A 100kg cannon at rest contains a 10kgcannonball. when fired,
the cannonball leaves the cannon with avelocity of 90m/s. what is
the magnitude of the recoil speed of thecannon.

Answer:

[tex]\theta=145[/tex]

Explanation:

The amplitude of he combined wave is:

[tex]B=2Acos(\theta/2)\\[/tex]

A, is the amplitude from the identical harmonic waves

B, is the amplitude of the resultant wave

θ, is the phase, between the waves

The amplitude of the combined wave must be 0.6A:

[tex]0.6A=2Acos(\theta/2)\\ cos(\theta/2)=0.3\\\theta/2=72.5\\\theta=145[/tex]

Answer:

The current through [tex]9 \Omega[/tex] is 0.297 A

Solution:

As per the question:

[tex]R_{5} = 5.0 \Omega[/tex]

[tex]R_{9} = 9.0 \Omega[/tex]

[tex]R_{4} = 5.0 \Omega[/tex]

V = 6.0 V

Now, from the given circuit:

[tex]R_{5}[/tex] and [tex]R_{9}[/tex] are in parallel

Thus

[tex]\frac{1}{R_{eq}} = \frac{1}{R_{5}} + \frac{1}{R_{9}}[/tex]

[tex]R_{eq} = \frac{R_{5}R_{9}}{R_{5} + R_{9}}[/tex]

[tex]R_{eq} = \frac{5.0\times 9.0}{5.0 + 9.0} = 3.2143 \Omega[/tex]

Now, the [tex]R_{eq}[/tex] is in series with [tex]R_{4}[/tex]:

[tex]R'_{eq} = R_{eq} + R_{4} = 3.2143 + 4.0 = 7.2413 \Omega[/tex]

Now, to calculate the current through [tex]R_{9}[/tex]:

[tex]V = I\times R'_{eq}[/tex]

[tex]I = {6}{7.2143} = 0.8317 A[/tex]

where

I = circuit current

Now,

Voltage across [tex]R_{eq}[/tex], V':

[tex]V' = I\times R_{eq}[/tex]

[tex]V' = 0.8317\times 3.2143 = 2.6734 V[/tex]

Now, current through [tex]R_{9}[/tex], I' :

[tex]I' = \frac{V'}{R_{9}}[/tex]

[tex]I' = \frac{2.6734}{9.0} = 0.297 A[/tex]

Final answer:

To find the current through the 9.0 ohm resistor in a series-parallel circuit, we can calculate the equivalent resistance of the parallel combination.

Explanation:

To find the current through the 9.0 ohm resistor, we need to first determine the equivalent resistance of the circuit. The two resistors of 5.0 and 9.0 ohms that are connected in parallel have an equivalent resistance given by the formula:

1/Req = 1/R1 + 1/R2

1/Req = 1/5.0 + 1/9.0

1/Req = (9.0 + 5.0)/(5.0 * 9.0)

1/Req = 14.0/45.0

Req = 45.0/14.0 ≈ 3.21 ohms

The equivalent resistance of the parallel combination is approximately 3.21 ohms.

Final answer:

The time it takes for the car to fall is 0.606 s and the horizontal velocity of the car is 0.497 m/s.

Explanation:

To find the time it takes for the toy car to fall, we can use the kinematic equation:

h = (1/2)gt^2

Where h is the height, g is the acceleration due to gravity, and t is the time. Rearranging the equation, we can solve for t:

t = sqrt(2h / g)

Plugging in the values, we have:

t = sqrt(2 * 1.807 / 9.8) = 0.606 s

To find the horizontal velocity of the car, we can use the equation:

v = d / t

Where v is the velocity, d is the horizontal distance, and t is the time. Plugging in the values, we have:

v = 0.3012 / 0.606 = 0.497 m/s

Answer:

a) 0 m/s

b) - 24 m/s

c)  - 68 m

Explanation:

Given:

Initial distance = - 4 m

Initial velocity, u = 4 m/s

1) acceleration, a = - 2 m/s² for time, t = 2 seconds

thus,

velocity after 2 seconds will be

from Newton's equation of motion

v = u + at

v = 4 + (-2) × 2

v = 0 m/s

2) Velocity after 2 second is the initial velocity for this case

given acceleration = - 6 m/s² for 4 seconds

thus,

final velocity, v = 0 + ( - 6 ) × 4 = - 24 m/s

here the negative sign depicts the velocity in opposite direction to the initial direction of motion

thus, velocity after 6 seconds = - 24 m/s

3) Now,

Total displacement in 6 seconds

= Displacement in 2 seconds + Displacement in 4 seconds

From Newton's equation of motion

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

where,  

s is the distance

u is the initial speed  

a is the acceleration

t is the time

thus,

= [tex]0\times2+\frac{1}{2}\times(-2)\times2^2[/tex]  + [tex]0\times4+\frac{1}{2}\times-6\times4^2[/tex]

