A man is dragging a trunk up the loading ramp of a mover’s truck. The ramp has a slope angle of 20.0°, and the man pulls upward with a force FS whose direction makes an angle of 30.0° with the ramp (Fig. E4.4). (a) How large a force FS is necessary for the component Fx parallel to the ramp to be 90.0 N?(b) How large will the component Fy perpendicular to the ramp be then?

Answer:

+Q

Explanation:

As no electric field can exist (in electrostatic condition) inside a conductor, if we apply Gauss 'Law to a spherical gaussian surface with a radius just a bit larger than the distance of the inner surface to the center (but less tah the distance of the outer surface), the net flux through this surface must be zero, due to E=0 at any point of the gaussian surface.

Therefore, as the net flux must  be proportional to the  charge enclosed  by the surface, it follows that Qenc = 0.

⇒ Qenc = Qc + Qin = -2Q + Qin = 0 ⇒ Qin = +2Q

So, if the net charge of  the conductor is + 3Q (which must remain the same due to the conservation of charge principle) and no charge can exist within the conductor (in electrostatic conditions), we have the following equation:

Qnet = Qin + Qou = +3Q ⇒ +2Q + Qou = +3Q

⇒ Qou = +Q

Answer:

Explanation:

Given

Wavelength of incoming light [tex]\lambda =425\ nm[/tex]

We know

[tex]speed\ of\ wave=frequency\times wavelength[/tex]

[tex]frequency=\frac{speed}{wavelength}[/tex]

[tex]\mu =\frac{3\times 10^8}{425\times 10^{-9}}[/tex]

[tex]\mu =7.058\times 10^{14}\ Hz[/tex]

Energy associated with this frequency

[tex]E=h\mu [/tex]

where h=Planck's constant

[tex]E=6.626\times 10^{-34}\times 7.058\times 10^{14}[/tex]

[tex]E=46.76\times 10^{-20}\ Hz[/tex]

Energy of one mole of Photon[tex]=N_a\times E[/tex]

[tex]=6.022\times 10^{23}\times 46.76\times 10^{-20}[/tex]

[tex]=281.58\times 10^{3}[/tex]

[tex]=281.58\ kJ[/tex]

To calculate the energy of a mole of photons of the emission at 425 nm, use the equation E = hc/λ, where E is the energy, h is Planck's constant, c is the speed of light, and λ is the wavelength. Convert the wavelength to meters, substitute the values into the equation, and calculate to find the energy of a single photon. Multiply this by Avogadro's number to find the energy of a mole of photons.

To calculate the energy of a mole of photons of the emission at 425 nm, we can use the equation E = hc/λ, where E is the energy, h is Planck's constant (6.63 x 10^-34 J·s), c is the speed of light (3.00 x 10^8 m/s), and λ is the wavelength (in meters).

Converting the wavelength to meters, we have 425 nm = 425 x 10^-9 m.

Substituting the values into the equation, we get E = (6.63 x 10^-34 J·s)(3.00 x 10^8 m/s) / (425 x 10^-9 m). Calculating this gives us the energy of a single photon of this emission. To find the energy of a mole of photons, we can multiply this value by Avogadro's number (6.02 x 10^23 photons/mol).

Therefore, the energy of a mole of photons of this emission is (6.63 x 10^-34 J·s)(3.00 x 10^8 m/s) / (425 x 10^-9 m) x (6.02 x 10^23 photons/mol).

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Answer:

Q = 177J

Explanation:

Specific heat capacity of lead=0.13J/gc

Q=MCΔT

ΔT=T2-T1,where T1=22degrees Celsius and T2=37degree Celsius.

ΔT=37 - 22 = 15

Q = Change in energy

M = mass of substance

C= Specific heat capacity

Q = (91g) * (0.13J/gc) * (15c)= 177.45J

Approximately, Q = 177J

Explanation:

The given data is as follows.

    Inner wheel Radius = 20 m,

   Distance between left and right wheel = 1.5m,

Let us assume speed of drive shaft is N rpm.

Formula to calculate angular velocity is as follows.

