Rosa goes into the hospital with her son.
Her son does not recall what happened so Rosa describes the following scene to the doctor.
“He was sitting on the couch watching a movie when his muscles got really tense. He started to drool and grunt and then started convulsing. He then urinated on himself. After a few seconds, he stopped convulsing and complained of being very tired.”
What most likely caused Rosa’s son to have that experience?
a stroke
a metallic taste
feeling anxious
a seizure

Answer:

magnitude: 21.6; direction: 33.7 degrees

Explanation:

When we multiply a vector by a scalar, we have to multiply each component of the vector by the scalar number. In this case, we have

vector: (-3,-2)

Scalar: -6

so the vector multiplied by the scalar will have components

[tex](-3\cdot (-6), -2 \cdot (-6))=(18,12)[/tex]

The magnitude is given by Pythagorean's theorem:

[tex]m=\sqrt{18^2+12^2}=21.6[/tex]

and the direction is given by the arctan of the ratio between the y-component and the x-component:

[tex]\theta = tan^{-1} (\frac{12}{18})=33.7^{\circ}[/tex]

(a) [tex]2.7\cdot 10^{25} kg[/tex]

The acceleration due to gravity on the surface of the planet is given by

[tex]g=\frac{GM}{R^2}[/tex] (1)

where

G is the gravitational constant

M is the mass of the planet

R is the radius of the planet

Here we know:

[tex]g=22.4 m/s^2[/tex]

[tex]d=1.8\cdot 10^7 m[/tex] is the diameter, so the radius is

[tex]R=\frac{d}{2}=\frac{1.8\cdot 10^7 m}{2}=9\cdot 10^6 m[/tex]

So we can re-arrange eq.(1) to find M, the mass of the planet:

[tex]M=\frac{gR^2}{G}=\frac{(22.4 m/s^2)(9\cdot 10^6 m)^2}{6.67\cdot 10^{-11}}=2.7\cdot 10^{25} kg[/tex]

(b) [tex]4.8\cdot 10^{31}kg[/tex]

The planet is orbiting the star, so the centripetal force is equal to the gravitational attraction between the planet and the star:

[tex]m\frac{v^2}{r}=\frac{GMm}{r^2}[/tex] (1)

where

m is the mass of the planet

M is the mass of the star

v is the orbital speed of the planet

r is the radius of the orbit

The orbital speed is equal to the ratio between the circumference of the orbit and the period, T:

[tex]v=\frac{2\pi r}{T}[/tex]

where

[tex]T=402 days = 3.47\cdot 10^7 s[/tex]

Substituting into (1) and re-arranging the equation

[tex]m\frac{4\pi r^2}{rT^2}=\frac{GMm}{r^2}\\\frac{4\pi r}{T^2}=\frac{GM}{r^2}\\M=\frac{4\pi r^3}{GMT^2}[/tex]

And substituting the numbers, we find the mass of the star:

[tex]M=\frac{4\pi^2 (4.6\cdot 10^{11} m)^3}{(6.67\cdot 10^{-11})(3.47\cdot 10^7 s)^2}=4.8\cdot 10^{31}kg[/tex]

Answer:

0.20 A

Explanation:

The current in the wire is given by Ohm's law:

[tex]I=\frac{V}{R}[/tex]

where

I is the current

V is the voltage

R is the resistance

In this problem,

V = 3.0 V is the voltage

[tex]R=15 \Omega[/tex] is the resistance

Substituting into the formula,

[tex]I=\frac{3.0 V}{15 \Omega}=0.20 A[/tex]

Final answer:

Using Ohm's Law, we calculate the current in a nichrome wire with 15Ω resistance connected to a 3.0 V battery to be 0.20 amps.

Explanation:

To determine the amount of current flowing through a nichrome wire when connected to a 3.0 V flashlight battery, we can use Ohm's Law, which states that the current (I) is equal to the voltage (V) divided by the resistance (R). Given that the resistance of the wire is 15Ω, we can calculate the current using the formula I = V/R.

Therefore, the current in the wire is as follows:

I = 3.0 V / 15Ω = 0.2 A

So, the current in the nichrome wire is 0.20 amps.

Answer:

Explanation:

The energy stored in the spring = the kinetic energy at the bottom of the incline

1/2 kx² = 1/2 mv²

kx² = mv²

(500 N/m) (0.75 m)² = (10 kg) v²

v ≈ 5.3 m/s

The energy stored in the spring = the initial potential energy - work done by friction

1/2 kx² = mgh - W

1/2 (500 N/m) (0.75 m)² = (10 kg) (9.8 m/s²) (2.0 m) - W

W ≈ 55 J

Since the horizontal surface is frictionless, the object will have the same speed at the bottom of the incline as before.

v ≈ 5.3 m/s

Answer:

In this problem we need to use the conservation of energy theorem, which is about to constant exchange between kinetic and potential energy. In this case, we have two important points to analyse. The first one is at the initial point, where the mass is just released at 2 m from the ground. The final point would be when the mass hits the massless spring.

So, in the initial point the total energy of the mass would be only potential energy due to its position and its lack of speed. Then, the mass will be released gaining speed, until reaches the spring, where all the kinetic energy is transformed in potential energy but due to elastic forces (spring). So, in the final point, both energy are the same, the kinetic and the elastic potential energy, which we express like this:

[tex]K_{f}=U_{f}[/tex]

But, the definition of each one is:

[tex]K=\frac{1}{2}mv^{2}  \\U=\frac{1}{2}kx^{2}[/tex]

Where k is the constant of the spring. So, we replace this in the initial equation and solve for v:

[tex]\frac{1}{2}mv_{f} ^{2}=\frac{1}{2}kx^{2}\\v_{f}=\sqrt{\frac{kx^{2}}{m}}\\v_{f}=\sqrt{\frac{500(0.75)^{2} }{10}}=5.30m/s[/tex]

Therefore, the speed in the spring point is 5.30 m/s.

Now, to calculate the work due to friction, we have to analyse first, because the problem doesn't offer a friction coefficient, but, we can analyse according to energies again, because work and energy are magnitudes close related, and they have the same dimension.

So, if we thought about it, we'll find that the work due to friction is what slows down the mass, subtracting its potential energy from the initial point, giving as a result the final potential energy:

[tex]U_{f}=U_{i}-W_{k}[/tex]

[tex]\frac{1}{2}kx^{2}  =mgy-W_{k} \\W_{k}=10(9.8)(2)-\frac{1}{2}(500)(0.75)^{2}\\W_{k}=196-140.625=55.38J[/tex]

Therefore, the word due to friction is 55.38 J.

Lastly, the option c is asking about the speed of the object when it reaches the vase of the incline, which is the horizontal ground. The problem specify that this portion of the trajectory is frictionless, which mean that nothing slows down the mass.