= - 16 - 48

= - 64 m

Hence, the final displacement = - 64 - 4 = - 68 m

Answer:

a) 38.27      b) 322.5°

c) 126.99    d) 1.17°

e) 62.27     e) 139.6°

Explanation:

First of all we have to convert the coordinates into rectangular coordinates, so:

a=( 43.3 , 25)

b=( -48.3 , -12.94)

c=( 35.36 , -35.36)

Now we can do the math easier (x coordinate with x coordinate, and y coordinate with y coordinate):

1.)  a+b+c=( 30.36 , -23.3) = 38.27 < 322.5°

2.)  a-b+c=( 126.96 , 2.6) = 126.99 < 1.17°

3.)  (a+b) - (c+d)=0   Solving for d:

     d=(a+b) - c = ( -40.36 , 47.42) = 62.27 < 139.6°

The acceleration experienced by the space travelers during their launch would be considered unrealistically large. It would exceed both the limits of human endurance and the acceleration experienced during free fall.

To calculate the acceleration experienced by the space travelers during their launch, we can use the formula:

Acceleration = (Final Velocity - Initial Velocity) / Time

In this case, the final velocity is given as 10.97 km/s, and the initial velocity is 0 m/s (since the spaceship starts from rest). The time is not provided, so we cannot calculate the exact acceleration. However, we can compare it to the acceleration that a human can withstand, which is 15g for a short time. One g is equivalent to the acceleration due to gravity, which is approximately 9.8 m/s².

So, 15g is equal to 15 * 9.8 m/s² = 147 m/s². Therefore, any acceleration larger than 147 m/s² would be unrealistically large for the space travelers during their launch.

Comparing this with the free-fall acceleration, which is approximately 9.8 m/s², we can see that the acceleration proposed by Jules Verne would be much larger than both the limits of human endurance and the acceleration experienced during free fall.

Answer:

a) [tex]y=16.53m[/tex]

b) [tex]t_{up}=1.83s[/tex]

c)[tex]t_{down}=1.84s[/tex]

d) [tex]v=-18m/s[/tex]

Explanation:

a) To find the highest point of the ball we need to know that at that point the ball stops going up and its velocity become 0

[tex]v^{2} =v^{2} _{o} +2g(y-y_{o})[/tex]

[tex]0=(18)^{2} -2(9.8)(y-0)[/tex]

Solving for y

[tex]y=\frac{(18)^{2} }{2(9.8)}=16.53m[/tex]

b) To find how long does it take to reach that point:

[tex]v=v_{o}+at[/tex]

[tex]0=18-9.8t[/tex]

Solving for t

[tex]t_{up} =\frac{18m/s}{9.8m/s^{2} }= 1.83s[/tex]

c) To find how long does it take to hit the ground after it reaches its highest point we need to find how long does it take to do the whole motion and then subtract the time that takes to go up

[tex]y=y_{o}+v_{o}t+\frac{1}{2}gt^{2}[/tex]

[tex]0=0+18t-\frac{1}{2}(9.8)t^{2}[/tex]

Solving for t

[tex]t=0 s[/tex] or [tex]t=3.67s[/tex]

Since time can not be negative, we choose the second option

[tex]t_{down}=t-t_{up}=3.67s-1.83s=1.84s[/tex]

d) To find the velocity when it returns to the level from which it started we need to use the following formula:

[tex]v=v_{o}+at[/tex]

[tex]v=18m/s-(9.8m/s^{2} )(3.67s)=-18m/s[/tex]

The sign means the ball is going down

Answer:

Explanation:

The total energy of an electron in an orbit consists of two components

1 ) Potential energy which is - ve because the field is attractive

2) Kinetic energy which represents moving electron having some velocity.

Kinetic is always positive.

3 ) In an orbit , The magnitude of potential energy is twice that of kinetic energy. So if -2E is the value of potential energy E wil be the value of kinetic energy.

4 ) Total energy will become some of potential energy and kinetic energy

-2E + E = -E

5 ) So total energy becomes equal to kinetic energy with only sign reversed.

In the given case total energy is -0.28 eV . Hence kinetic energy will be +0.28 eV.