    Angular velocity of automobile = w = [tex]\frac{V}{R}[/tex]

where,   V = linear velocity of automobile m/min,

              R = turning radius from automobile center in meter

In the given case, angular velocity remains same for inner and outer wheel but there is change in linear velocity of inner wheel and outer wheel.

Now, we assume that

         u = linear velocity of inner wheel

and,   u' = linear velocity of outer wheel.

Formula for angular velocity of inner wheel w = ,

Formula for angular velocity of outer wheel w =

Now, for inner wheels

                   w =

                      = [tex]\frac{u}{(R - d)}[/tex]

                  u = [tex]V \times \frac{(R - d)}{R}[/tex]

                    = [tex]V \times (1 - \frac{d}{R})[/tex]

If radius of wheel is r it will cover  distance in one min.

Since, velocity of wheel is u it will cover distance u in unit time(min)

Thus,             u = [tex]2\pi rn[/tex] = [tex]V \times (1 - \frac{d}{R})[/tex]

Now, rotation per minute of inner wheel is calculated as follows.

         n = [tex]\frac{V}{2 \pi r \times (1 - \frac{d}{R})}[/tex]

            = [tex]\frac{V}{2 \pi r \times (1 - \frac{0.75}{20})}[/tex] (since 2d = 1.5m given, d = 0.75m),

             = [tex]\frac{V}{r} \times 0.1532[/tex]

So, rotation per minute of outer wheel; n' =  

                   = [tex]\frac{V}{2 \pi r \times (1 + \frac{0.75}{20})}[/tex]

                   = [tex]\frac{V}{r} \times 0.1651[/tex]

In a sharp left turn, the car's right wheel rotates 1.0375 times the speed of the drive shaft, while the left wheel rotates at the speed of the drive shaft. This difference is due to the right wheel covering a larger distance than the left wheel.

The principle in operation here is that in a sharp turn, the outer wheel (in this case, the right wheel) has to cover a larger distance than the inner wheel (left wheel). Therefore, the right wheel will rotate faster than the left one.

Now, calculating the difference in rotation speeds of the wheels, we consider the radii of the paths of the two wheels. For the left wheel, the radius is 20 m and for the right wheel, the radius is greater by half the wheelbase distance, that is, 20.75 m.

The ratio of the speeds is equal to the ratio of the radii of the two paths. Therefore, the speed of the right wheel as a fraction of the drive shaft speed is 20.75/20 = 1.0375 and the speed of the left wheel as a fraction of the drive shaft speed is 20/20 = 1.

So, in conclusion, during a sharp turn to the left, the right wheel rotates 1.0375 times the speed of the drive shaft, while the left wheel rotates at the same speed as the drive shaft.

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To solve this problem it will be necessary to apply the interference principle. Under this principle interference is understood as a phenomenon in which two or more waves overlap to form a resulting wave of greater, lesser or equal amplitude. In this case, if both are at the same point, the result of the total displacement will be the sum of the individual displacements, therefore

[tex]x = \sum h_i[/tex]

[tex]x = 60cm + 80cm[/tex]

[tex]x =140cm[/tex]

Therefore the resulting displacement above equilibrium is 140cm

Final answer:

When two water waves meet at the same point, the resulting displacement above equilibrium can be calculated by adding the individual displacements together. In this case, the resulting displacement above equilibrium is 140 cm.

Explanation:

The resulting displacement above equilibrium when two water waves meet at the same point can be determined by adding the individual displacements together. In this case, one wave has a displacement above equilibrium of 60 cm and the other wave has a displacement above equilibrium of 80 cm. The resulting displacement is found by adding 60 cm and 80 cm, which gives a resulting displacement above equilibrium of 140 cm.

Answer:

B = 9.867 * 10^-8 T

Explanation:

Given:

- Rate of accumulating charge I = 37.0 mC/s

- radial distance from center of the plate r = 7.5 cm

- magnetic constant u_o = 4*pi *10^-7 H/m

Find:

The induced magnetic field strength 7.5 cm radially outward from the center of the plates?

