3 x 0.2 x 0.2 = 0.12 kg m^2

The relaton between the torque and teh moment off inertia is

τ = I α

Wheree, τ is torque and α be the angular acceleration and I be the moemnt of inertia

τ = 0.12 x 0.5 = 0.06 Nm

Answer:

The location of the center of gravity for the entire body, relative to the floor is 1.03 m

Explanation:

To find the center of gravity of a system of particles, we use that

[tex]R_{cog} = \frac{r_{1}*m_{1}+r_{2}*m_{2}+...+r_{n}*m_{n}}{M}[/tex]

where R is the vector center of gravity of the system, formed by n particles, and n masses.

In this case, for a person standing on the floor and being their body divided in three sectors, each one with a weight and an altitud in a specific point (center of gravity of the body sector), instead of mass we have every "particle" weight in Newtons (force instead of mass), being each "particle" in the formula, a sector of the body.

On the other hand, we use only magnitude for the calculation, because the gravity force is vertical to the floor, so instead of our vector formula, we use it in the vertical direction as a magnitude formula. Thus

[tex]Y_{cog} = \frac{y_{1}*W_{1}+y_{2}*W_{2}+y_{3}*W_{3}}{Wt}[/tex]

where Y is the center of gravity, y=1, 2, 3 is every "sector point" altitude from the floor, W=1, 2, 3 is every weight of a body "sector", and Wt is the sum of the three weights.

In this way we replace in our formula with the correspondent values

[tex]Y_{cog} = \frac{1.28m*438N+0.76m*144N+0.25m*87N}{669N}[/tex]

obtaining our result

[tex]Y_{cog}=1.03 m[/tex]

Answer:

[tex]v= v_{student} ( 1 + \frac{d_{bus \ to \ the \ student}}{d_{student \ to \ puddle}} )[/tex]

Explanation:

We know that speed is:

[tex]v= \frac{distance}{time}[/tex].

So, to find the speed for the bus, we need to know:

Lets call [tex]v_{student}[/tex] the speed of the student, and [tex]d_{student \ to \ puddle}[/tex] the distance from the student to the puddle.

We can obtain the time taking

[tex]v_{student}= \frac {d_student \ to \ puddle}{t}[/tex]

as t must be the time that the student will take to reach the puddle:

[tex]t= \frac {d_{student \ to \ puddle}}{v_{student}}[/tex]

The bus is at a distance [tex]d_{bus \ to \ the \ student}[/tex] behind the student, so, the total distance that the bus must travel to the puddle is:

[tex]d_{bus \ to \ the \ puddle} = d_{bus \ to \ the \ student} + d_{student \ to \ puddle}[/tex]

Taking all this togethes, the formula must be:

[tex]v= \frac{d_{bus \ to \ the \ puddle}}{t}[/tex].

[tex]v= \frac{d_{bus \ to \ the \ student} + d_{student \ to \ puddle}}{\frac {d_{student \ to \ puddle}}{v_{student}}}[/tex].

[tex]v= v_{student} \frac{d_{bus \ to \ the \ student} + d_{student \ to \ puddle}}{d_{student \ to \ puddle}}[/tex].

[tex]v= v_{student} (\frac{d_{bus \ to \ the \ student}}{d_{student \ to \ puddle}} + \frac{d_{student \ to \ puddle}}{d_{student \ to \ puddle}} )[/tex].

[tex]v= v_{student} ( 1 + \frac{d_{bus \ to \ the \ student}}{d_{student \ to \ puddle}} )[/tex].

And this is the formula we are looking for.

Taking the values from the problem, we find

[tex]v= 1 \frac{m}{s} ( 1 + \frac{2 m}{1 m})[/tex]

[tex]v= 1 \frac{m}{s} ( 1 + 2)[/tex]

[tex]v= 3 \frac{m}{s}[/tex]

Answer:

Explanation:

Knowing the magnitude and directions relative to the x axis, we can find the Cartesian representation of the vectors using the formula

[tex]\vec{A}= | \vec{A} | \ ( \ cos(\theta) \ , \ sin (\theta) \ )[/tex]

where [tex]| \vec{A} |[/tex] its the magnitude and θ.

So, for our vectors, we will have:

[tex]\vec{D}= 3.00 m \ ( \ cos(315) \ , \ sin (315) \ )[/tex]

[tex]\vec{D}=  ( 2.121 m , -2.121 m )[/tex]

and

[tex]\vec{E}= 4.50 m \ ( \ cos(53.0) \ , \ sin (53.0) \ )[/tex]

[tex]\vec{E}= ( 2.71 m , 3.59 m )[/tex]

Now, we can take the sum of the vectors

[tex]\vec{R} = \vec{D} + \vec{E}[/tex]

[tex]\vec{R} = ( 2.121 \ m , -2.121 \ m ) + ( 2.71 \ m , 3.59 \ m )[/tex]

[tex]\vec{R} = ( 2.121 \ m  + 2.71 \ m , -2.121 \ m + 3.59 \ m ) [/tex]