When kinetic energy is calculated as +.28 eV , the potential energy will be

- 2 x .28 or - 0.56  eV .

Answer:

The spring constant is [tex]1.09\times10^{9}\ N/m[/tex]

Explanation:

Given that,

length = 500 mm

Diameter = 2 cm

Young's modulus = 17.4 GPa

We need to calculate the young's modulus

Using formula of young's modulus

[tex]Y=\dfrac{\dfrac{F}{A}}{\dfrac{\Delta l}{l}}[/tex]....(I)

[tex]Y=\dfrac{Fl}{\Delta l A}[/tex]

From hook's law

[tex]F=kx[/tex]

[tex]k=\dfrac{F}{x}[/tex]

[tex]F=k\times\Delta l[/tex]....(II)

Put the value of F in equation

[tex]Y=\dfrac{k\times\Delta l\times l}{\Delta l A}[/tex]

[tex]Y=\dfrac{kl}{A}[/tex]

We need to calculate the spring constant

[tex]k = \dfrac{YA}{l}[/tex]....(II)

We need to calculate the area of cylinder

Using formula of area of cylinder

[tex]A=2\pi\times r\times l[/tex]

Put the value into the formula

[tex]A=2\pi\times 1\times10^{-2}\times500\times10^{-3}[/tex]

[tex]A=0.0314\ m^2[/tex]

Put the value of A in (II)

[tex]k=\dfrac{1.74\times10^{10}\times0.0314}{500\times10^{-3}}[/tex]

[tex]k=1.09\times10^{9}\ N/m[/tex]

Hence, The spring constant is [tex]1.09\times10^{9}\ N/m[/tex]

Answer:

Distance, d = 112.5 meters

Explanation:

Initially, the bicyclist is at rest, u = 0

Final speed of the bicyclist, v = 30 m/s

Acceleration of the bicycle, [tex]a=4\ m/s^2[/tex]

Let s is the distance travelled by the bicyclist. The third equation of motion is given as :

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

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

[tex]s=\dfrac{(30)^2}{2\times 4}[/tex]

s = 112.5 meters

So, the distance travelled by the bicyclist is 112.5 meters. Hence, this is the required solution.

Answer:

Explanation:

Given:

Length of each side of the cube, [tex]L=10\ cm[/tex]

The Elecric flux through one of the side of the cube is, [tex]\phi =-1 kNm^2/C.[/tex]

The net flux through a closed surface is defined as the total charge that lie inside the closed surface divided by [tex]\epsilon_0[/tex]

Since Flux is a scalar quantity. It can added to get total flux through the surface.

[tex]\phi_{total}=\dfrac{Q_{in}}{\epsilon_0}\\6\times {-1}\times 10^3=\dfrac{Q_{in}}{\epsilon_0}\\\\Q_{in}=-6}\times 10^3\epsilon_0\\Q_{in}=-6}\times 10^3\times8.85\times 10^{-12}\\Q_{in}=-53.1\times10^{-9}\ C[/tex]

So the the charge at the centre is calculated.

Answer:

Average velocity

[tex]v=\frac{d}{t}\\ v=\frac{10m}{70s}\\v=.1428 \frac{m}{s}[/tex]

Average speed,

[tex]S=\frac{D}{t}\\ S=\frac{90}{70}\\ S=1.29\frac{m}{s}[/tex]

Explanation:

(a)Average velocity

We have to find the average velocity. We know that velocity is defined as the rate of change of displacement with respect to time.

To find the average velocity we have to find the total displacement.

since displacement along east direction is 50m

and displacement along west=40m

so total displacement,

[tex]d=50m-40m\\d=10m[/tex]

total time,

[tex]t=28 s+42 s\\t=70 s[/tex]

therefore, average velocity

[tex]v=\frac{d}{t}\\ v=\frac{10m}{70s}\\v=.1428 \frac{m}{s}[/tex]

(b)Average Speed:

Average speed is defined as the ratio of total distance to the total time

it means

Average speed= total distance/total time

here total distance,

[tex]D= 50m+40m\\D=90m[/tex]

and total time,

[tex]t= 28s+40s\\t=70s[/tex]

therefore,

Average speed,

[tex]S=\frac{D}{t}\\ S=\frac{90}{70}\\ S=1.29\frac{m}{s}[/tex]

The total charge on a cloud when breakdown of the surrounding air begins is calculated using Gauss's Law. The breakdown field strength, radius of the cloud and permittivity of free space are needed. The number of excess electrons in the cloud is then found by dividing the total charge by the charge of an electron.

