According to Archimedes' principle, the mass of a floating object equals the mass of the fluid displaced by the object. A 150-lbm swimmer is floating in a nearby pool; 95% of his or her body's volume is in the water while 5% of his or her body's volume is above water. Determine the density of the swimmer's body. The density of water is 0.036lbm/in^3.

The distance of the image is 60 cm and the focal length of this concave mirror is 30 cm.

The focal length of the lens is length of the distance between the middle of the lens to the focal point.

It can be find out using the following formula as,

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

Here, (v)is the distance of the image, (u) is the distance of the object, and (f) is the focal length of the lens.

Here, the concave mirror produces a real image that is three times as large as the object. The object is 20 cm in front of the mirror, and the concave mirror produces a real image that is three times as large as the object.

Hence, the value of magnification of the mirror is 3.

The object distance is 20 cm thus the image distance, using the magnification formula, can be given as,

[tex]m=\dfrac{v}{u}\\3=\dfrac{v}{20}\\v=60\rm cm[/tex]

Put the values in the lens formula as,

[tex]\dfrac{1}{60}+\dfrac{1}{20}=\dfrac{1}{f}\\f=30 \rm \; cm[/tex]

Hence, the distance of the image is 60 cm and the focal length of this concave mirror is 30 cm.

Learn more about the focal length here;

https://brainly.com/question/25779311

Final answer:

Using the mirror equation and magnification formula, the image distance is found to be -60 cm and the focal length of the concave mirror is calculated to be 30 cm.

Explanation:

The question involves determining the image distance and the focal length of a concave mirror that produces a real image three times larger than the object. The object is placed 20 cm in front of the mirror. To find these values, we can use the mirror equation which is 1/f = 1/do + 1/di, where f is the focal length, do is the object distance, and di is the image distance.

Given that the magnification (m) is -3 (negative sign since the image is real and inverted), and magnification is also equal to -di/do, we can express di as -3do. We are told that do is 20 cm, so di is -3(20 cm) = -60 cm (negative because the image is on the same side as the object).

Now, we can plug these values into the mirror equation to find the focal length (f):

Therefore, the image distance is -60 cm and the focal length of the concave mirror is 30 cm.

Answer:

work done is = 0

Explanation:

given data

distance = 2 cm

potential = 1 V

charge with magnitude = 1 nC

to find out

work done by the electrostatic force

solution

we know that at equipotential surface is that surface which have equal potential at each every point that we say

work done will be

work done = ∫dw

∫dw = [tex]\int\limits^v1_v2 {q} \, dv[/tex]

here q is charge

so

net work done = q ( v2 - v1 )

and

so v2 = v1 = 0

so

work done is = 0

No work is done in moving a charge along an equipotential line because there is no difference in electric potential along that line, making the work done by the electrostatic force zero.

The question asks to calculate the work done by the electrostatic force when moving a charge along an equipotential line. By definition, the potential difference along an equipotential line is zero. Therefore, the work done W by the electric force to move a charge q in an electric potential V along an equipotential line is given by the equation W = qΔV, where ΔV is the change in electric potential. Since ΔV is zero along an equipotential line, the work done is also zero. No work is required to move a charge along an equipotential line because there is no change in electric potential energy.

Answer:

Mass of second object will be 15 kg

Explanation:

We have given mass of first object = 12 kg

Acceleration [tex]a=5m/sec^2[/tex]

According to second law of motion we know that force F = MA

So force [tex]F=12\times 5=60N[/tex]

As the same force is applied to the second object of acceleration [tex]a=4m/Sec^2[/tex]

So force = ma

[tex]60=m\times 4[/tex]

m = 15 kg

So mass of second object will be 15 kg

Answer:

Volume left, v = 67 mL

Explanation:

Given that,

Volume of soda contained in a can, V = 375 mL

We need to find the volume of soda left after 308 mL of soda is removed, V' = 308 mL

Let v is the left volume of soda. It can be calculated using simple calculations as :

v = V - V'

v = 375 mL - 308 mL

v = 67 mL

So, the left volume in the can is 67 mL. Hence, this is the required solution.

Answer:

Distance, d = 99990 meters

Explanation:

It is given that,

Speed of the train, v = 200 km/h = 55.55 m/s

Time taken, t = 1800 s

Let d is the distance covered by the train. We know that the speed of an object is given by total distance covered divided by total time taken. Mathematically, it is given by :

[tex]v=\dfrac{d}{t}[/tex]

[tex]d=v\times t[/tex]

[tex]d=55.55\times 1800[/tex]

d = 99990 m

So, the distance covered by the train is 99990 meters. Hence, this is the required solution.

The magnitude of [tex]\( q_2 \)[/tex] is [tex]\( 674.99 \, \mu \text{C} \)[/tex].