[tex]\vec{R} = ( 4.831 \ m , 1.469 \ m ) [/tex]

This is R in Cartesian representation, now, to find the magnitude we can use the Pythagorean theorem

[tex]|\vec{R}| = \sqrt{R_x^2 + R_y^2}[/tex]

[tex]|\vec{R}| = \sqrt{(4.831 m)^2 + (1.469 m)^2}[/tex]

[tex]|\vec{R}| = \sqrt{23.338 m^2 + 2.158 m^2}[/tex]

[tex]|\vec{R}| = \sqrt{25.496 m^2}[/tex]

[tex]|\vec{R}| = 5.049 m[/tex]

To find the direction, we can use

[tex]\theta = arctan(\frac{R_y}{R_x})[/tex]

[tex]\theta = arctan(\frac{1.469 \ m}{4.831 \ m})[/tex]

[tex]\theta = arctan(0.304)[/tex]

[tex]\theta = 16\°54'33''[/tex]

As we are in the first quadrant, this is relative to the x axis.

Answer:

Double the charge to 2Q

Explanation:

The magnitude of a electric field caused by a charge Q at a distance R can be find by the following expression:

[tex]E = K\frac{Q}{R^2}[/tex]

Where K is the Coulomb constant.

As the relationship between the charge and the electric fiel is proportional, by simply doubling the charge, you would double the magnitude of the electric field.

Notice that the distance affects the magnitude of the electric field by the inverse square. So if you half the distance, the magnitude of the field will quadruple.

To double the electric field's magnitude at point P from a point charge Q at distance R, double the charge to 2Q while keeping the distance R unchanged. Option B is the correct option.

To double the magnitude of the electric field at point P due to a point charge Q a distance R away from P, you can follow Coulomb's law. Electric field (E) is directly proportional to the charge (Q) and inversely proportional to the square of the distance (R) between the point charge and the observation point.

If you double the charge to 2Q while keeping the distance R the same, the new electric field magnitude at point P will be double the original E. This is because E ∝ Q. Other options like reducing the distance to R/2 or R/4, or doubling the distance to 2R, would not result in a doubling of the field magnitude but rather lead to different field strengths according to the inverse square law, and altering both the charge and distance simultaneously is not necessary to achieve this specific goal. Option B is the correct option.

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The appropriate question is :

The electric field at point P due to a point charge Q a distance R away from P has magnitude E. In order to double the magnitude of the field at P, you could The electric field at point P due to a point charge Q a distance R away from P has magnitude E. In order to double the magnitude of the field at P, you could double the charge to 2Q and at the same time reduce the distance to R/2 OR

A) reduce the distance to R/4.

B) double the charge to 2Q.

C)reduce the distance to R/2.

D) double the distance to 2R.

The Earth moves 9.48 × 10⁸ kilometers in a year as it revolves around the Sun, calculated by multiplying the average speed of 30 km/s by the time of one year, 3.16 × 10⁷ seconds.

To calculate the distance that Earth moves in a year as it revolves around the Sun, we can use the formula for distance traveled, which is distance = speed × time. Given that Earth orbits around the Sun at an average speed of 30 kilometers per second (km/s) and the time it takes for one orbit is 3.16 × 107 seconds, the calculation would be:

30 km/s × 3.16 × 107 s = distance

Distance = 30 × 3.16 × 107 km

Distance = 94.8 × 107 km

Distance = 9.48 × 108 km

The Earth moves 9.48 × 108 kilometers in a year as it revolves around the Sun, using the correct number of significant figures.

Answer:

Explanation:

Ball is thrown downward:

initial velocity, u = - 20 m/s (downward)

height, h = - 60 m

Acceleration due to gravity, g = - 9.8 m/s^2 (downward)

(a) Let the speed of the ball as it hits the ground is v.

Use third equation of motion

[tex]v^{2}=u^{2}+2as[/tex]

[tex]v^{2}=(-20)^{2}+2\times 9.8 \times 60[/tex]

v = 39.69 m/s

(b) Let t be the time taken

Use First equation of motion

v = u + a t

- 39.69 = - 20 - 9.8 t

t = 2 second

Now the ball is thrown upwards:

initial velocity, u = 20 m/s (upward)

height, h = - 60 m

Acceleration due to gravity, g = - 9.8 m/s^2 (downward)

(c) Let the speed of the ball as it hits the ground is v.