The charge q on a spherically symmetric object, such as our cloud, when an electric field E is induced on its surface can be calculated using Gauss's Law, which states that the total charge enclosed by a Gaussian surface is equal to the electric field times the surface area of the Gaussian surface divided by the permittivity of free space (ε0). In our case, E = Eb (the breakdown field strength), the radius of the sphere, r = 0.5km = 500m (half the diameter), and ε0 = 8.85 x 10^-12 C2/N·m2. Therefore, q = (4πr2Eb)ε0.

For question 2: To calculate the number of excess electrons in the cloud, we should recall that the charge of an electron is -1.602 x 10^-19 C. Therefore, the number of excess electrons, N, can be calculated using N = q / Charge of an electron.

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Answer:

magnitude is 2.52 m  with 66.15° north east

Explanation:

given data

displace 1 = 3.45 m north = 3.45 j

displace 2 = 1.53 m southeast = 1.53 cos(315) i + sin(315) j

displace 3 = 0.877 m southwest  = 0.877 cos(225) i- sin(225) j

to find out

displacement and direction

solution

we consider here direction i as east and direction j as north

so here

displacement = displace 1 + displace 2 + displace 3

displacement =  3.45 j + 1.53 cos(315) i + sin(315) j + 0.877 cos(225) i- sin(225) j

displacement =  3.45 j + 1.081 i -1.0818 j - 0.062 i -0.062 j

displacement = 1.019 i + 2.306 j

so magnitude

magnitude = [tex]\sqrt{1.019^{2} + 2.306^{2}}[/tex]

magnitude = 2.52

and

angle will be = arctan(2.306/1.019)

angle = 1.15470651 rad

angle is 66.15 degree

so magnitude is 2.52 m  with 66.15° north east

Answer:

The vector has a magnitude of 33.86 units and a direction of 121.64°.

Explanation:

To find the magnitude, you use the Pitagorean theorem:

[tex]||V|| = \sqrt{x^2 + y^2}= \sqrt{(-26.5)^2 + (43)^2} = 33.86 units[/tex]

In order to find the direction, you can use trigonometry. You have to keep in mind, that as the y component of the vector is positive and the x component is negative, the vector must have an angle between 90 and 180°, or in the second quadrant of the plane.:

[tex]tan(\alpha) = \frac{y}{x} \\\alpha = tan^{-1}(\frac{43}{-26.5})= 180 - tan^{-1}(\frac{43}{26.5}) = 121.64[/tex]° or, 58.36° in the second quadrant

Answer:

0.02442 × 10⁻⁹

Explanation:

Given:

Diameter of copper ball = 2.00 mm = 0.002 m

Charge on ball = 40 nC = 40 × 10⁻⁹ C

Density of copper = 8900 Kg/m³

Now,

The number of electrons removed, n = [tex]\frac{\textup{Charge on ball}}{\textup{Charge of an electron}}[/tex]

also, charge on electron = 1.6 × 10⁻¹⁹ C

Thus,

n = [tex]\frac{40\times10^{-9}}{1.6\times10^{-19}}[/tex]

or

n = 25 × 10¹⁰ Electrons

Now,

Mass of copper ball = volume × density

Or

Mass of copper ball =  [tex]\frac{4}{3}\pi(\frac{d}{2})^3[/tex]  × 8900

or

Mass of copper ball =  [tex]\frac{4}{3}\pi(\frac{0.002}{2})^3[/tex]  × 8900

or

Mass of copper ball = 0.03726 grams

Also,

molar mass of copper = 63.546 g/mol

Therefore,

Number of mol of copper in  0.03726 grams = [tex]\frac{ 0.03726}{63.546}[/tex]

or

Number of mol of copper in  0.03726 grams = 5.86 × 10⁻⁴ mol

and,

1 mol of a substance contains = 6.022 × 10²³ atoms

Therefore,

5.86 × 10⁻⁴ mol of copper contains = 5.86 × 10⁻⁴ × 6.022 × 10²³ atoms.

or

5.86 × 10⁻⁴ mol of copper contains = 35.88 × 10¹⁹ atoms

Now,

A neutral copper atom has 29 electrons.  

Therefore,

Number of electrons in ball = 29 × 35.88 × 10¹⁹ = 1023.37 × 10¹⁹ electrons.