Therefore, the speed in the horizontal part is the same 5.30 m/s.

Answer:

The percentage uncertainty in the average speed is 0.10% (2 sig. fig.)

Explanation:

Consider the formula for average speed [tex]\bar{v}[/tex].

[tex]\displaystyle \bar{v} = \frac{s}{t}[/tex],

where

The percentage uncertainty of a fraction is the sum of percentage uncertainties in

What are the percentage uncertainties in [tex]s[/tex] and [tex]t[/tex] in this question?

The unit of the absolute uncertainty in [tex]s[/tex] is meters. Thus, convert the unit of [tex]s[/tex] to meters:

[tex]s = \rm 42.238\;km = 42.238\times 10^{3}\;m[/tex].

[tex]\begin{aligned}\displaystyle \text{Percentage Uncertainty in }s &= \frac{\text{Absolute Uncertainty in } s}{\text{Measured Value of }s}\times 100\% \\ &=\rm\frac{29\; m}{42.238\times 10^{3}\;m}\times 100\%\\ &= 0.0687\%\end{aligned}[/tex].

The unit of the absolute uncertainty in [tex]t[/tex] is seconds. Convert the unit of [tex]t[/tex] to seconds:

[tex]t = \rm 2\times 3600 + 31\times 60 + 46 = 9106\;s[/tex]

Similarly,

[tex]\begin{aligned}\displaystyle \rm \text{Percentage Uncertainty in }t &= \frac{\text{Absolute Uncertainty in }t}{\text{Measured Value of }t}\times 100\% \\ &=\rm\frac{46\; s}{9106\;s}\times 100\%\\ &= 0.0329\%\end{aligned}[/tex].

The average speed [tex]\bar{v}[/tex] here is a fraction of [tex]s[/tex] and [tex]t[/tex]. Both [tex]s[/tex] and [tex]t[/tex] come with uncertainty. The percentage uncertainty in [tex]\bar{v}[/tex] will be the sum of percentage uncertainties in [tex]s[/tex] and [tex]t[/tex]. That is:

[tex]\text{Percentage Uncertainty in }\bar{v}\\=(\text{Percentage Uncertainty in } s) + (\text{Percentage Uncertainty in } t)\\ = 0.0687\% + 0.0329\%\\ = 0.010\%[/tex].

Generally, keep

The percentage uncertainty in [tex]\bar{v}[/tex] here is less than 2%. Thus, keep two significant figures. However, keep more significant figures than that in calculations to make sure that the final result is accurate.

Answer:

22.10 kg

Explanation:

The worker wants to recast the model from iron to gold, so the volume of the model made of iron and made of gold will be the same:

[tex]V_i = V_g[/tex]

the volume of each model can be rewritten as the ratio between the mass of the model, m, and the density of the material, d:

[tex]V=\frac{m}{d}[/tex]

so we can rewrite the first equation as

[tex]\frac{m_i}{d_i}=\frac{m_g}{d_g}[/tex]

where we have

[tex]m_i = 9.00 kg[/tex] is the mass of the model in iron

[tex]d_i = 7.86\cdot 10^3 kg/m^3[/tex] is the density of iron

[tex]m_g[/tex] is the mass of gold needed

[tex]d_g = 19.30\cdot 10^3 kg/m^3[/tex] is the density of gold

Solving for [tex]m_g[/tex], we find

[tex]m_g = d_g \frac{m_i}{d_i}=(19.30\cdot 10^3 kg/m^3) \frac{9.00 kg}{7.86\cdot 10^3 kg/m^3}=22.10 kg[/tex]

a. [tex]E=k\frac{e}{6.25}[/tex], [tex]53.1^{\circ}[/tex] above x-axis

The electric field produced by a single point charge is

[tex]E=k\frac{q}{r^2}[/tex]

where

q is the charge

r is the distance from the charge

In this case,

q = e

the charge is at the origin, so the distance of point A (1.50, 2.00) from the charge is

[tex]r=\sqrt{(1.50-0)^2+(2.00-0)^2}=2.5[/tex]

So the magnitude of the electric field at point A is

[tex]E=k\frac{e}{(2.5)^2}=k\frac{e}{6.25}[/tex]

The electric field produced by a single-point positive charge is radial, pointing away from the charge. So the direction of the field in this case is given by

[tex]\theta = tan^{-1} (\frac{A_y}{A_x})=tan^{-1} (\frac{2.00}{1.50})=53.1^{\circ}[/tex]

above the x-axis.

b. [tex]F=k\frac{e^2}{6.25}[/tex], [tex]53.1^{\circ}[/tex] above x-axis

The magnitude of the electric force exerted on a charge is given by

[tex]F=qE[/tex]

where

q is the charge magnitude

E is the electric field strength

In this case, we have a proton (charge e), and the electric field strength at point A is

[tex]E=k\frac{e}{6.25}[/tex]

So the force exerted on the proton is

[tex]F=(e)(k\frac{e}{6.25})=k\frac{e^2}{6.25}[/tex]

Moreover, the direction of the electric force exerted on a positive charge is equal to the direction of the electric field at that point, so the direction of the electric force is also [tex]53.1^{\circ}[/tex] above x-axis.

c. [tex]F=k\frac{e^2}{6.25}[/tex], [tex]233.1^{\circ}[/tex]

The electric force exerted on a charge is given by

[tex]F=qE[/tex]

In this case, we have an electron (charge e), and the electric field strength at point A is

[tex]E=k\frac{e}{6.25}[/tex]

So the force exerted on the electron is equal as the one for the proton:

[tex]F=(e)(k\frac{e}{6.25})=k\frac{e^2}{6.25}[/tex]

However, this time the direction is different: in fact, for a negative charge the electric force points in a direction opposite to the electric field. So, the direction of the electric force in this case is

[tex]\theta = 53.1^{\circ}+180^{\circ}=233.1^{\circ}[/tex]

d. [tex]E'=k\frac{e}{6.25}[/tex],  [tex]233.1^{\circ}[/tex]

In order for the electron at point A not to have acceleration, the net force on it should be zero:

[tex]F_{net} =0[/tex]

The net force is the algebraic sum of the force calculated at point c), due to the electric field produced by the proton, and the additional force F' due to the uniform electric field that we need to add:

[tex]F_{net}=F+F' = 0\\F' = -F[/tex]

Rewriting [tex]F'=eE'[/tex]

where E' is the uniform electric field that we add, we find

[tex]E' = -\frac{F}{e}=-\frac{1}{e}k\frac{e^2}{6.25}=-k\frac{e}{6.25}\\[/tex]

So the magnitude of the uniform electric field must be

[tex]E'=k\frac{e}{6.25}[/tex]

and the direction must be opposite to the electric field produced by the proton, so at an angle of [tex]233.1^{\circ}[/tex].

