An internal explosion breaks an object, initially at rest,into
two pieces, one of which has 1.5 times the mass of the other.If
7500 J were released in the explosion, how much kinetic energydid
each piece acquire?

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.

Answer:

It take 1.78033 second get away

Explanation:

We have given that a car takes 30 m to stop when its speed is 25 m/sec

As the car stops its final speed v = 0 m/sec

Initial speed u = 25 m/sec

Distance s = 30 m

From third law of motion [tex]v^2=u^2+2as[/tex]

So [tex]0^2=25^2+2\times a\times 30[/tex]

[tex]a=-10.4166m/sec^2[/tex]

Now in second case distance s = 28 m

So [tex]v^2=25^2+2\times -10.4166\times 28[/tex]

[tex]v^2=41.666[/tex]

v = 6.4549 m/sec

Now from first equation of motion v=u+at

So [tex]6.4549=25-10.4166\times t[/tex]

t = 1.78033 sec

Answer:

Zero

Explanation:

The overall work done by gravitational force on completion of a complete turn is zero.

Since the work done by gravitaional force is conservative and depends only on the initial and end position here height and the path followed does not matter.

Since, in a complete turn the wheel return to its final position as a result of which displacement is zero and as work is the dot product of Force exerted and displacement, the work done is zero.

Also the work done in half cycle by gravity is counter balance by the work which is done against the gravity in the other half cycle.

The net work done by the gravitational force when a person is riding on a Ferris Wheel and it makes one complete turn is zero.

To understand why the net work done by the gravitational force is zero, we need to consider the definition of work and the nature of the motion on a Ferris Wheel. Work done by a force is defined as the product of the force and the displacement in the direction of the force. Mathematically, this is expressed as:

[tex]\[ W = F \cdot d \cdot \cos(\theta) \][/tex]

where ( W ) is the work, ( F ) is the force, ( d ) is the displacement, and [tex]\( \theta \)[/tex] is the angle between the force and the displacement.

In the case of the Ferris Wheel, the gravitational force acts vertically downward towards the center of the Earth, while the displacement of the person on the Ferris Wheel is along the circumference of the wheel, which is horizontal at any given point. Since the force and displacement are perpendicular to each other at every point in the circle, the angle \( \theta \) between them is always 90 degrees. Therefore, the cosine of 90 degrees is zero, which means that the work done by the gravitational force at each point is zero:

[tex]\[ W = F \cdot d \cdot \cos(90^\circ) = F \cdot d \cdot 0 = 0 \][/tex]

Moreover, over one complete turn of the Ferris Wheel, the initial and final positions of the person are the same. This means that the total displacement over one complete cycle is zero. Even if we consider the components of the gravitational force along the direction of displacement during different parts of the cycle, the net effect is zero because the person is raised to a certain height and then lowered back to the starting point. The work done against gravity to raise the person is equal in magnitude and opposite in sign to the work done by gravity as the person descends, resulting in a net work of zero for the entire cycle.

Therefore, the gravitational force does no net work on the person over one complete turn of the Ferris Wheel.

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:

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:

(a) Maximum height = 53.88 meters

(b) Range of the ball = 924.36 meters

Explanation:

The ball has been launched at a speed = 44.4 meters per second

Angle of the ball with the horizontal = 25°

Horizontal component of the speed of the ball = 44.4cos25° = 40.24 meters per second

Vertical component = 44.4sin25° = 18.76 meters per second

We know vertical component of the speed decides the height of the ball so by the law of motion,

v² = u² - 2gh

where v = velocity at the maximum height = 0

u = initial velocity = 18.76 meter per second

g = gravitational force = 9.8 meter per second²

Now we plug in the values in the given equation

0 = (18.76)² - 2(9.8)(h)

19.6h = 352.10

h = [tex]\frac{352.10}{19.6}[/tex]

h = 17.96 meters

By another equation,

[tex]v=ut-\frac{1}{2}gt^{2}[/tex]

Now we plug in the values again

[tex]0=(18.76)t-\frac{1}{2}(9.8)t^{2}[/tex]

18.76t = 4.9t²

18.76 = 4.9t

t = [tex]\frac{18.76}{4.9}=3.83[/tex]seconds

Since time t is the time to cover half of the range.

Therefore, time taken by the ball to cover the complete range = 2×3.83 = 7.66 seconds

  So the range of the ball = Horizontal component of the velocity × time

                                           = 40.24 × 7.66

                                           = 308.12 meters

This we have calculated all for our planet.

Now we take other planet.

(a) Since the golfer drives the ball 3 times as far as he would have on earth then maximum height achieved by the ball = 17.96 × 3 = 53.88 meters

(b) Range of the ball = 3×308.12 = 924.36 meters

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:

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:

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.