Hence,

The fraction of electrons removed = [tex]\frac{25\times10^{10}}{1023.37\times10^{19}}[/tex]

or

The fraction of electrons removed = 0.02442 × 10⁻⁹

The fraction of electrons removed from the copper ball can be calculated by comparing its net charge to the charge of a single electron. Using the provided information, we can find that approximately 2.5 x 10^10 electrons have been removed from the ball. To determine the fraction, we need to compare this number to the total number of electrons in the ball, which can be calculated using the mass, density, and atomic mass of copper. By plugging in the values, we can find the fraction of electrons removed.

To determine the fraction of electrons that have been removed from the copper ball, we need to compare the net charge of the ball to the charge of a single electron. The net charge of the ball is 40 nC, which is equivalent to 40 x 10^-9 C. The charge of a single electron is 1.60 x 10^-19 C.

We can calculate the number of electrons that have been removed using the formula:

Number of electrons removed = Net charge of the ball / Charge of a single electron

Number of electrons removed = (40 x 10^-9 C) / (1.60 x 10^-19 C) = 2.5 x 10^10 electrons

To find the fraction of electrons removed, we need to compare the number of electrons removed to the total number of electrons in the ball. The total number of electrons in the ball can be calculated using the formula:

Total number of electrons = Number of copper atoms x Number of electrons per copper atom

The number of copper atoms can be calculated using the formula:

Number of copper atoms = Mass of the ball / Atomic mass of copper

The mass of the ball can be calculated using the formula:

Mass of the ball = Volume of the ball x Density of copper

Given that the diameter of the ball is 2.0 mm, the volume of the ball can be calculated using the formula for the volume of a sphere:

Volume of the ball = (4/3) x pi x (radius)^3

As the ball is a sphere, the radius is half the diameter, so the radius is 1.0 mm or 1 x 10^-3 m.

Using the given density of copper (8900 kg/m^3) and atomic mass of copper (63.5 g/mol), we can now calculate the fraction of electrons removed:

Fraction of electrons removed = Number of electrons removed / Total number of electrons = (2.5 x 10^10 electrons) / (Number of copper atoms x Number of electrons per copper atom)

Answer:

The final velocity of the car is 30 m/s.

Explanation:

Given that,

Initial speed of the car, u = 0

Acceleration of the car, [tex]a=5\ m/s^2[/tex]

Time taken, t = 6 s

Let v is the final velocity of the car. It can be calculated using first equation of kinematics as :

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

[tex]v=at[/tex]

[tex]v=5\ m/s^2\times 6\ s[/tex]

v = 30 m/s

So, the final velocity of the car is 30 m/s. Hence, this is the required solution.

Answer:

The resultant field will have a magnitude of 241.71 V/m, 30.28° to the left of E1.

Explanation:

To find the resultant electric fields, you simply need to add the vectors representing both electric field E1 and electric field E2. You can do this by using the component method, where you add the x-component and y-component of each vector:

E1 = 99 V/m, 0° from the y-axis

E1x = 0 V/m  

E1y = 99 V/m, up

E2 = 164 V/m, 48° from y-axis

E2x = 164*sin(48°) V/m, to the left

E2y = 164*cos(48°) V/m, up

[tex]Ex: E_{1_{x}} + E_{2_{x}} = 0 V/m - 164 *sin(48) V/m= -121.875 V/m\\Ey: E_{1_{y}} + E_{2_{y}} = 99 V/m + 164 *cos(48) V/m = 208.74 V/m\\[/tex]

To find the magnitude of the resultant vector, we use the pythagorean theorem. To find the direction, we use trigonometry.

[tex]E_r = \sqrt{E_x^2 + E_y^2}= \sqrt{(-121.875V/m)^2 + (208.74V/m)^2} = 241.71 V/m[/tex]

The direction from the y-axis will be:

[tex]\beta = arctan(\frac{-121.875 V/m}{208.74 V/m}) = 30.28[/tex]° to the left of E1.

the resultant electric field has a magnitude of approximately 247.8 V/m and is directed at an angle of approximately 63.5 degrees above the horizontal, not the vertical.

To find the resultant electric field, you should add the horizontal and vertical components of E1 and E2 separately:

Vertical Component:

E1y = 99 V/m (vertical component of E1)

E2y = 164 V/m * sin(48°) ≈ 123.6 V/m (vertical component of E2)

Horizontal Component:

E2x = 164 V/m * cos(48°) ≈ 109.8 V/m (horizontal component of E2)

Add the vertical components:

Ey = E1y + E2y

Ey = 99 V/m + 123.6 V/m

Ey ≈ 222.6 V/m

Add the horizontal components:

Ex = E2x

Ex ≈ 109.8 V/m

Calculate the magnitude of the resultant electric field (E) using the Pythagorean theorem:

E = √(Ex² + Ey²)

E = √((109.8 V/m)² + (222.6 V/m)²)

E ≈ 247.8 V/m

So, the resultant electric field has a magnitude of approximately 247.8 V/m and is directed at an angle of approximately 63.5 degrees above the horizontal, not the vertical, as previously stated.