e. [tex]V=k \frac{e}{2.5}[/tex]

The electric potential generated by a single-point positive charge is

[tex]V=k \frac{q}{r}[/tex]

where here we have

q = e  is the charge of the proton

r = 2.5 is the distance of point A from the proton

Substituting into the formula, we find

[tex]V=k \frac{e}{2.5}[/tex]

f. [tex]k\frac{e^2}{2.5}[/tex]

The electric potential energy of a charge in an electric field is

[tex]U=qV[/tex]

where q is the charge and V is the electric potential

Here we have a proton (charge q=e) located at point A, where the potential is

[tex]V=k \frac{e}{2.5}[/tex]

So its electric potential energy is

[tex]U=(e)(k\frac{e}{2.5})=k\frac{e^2}{2.5}[/tex]

g. [tex]-k\frac{e^2}{2.5}[/tex]

As before, electric potential energy of a charge in an electric field is

[tex]U=qV[/tex]

Here we have an electron (charge q=-e) located at point A, where the potential is

[tex]V=k \frac{e}{2.5}[/tex]

So its electric potential energy this time is

[tex]U=(-e)(k\frac{e}{2.5})=-k\frac{e^2}{2.5}[/tex]

So it's the same as the one calculated at part f), but with a negative sign, since the charge is negative.

Answer:

83.2 W/m^2

Explanation:

The radiation per unit area of a star is directly proportional to the power emitted, which is given by Stefan-Boltzmann law:

[tex]P=\sigma A T^4[/tex]

where

[tex]\sigma[/tex] is the Stefan-Boltzmann constant

A is the surface area

T is the surface temperature

So, we see that the radiation per unit area is proportional to the fourth power of the temperature:

[tex]I \propto T^4[/tex]

So in our problem we can write:

[tex]I_1 : T_1^4 = I_2 : T_2^4[/tex]

where

[tex]I_1 = 1400 W/m^2[/tex] is the power per unit area of the present sun

[tex]T_1 = 5800 K[/tex] is the temperature of the sun

[tex]I_2[/tex] is the power per unit area of sun X

[tex]T_2 = 2864 K[/tex] is the temperature of sun X

Solving for I2, we find

[tex]I_2 = \frac{I_1 T_2^4}{T_1^4}=\frac{(1400 W/m^2)(2864 K)^4}{(5800 K)^4}=83.2 W/m^2[/tex]

Answer:

[tex]0.116 m^3/s[/tex]

Explanation:

The volume flow rate of a fluid in a pipe is given by:

[tex]Q=Av[/tex]

where

A is the cross-sectional area of the pipe

v is the speed of the fluid

In this problem, at the initial point we have

v = 0.84 m/s is the speed of the water

r = 0.21 m is the radius of the pipe, so the cross-sectional area is

[tex]A=\pi r^2 = \pi (0.21 m)^2 =0.138 m^2[/tex]

So, the volume flow rate is

[tex]Q=(0.138 m^2)(0.84 m/s)=0.116 m^3/s[/tex]

Answer:

[tex]d=\frac{V}{E}[/tex]

Explanation:

The electric field strength between the plates of a parallel plate capacitor is given by:

[tex]E=\frac{V}{d}[/tex]

where

V is the potential difference

d is the distance between the plates

If we re-arrange the formula, we find a new equation that lets us calculate d:

[tex]d=\frac{V}{E}[/tex]

We can at this point substitute the values:

V = 112 V

E = 1.1 kV/cm = 1100 V/cm = 110,000 V/m

To find the distance:

[tex]d=\frac{112 V}{110,000 V/m}=1.02\cdot 10^{-3} m[/tex]

The distance between capacitor plates, using the formula d = V/E, is approximately 1 cm

To calculate the distance d between the plates of a parallel plate capacitor, we use the relationship between the electric field strength E, the voltage V, and the distance d: E = V/d. Given:

V = 112 V (voltage across the plates)

E = 1.1 kV/cm = 1.1 * 10⁵ V/m (electric field strength)

The formula can be rearranged to solve for d:

d = V/E

Where:

Substituting the given values into the equation will allow us to calculate the distance between the plates.

Given:

Voltage (V) = 112 V

Electric field strength (E) = 1.1 kV/cm = 1.1 × 10⁴ V/m

Distance (d) = ?

Formula: E = V/d

Substitute: 1.1 × 10⁴ = 112/d

Solve for d: d = 112 / (1.1 × 10⁴)

[tex]\approx[/tex] 0.01 m [tex]\approx[/tex] 1 cm

The distance between the capacitor plates is thus approximately 1 cm.

Answer:

Explanation:

The rate of conductive heat transfer in watts is:

q = (k/s) A ΔT

where k is the heat conductivity, s is the thickness, A is the area, and ΔT is the temperature difference.

a)

Given k = 52.4 W/m/K, s = 0.014 m, and ΔT = 350-140 = 210 K, we can find q/A:

q/A = (52.4 / 0.014) (210)

q/A = 786,000 W/m²

b)

Given that A = 0.42 m², we can find q:

q = (0.42 m²) (786,000 W/m²)

q = 330,120 W

A watt is a Joule per second.  Convert to Joules per hour:

q = 330,120 J/s * 3600 s/hr

q = 1.19×10⁹ J/hr

c)

If we change k to 1.8 W/m/K:

q = (k/s) A ΔT

q = (1.8 / 0.014) (0.42) (210)

q = 11,340 J/s

q = 4.08×10⁷ J/hr

d)

If k is 52.4 W/m/K and s is 0.024 m:

q = (k/s) A ΔT

q = (52.4 / 0.024) (0.42) (210)

q = 192,570 J/s

q = 6.93×10⁸ J/hr

The heat flux is 785,600 W/m^2, and the heat loss per hour is 1,110,912,000 J/h for the first metal. Using a material with a thermal conductivity of 1.8 W/m-K, the heat loss is 42,504,000 J/h. With increased thickness to 24 mm, the heat loss is 491,820,000 J/h.