To find the magnitude of [tex]\( q_2 \)[/tex], we can use Coulomb's Law and the principle of superposition. Coulomb's Law states that the force between two point charges is given by:

[tex]\[ F = \frac{k \cdot |q_1 \cdot q|}{r_1^2} + \frac{k \cdot |q_2 \cdot q|}{r_2^2} \][/tex]

where ( k ) is Coulomb's constant [tex](\( 8.99 \times 10^9 \, \text{N}\cdot\text{m}^2/\text{C}^2 \))[/tex], [tex]\( q_1 \)[/tex] and [tex]\( q_2 \)[/tex] are the charges, [tex]\( q \)[/tex] is the test charge, and [tex]\( r_1 \)[/tex] and [tex]\( r_2 \)[/tex] are the distances from [tex]\( q \)[/tex] to the charges [tex]\( q_1 \)[/tex] and [tex]\( q_2 \)[/tex], respectively.

Given the problem, we have [tex]\( F = 23 \, \text{N} \)[/tex], [tex]\( q_1 = -27 \times 10^{-6} \, \text{C} \)[/tex], [tex]\( q = 9 \times 10^{-6} \, \text{C} \)[/tex], [tex]\( r_1 = 0.21 \, \text{m} \)[/tex], [tex]\( r_2 = 0.35 \, \text{m} \)[/tex], and we need to find [tex]\( q_2 \)[/tex].

First, let's calculate the force due to [tex]\( q_1 \)[/tex]:

[tex]\[ F_1 = \frac{k \cdot |q_1 \cdot q|}{r_1^2} = \frac{8.99 \times 10^9 \cdot |-27 \times 10^{-6} \cdot 9 \times 10^{-6}|}{(0.21)^2} \][/tex]

[tex]\[ F_1 = \frac{8.99 \times 10^9 \cdot 243 \times 10^{-12}}{0.0441} = \frac{2185.57}{0.0441} = 49536.73 \, \text{N} \][/tex]

Now, the force due to [tex]\( q_2 \)[/tex]:

[tex]\[ F_2 = \frac{k \cdot |q_2 \cdot q|}{r_2^2} \][/tex]

We know the net force is in the positive y-direction, so [tex]\( F_2 \)[/tex] must be positive.

Now, sum the forces and equate to the given net force:

[tex]\[ F = F_1 + F_2 \][/tex]

[tex]\[ 23 = 49536.73 + F_2 \][/tex]

[tex]\[ F_2 = 23 - 49536.73 = -49513.73 \, \text{N} \][/tex]

Now, plug in [tex]\( F_2 \)[/tex] and solve for [tex]\( q_2 \)[/tex]:

[tex]\[ -49513.73 = \frac{8.99 \times 10^9 \cdot |q_2 \cdot 9 \times 10^{-6}|}{(0.35)^2} \][/tex]

[tex]\[ -49513.73 = \frac{8.09 \times 10^9 \cdot |q_2|}{0.1225} \][/tex]

[tex]\[ |q_2| = \frac{-49513.73 \cdot 0.1225}{8.99 \times 10^9} \][/tex]

[tex]\[ |q_2| = -674.99 \times 10^{-6} \][/tex]

Since [tex]\( q_2 \)[/tex] is positive, [tex]\( q_2 = 674.99 \times 10^{-6} \, \text{C} \)[/tex].

Thus, the magnitude of [tex]\( q_2 \)[/tex] is [tex]\( 674.99 \, \mu \text{C} \)[/tex].

Answer:

a) To the nearest tenth the average value of the measures is 217.4cm

b) The standard deviation for the four measurements is 0.415

Explanation:

The average is the result of adding all the values and dividing by the number of measures, in this case 4 then

[tex]Average = \frac{217.6 + 217.2 + 216.8 + 217.9}{4} \\[/tex]

[tex]Average = \frac{869.5}{4} \\[/tex]

[tex]Average = 217.375 [/tex]

a) To the nearest tenth the average value of the measures is 217.4cm

The standard deviation equals to the square root of the variance, and the variance is the adition of all the square of the difference between the measure and the average for each value divided by the number of measurements

So first, we must calculate the variance:

[tex]Variance = \frac{(217.6-217.4)^2+(217.2-217.4)^2+(216.8-217.4)^2+(217.9-217.4)^2}{4}[/tex].

[tex]Variance = \frac{(0.2)^2+(-0.2)^2+(-0,6)^2+(0.5)^2}{4}[/tex].

[tex]Variance = \frac{0.04+0.04+0.36+0.25}{4}[/tex].

[tex]Variance = \frac{0.69}{4}[/tex].

[tex]Variance = 0.1725[/tex].

This represent the difference between the average and the measurements.

Now calculate the standard deviation

Standard deviation  = [tex]\sqrt{Variance}[/tex]

Standard deviation  = [tex]\sqrt{0.1725}[/tex]

Standard deviation  = 0.415

b) The standard deviation for the four measurements is 0.415

This measure represent a standard way to know what is normal in this sample. so the differences between the average should be of ±0.415

The average value of the four door height measurements is 217.4 cm, and the standard deviation of these measurements is approximately 0.4 cm.