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Answer:

A. [tex]H=\frac{v_{0}^{2}Sin^{2}\theta }{2g}[/tex]

B. 5556.1 m

Explanation:

A.

Launch speed, vo

Angle of projection = θ

The value of vertical component of velocity at maximum height is zero. Let the maximum height is H.

Use third equation of motion in vertical direction

[tex]v_{y}^{2}=u_{y}^{2}+2a_{y}H[/tex]

[tex]0^{2}=\left (v_{0}Sin\theta  \right )^{2}-2gH[/tex]

[tex]H=\frac{v_{0}^{2}Sin^{2}\theta }{2g}[/tex]

B.

u = 330 m/s

Let it goes upto height H.

V = 0 at maximum height

Use third equation of motion in vertical direction

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

[tex]0^{2}=330^{2}-2\times 9.8\times H[/tex]

H = 5556.1 m

Answer:

Explanation:

Total volume of the shell on which charge resides

= 4/3 π ( R₁³ - R₂³ )

= 4/3 X 3.14 ( 33³ - 20³) X 10⁻⁶ m³

= 117 x 10⁻³ m³

Charge inside the shell

-117 x 10⁻³ x 1.3 x 10⁻⁶

= -152.1 x 10⁻⁹ C

Charge at the center

= - 60 x 10⁻⁹ C

Total charge inside the shell

= - (152 .1 + 60 ) x 10⁻⁹ C

212.1 X 10⁻⁹C

Force between - ve charge and proton

F = k qQ / R²

k = 9 x 10⁹ .

q = 1.6 x 10⁻¹⁹ ( charge on proton )

Q = 212.1 X 10⁻⁹ ( charge on shell )

R = 33 X 10⁻² m ( outer radius )

F = [tex]\frac{9\times10^9\times1.6\times10^{-19}\times212.1\times 10^{-9}}{(33\times10^{-2})^2}[/tex]

F = 2.8 X 10⁻¹⁵ N

This force provides centripetal force for rotating proton

mv² / R = 2.8 X 10⁻¹⁵

V² = R X 2.8 X 10⁻¹⁵ / m

= 33 x 10⁻² x 2.8 x 10⁻¹⁵ /(  1.67 x 10⁻²⁷ )

[ mass of proton = 1.67 x 10⁻²⁷ kg)

= 55.33 x 10¹⁰

V = 7.44 X 10⁵ m/s

Answer:

Average speed, v = 48.01 km/h

Explanation:

Given that,

Distance covered by the walker, d = 4 km

Tim taken to cover that distance, t = 53 min = 0.0833 hours

We need to find the average speed of the walker for this distance. It is given by :

[tex]speed=\dfrac{distance}{time}[/tex]

[tex]speed=\dfrac{4\ km}{0.0833\ h}[/tex]  

Speed, v = 48.01 km/h

So, the average speed of the walker is 48.01 km/h. Hence, this is the required solution.

Answer:

a) Focal length of the lens is 8 cm which is a convex lens

b) 6 cm

c) The lens is a convex lens and produces a virtual image which is upright and two times larger than the object.

Explanation:

u = Object distance =  4 cm

v = Image distance = -8 cm

f = Focal length

Lens Equation

[tex]\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\\\Rightarrow \frac{1}{f}=\frac{1}{4}+\frac{1}{-8}\\\Rightarrow \frac{1}{f}=\frac{1}{8}\\\Rightarrow f=\frac{8}{1}=-8\ cm[/tex]

a) Focal length of the lens is 8 cm which is a convex lens

Magnification

[tex]m=-\frac{v}{u}\\\Rightarrow m=-\frac{-8}{4}\\\Rightarrow m=2[/tex]

b) Height of image is 2×3 = 6 cm

Since magnification is positive the image upright

c) The lens is a convex lens and produces a virtual image which is upright and two times larger than the object.

(a) The focal length of the lens is -2.67 cm.

(b) The magnification of the image is 2 and the height is 6 cm.

(c) The imaged formed by the lens is upright, virtual and magnified.