Firstly, you would calculate the heat flux through the metal sheet using Fourier's Law of Heat Conduction, which is expressed as:

q = -k * (T_2 - T_1) / d

In this case, the temperature difference (T_2 - T_1) is (350°C - 140°C), the thickness (d) of the metal sheet is 14 mm (which is 0.014 meters), and the thermal conductivity (k) is given as 52.4 W/m-K. Substituting these values in:

q = -52.4 * (350 - 140) / 0.014 = 785,600 W/m^2

For part (b), to calculate the heat loss per hour, we multiply the heat flux by the area of the sheet and convert seconds to hours:

Q/t = q * A = 785,600 * 0.42 m^2 * 3600 seconds = 1,110,912,000 J/h

For part (c), using a material with a thermal conductivity of 1.8 W/m-K, the process is similar, and the heat loss per hour would be:

Q/t = -1.8 * (350 - 140) / 0.014 * 0.42 * 3600 = 42,504,000 J/h

For part (d), with the increased thickness of 24 mm (0.024 meters), the heat loss per hour would be:

Q/t = -52.4 * (350 - 140) / 0.024 * 0.42 * 3600 = 491,820,000 J/h

1. [tex]1.69\cdot 10^{11} m[/tex]

The Schwarzschild radius of an object of mass M is given by:

[tex]r_s = \frac{2GM}{c^2}[/tex] (1)

where

G is the gravitational constant

M is the mass of the object

c is the speed of light

The black hole in the problem has a mass of

[tex]M=5.7\cdot 10^7 M_s[/tex]

where

[tex]M_s = 2.0\cdot 10^{30} kg[/tex] is the solar mass. Substituting,

[tex]M=(5.7\cdot 10^7)(2\cdot 10^{30}kg)=1.14\cdot 10^{38} kg[/tex]

and substituting into eq.(1), we find the Schwarzschild radius of this black hole:

[tex]r_s = \frac{2(6.67\cdot 10^{-11})(1.14\cdot 10^{38} kg)}{(3\cdot 10^8 m/s)^2}=1.69\cdot 10^{11} m[/tex]

2) 242.8 solar radii

We are asked to find the radius of the black hole in units of the solar radius.

The solar radius is

[tex]r_S = 6.96\cdot 10^5 km = 6.96\cdot 10^8 m[/tex]

Therefore, the Schwarzschild radius  of the black hole in solar radius units is

[tex]r=\frac{1.69\cdot 10^{11} m}{6.96\cdot 10^8 m}=242.8[/tex]

The balloon was 296.4 m high when the rock was thrown out; b) The balloon is 176.4 m high when the rock hits the ground; c) The distance between the rock and the basket when the rock hits the ground is 198.03 m; d) The horizontal velocity of the rock as measured by an observer at rest in the basket and ground is 15.0 m/s.

a) The initial velocity of the balloon is 20.0 m/s downward. The initial velocity of the rock is 15.0 m/s perpendicular to the path of the descending balloon. The person in the basket sees the stone hit the ground 6.00 s after being thrown.

Let the height of the balloon when the rock was thrown be h. The time it takes the rock to hit the ground is:

t = h / 15.0 m/s

Plugging in the value of h, we get:

t = 296.4 m / 15.0 m/s = 19.76 s

The total time it takes the rock to hit the ground is 6.00 s + 19.76 s = 25.76 s.

The vertical velocity of the rock is constant. The vertical velocity of the rock is:

v_y = 15.0 m/s

The horizontal velocity of the rock is also constant. The horizontal velocity of the rock is:

v_x = 15.0 m/s * (6.00 s / 25.76 s) = 3.53 m/s

The horizontal distance traveled by the rock is:

d_x = v_x * t

d_x = 3.53 m/s * 25.76 s = 90.36 m

The height of the balloon, when the rock was thrown, is:

h = d_x + v_y t

h = 90.36 m + 15.0 m/s * 25.76 s = 296.4 m

Therefore, the balloon was 296.4 m high when the rock was thrown out.

(b)The vertical velocity of the rock is constant. The vertical velocity of the rock is:

v_y = 15.0 m/s

The time it takes the rock to hit the ground is:

t = v_y / g

t = 15.0 m/s / 9.8 m/s² = 1.53 s

The height of the balloon when the rock hits the ground is:

h = v_y t

h = 15.0 m/s * 1.53 s = 22.95 m

Therefore, the balloon is 22.95 m high when the rock hits the ground.

(c) The horizontal velocity of the rock is constant. The horizontal velocity of the rock is:

v_x = 3.53 m/s

The time it takes the rock to hit the ground is:

t = 1.53 s

The horizontal distance traveled by the rock is:

d_x = v_x * t

d_x = 3.53 m/s * 1.53 s = 5.40 m

The distance between the rock and the basket when the rock hits the ground is:

d = d_x + h

d = 5.40 m + 22.95 m = 28.35 m

Therefore, the distance between the rock and the basket when the rock hits the ground is 28.35 m.

(d) i) The horizontal velocity of the rock as measured by an observer at rest in the basket is 15.0 m/s.

ii) The vertical velocity of the rock as measured by an observer at rest on the ground is 15.0 m/s.

To know more about velocity:

https://brainly.com/question/34025828

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a. 37500;  299,792,457.9 m/s

The formula for the length contraction for a particle moving close to the speed of light is:

[tex]L' = \frac{L}{\gamma}[/tex]

where

L' is the length observed by the particle moving

L is the length observed by an observer at rest

In this problem, we have

[tex]L = 3 km = 3000 m[/tex] is the length of the SLAC measured by an observer at rest

[tex]L' = 8cm = 0.08 m[/tex] is the length measured by the electrons moving

Substituting into the formula, we find the gamma factor

[tex]\gamma = \frac{L}{L'}=\frac{3000 m}{0.08 m}=37,500[/tex]

The formula for the gamma factor is

[tex]\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}[/tex]

where v is the electron's speed and c is the speed of light. Re-arranging the equation, we find v:

[tex]1-\frac{v^2}{c^2}=\frac{1}{\gamma^2}\\v=c \sqrt{ 1-\frac{1}{\gamma^2}}=(299 792 458 m/s)\sqrt{1-\frac{1}{(37500)^2}}=299,792,457.9 m/s[/tex]

b. [tex]1.0\cdot 10^{-5}s[/tex]

For an observer at rest in the laboratory, the electron is moving at a speed of

v = 299,792,457.9 m/s

and it covers a total distance of

L = 3000 m

which is the length of the SLAC measured by the observer. Therefore, the time it takes for the electron to travel down the tube is

[tex]t=\frac{L}{v}=\frac{3000 m}{299,792,457.9 m/s}=1.0\cdot 10^{-5}s[/tex]

c. [tex]2.67\cdot 10^{-10}s[/tex]

From the electron's point of view, the length of the SLAC is actually contracted, so the electron "sees" a total distance to cover of

[tex]L' = 0.08 m[/tex]

And this means that the total time of travel of the electron, in its frame of reference will be shorter; in particular it is given by the formula:

[tex]t' = \frac{t}{\gamma}[/tex]

where

[tex]t=1.0\cdot 10^{-5}s[/tex] is the time measured by the observer at rest

[tex]\gamma=37500[/tex]

Substituting,

[tex]t' = \frac{1.0\cdot 10^{-5}s}{37500}=2.67\cdot 10^{-10}s[/tex]

I don’t know what the answer is I Wish I could help

Both light and matter can possess particle properties and wave properties at the same time.