Use third equation of motion

[tex]v^{2}=u^{2}+2as[/tex]

[tex]v^{2}=(-20)^{2}+2\times 9.8 \times 60[/tex]

v = 39.69 m/s

(d) Let t be the time taken

Use First equation of motion

v = u + a t

- 39.69 = + 20 - 9.8 t

t = 6.09 second

Answer:

a) 5.65 s

b) 5cm/s

c) They will pass each other at both 1.1168 s and 5.84s

d)15.084cm and 38.7 cm

Explanation:

For part A, you need to keep in mind that acceleration is the rate of change of velocity per unit of time. For a constant acceleration, this can be told in this way:

[tex]a = \frac{v - v_o}{t}[/tex]

Reordering this equation, we can get v in terms of the initial velocity, the acceleration, and the time elapsed:

[tex]v = at + v_o[/tex]

Now, we can get the expressions for velocity of each toy car, and equalize them:

[tex]v_1 =a_1t + v_o_1\\v_2 =a_2t + v_o_2\\v_1 = v2\\a_1t +v_o_1 =a_2t + v_o_2\\(a_1 - a_2)t = v_o_2 - v_o_1\\t = \frac{v_o_2 - v_o_1}{a_1 - a_2} = \frac{5cm/s - (-3cm/s)}{2.3 cm/s^2 - 0 cm/s^2}= 3.47 s[/tex]

As toy car has no acceleration and, therefore, constant speed, both car will have the same speed when toy car 1 reaches this velocity = 5cm/s

c) The position of car 1, as it follows a constant acceleration motion, is given by this equation:

[tex]x_1 = \frac{1}{2}a_1t^2 + v_o_1t + x_o_1[/tex]

The position for car 2, as it has constant velocity, is given by this equation:

[tex]x_2 = v_2t + x_o_2[/tex]

We equalize both equation to find the time where the cars pass each other:

[tex]x_1 = x_2\\\frac{1}{2}a_1t^2 + v_o_1t + x_o_1 = v_2t+x_o_2\\\frac{1}{2} a_1t^2 + (v_o_1 - v_2)t + x_o_1 - x_o_2 = 0\\\frac{1}{2}2.3m/s^2t^2 +(-3cm/s-5cm/s)t+ 17cm - 9.5cm = 0\\1.15t^2 -8t + 7.5 = 0 | a = 1.15, b = -8, c = 7.5\\t = \frac{-b +-\sqrt{b^2 - 4ac}}{2a} = 5.84s | 1.1168 s[/tex]

The car will pass each other at both 1.1168s and 5.84s.

For the positions, we solve any of the position equation with the solutions:

[tex]x = v_2*t + x_o_2 = 5cm/s *5.84s + 9.5cm = 38.7 cm\\x = 5cm/s * 1.1168s + 9.5cm = 15.084 cm[/tex]

The two toy cars have equal speeds at approximately t = 3.48s, with both traveling at 5 cm/s. They pass each other at t ≈ 1.99s, with their location at approximately 15.4 cm.

Part A: Equal Speeds:

We know the first car has an initial velocity of -3 cm/s and an acceleration of 2.30 cm/s². Its velocity at any time t can be given by v1 = v0 + at, that is -3 + 2.30t.

The second car has a constant velocity of 5.0 cm/s.

For them to have equal speeds, v1 = v2,

so -3 + 2.30t = 5.

Solving for t gives t ≈ 3.48 s.

Part B: Speeds at Equal Times:

The common speed when both cars have equal speeds can be found by substituting the time back into either equation for velocity.

Doing so for the first car gives

-3 + 2.30(3.48) ≈ 5 cm/s.

Part C: Cars Passing Each Other:

The cars pass each other when their positions are equal. Using the equations for position x1 = x01 + v01t + (1/2)at² and x2 = x02 + v02t,

we get 17 + (-3)t + (1/2)(2.30)t² = 9.5 + (5)t.

Solving for t gives two possible times, but only the positive one is relevant, t ≈ 1.99 s.

Part D: Location at Passing Time:

Substituting the time when they pass each other back into the position equations, for the first car we have

17 + (-3)(1.99) + (1/2)(2.30)(1.99)² ≈ 15.4 cm.

The second car will be at the same position since they are passing each other.

Answer:

Part A: [tex]8.06\times10^{-3}\ m[/tex]

Part B: [tex]6.36\times 10^{-9}\ s[/tex]

Explanation:

Given:

Assumptions:

Part A:

Since the electron moves in the direction of the electric field, the electric force will act on it in the direction opposite to electric field. This electric force does work on it to make the electron come to rest.

Using the work-energy theorem, the work done by the electric field will be equal to the kinetic energy change of the electron.

[tex]\therefore -eEx = \dfrac{1}{2}m(v^2-v_o^2)\\\Rightarrow -eEx=-\dfrac{1}{2}mv_o^2\\\Rightarrow x=\dfrac{mv_o^2}{2eE}\\\Rightarrow x=\dfrac{9.1\times 10^{-31}\times (5.08\times 10^{6})^2}{2\times 1.6\times10^{-19}\times 9100}\\\Rightarrow x=8.06\times 10^{-3}\ m[/tex]

Hence, the electron comes to rest by travelling a distance of [tex]8.06\times 10^{-3}\ m[/tex].

Part B:

In this part, let us first find out the acceleration of the electron due to the electric force.