The focal length of the lens is determined by using lens formulas as given below;

[tex]\frac{1}{f} = \frac{1}{v} - \frac{1}{u} \\\\\frac{1}{f} = \frac{-1}{8} - \frac{1}{4} \\\\\frac{1}{f} = \frac{-1 -2}{8} = \frac{-3}{8} \\\\f = -2.67 \ cm[/tex]

The magnification of the image is calculated as follows;

[tex]m = \frac{v}{u} \\\\m = \frac{8}{4} \\\\m = 2\\\\[/tex]

The height of the image is calculated as follows;

[tex]H = mu\\\\H = 2(3cm) = 6\ cm[/tex]

The imaged formed by the lens is;

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Answer:

The answer is [tex] 1.956 \times 10^7\ m/s[/tex]

Explanation:

The amount of energy is not enough to apply the relativistic formula of energy [tex]E = mc^2[/tex], so the definition of energy in this case is

[tex]E = \frac{1}{2}m v^2[/tex].

From the last equation,

[tex]v= \sqrt{2E/m}[/tex]

where

[tex]E = 2 MeV = 3.204 \times 10^{-13} J[/tex]

and the mass of the neutron is

[tex]m = 1.675\times 10^{-27}\ Kg[/tex].

Then

[tex]v = 1.956 \times 10^7\ m/s[/tex]

the equivalent of [tex]0.065[/tex] the speed of light.

By installing motion sensors, the storage room would save $1.27 per day in energy costs. The simple payback period for installing motion sensors is approximately 77.17 days.

To determine the amount of energy and money that will be saved as a result of installing motion sensors, we need to compare the energy consumption and cost of the current lighting system with that of the motion sensor system. Currently, the storage room uses six fluorescent light fixtures, each containing four lamps rated at 60 W each. The lamps are on for 12 hours a day. So, the total energy consumed by the current system per day is 6 * 4 * 60 W * 12 hours = 17,280 W. With an electricity price of $0.11/kWh, the cost of energy consumed per day is 17,280 W / 1000 * $0.11 = $1.90.

Now, let's consider the motion sensor system. If the room is only used for an average of 3 hours a day, the total energy consumed by the motion sensor system per day would be 6 * 4 * 60 W * 3 hours = 5,760 W. This results in a cost of 5,760 W / 1000 * $0.11 = $0.63 per day. Therefore, by installing motion sensors, the storage room would save $1.90 - $0.63 = $1.27 per day in energy costs.

To determine the simple payback period, we need to calculate the cost of the motion sensor system and the savings achieved per day. The total cost of the motion sensor system is $32 for the purchase price + $66 for installation, which equals $98. With daily savings of $1.27, the payback period can be calculated as $98 / $1.27 = 77.17 days. Therefore, the simple payback period for installing motion sensors in the storage room is approximately 77.17 days.

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The relativistic factor y (gamma) quantifies the relativistic effects based on the speed of an object. In your given case with v = 0.932c, gamma would be approximately 2.61, indicating significant relativistic effects. The provided options do not include this value, suggesting some error.

The dimensionless parameter used in relativity is denoted as y (gamma), and it is referred to as the relativistic factor. It is associated with the relative speed of an object and is defined by the equation y = 1/√(1 − (v²/c²)), where v is the object's velocity and c is the speed of light.

For your case where v = 0.932c, we'll substitute this into the formula and solve for y. This results in y being approximately equal to 2.61. This isn't available in the provided options, which would indicate an error in the question or provided choices.

Without a doubt, as y approaches and surpasses 1, the relativistic effects become more noticeable in an object's behavior. Significant relativistic effects mean that the classical interpretation, where we neglect these effects, becomes increasingly inaccurate.

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Answer:

The shell hit at a distance of 1.9 x 10² km

The time of flight of the shell was 5.3 x 10² s

Explanation:

The position of the shell is given by the vector "r":

r  = (x0 + v0 * t * cos α ; y0 + v0 * t * sin α + 1/2 g t²)

where:

x0 = initial horizontal position

v0 = magnitude of the initial velocity

t = time

α = launching angle

y0 = initial vertical position

g = acceleration of gravity

When the shell hit, the vertical component (ry) of the vector position r is 0. See figure.

Then:

ry = 0 =  y0 + v0 * t * sin α + 1/2 g t²

Since the gun is at the center of our system of reference, y0 and x0 = 0

0 = t (v0 sin α + 1/2 g t)

t= 0 is discarded as solution

v0 sin α + 1/2 g t = 0

t = -2v0 sin α / g

t = (-2 * 2610 m/s * sin 81.9°)/ (-9.8 m/s²) = 5.3 x 10² s. This is the time of flight of the shell until it hit.