To find the average value of the four measurements: 217.6 cm, 217.2 cm, 216.8 cm, and 217.9 cm, you add them up and then divide by the number of measurements, which is four:

Average = (217.6 + 217.2 + 216.8 + 217.9) / 4 = 217.375

Rounded to the nearest tenths place, the average is 217.4 cm.

For the standard deviation, we first calculate the variance. First, find the difference of each measurement from the mean, square that difference, and then find the average of those squared differences. Finally, take the square root of that average to find the standard deviation:

Substituting the values and performing the calculations gives us a standard deviation of approximately 0.4 cm.

Answer:

The answer is 9 m.

Explanation:

Using the kinematic equation for an object in free fall:

[tex]y = y_o - v_o-\frac{1}{2}gt^{2}[/tex]

In this case:

[tex]v_o = \textrm{Initial velocity} = 1.5[m/s]\\t = \textrm{air time} =  1.2 [s][/tex][tex]y_o = 0[/tex]

[tex]g = \textrm{gravity} = 9.8 [m/s^{2} ][/tex]

Plugging those values into the previous equation:

[tex]y = 0 - 1.5*1.2-\frac{1}{2}*9.8*1.2^{2} \\y = -8.85 [m] \approx -9 [m][/tex]

The negative sign is because the reference taken. If I see everything from the rescuer point of view.

Answer:

3.122×10¹⁴ g/cm³

Explanation:

Diameter of neutron star = 23 km = 2300000 cm

Radius of neutron star = 2300000/2 = 1150000 cm = r

Mass of neutron star = 1.989 × 10³⁰ kg = 1.989 × 10³³ g = m

Volume of neutron star

[tex]v=\frac{4}{3}\pi r^3\\\Rightarrow v=\frac{4}{3}\pi 1150000^3[/tex]

Density = Mass / Volume

[tex]\rho=\frac{m}{v}\\\Rightarrow \rho=\frac{1.989\times 10^{33}}{\frac{4}{3}\pi 1150000^3}\\\Rightarrow \rho=3.122\times 10^{14}\ g/cm^3[/tex]

∴ Density of neutron star is 3.122×10¹⁴ g/cm³

Final answer:

The typical density of a neutron star is about 10¹⁴g/cm³. This is calculated by dividing the mass of the star in grams by its volume in cubic centimeters, and taking into account that a neutron star's mass is typically 1.4 solar masses and its diameter is about 20 kilometers.

Explanation:

To calculate the density of a neutron star, we consider it as a sphere with a typical mass of 1.4 solar masses and a diameter of about 20 kilometers. The formula for density (d) is mass (m) divided by volume (V), and the volume of a sphere is given by the formula V = ⅔πr3, where r is the radius of the sphere.

First, we convert the solar mass to kilograms (1 solar mass = 1.99 × 1030 kg) and 1.4 solar masses to kg gives us 2.786 × 1030 kg. Next, we convert the diameter to radius in centimeters (10,000 cm), then calculate the volume. Now we can find the density:

Mass: 2.786 × 1030 kg
Volume: ⅔π(105 cm)³= 4.18879 × 1015 cm³
Density = Mass/Volume
Density = 2.786 × 1030 kg / 4.18879 × 1015 cm³

The density, in grams per cubic centimeter (g/cm³), is calculated by converting the mass from kilograms to grams. This gives us a typical neutron star density of about 1014 g/cm³, which is exceedingly high compared to materials we experience on Earth.

Answer:3.6

Explanation:

Given

Architect decide to double the width

length increases by 50 percent

height increases by 20 percent

Let L,B,H be length,height and width of rectangular room

L'=1.5L

B'=2B

H'=1.2H

therefore new volume V'=L'\times B'\times H'

[tex]V'=1.5L\times 2B\times 1.2H[/tex]

V'=3.6LBH

new volume is 3.6 times the original

Answer:

time will elapse before it return to  its staring point is 23.6 ns

Explanation:

given data

speed u = 2.45 × [tex]10^{7}[/tex] m/s

uniform electric field E = 1.18 × [tex]10^{4}[/tex] N/C

to find out

How much time will elapse before it returns to its starting point

solution

we find acceleration first by electrostatic force that is

F = Eq

here

F = ma by newton law

so

ma = Eq

here m is mass , a is acceleration and E is uniform electric field and q is charge of electron

so

put here all value

9.11 × [tex]10^{-31}[/tex] kg ×a = 1.18 × [tex]10^{4}[/tex] × 1.602 × [tex]10^{-19}[/tex]

a = 20.75 × [tex]10^{14}[/tex] m/s²

so acceleration is 20.75 × [tex]10^{14}[/tex] m/s²

and

time required by electron before come rest is

use equation of motion

v = u + at

here v is zero and u is speed given and t is time so put all value

2.45 × [tex]10^{7}[/tex] = 0 + 20.75 × [tex]10^{14}[/tex] (t)

t = 11.80 × [tex]10^{-9}[/tex] s

so time will elapse before it return to  its staring point is

time = 2t

time = 2 ×11.80 × [tex]10^{-9}[/tex]

time is 23.6 × [tex]10^{-9}[/tex] s

time will elapse before it return to  its staring point is 23.6 ns

Answer:

Frequency of the wave will be 31.84 Hz

Explanation:

We have given the equation [tex]y=0.02sin(30x-200t)[/tex]-----eqn 1

The standard equation of sine wave is given by [tex]y=Asin(kx-\omega t)[/tex]----eqn 2

On comparing eqn 1 and eqn 2

[tex]\omega =200[/tex]

Angular frequency [tex]\omega[/tex] is given by [tex]\omega =2\pi f[/tex]

So [tex]200=2\times 3.14\times f[/tex]

[tex]f=31.84Hz[/tex]

Answer:

De broglie wavelength = 4.17 pm

Explanation:

We have given mass of proton [tex]m=1.67\times 10^{-27}kg[/tex]

Speed of proton v [tex]v=9.50\times 10^{6}m/sec[/tex]

Plank's constant [tex]h=6.63\times 10^{-34}J-s[/tex]

We have to find de broglie wavelength

De broglie wavelength is given by [tex]\lambda =\frac{h}{mv}=\frac{6.63\times 10^{-34}}{1.67\times 10^{-27}\times 9.5\times 10^6}=4.17\times 10^{-12}m=4.17pm[/tex]

Answer:

The speed of light remains the same.

Explanation:

(a)  Time : second, hour minutes are example of time.

(b) Distance : meter, kilometers etc are some example of distance.

(c) The speed of light : It always remains constant. It is equal to [tex]3\times 10^8\ m/s[/tex].

(d) The weight of an object is given by the product of mass of an object and the acceleration due to gravity. As the value of g is not same everywhere, its weight varies.

So, the speed of light is always remains the same. Hence, the correct option is (c).                                                                

To find the charge per unit length on the copper rod, use the formula E = k * λ / r. Plug in the given values to find the charge per unit length. The electric flux through the cube can be calculated by multiplying the electric field by the area of one face of the cube.

(a) To find the charge per unit length on the copper rod, we can use the formula for an electric field (E = k * λ / r), where E is the electric field, k is the electrostatic constant, λ is the charge per unit length, and r is the distance from the center of the rod. Rearranging the formula, we can solve for λ: (λ = E * r / k). Plugging in the given values, we have: λ = (4 N/C) * (0.04 m) / (9 x 10^9 Nm^2/C^2) = 1.78 x 10^-10 C/m.

(b) The electric flux through a surface is given by the formula (Φ = E * A), where Φ is the electric flux, E is the electric field, and A is the area of the surface. In this case, the electric flux through the cube can be calculated by multiplying the electric field (4 N/C) by the area of one face of the cube (3 cm)^2 = (0.03 m)^2 = 9 x 10^-4 m^2. Therefore, the electric flux is Φ = (4 N/C) * (9 x 10^-4 m^2) = 3.6 x 10^-3 Nm^2/C.

(a) The charge per unit length on the copper rod is [tex]\( {8.90 \times 10^{-10} \, \text{C/m}} \).[/tex]

(b) The electric flux through the cube is [tex]\( {0.0036 \, \text{N} \cdot \text{m}^2/\text{C}} \).[/tex]

(a): Charge per Unit Length on the Copper Rod

The electric field ( E ) at a distance ( r ) from the axis of a long charged rod is given by:

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

From the given data:

[tex]\[4 = \frac{2 \cdot 8.99 \times 10^9 \cdot \lambda}{0.04}\][/tex]

Solving for [tex]\( \lambda \):[/tex]

[tex]\[\lambda = \frac{4 \cdot 0.04}{2 \cdot 8.99 \times 10^9}\][/tex]

[tex]\[\lambda = \frac{0.16}{1.798 \times 10^9}\][/tex]

[tex]\[\lambda \approx 8.90 \times 10^{-10} \, \text{C/m}\][/tex]

(b): Electric Flux through a Cube

To find the electric flux [tex]\( \Phi_E \)[/tex] through the cube, we use Gauss's law, which states:

[tex]\[\Phi_E = \oint_S \mathbf{E} \cdot d\mathbf{A} = \frac{q_{\text{enclosed}}}{\epsilon_0}\][/tex]

Since the rod is long and the cube is oriented such that the rod passes through opposite sides perpendicularly, the electric flux [tex]\( \Phi_E \)[/tex] through the cube is:

[tex]\[\Phi_E = E \cdot A\][/tex]

Calculate ( A ):