Solution:

- Apply Gaussian Law on the surface:

                            B.(dA) = u_o*I

- The surface integral is dA = (2*pi*r):

                            B.(2*pi*r) = u_o*I

                            B =  u_o*I / (2*pi*r)

- Plug values in:

                            B = (4*pi *10^-7)*(37*10^-3) / (2*pi*0.075)

                            B = 9.867 * 10^-8 T

Answer:

answer is B on Plato if you have it

Explanation:

Answer:

[tex]\omega = \frac{mv\frac{L}{2}\sin(29^\circ)}{\frac{1}{3}ML^2 + m(\frac{L}{2})^2} = 0.91~{\rm rad/s}[/tex]

Explanation:

The angular speed of the system can be found by conservation of angular momentum. Since the ball and the rod are stick together, the collision is completely inelastic, ergo kinetic energy is not conserved.

[tex]L_{initial} = L_{final}\\m\vec{v}\times \vec{r} = I\omega[/tex]

The moment of inertia of the combined objects is equal to the sum of moment of inertia of the separate objects.

[tex]I = \frac{1}{3}ML^2 + m(\frac{L}{2})^2[/tex]

The cross product in the left-hand side can be written as a sine of the angle.

Therefore;

[tex]mv\frac{L}{2}\sin(29^\circ) = (\frac{1}{3}ML^2 + m(\frac{L}{2})^2)\omega\\(19\times 10^{-3})(5)(\frac{47\times 10^{-2}}{2})\sin(29^\circ) = (\frac{1}{3}(146\times 10^{-3})(47\times 10^{-2})^2 + (19\times 10^{-3})(\frac{47\times 10^{-2}}{2})^2)\omega\\\omega = 0.91~{\rm rad/s}[/tex]In terms of system parameters:

[tex]\omega = \frac{mv\frac{L}{2}\sin(29^\circ)}{\frac{1}{3}ML^2 + m(\frac{L}{2})^2}[/tex]

Answer:

8.07 m/s, 81.7º NE.

Explanation:

        [tex]vocx = voc* cos 40 = 1.53 m/s * 0.766 = 1.17 m/s\\vocy = voc* sin 40 = 1.53 m/s * 0.643 = 0.98 m/s[/tex]

        [tex]vshy = vsw + vwy = 7.00 m/s + 0.98 m/s = 7.98 m/s[/tex]

        vshx = 1.17 m/s

        [tex]v = \sqrt{vshx^{2} + vshy^{2} } =\sqrt{(7.98m/s)^{2} +(1.17m/s)^{2} } =8.07 m/s[/tex]

        [tex]tg \theta = \frac{vshy}{vshx} = \frac{7.98}{1.17} = 6.82 \\ \theta = tg^{-1} (6.82)\\ \theta= 81.7\deg[/tex]

Answer:

(a) the electric field of the Earth will be directed towards the negatively charged water droplet.

(b) E = (9.8*m)/q

Explanation:

Part (a) the direction of the Earth’s electric field

Electric field is always directed towards negatively charged objects.

The water droplet has a negative charge ‘-q’, therefore the electric field of the Earth will be directed towards the negatively charged water droplet.

Part (b) an expression for the Earth’s electric field in terms of the mass and charge of the droplet

The magnitude of Electric field is given as;

E = F/q

where;

f is force and q is charge

Also from Newton's law, F = mg

where;

m is mass and g is acceleration due to gravity = 9.8 m/s²

E =  F/q = mg/q

E = (9.8*m)/q, Electric field in terms of mass and charge of the droplet.

Answer:

Explanation:

Given

mass of rock [tex]m=40\ kg[/tex]

Elevation of Rock [tex]h=10\ m[/tex]

Distance traveled by rock with time

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

where, u=initial velocity

t=time

a=acceleration

here initial velocity is zero

when rock is 5 m from ground then it has traveled a distance of 5 m from top because total height is 10 m

[tex]5=0\times t+\frac{1}{2}(9.8)(t^2)[/tex]

[tex]t^2=\frac{10}{9.8}[/tex]

[tex]t=1.004\approx 1\ s[/tex]

velocity at this time

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

[tex]v=0+9.8\times 1.004[/tex]

[tex]v=9.83\ m/s[/tex]

Final answer:

To calculate the velocity of a 40-kg rock when it is 5 meters from the ground, after being dropped from 10 meters, we use the kinematic equation, leading to a final velocity of approximately 9.9 m/s.