Then, the distance at which the shell hit is:

Distance = Module of r = ( x0 + v0 * t * cos α; 0) = x0 + v0 * t * cos α  

Distance = 2.61 km/s * 5.3 x 10² s * cos 81.9 = 1.9 x 10² km

The car strikes the tree at approximately 10.7 m/s after slowing down with a uniform acceleration of -5.75 m/s^2 for 60.0 m.

To calculate the speed with which the car strikes the tree, we'll use the kinematic equations for uniformly accelerated motion. Given the uniform acceleration of -5.75 m/s2 and the time of 4.40 s, we can use the following equation to find the initial velocity (vi) before the car started braking:

v = vi + at
Where:

By rearranging the equation to solve for the initial velocity (vi), we get:

vi = v - at

Substituting the known values:
vi = 0 - (-5.75 m/s2 * 4.40 s)
vi = 25.3 m/s

This is the speed with which the car was travelling before it hit the brakes. However, to find the speed at which the car strikes the tree, we must consider the distance of the skid marks. Using the kinematic equation for distance (d), where d equals the initial velocity times time plus half the acceleration times time squared:

d = vit +  rac{1}{2}at2

Since the distance to the tree is 60.0 m and we're looking for the speed at the end of this distance, we rearrange the equation to solve for final velocity (v). But first, we need to calculate the time it takes to stop over this distance. We can use the formula:

d =  rac{vi2 - v2}{2a}

By solving for v we find:

v = [tex]\sqrt{x}[/tex]{vi2 - 2ad}

v = [tex]\sqrt{x}[/tex]{(25.3 m/s)2 - 2(-5.75 m/s2)(60.0 m)}

v = [tex]\sqrt{x}[/tex]{(640.09) - (-690)}

v = 10.7 m/s

Therefore, the car strikes the tree at approximately 10.7 m/s.

The car strikes the tree at a speed of 0.98 m/s.

To determine the speed at which the car strikes the tree, we can utilize the kinematic equations of motion. We are given the following data:

First, we calculate the initial velocity using the equation:

Substituting the given values:

Solving this:

Now, to find the final velocity when the car strikes the tree, we use the kinematic equation:

Substituting the values:

So, the car strikes the tree at a speed of 0.98 m/s.

Answer:

Explanation:

mass, m = 1 kg

Position (2, 3 ) m

height, h = 2 m

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

Here, no force is acting in horizontal direction, the force of gravity is acting in vertical direction, so the work done by the gravitational force is to be calculated.

Force  mass x acceleration due to gravity

F = 1 x 9.8 = 9.8 N

Work = force x displacement x CosФ

Where, Ф be the angle between force vector and the displacement vector.

Here the value of Ф is 180° as the force acting vertically downward and the displacement is upward

So, W = 9.8 x 2 x Cos 180°

W = - 19.6 J

Thus, option (A) is correct.

To determine the magnitude of acceleration, use the formula (final velocity - initial velocity) / time. To calculate the distance traveled during the braking period, use the formula (initial velocity + final velocity) / 2 * time.

The magnitude of acceleration can be calculated using the formula:

acceleration = (final velocity - initial velocity) / time

acceleration = (50 mi/h * 1.46667 ft/s - 74 mi/h * 1.46667 ft/s) / 3 s

The distance traveled during the braking period can be calculated using the formula:

distance = (initial velocity + final velocity) / 2 * time

distance = (74 mi/h * 1.46667 ft/s + 50 mi/h * 1.46667 ft/s) / 2 * 3 s

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Answer:

[tex]F_1+F_2= (28.26, 9.05) N[/tex]

[tex]\alpha = 17.7\º[/tex]

[tex]F = 29.67 N[/tex]

Explanation:

Hi!

In a (x, y) coordinate representation, the two forces are:

[tex]F_1=(12.7N, 0)\\F_2=(18.1N\cos(30\º), 18.1N \sin(30\º) )\\\cos(\º30)=0.86\\\sin(\º30)= 0.5[/tex]

The sum of the two forces is:

[tex]F_1 + F_2 = ( 12.7 + 0.86*18.1, 18.1*0.5) N[/tex]

[tex]F_1+F_2= (28.26, 9.05) N[/tex]

The angle to x-axis is calculated using arctan:

[tex]\alpha = \arctan(\frac{F_y}{F_x}) = \arctan(\frac{9.05}{28.26} = 17.7\º[/tex]

The magnitude is:

[tex]F = \sqrt {F_x^2 + F_y^2}= \sqrt{798.6 + 81.9} = 29.67 N[/tex]