[tex]a = -\dfrac{eE}{m}\\\Rightarrow a= -\dfrac{1.6\times10^{-19}\times 9100}{9.1\times 10^{-31}}\\\Rightarrow a= -1.6\times 10^{15}\ m/s^2\\[/tex]

The electron moves with the above acceleration constantly as it moves in the uniform electric field.

Since the electron is supposed to move from a point and then again move back to the same point. This means the displacement of the electron is zero.

[tex]i.e.,\ s=0\\\Rightarrow v_ot+\dfrac{1}{2}at^2=0\\\Rightarrow (v_o+\dfrac{1}{2}at)t=0\\\Rightarrow \dfrac{(2v_o+at)}{2}t=0\\\Rightarrow t = 0\,\,\, or\,\,\, (2v_o+at)=0\\\Rightarrow t = 0\,\,\, or\,\,\, t=\dfrac{-2v_o}{a}\\[/tex]

Since the electron starts moving at t = 0 s.

[tex]\therefore t = \dfrac{-2v_o}{a}\\\Rightarrow t=\dfrac{-2\times 5.08\times 10^6}{-1.6\times 10^{15}}\\\Rightarrow t= 6.36\times 10^{-9}\ s[/tex]

Hence, the electron returns to the starting position after [tex]6.36\times 10^{-9}\ s[/tex].

Answer:

[tex]Volume=160 cm^3[/tex]

Explanation:

The volume of the cylinder is given by the area of the base multiplied by the height.

In this case:

Height:

[tex]h = 10 cm\\[/tex]

Base area = area of the square ([tex]area=side*side=side^2[/tex])

the side of the square is:

[tex]side=4cm[/tex]

thus, the area of the base:

[tex]area=(4cm)^2 = 16 cm^2[/tex]

Now we multiply this quantities, to find the volume:

[tex]Volume= 16 cm^2*10cm=160 cm^3[/tex]

Answer:

[tex]v_{max}=52.38\frac{m}{s}[/tex]

[tex]v_{100}=33.81[/tex]

Explanation:

the maximum speed is reached when the drag force and the weight are at equilibrium, therefore:

[tex]\sum{F}=0=F_d-W[/tex]

[tex]F_d=W[/tex]

[tex]kv_{max}^2=m*g[/tex]

[tex]v_{max}=\sqrt{\frac{m*g}{k}} =\sqrt{\frac{70*9.8}{0.25}}=52.38\frac{m}{s}[/tex]

To calculate the velocity after 100 meters, we can no longer assume equilibrium, therefore:

[tex]\sum{F}=ma=W-F_d[/tex]

[tex]ma=W-F_d[/tex]

[tex]ma=mg-kv_{100}^2[/tex]

[tex]a=g-\frac{kv_{100}^2}{m}[/tex] (1)

consider the next equation of motion:

[tex]a = \frac{(v_{x}-v_0)^2}{2x}[/tex]

If assuming initial velocity=0:

[tex]a = \frac{v_{100}^2}{2x}[/tex] (2)

joining (1) and (2):

[tex]\frac{v_{100}^2}{2x}=g-\frac{kv_{100}^2}{m}[/tex]

[tex]\frac{v_{100}^2}{2x}+\frac{kv_{100}^2}{m}=g[/tex]

[tex]v_{100}^2(\frac{1}{2x}+\frac{k}{m})=g[/tex]

[tex]v_{100}^2=\frac{g}{(\frac{1}{2x}+\frac{k}{m})}[/tex]

[tex]v_{100}=\sqrt{\frac{g}{(\frac{1}{2x}+\frac{k}{m})}}[/tex] (3)

[tex]v_{100}=\sqrt{\frac{9.8}{(\frac{1}{2*100}+\frac{0.25}{70})}}[/tex]

[tex]v_{100}=\sqrt{\frac{9.8}{(\frac{1}{200}+\frac{1}{280})}}[/tex]

[tex]v_{100}=\sqrt{\frac{9.8}{(\frac{3}{350})}}[/tex]

[tex]v_{100}=\sqrt{1,143.3}[/tex]

[tex]v_{100}=33.81[/tex]

To plot velocity as a function of distance, just plot equation (3).

To plot velocity as a function of time, you have to consider the next equation of motion:

[tex]v = v_0 +at[/tex]

as stated before, the initial velocity is 0:

[tex]v =at[/tex] (4)

joining (1) and (4) and reducing you will get:

[tex]\frac{kt}{m}v^2+v-gt=0[/tex]

solving for v:

[tex]v=\frac{ \sqrt{1+\frac{4gk}{m}t^2}-1}{\frac{2kt}{m} }[/tex]

Plots:

Answer:

V1=2.5ft3

V2=1ft3

n=1.51

Explanation:

PART A:

the volume of each state is obtained by multiplying the mass by the specific volume in each state

V=volume

v=especific volume

m=mass

V=mv

state 1

V1=m.v1

V1=2lb*1.25ft3/lb=2.5ft3

state 2

V2=m.v2

V2=2lb*0.5ft3/lb=   1ft3

PART B:

since the PV ^ n is constant we can equal the equations of state 1 and state 2

P1V1^n=P2V2^n

P1/P2=(V2/V1)^n

ln(P1/P2)=n . ln (V2/V1)

n=ln(P1/P2)/ ln (V2/V1)

n=ln(15/60)/ ln (1/2.5)

n=1.51

Answer:

Option d

Explanation:

When we throw an object in the upward direction, we provide it with certain initial velocity due to which it covers a certain distance up to the maximum height.