Explanation:

The question concerns calculating the velocity of a 40-kg rock when it is 5 meters from the ground, after being dropped from an elevation of 10 meters. This problem is fundamentally based on the principles of kinematic equations that describe the motion of objects under the influence of gravity. Assuming the acceleration due to gravity (g) is 9.8 m/s2, and ignoring air resistance, we can use the equation of motion v² = u² + 2as, where v is the final velocity, u is the initial velocity (0 m/s, since the rock is dropped), a is the acceleration due to gravity, and s is the distance covered.

To find out how fast the rock is moving when it is 5 meters from the ground, we first calculate the distance it has fallen, which is 10 meters - 5 meters = 5 meters. Plugging the values into the equation, we have v² = 0 + 2(9.8)(5) = 98. Therefore, v = √98 = approximately 9.9 m/s.

Answer:

E = 39.68 V

Explanation:

EMF ( Electromotive force) : This is defined as the magnitude of the potential difference of both the external circuit and the inside of the cell. The S.I unit is Volt.

The Expression for E.M.F is given as,

E = I(R+r) ................... Equation 1

Where E = EMF, I = current, R = External Resistance, r = internal resistance.

Also,

P = I²R

I = √(P/R) ..................... Equation 2

Where P = power, R = External resistance.

Given: P = 92 J/s, R = 14 Ω.

Substitute into equation 2

I = √(92/14)

I = √(6.57)

I = 2.56 A.

Also Given: r = 1.5 Ω.

Substitute into equation 1

E = 2.56(1.5+14)

E = 2.56(15.5)

E = 39.68 V.

A battery has emf E and internal resistance r = 2.00 Ω. A 11.5 Ω resistor is connected to the battery, and the resistor consumes electrical power at a rate of 96.0 J/s.

Part APart complete

What is the emf of the battery?

Express your answer with the appropriate units.

E =

39.0 V

3 trams must be added

Explanation:

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

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

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

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

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

And the number of trams will become

[tex]12+n[/tex]

So eq.(1) will become

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

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

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

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Answer: 3

Explanation:

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

Answer:

a. 6.69m/s

b. y=4.48m

c. t=1.43secs

Explanation:

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

distance covered,d=64cm=0.64m,

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

initial velocity,u=0m/s

if we substitute values we arrive at

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

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

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

Hence we can write the equation above again

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

if we substitute values we have

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

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

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

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

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

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

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

The given parameters;

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

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

The height risen by the shot is calculated as follows;

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

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

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

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

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

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

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

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

The total time spent in air;

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

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Answer:

Explanation:

A ball is thrown vertically upward with a certain Kinetic Energy in the absence of air resistance and while returning it experiences air resistance.

Air resistance causes the ball to lose its kinetic energy as it provides resistance which will convert some of its kinetic energy to heat energy.

So in a way total energy is conserved but not kinetic energy as some portion of it is lost in the form of heat.              

Final answer:

A ball thrown upward with air resistance will return with less kinetic energy due to the negative work done by air resistance, which transforms some of its kinetic energy into heat. This does not violate the conservation of energy as the energy is still conserved but in different forms. Mechanical energy decreases but is compensated by the increase in thermal energy in the air.

Explanation:

When a ball is thrown vertically upward in the absence of air resistance, it will return to its original level with the same kinetic energy (KE) because energy is conserved in a system where no external forces do work. However, when air resistance is a factor, it will return with less kinetic energy. This is because air resistance does negative work on the ball, converting some of its Kinetic Energy into thermal energy as it dissipates heat into the air.

The conservation of energy principle is not contradicted by this scenario. The total energy (kinetic plus potential plus any energy converted into heat due to air resistance) is still conserved. However, the mechanical energy of the ball (the sum of its kinetic and potential energy) decreases due to the work done by air resistance. This reduction in mechanical energy is exactly balanced by the increase in thermal energy of the air.