[tex]\[A = a^2 = (0.03)^2 = 0.0009 \, \text{m}^2\][/tex]

Now, calculate [tex]\( \Phi_E \):[/tex]

[tex]\[\Phi_E = 4 \times 0.0009\][/tex]

[tex]\[\Phi_E = 0.0036 \, \text{N} \cdot \text{m}^2/\text{C}\][/tex]

Answer:

1.35×10⁻⁷ m

37.278 mi/My

Explanation:

Speed of the tectonic plate= 6 cm/yr

Converting to seconds

[tex]6=\frac{6}{365.25\times 24\times 60\times 60}[/tex]

So in one second it will move

[tex]\frac{6}{365.25\times 24\times 60\times 60}[/tex]

In 71 seconds

[tex]71\times \frac{6}{365.25\times 24\times 60\times 60}=1.35\times 10^{-5}\ cm[/tex]

The tectonic plate will move 1.35×10⁻⁵ cm or 1.35×10⁻⁷ m

Convert to mi/My

1 cm = 6.213×10⁻⁶ mi

1 M = 10⁶ years

[tex]6\times 6.213\times 10^{-6}\times 10^6=37.278\ mi/My[/tex]

Speed of the tectonic plate is 37.278 mi/My

Answer:

[tex]E=0.697*10^{6}V/m\\[/tex]

Explanation:

If the voltage is constant, the relation between the electric field E and the voltage V , for two plates separated by a distance d, is:

[tex]V=E*d\\[/tex]

d=6cm=0.06m

V=4.18×10^4 V

We solve to find E:

[tex]E=V/d=4.18*10^{4}/0.06=6.97*10^{5}=0.697*10^{6}V/m\\[/tex]

Answer:

The net force on X is Fx=75.4N

The net force on Y is FY=25.17N

Explanation:

This is an electrostatic problem, we can calculate de force applying the formula:

[tex]F=k*\frac{Q*Q'}{r^2}\\where:\\k=coulomb constant\\r=distance\\Q=charge[/tex]

the force because of charge at 1cm on the X axis, will only have an X component of force, so:

[tex]Fx1=8.98755*10^9*\frac{(9\µC)(7\µC)}{(5*10^{-2}m)^2}\\Fx1=226.48N[/tex]

For the force because of the charge of the Y axis we have to find the distances usign pitagoras, and the angle:

[tex]r=\sqrt{(-1*10^{-2}m)^2+(6*10^{-2}m)^2} \\r=6.08cm=0.0608m[/tex]

we can find the angle with:

[tex]\alpha = arctg(\frac{1cm}{6cm})=9.46^o[/tex]

We now can calculate the force of the X axis because of the second charge:

[tex]Fx2=8.98755*10^9*\frac{(-9\µC)(7\µC)}{(6.08*10^{-2}m)^2}*cos(9.46 )\\Fx2=-151.08N[/tex]

and for the Force on Y axis:

[tex]Fy2=8.98755*10^9*\frac{(-9\µC)(7\µC)}{(6.08*10^{-2}m)^2}*sin(9.46 )\\Fy2=-25.17N[/tex]

The net force on X axis is:

Fx=226.48N-151.08N=75.4N

Fy=-25.17N

Answer:

The magnitude of the resultant force is equal to 0.0216 N

The direction is along the negative y axis

Explanation:

According to the exercise data:

q1 = 9x10^-6 C (0,1)

q2 = -9x10^-6 C (0,-1)

q = 7x10^6 C (6,0)

the distance between load q1 and load q will be equal to:

[tex]r_{1} = \sqrt{(0-1)^{2}+(6-0)^{2} }= 6.08 m[/tex]

The same way to calculate the distance between q2 and q:

[tex]r_{2} =\sqrt{(0-(-1))^{2}+(6-0)^{2} } =6.08 m[/tex]

The force on q due to the load q1, is calculated with the following equation:

[tex]F_{1}=\frac{K*q_{1}*q }{r1^{2} } *(cos45i-sin45j)=\frac{8.987559x10^{9}*9x10^{-6} *7x10^{-6} }{6.08^{2} }*(cos45i-sin45j)=0.0153*(0.707i-0.707j)=0.0108N(i-j)[/tex]

The same way to force F2:

F2 = (8.987559x10^9*-9x10^-6*7x10^-6))/(6.08^2)*(cos45i-sin45j) = 0.0108 N(i-j)

The resultant force:

F = F1 + F2 = 0.0108 N *(-2j) = - 0.0216 N j

The magnitude is equal to 0.0216 N

The direction is along the negative y axis

Answer:

The density of nuclear matter is [tex]10^{6}\ Mg/\mu L[/tex]

Explanation:

Given that,

Density [tex]\rho= 10^{18}\ kg/m^3[/tex]