While the object is moving in the upward direction, its velocity keeps on reducing due to the acceleration due to gravity which acts vertically downwards in the opposite direction thus reducing its velocity.

So, the maximum height attained by the object is the point where this upward velocity of the body becomes zero and after that the object starts to fall down.

Answer:

The gauge pressure is [tex]1.8\times10^{5}\ N/m^2[/tex]

Explanation:

Given that,

Gauge pressure of car tires [tex]P_{1}=2.40\times10^{5}\ N/m^2[/tex]

Temperature [tex]T_{1}=35.0^{\circ}C = 35.0+273=308 K[/tex]

Dropped temperature [tex]T_{2}= -42.0^{\circ}C=273-42=231 K[/tex]

We need to calculate the gauge pressure P₂

Using relation pressure and temperature

[tex]\dfrac{P_{1}}{T_{1}}=\dfrac{P_{2}}{T_{2}}[/tex]

Put the value into the formula

[tex]\dfrac{2.40\times10^{5}}{308}=\dfrac{P_{2}}{231}[/tex]

[tex]P_{2}=\dfrac{2.40\times10^{5}\times231}{308}[/tex]

[tex]P_{2}=180000 = 1.8\times10^{5}\ N/m^2[/tex]

Hence, The gauge pressure is [tex]1.8\times10^{5}\ N/m^2[/tex]

Answer:

The electric field at a distance r = 0.02 m is 14062.5 N/C.

Solution:

Refer to fig 1.

As per the question:

Radius of sphere, R = 0.04 m

Charge, Q = [tex]5.0\times 10^{- 9} C[/tex]

Distance from the center at which electric field is to be calculated, r = 0.02 m

Now,

According to Gauss' law:

[tex]E.dx = \frac{Q_{enclosed}}{\epsilon_{o}}[/tex]

Now, the charge enclosed at a distance r is given by volume charge density:

[tex]\rho = \frac{Q_{enclosed}}{area}[/tex]

[tex]\rho = \frac{Q_{enclosed}}{\frac{4}{3}\pi R^{3}}[/tex]

Also, the charge enclosed Q' at a distance r is given by volume charge density:

[tex]\rho = \frac{Q'_{enclosed}}{\frac{4}{3}\pi r^{3}}[/tex]

Since, the sphere is no-conducting, Volume charge density will be constant:

Thus

[tex]\frac{Q_{enclosed}}{\frac{4}{3}\pi R^{3}} = \frac{Q'_{enclosed}}{\frac{4}{3}\pi r^{3}}[/tex]

Thus charge enclosed at r:

[tex]Q'_{enclosed} = \frac{Q_{enclosed}}{\frac{r^{3}}{R^{3}}[/tex]

Now, By using Gauss' Law, Electric field at r is given by:

[tex]4\pi r^{2}E = \frac{Q_{enclosed}r^{3}}{\epsilon_{o}R^{3}}[/tex]

Thus

[tex]E = \frac{Q_{enclosed}r}{4\pi\epsilon_{o}R^{3}}[/tex]

[tex]E = \frac{(9\times 10^{9})\times 5.0\times 10^{- 9}\times 0.02}{0.04^{3}}[/tex]

E = 14062.5 N/C

Answer:

total number of electron in 1 litter is 3.34 × [tex]10^{26}[/tex] electron

Explanation:

given data

mass per mole = 18 g/mol

no of electron = 10

to find out

how many electron in 1 liter of water

solution

we know molecules per gram mole is 6.02 ×[tex]10^{23}[/tex] molecules

no of moles is 1

so

total number of electron in water is = no of electron ×molecules per gram mole × no of moles

total number of electron in water is = 10 × 6.02 ×[tex]10^{23}[/tex] × 1

total number of electron in water is = 6.02×[tex]10^{24}[/tex] electron

and

we know

mass = density × volume    ..........1

here we know density of water is 1000 kg/m

and volume = 1 litter = 1 × [tex]10^{-3}[/tex] m³

mass of 1 litter = 1000 × 1 × [tex]10^{-3}[/tex]

mass = 1000 g

so

total number of electron in 1 litter =  mass of 1 litter × [tex]\frac{molecules per gram mole}{mass per mole}[/tex]

total number of electron in 1 litter =  1000 × [tex]\frac{6.02*10{24}}{18}[/tex]

total number of electron in 1 litter is 3.34 × [tex]10^{26}[/tex] electron

Using the formula for acceleration, a = (v^2) / 2d, and the given velocity and acceleration distance, the acceleration of the tennis ball during Lisicki's serve was approximately 11368 m/s². The force exerted by the racket is generally higher than the force due to gravity during this action.