If an object like a feather is thrown upward, it will experience significant air resistance, and it will definitely return with less kinetic energy than it had when it was thrown up. Again, this doesn't violate the conservation of energy; instead, the energy is simply transformed into different forms, primarily heat, due to interactions with the air.

Answer:

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

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

Explanation:

We are given that

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

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

a.We have to find its velocity

We know that

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

Using the formula

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

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

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

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

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

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

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

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

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

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Answer

given,

Mean of temperature in Celsius = 37    Standard deviation in Celsius = 0.5

relation between Fahrenheit and Celsius

[tex]F = \dfrac{9}{5}C + 32[/tex]

Mean of temperature in Fahrenheit

[tex]Mean_F =\dfrac{9}{5}\times Mean_C + 32[/tex]

[tex]Mean_F =\dfrac{9}{5}\times 37 + 32[/tex]

[tex]Mean_F = 98.6\ F[/tex]

Standard deviation of Fahrenheit.

[tex]SD_F = \dfrac{9}{5}\times SD_C [/tex]

addition of 32 will not change the Standard deviation.

[tex]SD_F = \dfrac{9}{5}\times 0.5[/tex]

[tex]SD_F = 0.9[/tex]

Answer:

Explanation:

Given

Initial Volume [tex]v_1=1\ m^3[/tex]

final Volume [tex]v_2=3\ m^3[/tex]

Heat added at constant Pressure [tex]Q=200\ kJ[/tex]

Decrease in Energy of System [tex]\Delta U=-340\ kJ[/tex]

According to First law of thermodynamics

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

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

[tex]W=200-(-340)[/tex]

[tex]W=540\ kJ[/tex]

Work done in a constant Pressure Process is given by

[tex]W=P\Delta V[/tex]

where P is the constant Pressure

[tex]540=P\times (3-1)[/tex]

[tex]P=270\ kPa[/tex]                          

Answer:

d. the same within the uncertainty of each measurement method

Explanation:

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

The magnitude of the gravitational force exerted by the Sun on Mars is approximately 1.49x10^22 N, while the magnitude of the gravitational force exerted by Mars on the Sun is approximately 5.92x10^15 N.

To calculate the magnitude of the gravitational force exerted by the Sun on Mars, we can use Newton's law of gravitation. The formula is given by F = G * (m1 * m2) / r^2, where F is the magnitude of the force, G is the gravitational constant, m1 and m2 are the masses of the objects, and r is the distance between them.

Plugging in the values for the Sun's mass (2x10^30 kg), Mars' mass (6.4x10^23 kg), and the distance between them (2.3x10^11 m), we get

F = (6.673x10^-11 N·m²/kg²) * ((2x10^30 kg) * (6.4x10^23 kg)) / (2.3x10^11 m)^2

Simplifying the equation and performing the calculations, the magnitude of the gravitational force exerted by the Sun on Mars is approximately 1.49x10^22 N.

Similarly, to calculate the magnitude of the gravitational force exerted by Mars on the Sun, we can use the same formula with the masses and distance reversed. Plugging in the values, we get

F = (6.673x10^-11 N·m²/kg²) * ((6.4x10^23 kg) * (2x10^30 kg)) / (2.3x10^11 m)^2

Simplifying and calculating the equation, the magnitude of the gravitational force exerted by Mars on the Sun is approximately 5.92x10^15 N.

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Answer:

T2=336K

Explanation:

Clausius-Clapeyron equation is used to determine the vapour pressure at different temperatures:

where:

In(P2/P1) = ΔvapH/R(1/T1 - 1/T2)

p1 and p2 are the vapour pressures at temperatures 

T1 and T2

ΔvapH = the enthalpy  of vaporization of the liquid

R = the Universal Gas Constant

p1=p1, T1=307K

p2=3.50p1; T2=?