Using unit conversation,

[tex]\Rightarrow \dfrac{10^{18}\ kg}{1\ m^3}\times\dfrac{1 Mg}{10^{6}g}\times\dfrac{1000\ g}{1\ kg}\times\dfrac{1\ m^3}{10^6\ cm^3}\times\dfrac{1\ cm^3}{1\ mL}\times\dfrac{1000\ mL}{1\ L}\times\dfrac{10^{-6}\ L}{1\ \muL}[/tex]

[tex]\Rightarrow \dfrac{10^{18}\times10^{3}\times10^{3}\times10^{-6}}{10^{6}\times10^{6}\times10^{6}}[/tex]

The density of nuclear matter is

[tex]\rho=10^{6}\ Mg/\mu L[/tex]

Hence, The density of nuclear matter is [tex]10^{6}\ Mg/\mu L[/tex]

Answer :

(a). The wavelength of electron is 26.22 μm.

(b).The wavelength of car is [tex]2.38\times10^{-38}\ m[/tex]

Explanation :

Given that,

Speed = 100 km/hr

Mass of car = 1 ton

(a). We need to calculate the wavelength of electron

Using formula of wavelength

[tex]\lambda_{e}=\dfrac{h}{p}[/tex]

[tex]\lambda_{e}=\dfrac{h}{mv}[/tex]

Put the value into the formula

[tex]\lambda_{e}=\dfrac{6.63\times10^{-34}}{9.1\times10^{-31}\times100\times\dfrac{5}{18}}[/tex]

[tex]\lambda=0.00002622[/tex]

[tex]\lambda=26.22\times10^{-6}\ m[/tex]

[tex]\lambda=26.22\ \mu m[/tex]

(II).  We need to calculate the wavelength of car

Using formula of wavelength again

[tex]\lambda_{e}=\dfrac{6.63\times10^{-34}}{1000\times100\times\dfrac{5}{18}}[/tex]

[tex]\lambda=2.38\times10^{-38}\ m[/tex]

The wavelength of the electron is greater than the dimension of electron and the wavelength of car is less than the dimension of car.

Therefore, electron is quantum particle and car is classical.

Hence, (a). The wavelength of electron is 26.22 μm.

(b).The wavelength of car is [tex]2.38\times10^{-38}\ m[/tex].

Answer:

the square of the average atomic velocity.

Explanation:

From the formulas for kinetic energy and temperature for a monoatomic gas, which has three translational degrees of freedom, the relationship between root mean square velocity and temperature is as follows:

[tex]v_{rms}=\sqrt{\frac{3RT}{M}}[/tex] (1)

Where  [tex]v_{rms}[/tex] is the root mean square velocity, M is the molar mass of the gas, R is the universal constant of the ideal gases and T is the temperature.

The root mean square velocity is a measure of the velocity of the particles in a gas. It is defined as the square root of the mean square velocity of the gas molecules:

[tex]v_{rms}=\sqrt{<v^2>}[/tex] (2)

substituting 2 in 1, we find the relationship between mean square speed and temperature:

[tex]\sqrt{<v^2>}=\sqrt{\frac{3RT}{M}}\\T=\frac{M<v^2>}{3R}\\\\T\sim  <v^2>[/tex]

Final answer:

The temperature of a monatomic ideal gas is directly proportional to the average atomic velocity.

Explanation:

The temperature of a monatomic ideal gas is directly proportional to the average atomic velocity. As the temperature increases, the average atomic velocity also increases. This relationship is a result of the fact that temperature is a measure of the kinetic energy of the gas particles, and the average velocity is related to the kinetic energy. Therefore, temperature and average atomic velocity are directly proportional in a monatomic ideal gas.

Answer:

(a) 82 Km

(b) 32°

Explanation:

First you should draw the vectors in the cartesian plane (please see the picture below).

As the car is driven east for a distance of 47 Km, your first vector should be drawn from the origin (0,0) and on the x axis

Then the car is driven north for a distance of 28 Km, so your second vector should be drawn from the origin and on the y axis.

And the car finally goes east of north for 27Km, so the third vector should be drawn from the origin east of north forming an 35° angle with x axis.

Then you should find the components of the vector in x and y:

For Vector 1 ([tex]V_{1}[/tex])

[tex]V_{1x}=47Km[/tex]

[tex]V_{1y}=0[/tex]

For Vector 2 ([tex]V_{2}[/tex])

[tex]V_{2x}=0[/tex]

[tex]V_{2y}=28Km[/tex]

For Vector 3 ([tex]V_{3}[/tex])

[tex]V_{3x}=27cos37^{o}[/tex]

[tex]V_{3x}=21.56[/tex]

[tex]V_{3y}=27sin37^{o}[/tex]

[tex]V_{3y}=16.25[/tex]

To find the magnitud of the car´s total displacement, (R) you should add up all the x and y components.