This quantum behavior was proposed by the French physicist Louis De Broglie in 1924 in his doctoral thesis when proposing that all matter has an associated wave, and light did not escape from that behavior. Then this was empirically tested in various experiments where light behaved as a photon (particle) and in others as particles.

Einstein also demonstrated this fact, however, in spite of being observed this behavior several times, it could not have been done simultaneously, until in 2015 a group of scientists from the Federal Polytechnic School of Lausanne (Switzerland), succeeded to perform an experiment and capture for the first time the image of the wave-particle duality of light.

1. 13,500 cal

First of all, we need to find the amount of heat needed to raise the temperature of the ice from -20°C to 0°C. This is given by

[tex]Q_1 = m C_i \Delta T[/tex]

where

m = 150 g is the mass of the ice

C_i = 0.5 cal/g·C° is the specific heat capacity of the ice

[tex]\Delta T=0 C-(-20 C)=20^{\circ}C[/tex] is the change in temperature of the ice

Substituting,

[tex]Q_1 = (150 g)(0.5 cal/gC)(20 C)=1500 cal[/tex]

Now we have to find the amount of heat needed to melt the ice, which is

[tex]Q_2 = m \lambda_f[/tex]

where

m = 150 g is the mass of the ice

[tex]\lambda_f = 80 cal/g[/tex] is the latent heat of fusion

Substituting,

[tex]Q_2 = (150 g)(80 cal/g)=12,000 cal[/tex]

So the total heat required is

[tex]Q_3 = 1500 cal + 12,000 cal = 13,500 cal[/tex]

2. 3750 cal

The additional amount of heat required to heat the water to 25°C is

[tex]Q_4 = m C_w \Delta T[/tex]

where

m = 150 g is the mass of water

C_w = 1 cal/g·C is the speficic heat capacity of water

[tex]\Delta T=25 C-0 C=25^{\circ}C[/tex] is the change in temperature

Substituting,

[tex]Q_4 = (150 g)(1 cal/gC)(25 C)=3,750 cal[/tex]

3. 9200 cal

First of all, we need to find the amount of heat needed to raise the temperature of the ice from -20°C to 0°C. As at point 1., this is given by

[tex]Q_1 = m C_i \Delta T[/tex]

where

m = 80 g is the mass of the ice

C_i = 0.5 cal/g·C° is the specific heat capacity of the ice

[tex]\Delta T=0 C-(-20 C)=20^{\circ}C[/tex] is the change in temperature of the ice

Substituting,

[tex]Q_1 = (80 g)(0.5 cal/gC)(20 C)=800 cal[/tex]

Now we have to find the amount of heat needed to melt the ice:

[tex]Q_2 = m \lambda_f[/tex]

where

m = 80 g is the mass of the ice

[tex]\lambda_f = 80 cal/g[/tex] is the latent heat of fusion

Substituting,

[tex]Q_2 = (80 g)(80 cal/g)=6,400 cal[/tex]

Finally, the amount of heat required to heat the water to 25°C is

[tex]Q_3 = m C_w \Delta T[/tex]

where

m = 80 g is the mass of water

C_w = 1 cal/g·C is the speficic heat capacity of water

[tex]\Delta T=25 C-0 C=25^{\circ}C[/tex] is the change in temperature

Substituting,

[tex]Q_3 = (80 g)(1 cal/gC)(25 C)=2,000 cal[/tex]

So the total heat required is

[tex]Q=Q_1+Q_2+Q_3=800 cal+6,400 cal+2,000 cal=9,200 cal[/tex]

4. No

Explanation:

The total heat required for this process consists of 3 different amounts of heat:

1- The heat required to bring the ice at melting temperature

2- The heat required to melt the ice, while its temperature stays constant

3- The heat required to raise the temperature of the water

However, computing how much heat is required to melt the ice and adding the amount of heat required to raise the temperature of 80 g of water by 45°C is not equivalent: in fact, the calculation of point 1) requires to use the specific heat capacity of ice, not that of water, therefore the two are not equivalent.

Answer:

0.0835 m

Explanation:

The magnetic force exerted on the deuteron is equal to the centripetal force that keeps it in circular motion:

[tex]qvB = m\frac{v^2}{r}[/tex]

where

[tex]q=+e = 1.6\cdot 10^{-19} C[/tex] is the charge

[tex]v=8.0\cdot 10^5 m/s[/tex] is the speed of the deuteron

[tex]B=0.20 T[/tex] is the magnetic field strength

[tex]m=3.34\cdot 10^{-27} kg[/tex] is the mass of the deuteron

r is the radius of the orbit

re-arranging the equation, we find:

[tex]r=\frac{mv}{qB}=\frac{(3.34\cdot 10^{-27}kg)(8.0\cdot 10^5 m/s)}{(1.6\cdot 10^{-19} C)(0.20 T)}=0.0835 m[/tex]

1) Static friction coefficient: 0.355

The crate is initially at rest. The crate remains at rest until the horizontal pushing force is less than the maximum static frictional force.

The maximum static frictional force is given by

[tex]F_s = \mu_s mg[/tex]

where

[tex]\mu_s[/tex] is the static coefficient of friction

m = 21 kg is the mass of the crate

g = 9.8 m/s^2 is the acceleration due to gravity

The horizontal force required to set the crate in motion is 73 N: this means that this is the value of the maximum static frictional force. So we have

[tex]F_s=73 N[/tex]

Using this information into the previous equation, we can find the coefficient of static friction:

[tex]\mu_s = \frac{F}{mg}=\frac{73 N}{(21 kg)(9.8 m/s^2)}=0.355[/tex]

2) Kinetic friction coefficient: 0.267

Now the crate is in motion: this means that the kinetic friction is acting on the crate, and its magnitude is

[tex]F_k = \mu_k mg[/tex] (1)

where

[tex]\mu_k[/tex] is the coefficient of kinetic friction

There is a horizontal force of

F = 55 N

pushing the crate. Moreover, the speed of the crate is constant: this means that the acceleration is zero, a = 0.

So we can write Newton's second law as

[tex]F-F_k = ma = 0[/tex]

And by substituting (1), we can find the value of the coefficient of kinetic friction:

[tex]F-\mu_k mg = 0\\\mu_k = \frac{F}{mg}=\frac{55 N}{(21 kg)(9.8 m/s^2)}=0.267[/tex]

The coefficients of static and kinetic friction between the crate and floor are;

μ_s = 0.3547

μ_k = 0.2672

1) To find the coefficient of static friction, is given by the formula;

F_s = μ_s*mg

We are given;

Mass; m = 21 kg

Horizontal force to set it to motion; F_h = 73 N

Now, in this static friction equation, we are using this force of 73 N because static friction is friction at rest and 73 N is the force to pull the object from rest. Thus;

73 = μ_s(21 × 9.8)

μ_s = 73/(21 × 9.8)

μ_s = 0.3547

2) To find the coefficient of kinetic friction, is given by the formula;

F_k = μ_k*mg

We will make use of the force of 55N because we are dealing with friction associated with motion and 55N is the force that keeps the object in motion. Thus;

μ_k = F_k/(mg)

μ_k = 55/(21 × 9.8)

μ_k = 0.2672

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The average speed and kinetic energy of gas molecules decrease with decreasing temperature, and there are gas molecules that move slower than the average.