ΔvapH=37.51kJ/mol=37510J/mol

R=8.314J.K^-1moL^-1

In(3.50P1/P1)= (37510J/mol)/(8.314J.K^-1)*(1/307 - 1/T2)

P1 and P1 cancelled out:

In(3.50)=4511.667(T2 - 307/307T2)

1.253=14.696(T2 - 307/T2)

1.253=(14.696T2) - (14.696*307)/T2

1.253T2=14.696T2 - 4511.672

Therefore,

4511.672=14.696T2 - 1.253T2

4511.672=13.443T2

So therefore, T2=4511.672/13.443=335.61

Approximately, T2=336K

The electric flux through a spherical shell in a uniform electric field is calculated using Gauss's Law. The physical principles of uniform electric fields and spherical symmetry are applied to determine the flux, emphasizing the concept of electric flux through closed surfaces.

The question involves calculating the electric flux through a spherical shell placed in a uniform electric field. According to Gauss's Law, the electric flux (ΦE) through a closed surface surrounding a charge is proportional to the enclosed charge (ΦE = q/ε0), regardless of the shape of the surface. In a uniform electric field, the electric flux through a closed surface, like a spherical shell, can be derived from the formula ΦE = E⋅A, where E is the electric field strength and A is the area of the spherical shell.

For a spherical shell of radius 9.7 m in a uniform electric field of 1310 N/C, the area of the shell (A) is 4πr2, resulting in A = 4π(9.72) square meters. Substituting the values into ΦE = E⋅A gives us the total electric flux through the shell.

Answer: Option (b) is the correct answer.

Explanation:

Since, there is a negative charge present on the ball and a positive charge present on the rod. So, when the negatively charged metal ball will come in contact with the rod then positive charges from rod get conducted towards the metal ball.

Hence, the rod gets neutralized. But towards the metal ball there is a continuous supply of negative charges. Therefore, after the neutralization of positive charge from the rod there will be flow of negative charges from the metal ball towards the rod.

Thus, we can conclude that negative charge spread evenly on both ends.

Option (b) is the correct answer.

b. There is negative charge spread evenly on both ends.

The following information should be considered:

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Answer:

[tex] f_s = 640 Hz[/tex]

Explanation:

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

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

And we have the following info provided:

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

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

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

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

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

And replacing we got:

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

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

[tex] f_s = 640 Hz[/tex]

Answer:

B. 0.0254m

Explanation:

A is the amplitude of the oscillation, i.e. the

maximum displacement of the object from

equilibrium, either in the positive or negative x-direction. Simple harmonic motion is repetitive.

The period T is the time it takes the object tocomplete one oscillation and return to the startingposition.

d = 2A = 2×0.0127

Answer:

884 balloons

Explanation:

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

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

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

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

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

The number of balloons that 12.5 m3 can fill in is

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

Final answer:

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

Explanation:

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

Given:

The volume of the tank is 0.150 m³

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

The pressure of the helium in the tank is 100 atm

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

The temperature of the filled balloon is 27.0°C

The absolute pressure of the filled balloon is 1.20 atm

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

0.150 m³ * 1000 L/m³ = 150 L

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

P * V = n * R * T

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

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

Using the given values:

Pressure (P) = 100 atm

Volume (V) = 150 L

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

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

Calculating the number of moles:

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

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

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

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

P * V = n * R * T

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

Using the given values:

Pressure (P) = 1.20 atm

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

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

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

Calculating the number of moles:

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

Finally, calculating the number of balloons:

Number of balloons = 6.19 moles / 0.006 moles = 1031

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

Answer:

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

Explanation:

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

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

This is equal to the Coulomb's Force.

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

Answer:

It's true. Potential energy is actually the concentrations of different elements. Diffusion is the process of moving the elements or materials from the area of high concentration (high potential energy) to the low concentration. So in bot facilitated and simple types of diffusion the level of potential energy decreases across the membrane.