For the x component:

[tex]R_{x}=V_{1x}+V_{2x}+V_{3x}[/tex]

[tex]R_{x}=47+0+21.56[/tex]

[tex]R_{x}=68.56Km[/tex]

For the x component:

[tex]R_{y}=V_{1y}+V_{2y}+V_{3y}[/tex]

[tex]R_{y}=0+28+16.25[/tex]

[tex]R_{y}=44.25Km[/tex]

Now please see the second picture that is showing the components x and y as the sides of a right triangle, and we are going to use the Pythagorean theorem to find the resultant, R.

[tex]R=\sqrt{R_{x}^{2}+R_{y}^{2}[/tex]

[tex]R=\sqrt{68.56^{2}+44.25^{2}}[/tex]

[tex]R=82Km[/tex]

And to find the angle of the car´s total displacement (α), we use the same right triangle with the relationship between its legs.

[tex]tan(\alpha)=\frac{44.25}{68.56}[/tex]

[tex]tan(\alpha)=0.64[/tex]

[tex]\alpha=tan^{-1}(0.64)[/tex]

[tex]\alpha =32^{o}[/tex]

Answer:

a) [tex]v=230 m/s[/tex]

b) [tex]v=196.5 m/s[/tex]

Explanation:

a) The formula for average velocity is

[tex]v=\frac{y_{2}-y_{1}  }{t_{2}-t_{1}  }[/tex]

For the first Δt=4.7s

[tex]v=\frac{(1150-70)m}{(4.7)s} =230 m/s[/tex]

b) For the secont  Δt=5.85s we know that the displacement is 1150m. So, the average velocity is:

[tex]v=\frac{(1150)m}{(5.85)s}=196.5m/s[/tex]

Explanation:

Let [tex]q_1[/tex] is the first charge and [tex]q_2[/tex] is the second charge.

Force between them, F = 0.12 N

Distance between charges, d = 0.46 m

(a) Force acting between two point charges is given by :

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

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

[tex]q_1q_2=\dfrac{0.12\times (0.46)^2}{9\times 10^9}[/tex]

[tex]q_1q_2=2.82\times 10^{-12}[/tex]..............(1)

Also,

[tex]q_1+q_2=-8.9\ \mu C=-8.9\times 10^{-6}\ C[/tex]............(2) (both charges are negative)

On solving equation (1) and (2) :

[tex]q_1=-8.571\ C[/tex]

and

[tex]q_2=-0.329\ C[/tex]

(b) If the force is attractive, F = -0.12 N

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

[tex]q_1q_2=\dfrac{-0.12\times (0.46)^2}{9\times 10^9}[/tex]

[tex]q_1q_2=-2.82\times 10^{-12}[/tex]..............(3)

[tex]q_1+q_2=-8.9\ \mu C=-8.9\times 10^{-6}\ C[/tex]............(4)    

Solving equation (3) and (4) we get :

[tex]q_1=-0.306\ C[/tex]

[tex]q_2=9.206\ C[/tex]      

Hence, this is the required solution.                                  

Using Coulomb's law, set up a system of equations with q1 and q2 as unknowns and solve to find the values of the individual charges for the case of repulsive force. Repeat the process for attractive forces, keeping in mind that charges will have the opposite sign.

This problem can be solved using the formula for Coulomb's law, which states that the force between two charges is equal to the product of the charges divided by the distance squared, times the Coulomb constant: F = k*q1*q2/r², where F is the force, k is the Coulomb constant (8.99 * 10^9 N.m²/C²), q1 and q2 are the charges, and r is the distance between them.

For repulsive force, both charges have the same sign. But in this particular problem, we are given that the total charge is 8.90 μC, so let's take q1 and q2 as unknowns. Now q1 + q2 = 8.90 μC and using the above formula we get another equation. Now you have two equations to solve the unknown charges. Same procedure applies for the attractive force, but know that charges are of opposite sign for attractive force.

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Answer:0.10283 J

Explanation:

Given

mass of block(m)=0.0475 kg

radius of track (r)=0.425 m

when the Block is at Bottom Normal has a magnitude of 3.95 N

Force acting on block at bottom

[tex]N-mg=\frac{mu^2}{r}[/tex]

[tex]N=mg+\frac{mu^2}{r}[/tex]

[tex]3.95=0.0475\timess 9.81+\frac{0.0475\times u^2}{0.425}[/tex]

[tex]u^2=31.172[/tex]

[tex]u=\sqrt{31.172}[/tex]

u=5.583 m/s

At top point

[tex]N+mg=\frac{mv^2}{r}[/tex]

[tex]0.670+0.0475\timess 9.81=\frac{mv^2}{r}[/tex]

[tex]v^2=10.1639[/tex]

v=3.188 m/s

Using Energy conservation

[tex]\frac{mu^2}{2}=\frac{mv^2}{2}+mg\left ( 2r\right )+W_f[/tex]

[tex]W_f=0.4989-0.39607[/tex]

[tex]W_f=0.10283 J[/tex]

i.e. 0.10283  J of energy is wasted while moving up.