The correct statements are:

The average speed of gas molecules decreases with decreasing temperature. This is because at lower temperatures, the molecules have less kinetic energy and move slower.

The average kinetic energy of gas molecules decreases with decreasing temperature. As temperature decreases, the average kinetic energy of the gas molecules also decreases.

There are gas molecules that move slower than the average. In a sample of gas, there will always be molecules that have speeds below the average.

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Several statements about gas molecules and temperature are true, including the decrease in average speed and kinetic energy with decreasing temperature. Molecular speeds and kinetic energies vary, ensuring not all molecules have the same energy and some move slower than average. The kinetic energy of a molecule can determine its speed.

Let's analyze each statement one by one to determine which are true:

A) [tex]E=\frac{4.50\cdot 10^{10}}{r^2} V/m[/tex]

r < a

We can find the magnitude of the electric field by using Gauss theorem. Taking a Gaussian spherical surface of radius r centered in the centre of the sphere, the electric flux through the surface of the sphere is equal to the ratio between the charge contained in the sphere and the vacuum permittivity:

[tex]E\cdot 4 \pi r^2 = \frac{q}{\epsilon_0}[/tex]

For r < a, the charge contained in the gaussian sphere is the point charge:

[tex]q=5.00 C[/tex]

So the electric field in this region is

[tex]E=\frac{q}{4\pi \epsilon_0 r^2}=\frac{5.00 C}{4\pi (8.85\cdot 10^{-12} F/m)}\frac{1}{r^2}=\frac{4.50\cdot 10^{10}}{r^2} V/m[/tex]

B) E = 0

a < r < b

The region a < r < b is the region between the inner and the outer surface of the shell. We have to keep in mind that the presence of the single point charge +q = 5.00 C at the center of the sphere induces an opposite charge -q on the inner surface (r=a), and a charge of +q at the outer surface (r=b).

Using again Gauss theorem

[tex]E\cdot 4 \pi r^2 = \frac{q'}{\epsilon_0}[/tex]

this time we have that the gaussian sphere contains both the single point charge +q and the negative charge -q induced at r=a, so the net charge contained in the sphere is

q' = +q - q = 0

And so, the electric field in this region is zero.

C) [tex]E=\frac{4.50\cdot 10^{10}}{r^2} V/m[/tex]

r > b

Here we are outside of the sphere. Using Gauss theorem again

[tex]E\cdot 4 \pi r^2 = \frac{q'}{\epsilon_0}[/tex]

this time we have that the gaussian sphere contains the single point charge +q, the negative charge -q induced at r=a, and the positive charge +q induced at r=b, so the net charge contained in the sphere is

q' = +q - q +q = q

And so the electric field is identical to the one inside the sphere:

[tex]E=\frac{q}{4\pi \epsilon_0 r^2}=\frac{5.00 C}{4\pi (8.85\cdot 10^{-12} F/m)}\frac{1}{r^2}=\frac{4.50\cdot 10^{10}}{r^2} V/m[/tex]

D) -12.29 C/m^2

We said that the charge induced at the inner surface r=a is

-q = -5.00 C

The induced charge density is

[tex]\sigma = \frac{-q}{A}[/tex]

where A is the area of the inner surface of radius r = a = 0.18 m, so it is

[tex]A=4\pi a^2 = 4 \pi (0.18 m)^2=0.407 m^2[/tex]

So the induced charge density is

[tex]\sigma = \frac{-5.00 C}{0.407 m^2}=-12.29 C/m^2[/tex]

E) +3.65 C/m^2

We said that the charge induced at the outer surface r=b is

+q = +5.00 C

The induced charge density is

[tex]\sigma = \frac{+q}{A}[/tex]

where A is the area of the outer surface of radius r = b = 0.33 m, so it is

[tex]A=4\pi b^2 = 4 \pi (0.33 m)^2=1.368 m^2[/tex]

So the induced charge density is

[tex]\sigma = \frac{+5.00 C}{1.368 m^2}=+3.65 C/m^2[/tex]

Answer:

kinetic energy from wind turns blades of wind turbines → wind turbine blades turn generator → generator changes kinetic energy to electricity

Answer:

kinetic energy from wind turns blades of wind turbines → wind turbine blades turn generator → generator changes kinetic energy to electricity

Explanation:

Wind energy is the energy harnessed from the velocity of wind. It is obtained with help of wind mills that have large fan type structure. The fan type structure has blades fixed on frame. When the high velocity wind blows, it cause the fan blades to rotate. The rotating fan causes the generator to turn. This rotating generator produces electricity.  Now according to the explanation second option is the correct answer that is kinetic energy from wind turns blades of wind turbines → wind turbine blades turn generator → generator changes kinetic energy to electricity

a) See part a of the attached picture.

There are two forces involved here:

- The weight of the block attached to the spring, of magnitude

W = mg

where m is the mass of the block and g = 9.8 m/s^2 is the acceleration due to gravity, pointing downward

- The restoring force of the spring,

F = kx

with k being the spring constant and x the displacement of the spring with respect to the equilibrium position, pointing upward

b) [tex]L' = L+\frac{mg}{k}[/tex]

The spring is in equilibrium when the restoring force of the spring is equal to the weigth of the block:

[tex]kx = mg[/tex]

where

x is the displacement of the spring with respect to the natural length of the spring, L. Solving for x,

[tex]x=\frac{mg}{k}[/tex]

And since the natural length is L, the equilibrium length of the spring is

[tex]L' = L+x=L+\frac{mg}{k}[/tex]

c) [tex]y(t) = y cos(\sqrt{\frac{k}{m}} t + \pi)[/tex]

Let's assume that

y = 0

corresponds to the equilibrium position of the spring (which is stretched by an amount [tex]L+\frac{mg}{k}[/tex]). If doing so, the vertical position of the mass at time t is given by

[tex]y(t) = y cos(\omega t + \pi)[/tex]

where

[tex]\omega = \sqrt{\frac{k}{m}}[/tex] is the angular frequency

t is the time

y is the initial displacement with respect to the equilibrium position

[tex]\pi[/tex] is the phase shift, so that the position at time t=0 is negative:

y(0) = -y

So rewriting the angular frequency:

[tex]y(t) = y cos(\sqrt{\frac{k}{m}} t + \pi)[/tex]

d) [tex]T=2 \pi \sqrt{\frac{m}{k}}[/tex]

The period of oscillation is given by

[tex]T=\frac{2\pi}{\omega}[/tex]

where

[tex]\omega = \sqrt{\frac{k}{m}}[/tex] is the angular frequency

Substituting [tex]\omega[/tex], we find an expression for the period

[tex]T=2 \pi \sqrt{\frac{m}{k}}[/tex]

e) k' = 2k (see part e of attached picture)

Here the two half springs of spring constant k' are connected in series, so the sum of the stretchings of the two springs is equal to the total stretching of the spring, x:

[tex]x=x_1 + x_2[/tex]

Also, since the two springs are identical, their stretching will be the same:

[tex]x_1 = x_2 = x'[/tex]

so we have

[tex]x=x'+x'=2x'[/tex]

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

Substituting x find in part (b),

[tex]x'=\frac{mg}{2k}[/tex] (1)

Hooke's law for each spring can be written as

[tex]F=k' x'[/tex] (2)

where

F = mg (3)

is still the weight of the block

Using (1), (2) and (3) together, we find an expression for the spring constant k' of each spring

[tex]mg=k'\frac{mg}{2k}\\k' =2k[/tex]

So, the spring constant of each half-spring is twice the spring constant of the original spring.

fa) See part f) of attached picture

This time we have the block hanging from both the two half-springs, each with spring constant k'. So at equilibrium, the weight of the block is equal to the sum of the restoring forces of the two springs:

[tex]k' x_1 + k' x_2 = mg[/tex]

fb) [tex]x'=\frac{L}{2}+\frac{mg}{2k'}[/tex]

For the two springs in parallel, the sum of the restoring forces of the two springs must be equal to the weight of the block:

[tex]F_1 + F_2 = mg\\k_1 x_1 + k_2 x_2 = mg[/tex]

The two springs are identical, so they have same spring constant:

[tex]k_1 = k_2 = k'[/tex]

So (1) can be rewritten as

[tex]k' (x_1 + x_2) = mg[/tex]

And since the two springs are identical, their stretching x' is the same:

[tex]x_1 = x_2 = x'[/tex]

so we can rewrite this as

[tex]k' (2x') = mg[/tex]

[tex]x'=\frac{mg}{2k'}[/tex]

and so the equilibrium length of each spring will be

[tex]x'=\frac{L}{2}+\frac{mg}{2k'}[/tex]

fc) [tex]y(t) = y cos(\sqrt{\frac{2k'}{m}} t + \pi)[/tex]

The system of two springs in parallel can be treated as a system of a single spring with equivalent spring constant given by

[tex]k_{eq}=k_1 + k_2 = 2k'[/tex]

where k' is the spring constant of each spring.

So, let's assume again that

y = 0

corresponds to the equilibrium position as calculated in the previous part. If doing so, the vertical position of the mass at time t is given by:

[tex]y(t) = y cos(\omega t + \pi)[/tex]

where this time we have

[tex]\omega = \sqrt{\frac{k_{eq}}{m}}[/tex] is the angular frequency of the system

t is the time

y is the initial displacement with respect to the equilibrium position

[tex]\pi[/tex] is the phase shift, which we put so that the position at time t=0 is negative:

y(0) = -y

If we rewrite the angular frequency,

[tex]\omega = \sqrt{\frac{2k'}{m}}[/tex]

the position of the mass is

[tex]y(t) = y cos(\sqrt{\frac{2k'}{m}} t + \pi)[/tex]

fd) [tex]T=2 \pi \sqrt{\frac{m}{2k'}}[/tex]

Similarly to part d), the period of oscillation is

[tex]T=\frac{2\pi}{\omega}[/tex]

where

[tex]\omega = \sqrt{\frac{k_{eq}}{m}}=\sqrt{\frac{2k'}{m}}[/tex] is the angular frequency

Substituting [tex]\omega[/tex], we find

[tex]T=2 \pi \sqrt{\frac{m}{2k'}}[/tex]

Answer:magnitude -5; angle 160°

Explanation:

Vector A is described as having magnitude 5 and angle -20°.

To get an equivalent vector, we either leave the magnitude at 5 and add 360° to the angle, or we reverse the magnitude to -5 and add 180° to the angle.

5 @ -20° = 5 @ 340°

5 @ -20° = -5 @ 160°

The third one is the answer.

(a) 5.35 s, 84.65 s

The distance covered by the rocket during the first part of the motion (accelerated motion) is

[tex]d_1 = \frac{1}{2}at_1 ^2[/tex] (1)

where

a = +35 ft/s^2 is the acceleration

While the distance covered during the second part of the motion (uniform motion) is

[tex]d_2 = v t_2[/tex]

where v is the constant speed reached after the first part of the motion, which is given by

[tex]v=at_1[/tex]

So we can rewrite the equation as

[tex]d_2 = a t_1 t_2[/tex] (2)

We also know:

[tex]d= d_1 + d_2 = 16350[/tex] (3) is the total distance

[tex]t= t_1 +t_2 = 90[/tex] (4) is the total time

We can rewrite (3) as

[tex]\frac{1}{2}at_1^2 + a t_1 t_2 = 16350[/tex] (5)

And (4) as

[tex]t_2 = 90-t_1[/tex]

Substituting into (5), we get

[tex]\frac{1}{2}at_1^2 + a t_1 (90-t_1) = 16350\\0.5 at_1^2 + 90at_1 - at_1^2 = 16350\\-0.5at_1^2 +90at_1 = 16350[/tex]

Substituting a=+35 ft/s^2, we have

[tex]-17.5 t_1^2 +3150 t_1 - 16350 = 0[/tex]

which has two solutions:

[tex]t_1 = 174.65 s[/tex] --> discarded, since the total time cannot be greater than 90 s

[tex]t_1 = 5.35 s[/tex] --> this is the correct solution

And therefore [tex]t_2[/tex] is

[tex]t_2 = 90-t_1 = 90 -5.35 s=84.65 s[/tex]

(b) 187.25 ft/s

The velocity v is the velocity at the end of the accelerated motion of the rocket, so after

[tex]t_1 = 5.35 s[/tex]

The acceleration is

a = +35 ft/s^2

Therefore, the velocity after the first part is:

[tex]v=at_1 = (+35 ft/s^2)(5.35 s)=187.25 ft/s[/tex]

(c) 16,585 ft

At the 15750 ft mark, the velocity of the sled is

[tex]v=187.25 ft/s[/tex]

Then, it starts decelerating with acceleration

a = -21 ft/s^2

So, its distance covered during this part of the motion will be given by:

[tex]v_f ^ 2 -v^2 = 2ad_3[/tex]

where

vf = 0 is the final speed

v = 187.25 ft/s is the initial speed

a = -21 ft/s^2 is the acceleration

d3 is the distance covered during this part

Solving for d3,

[tex]d_3 = \frac{v_f^2 -v^2}{2a}=\frac{0-(187.25 ft/s)^2}{2(-21 ft/s^2)}=835 ft[/tex]

So, the final position of the sled will be

[tex]x_3 = 15750 ft + 835 ft =16,585 ft[/tex]

(d) 93.92 s

The duration of the first part of the motion is

[tex]t_1 = 5.35 s[/tex]

The distance covered during this part is

[tex]d_1 = \frac{1}{2}at_1^2=\frac{1}{2}(+35 ft/s^2)(5.35 s)^2=501 ft[/tex]

In the second part (uniform motion), the sled continues with constant speed until the 15750 ft mark, so the distance covered is

[tex]d_2 = 15750- 835 =14915[/tex]

And so the duration of the second part is

[tex]t_2 =\frac{d_2}{v}=\frac{14915}{187.25}=79.65 s[/tex]

While the duration of the third part (decelerated motion) is

[tex]a=\frac{v_f-v}{t_3}\\t_3 = \frac{v_f-f}{a}=\frac{0-(187.25 ft/s)}{-21 ft/s^2}=8.92 s[/tex]

So the total duration is

t = 5.35 s+79.65 s+8.92 s=93.92 s

This physics problem involving kinematics can be solved returning to the kinematic equations and applying them to the scenarios described. Acceleration, constant velocity and deceleration periods of motion each have specific times and these can be used to derive the specifics of the journey such as times, velocity and final position.

The subject of this question is Physics, specifically, it is related to the concept of Kinematics which falls under Classical Mechanics.

In the first scenario, the sled accelerates from rest under the power of the rocket engine. We can use the kinematic equation: d = (1/2)at2 to solve for time t1, where a is the acceleration and d is the distance. The distance travelled under the acceleration is total distance minus the distance it goes with constant speed.

After finding t1, the time the sled moves with constant speed t2 is total time minus t1.

The constant speed v can be found by the distance traveled under constant speed divided by t2. When the sled hits the 15750ft mark, it starts to decelerate at -21 ft/s2. We can use the equation v2 = u2 + 2ad to get the distance it travels before it comes to rest.

Add this distance to 15750ft to get the final position of the sled when it comes to rest. Since we have the acceleration and final velocity is zero, one may get the time of this stage by using equation v = u + at . And the total duration is the sum of all time segments.

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Answer:

6.214g/cm³

Explanation:

The question is on density of a material

Density=mass/volume

Given, mass=87grams   and volume= 14 cm³  density=?

Density=m/v 87/14 =6.214g/cm³

Answer:

93 m

Explanation:

We need to find the resultant displacement.

Let's take north as positive y-direction and east as positive direction. We have

- Displacement 1 is 51 m to the north, so the two components are

[tex]x_1 =0\\y_1 = 51 m[/tex]

- Displacement 2 is 45 m [tex]60^{\circ}[/tex] north of east, so the two components are

[tex]x_1 = (45)cos 60^{\circ}=22.5 m\\y_1 = (45) sin 60^{\circ}=39.0 m[/tex]

So to find the resultant displacement we have to sum the components along each direction:

[tex]x=x_1 + x_2 = 0+22.5 m = 22.5 m\\y = y_1 + y_2 = 51 m +39.0 m = 90.0 m[/tex]

And the magnitude of the resultant displacement is

[tex]m=\sqrt{x^2+y^2}=\sqrt{(22.5 m)^2+(90.0 m)^2}=92.8 m \sim 93 m[/tex]

Answer:

Move further apart

Explanation:

The negative charge on the rod repels the electrons on the cap thus they concentrate at the leaves causing an increase in the amount of charge at the leaves. Since like charges repel, the two leaves of the electroscope move further apart.

Final answer:

When a negatively-charged rod is brought near a negatively-charged electroscope, the leaves of the electroscope will move farther apart due to increased electrostatic repulsion.

Explanation:

When a negatively-charged rod is brought close to an already negatively-charged electroscope, the electrostatic repulsion between the like charges will cause the two leaves of the electroscope to move farther apart. This is due to an increase in the density of negative charges within the leaves of the electroscope, which causes the leaves to repel each other even more.

Charging by induction occurs when a charged object is brought near, but not in contact with, a conductor. However, since the electroscope in this case is already negatively charged, bringing another negatively charged object close will simply enhance the repulsion between charges.

(a) [tex]1.24\cdot 10^{-11} F[/tex]

The general formula for the capacitance of a capacitor is:

[tex]C=\frac{Q}{V}[/tex]

where

Q is the charge stored on the capacitor

V is the potential difference across the capacitor

In this problem, we have

[tex]Q=3.10 nC = 3.10\cdot 10^{-9} C[/tex] is the charge stored

V = 250 V is the potential difference

Substituting, we find

[tex]C=\frac{3.10\cdot 10^{-9}C}{250 V}=1.24\cdot 10^{-11} F[/tex]

(b) 0.0294 m

The capacitance of a spherical capacitor is given by

[tex]C=\frac{4\pi \epsilon_0}{\frac{1}{a}-\frac{1}{b}}[/tex]

where

a is the radius of the inner shell

b is the radius of the outer shell

Here we have

b = 4.00 cm = 0.04 m

and the capacitance is

[tex]C=1.24\cdot 10^{-11} F[/tex]

So we can re-arrange the equation to find a, the radius of the inner sphere:

[tex]a=(\frac{4\pi \epsilon_0}{C}+\frac{1}{b})^{-1} =(\frac{4\pi (8.85\cdot 10^{-12}F/m)}{1.24\cdot 10^{-11}F}+\frac{1}{0.04 m})^{-1}=0.0294 m[/tex]

(c) [tex]3.225 \cdot 10^4 V/m[/tex]

The electric field just outside the surface of the inner sphere with charge Q is equal to the electric field produced by a single point charge Q at a distance of r = a:

[tex]E=\frac{Q}{4\pi \epsilon_0 a^2}[/tex]

where

[tex]Q=3.10 \cdot 10^{-9} C[/tex] is the charge on the sphere

a = 0.0294 m is the radius of the inner sphere

Substituting the data into the formula, we find:

[tex]E=\frac{3.10\cdot 10^{-9}C}{4\pi (8.85\cdot 10^{-12}F/m) (0.0294 m)^2}=3.225 \cdot 10^4 V/m[/tex]

Answer:

no

Explanation:

because will the cool air will be leaving the fridge the motor in the bottom will be running and eventually the mixture of the cool and hot air will balance out

'a' is the correct statement.