Explanation:

Answer:

6908316.619 N/C

25087609.3949 N/C

6652357.02259 N/C

690831.6619 N/C

Explanation:

x = Distance from the ring

R = Radius of ring = 10 cm

q = Charge = 78 μC

k = Coulomb constant = [tex]8.99\times 10^{9}\ Nm^2/C^2[/tex]

Electric field at a point x is from a ring given by

[tex]E=\dfrac{kqx}{(x^2+r^2)^{1.5}}[/tex]

For 1 cm

[tex]E=\dfrac{kqx}{(x^2+r^2)^{1.5}}\\\Rightarrow E=\dfrac{8.99\times 10^9\times 78\times 10^{-6}\times 0.01}{(0.01^2+0.1^2)^{1.5}}\\\Rightarrow E=6908316.619\ N/C[/tex]

The electric field is 6908316.619 N/C

For 5 cm

[tex]E=\dfrac{kqx}{(x^2+r^2)^{1.5}}\\\Rightarrow E=\dfrac{8.99\times 10^9\times 78\times 10^{-6}\times 0.05}{(0.05^2+0.1^2)^{1.5}}\\\Rightarrow E=25087609.3949\ N/C[/tex]

The electric field is 25087609.3949 N/C

For 30 cm

[tex]E=\dfrac{kqx}{(x^2+r^2)^{1.5}}\\\Rightarrow E=\dfrac{8.99\times 10^9\times 78\times 10^{-6}\times 0.3}{(0.3^2+0.1^2)^{1.5}}\\\Rightarrow E=6652357.02259\ N/C[/tex]

The electric field is 6652357.02259 N/C

For 100 cm

[tex]E=\dfrac{kqx}{(x^2+r^2)^{1.5}}\\\Rightarrow E=\dfrac{8.99\times 10^9\times 78\times 10^{-6}\times 1}{(1^2+0.1^2)^{1.5}}\\\Rightarrow E=690831.6619\ N/C[/tex]

The electric field is 690831.6619 N/C

When substances with different heat capacities are brought into thermal contact, heat transfers until they reach thermal equilibrium. In this case, the final temperature will be closer to the initial temperature of substance A.

When two substances with different heat capacities are brought into thermal contact, heat will transfer from the substance with a higher initial temperature to the substance with a lower initial temperature until they reach thermal equilibrium. In this case, since substance A has a much greater heat capacity than substance B, it can absorb and transfer a larger amount of heat. Therefore, the final temperature of the two substances will likely be closer to the initial temperature of substance A and lower than the initial temperature of substance B.

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Answer:

The Double Slits Experiment and Michelson Interferometer

The questions will be answered for the double slits experiment.

(b) Mathematically, the double slits experiment equation can be given as: d sin 0 = m(Wavelength). d is the separation distance, 0 is the diffraction angle, m = 1,2,3,...

(c) assumptions

The width of the slits is lesser than the microwave's wavelength. This is to set up a standing wave between the microwave source and detector.

m is an positive integer. To obtain a constructive interference of the E-M wave(microwave)

(d) The uncertainties are:

(i)The zero error in the reading of the multiplier will disrupt the value of the wavelength by small percentage. This can be adjusted to obtain more accurate result.

(ii) The angle as obtained by the gionometer can not be measured to highest level of accuracy, as there are some approximations. High sensitive equipment should be used to obtain accurate result

Explanation:

Double Slits Experiment.

The double slits is a pair of 2cm wide slits with a 6cm separation, cut in a metal foil, with the double slits at the centre of turnable with the microwave source and microwave detector through the slits. The shunt on the microammeter is adjusted to

so that a large scale deflection is obtained. The interference pattern may then be explored by moving the receiver arm, whilst keeping the transmitter arm fixed. A graph of intensity (current) versus θ(Measured through the gionometer) should be plotted, and an estimate made of the microwave wavelength by measuring the angular separation of adjacent maxima in the interference pattern and the separation of the two slits.

Michelson Interferometer

Setting up the interferometer, the mirrors are metal sheets and the beamsplitter is a hardboard sheet. The metal sheets should be carefully positioned to be perpendicular to the microwave beam, and the beamsplitter positioned at the centre of the turntable and at 45° to the beam. Any slight change in the path length of either of the beams leaving the beamsplitter changes the interference pattern at the receiver. One of the metal sheets is mounted in a frame which slides along rails. If this sheet is slowly moved, maxima and minima of current will be recorded by the receiver and a plot of “intensity” against distance can be made. The distance between successive maxima (or minima) is one half of the microwave wavelength which can thus be measured.