To calculate the acceleration of the tennis ball during Sabine Lisicki's serve, we can use the formula for acceleration: a = (v^2) / 2d, where 'v' corresponds to the final velocity and 'd' represents the distance over which the ball accelerated.

In this case, the final velocity 'v' is 211 km/h (converted to m/s gives us approximately 58.6 m/s), and Lisicki's racquet was in contact with the ball, causing it to accelerate over a distance 'd' of 0.15 m. Plugging these values into the formula gives us: a = (58.6 m/s)^2 / 2(0.15 m), which equals about 11368 m/s².

The average force exerted by the racket can be understood through the equation F = ma, a case of Newton's second law of motion. However, in this scenario, the force due to gravity is negligible as it's much smaller than the force exerted by the racket. The main focus here is the force exerted by the racket which made such a high acceleration possible.

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Answer:

Total distance covered during the journey is 235 m

Solution:

As per the question:

Initial velocity, v = 0 m/s

Acceleration, a = [tex]2 m/s^{2}[/tex]

Time, t = 5 s

Now,

For this, we use eqn 2 of motion:

[tex]d = vt + \farc{1}{2}at^{2}[/tex]

[tex]d = 0.t + \farc{1}{2}\times 2\times 5^{2} = 25 m[/tex]

The final speed of car after t = 5 s is given by:

v' = v + at

v' = 0 + 2(5) = 10 m/s

Now, the car travels at constant speed of 10 m/s for t' = 20 s with a = 0:

[tex]d' = vt + \farc{1}{2}at^{2}[/tex]

[tex]d' = 10\times 20 + \farc{1}{2}\times 0\times 20^{2} = 25 m[/tex]

d' = 200 m

Now, the car accelerates at a= - 5 [tex]m/s^{2}[/tex] until its final speed, v" = 0 m/s:

[tex]v"^{2} = v'^{2} + 2ad"[/tex]

[tex]0 = {10}^{2} + 2\times (- 5)d"[/tex]

[tex]100 = 10d"[/tex]

d" = 10 m

Total distance covered = d + d' + d"  = 25 + 200 + 10 = 235 m

Answer:

The acceleration is -9.8 m/s²

Explanation:

Hi there!!

When you throw a ball upward, there is a downward acceleration that makes the ball return to your hand. This acceleration is produced by gravity.

The average acceleration is calculated as the variation of the speed over time. In this case, we know the time and the initial and final speed. Then:

acceleration = final speed - initial speed/ elapsed time

acceleration = -4.3 m/s - 4.3 m/s / 0.88 s

acceleration = -9.8 m/s²  

The average acceleration vector would be 0 m/s² due to the upward (negative) and downward (positive) accelerations cancelling. However, the average magnitude of acceleration regardless of direction is 9.8 m/s², which is simply the acceleration due to gravity.

To answer your question about the average acceleration vector of a ball that was thrown upward, we can use the concept of free fall in physics. When you throw a ball upward, it initially slows down due to earth's gravitational pull until it stops at its highest point, then it starts accelerating downward due to gravity, until it reaches your hand again.

The acceleration when the ball is going upward will be the opposite of the acceleration when it's coming downward because they are in opposite directions but magnitude will be the same. We can assume the acceleration due to gravity as -9.8 m/s² (negative indicating upward direction) and when it's coming down it will be 9.8 m/s² (positive indicating downward direction).

So over the course of its flight (0.88s), the average acceleration would be ((-9.8)+(9.8))/2 = 0 m/s². However, this may not be what you're looking for as this is the averaged vector sum of the accelerations upward and downward.

If you are asking about the average magnitude of acceleration regardless of direction (in magnitude) it would be the average of the absolute values of the accelerations, which would be the gravitational acceleration g = 9.8 m/s².

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Final answer:

The electron's average velocity is found to be (0 m/s, 316,000 m/s, -146,000 m/s), and after another 9 microseconds, it will be at the position (0.02 m, 4.464 m, -2.104 m).

Explanation:

To calculate the average velocity of an electron, we use the formula:

Vavg = (rf - ri) / Δt, where rf is the final position, ri is the initial position, and Δt is the time interval between the positions.

Given the initial position (0.02 m, 0.04 m, -0.06 m) and the final position (0.02 m, 1.62 m, -0.79 m), with a time difference of 5 microseconds (μs), which is 5 x 10-6 seconds:

The position change in vector form is

Δr = (0.02 m - 0.02 m, 1.62 m - 0.04 m, -0.79 m - (-0.06 m))

= (0 m, 1.58 m, -0.73 m).