The work done on the block by friction during the motion of the block from point A to point B is 0.102 J.

The velocity of the block at the top of the vertical circle is calculated as follows;

[tex]W + mg = \frac{mv^2}{r} \\\\0.67 + 0.0475(9.8)= \frac{0.0475v^2}{0.425} \\\\1.136 = 0.112 v^2\\\\v^2 = 10.14\\\\v = 3.18 \ m/s[/tex]

The velocity of the block at the bottom of the vertical circle is calculated as follows;

[tex]W - mg = \frac{mv^2}{r} \\\\3.95 - 0.0475(9.8)= \frac{0.0475v^2}{0.425} \\\\3.485 = 0.112 v^2\\\\v^2 = 31.12\\\\v = 5.58 \ m/s[/tex]

The work done by friction is the change in the kinetic energy of the block.

[tex]W _f =P.E_f - \Delta K.E \\\\W_f = mgh - \frac{1}{2} m(v_f^2 - v_i^2)\\\\Wf = 0.0475\times 9.8(2 \times 0.425) - \frac{1}{2} \times 0.0475(5.58^2 - 3.18^2)\\\\W_f = 0.396 - 0.498\\\\W_f = -0.102 \ J[/tex]

Thus, the work done on the block by friction during the motion of the block from point A to point B is 0.102 J.

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Answer:

xf = - 3.16 m

Explanation:

the squirrel was initially in the position xi = 3.65 m, then it had a displacement of Δx = -6.81 m.

The negative sign indicates that it moved in the opposite direction, so we must subtract this displacement Δx = -6.81 m to the initial position xi = 3.65 m, to find its final position.

3.65 m - 6.81 m = - 3.16 m

Answer:

Option d)

Explanation:

As the Kinetic energy of the body ids due to the speed of the body and depends on it, it does not depend on the height of the body.

The kinetic energy is given by:

KE = [tex]\frac{1}{2}mv^{2}[/tex]

where

m = mass of the body

v = kinetic energy of the body

Now, as per the question:

Initial velocity is u

and final velocity, v = 2u

Thus

Initial KE = [tex]\frac{1}{2}mu^{2}[/tex]               (1)

FInal KE = [tex]\frac{1}{2}mv^{2}[/tex] = [tex]\frac{1}{2}m(2u)^{2}[/tex]          

FInal KE = 4[tex]\frac{1}{2}mu^{2}[/tex]               (2)  

Thus from eqn (1) and (2):

Final KE = 4(Initial KE)

The Kinetic Energy has changed by a factor of 4

Answer:

h/(m*c)

Explanation:

Hi!

Lets denote the units of X as [X]

Since the dimentions of h are:

[tex][h] = \frac{ML^{2}}{T}[/tex]

If we divide [h] by the units of mass, we get:

[tex]\frac{[h]}{[m]} = \frac{L^{2}}{T}}[/tex]

Also we know that:

[tex][c] = \frac{L}{T}[/tex]

So:

[tex][\frac{h}{mc}] = \frac{L^{2}}{T}}*\frac{T}{L}=L[/tex]

Therefore

 h/(mc) has dimentiosn of length

To construct a quantity with the dimensions of length using Planck's constant (h), mass (m), and the speed of light (c), the formula L = h/(mc) can be used. This length scale is relevant in the realms of quantum mechanics and high-energy physics.

The student has asked how to construct a quantity with the dimensions of length using Planck's constant (h), a mass (m), and the speed of light (c). To achieve this, we can use the formula for the Planck length given by:

Lp = √hG/c³

Where G is Newton's gravitational constant. However, since we need to construct a length using just h, m, and c without G, we can rearrange the Planck length equation to:

L = h/(mc)

This gives a length L which is dependent on the mass m in addition to Planck's constant h and the speed of light c. This length scale is significant in quantum mechanics and high-energy physics, where extremely small distances are relevant.

Answer:

Insufficient data.

Explanation:

Hi!

The data you have is only potential diference, so you know that the variation in potential energy of the electron when moving from -1.5V to 3.0V, is 4.5 V.

But you cannot know the acceleration. For that you need to know the electric field, so you can calculate force.

The tension in the rope must equal the weight of the supported mass. For the given situation, the tension can be calculated by multiplying the mass by the acceleration due to gravity.

Tension in the rope must equal the weight of the supported mass, as we can prove using Newton's second law. If the 5.00-kg mass in the figure is stationary, then its acceleration is zero, and thus Fnet = 0. The only external forces acting on the mass are its weight w and the tension T supplied by the rope. Thus, T - w = 0.

Thus, for a 5.00-kg mass (neglecting the mass of the rope), we can find that T = mg = (5.00 kg) (9.80 m/s²) = 49.0 N.

Thus, Tension in the rope equals the weight of the supported mass (5.00 kg) due to Newton's second law, making T = mg = 49.0 N.

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