Thus, the average velocity is

Vavg = Δr / Δt

= (0 m, 1.58 m, -0.73 m) / (5 x 10-6 s)

= (0 m/s, 316,000 m/s, -146,000 m/s)

The electron's new position after another 9 μs, moving with the same average velocity, is calculated by:

rnew = rf + Vavg × Δtnew

Here, Δtnew is 9 μs, which is 9 x 10-6 seconds, so:

rnew = (0.02 m, 1.62 m, -0.79 m) + (0 m/s, 316,000 m/s, -146,000 m/s) × (9 x 10-6 s)

= (0.02 m, 1.62 m + (2.844 m), -0.79 m - (1.314 m))

= (0.02 m, 4.464 m, -2.104 m).

Answer:

h = 4271.43 m

Explanation:

given,

Volume of the water = 1 m³

temperature decrease by = 10°C

heat removed from water

Q = m c ΔT                            

Q = ρ V c ΔT                            

   = 1000 × 1 × 4186 × 10

   = 4.186 × 10⁷ J

energy is used to do work to move the water against its weight

Q = force  × displacement

4.186 × 10⁷ J =  m g × h                    

4.186 × 10⁷ J =  1000 × 1 × 9.8 × h                

h = 4271.43 m                                

hence, the change in height of is equal to h = 4271.43 m

Answer:

n = 5 approx

Explanation:

If v be the velocity before the contact with the ground and v₁ be the velocity of bouncing back

[tex]\frac{v_1}{v}[/tex] = e ( coefficient of restitution ) = [tex]\frac{1}{\sqrt{10} }[/tex]

and

[tex]\frac{v_1}{v} = \sqrt{\frac{h_1}{6.1} }[/tex]

h₁ is height up-to which the ball bounces back after first bounce.

From the two equations we can write that

[tex]e = \sqrt{\frac{h_1}{6.1} }[/tex]

[tex]e = \sqrt{\frac{h_2}{h_1} }[/tex]

So on

[tex]e^n = \sqrt{\frac{h_1}{6.1} }\times \sqrt{\frac{h_2}{h_1} }\times... \sqrt{\frac{h_n}{h_{n-1} }[/tex]

[tex](\frac{1}{\sqrt{10} })^n=\frac{2.38}{6.1}[/tex]= .00396

Taking log on both sides

- n / 2 = log .00396

n / 2 = 2.4

n = 5 approx

The ball can bounce approximately 8 times before it reaches the height of 2.38 meters after accounting for a 10% energy loss each collision.

This problem involves a ball undergoing inelastic collisions, losing 10% of its kinetic energy with each bounce, and determining how many times it can bounce to still reach a windowsill 2.38 m high.

First, let's calculate the initial potential energy (PEinitial) of the ball when it is dropped from a height (hinitial) of 6.10 m:

PEinitial = mghinitial

where, g = 9.8 m/s² (acceleration due to gravity)

As it falls, this entire potential energy converts into kinetic energy (KEinitial) at the ground:

KEinitial = PEinitial = mghinitial

Upon each bounce, the ball loses 10% of its kinetic energy. Therefore, it retains 90% of its kinetic energy:

KEnew = 0.9 × KEprevious

To find out how high it can bounce after each loss of energy, convert kinetic energy back into potential energy:

PE = KE = mgh

After each bounce, the height the ball can reach is calculated by applying the 90% retention factor:

hnew = 0.9 × hprevious

Starting with h0 = 6.10 m:

The number of bounces can be calculated using the formula: hn = 6.10 × (0.9)n, where n is the number of bounces. Set hn = 2.38 m and solve for n:

2.38 = 6.10 × (0.9)n

Divide both sides by 6.10:

0.39 ≈ (0.9)n

Taking the natural log (ln) of both sides:

ln(0.39) = n × ln(0.9)

Finally, solving for n:

n ≈ ln(0.39) / ln(0.9) ≈ 8

Therefore, the ball can bounce approximately 8 times and still reach the windowsill that is 2.38 m above the ground.

Answer:

1497×10⁵ km

Explanation:

Speed of light in vacuum = 3×10⁵ km/s

Time taken by the light of the Sun to reach the Earth = 8 min and 19 s

Converting to seconds we get

8×60+19 = 499 seconds

Distance = Speed × Time

[tex]\text{Distance}=3\times 10^5\times 499\\\Rightarrow \text{Distance}=1497\times 10^5\ km[/tex]

1 AU = 1497×10⁵ km

The Sun is 1497×10⁵ km from Earth

Answer:

potential energy of net is 8.58 kJ

Explanation:

given data

mass m = 70 kg

height h = 11 m

stop distance x = 1.50 m

to find out

potential energy of net

solution

we know here man eventually stop

so elastic potential energy of net = change in potential energy of man body

so equation will be

potential energy of net = m×g×h    ...................1

here m is mass and h is total height = ( h+ x)  

height = 11 + 1.5 = 12.5 m and g is 9.8

so put here value in equation 1

potential energy of net = m×g×h

potential energy of net = 70×9.8×12.5

potential energy of net = 8575 J

so potential energy of net is 8.58 kJ

Answer: a) 8.2 * 10^-8 N or 82 nN and b) is repulsive

Explanation: To solve this problem we have to use the Coulomb force for two point charged, it is given by:

[tex]F=\frac{k*q1*q2}{d^{2}}[/tex]

Replacing the dat we obtain F=82 nN.

The force is repulsive because the points charged have the same sign.

The spring constant of the spring is -1.376 N/m.

To find the spring constant of the spring, we can use Hooke's Law, which states that the force exerted by a spring is directly proportional to its displacement from its equilibrium position. In this case, the displacement of the spring is 1.2 cm, and the charges are 2.2 cm apart. The force exerted by the spring can be calculated using the equation F = kx, where F is the force, k is the spring constant, and x is the displacement.

Given that the displacement (x) is 1.2 cm and the force exerted is caused by the electric force between the charges, which is given by Coulomb's Law, we can write the equation:

F = kx = k(1.2 cm) = (k/100 cm) * 1.2 cm = k/83.33

Similarly, the electric force between the charges is given by Coulomb's Law: F = kq1q2/r^2, where k is the electrostatic constant, q1 and q2 are the charges, and r is the distance between them. In this case, the charges are 2.2 cm apart and have magnitudes of 2.0 μC and -4.0 μC, respectively.

When the charges are 2.2 cm apart, we can calculate the electric force:

F = k(2.0 μC)(-4.0 μC)/(2.2 cm)^2 = (-8.0 kμC^2)/(4.84 cm^2) = (-8.0 k)/(4.84 cm^2) μC^2

Equating the force exerted by the spring to the electric force, we have:

k/83.33 = (-8.0 k)/(4.84 cm^2) μC^2

Solving for k:

k/83.33 = (-8.0 k)/(4.84 cm^2) μC^2

k * (4.84 cm^2)/(83.33) = -8.0 k * μC^2

(4.84 cm^2)/(83.33) = -8.0 μC^2

k = (-8.0 μC^2) * (83.33)/(4.84 cm^2) = -137.6 μC^2/cm^2 = -1.376 N/m

Therefore, the spring constant of the spring is -1.376 N/m.

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Answer:

a) m = 993 g

b) E = 6.50 × 10¹⁴ J

Explanation:

atomic mass of hydrogen = 1.00794

4 hydrogen atom will make a helium atom = 4 × 1.00794 = 4.03176

we know atomic mass of helium = 4.002602

difference in the atomic mass of helium = 4.03176-4.002602 = 0.029158

fraction of mass lost = [tex]\dfrac{0.029158}{4.03176}[/tex]= 0.00723

loss of mass for 1000 g = 1000 × 0.00723 = 7.23

a) mass of helium produced = 1000-7.23 = 993 g (approx.)

b) energy released in the process

E = m c²

E = 0.00723 × (3× 10⁸)²

E = 6.50 × 10¹⁴ J

Answer:

(a) 992.87 g

(b) [tex]6.419\times 10^{14} J[/tex]

Solution:

As per the question:

Mass of Hydrogen converted to Helium, M = 1 kg = 1000 g

(a) To calculate mass of He produced:

We know that:

Atomic mass of hydrogen is 1.00784 u

Also,

4 Hydrogen atoms constitutes 1 Helium atom

Mass of Helium formed after conversion:

[tex]4\times 1.00784 = 4.03136 u[/tex]

Also, we know that:

Atomic mass of Helium is 4.002602 u

The loss of mass during conversion is:

4.03136 - 4.002602 = 0.028758 u

Now,

Fraction of lost mass, M' = [tex]\frac{0.028758}{4.03136} = 0.007133 u[/tex]

Now,

For the loss of mass of 1000g = [tex]0.007133\times 1000[/tex] = 7.133 g

Mass of He produced in the process:

[tex]M_{He} = 1000 - 7.133 = 992.87 g[/tex]

(b) To calculate the amount of energy released:

We use Eintein' relation of mass-enegy equivalence:

[tex]E = M'c^{2}[/tex]

[tex]E = 0.007133\times (3\times 10^{8})^{2} = 6.419\times 10^{14} J[/tex]

Answer:

b) 2.2 kg

Explanation:

Net force acting on an object is the sum of the  two forces acting on the body.

The net force is calculated using the parallelogram law of vectors.

F =[tex]\sqrt{{A^{2}} + B^{2}+2 A B cos \theta}[/tex]

Here A = 20 N , B = 35 N and θ =80°

Net Force = F = 43.22 N

Acceleration = a = 20 m/s/s

Since F = ma, m = F/a = 43.22 / 20 = 2.161 kg = 2.2 kg

Answer:

It would have been longer.

Explanation:

Lets assume the Sun angle = θ

Distance between Syene and Alexandria = D

Circumference of Earth = C

As per Eratosthenes' calculations,

[tex]\frac{\theta}{360} =\frac{D}{C}[/tex]

From the above equation it is evident that if the circumference decreases value of θ will increase which implies that the shadow length would be longer as compared to that on the Earth.

Answer:

No

Explanation:

As we know that the electric field nullity does not define that the electric potential will be zero at that point.