Answer:42 units

Explanation:

Given

Price-production relationship=-0.7D+300

Total cost=Fixed cost+ variable cost

Total cost=10,287+25D

where D is the units produced

Total revenue[tex]=\left ( -0.7D+300\right )D[/tex]

Total revenue[tex]=-0.7D^2+300D[/tex]

For Break even point

Total revenue=Total cost

[tex]10,287+25D=-0.7D^2+300D[/tex]

[tex]7D^2-2750D+102870=0[/tex]

[tex]D=\frac{2750\pm \sqrt{2750^2-4\times 7\times 102870}}{2\times 7}[/tex]

[tex]D=41.869\approx 42[/tex] units

The velocity of the cart relative to the ground is approximately 89.44 km/h in a direction about 63.43 degrees north of west.

To find the velocity of the cart relative to the ground when a plane is traveling north at 100.0 km/h with a 40.0 km/h crosswind blowing west and the cart is rolling at 20.0 km/h towards the tail of the plane, we can use vector addition to determine the resultant velocity.

1. First, break down the velocities into their horizontal (west-east) and vertical (north-south) components:

- Plane's velocity (north): 100.0 km/h

- Crosswind velocity (west): 40.0 km/h

- Cart's velocity (towards tail): 20.0 km/h

2. The horizontal component of the cart's velocity is the crosswind velocity (40.0 km/h), and the vertical component is its velocity towards the tail of the plane (20.0 km/h).

3. To find the resultant velocity, we can use vector addition by adding the horizontal and vertical components of the velocities separately:

Horizontal component: 40.0 km/h (west) - 0 km/h (east) = 40.0 km/h (west)

Vertical component: 100.0 km/h (north) - 20.0 km/h (south) = 80.0 km/h (north)

4. Now, we can use the Pythagorean theorem to find the magnitude of the resultant velocity:

Resultant velocity = √(40.0^2 + 80.0^2)

Resultant velocity = √(1600 + 6400)

Resultant velocity = √8000

Resultant velocity ≈ 89.44 km/h

5. To find the direction of the resultant velocity, we can use trigonometry:

Direction = arctan(vertical component / horizontal component)

Direction = arctan(80.0 / 40.0)

Direction = arctan(2)

Direction ≈ 63.43 degrees north of west

6. Therefore, the velocity of the cart relative to the ground is approximately 89.44 km/h in a direction about 63.43 degrees north of west.

Answer:

b. the air

Explanation:

David Scott's experiment was performed on the moon, where there is gravity but there is no air. This experiment consisted of letting a hammer and a feather fall at the same time. The result was that the two objects touch the ground simultaneously.

Since these two objects obviously have a different mass, the experiment shows that in a vacuum, objects fall with the same acceleration regardless of their mass.

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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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:

[tex]v=0[/tex]

Explanation:

Knowing that the formula for average velocity is:

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

Being said that, we know that the person's displacement is zero because it returns to its starting point

[tex]x_{2}=x_{1}[/tex]

That means [tex]x_{2}-x_{1}=0[/tex]

[tex]v=\frac{0}{t_{2}-t_{1}}=0[/tex]

Answer:

44.755 m

Explanation:

Given:

Height of the movie star, H = 1.75 m = 1750 mm

Height of the image, h = - 8.25 mm

Focal length of the camera = 210 mm

Let the distance of the object i.e the distance between camera and the movie star be 'u'

and

distance between the camera focus and image be 'v'

thus,

magnification, m = [tex]\frac{\textup{h}}{\textup{H}}[/tex]

also,

m = [tex]\frac{\textup{-v}}{\textup{u}}[/tex]

thus,

[tex]\frac{\textup{-v}}{\textup{u}}=\frac{\textup{h}}{\textup{H}}[/tex]

or

[tex]\frac{\textup{-v}}{\textup{u}}=\frac{\textup{-8.25}}{\textup{1750}}[/tex]

or

[tex]\frac{\textup{1}}{\textup{v}}=-\frac{\textup{1750}}{\textup{-8.25}}\times\frac{1}{\textup{u}}[/tex]  ....................(1)

now, from the lens formula

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

on substituting value from (1)

[tex]\frac{\textup{1}}{\textup{210}}=\frac{\textup{1}}{\textup{u}}+-\frac{\textup{1750}}{\textup{-8.25}}\times\frac{1}{\textup{u}}[/tex]

or

[tex]\frac{\textup{1}}{\textup{210}}=\frac{\textup{1}}{\textup{u}}(1 -\frac{\textup{1750}}{\textup{-8.25}})[/tex]

or

u = 210 × ( 1 + 212.12 )

or

u = 44755.45 mm

or

u = 44.755 m

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:

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:

d = 3.44 *10^{-7} m  

Explanation:

given data:

length of metal plates = 16.50 cm

capacitor charge = 18.5 nC

potential difference = 37.8 V

capacitance of parallel plate capacitor

[tex]C = \frac{A\epsilon _{0}}{d}[/tex]

area of the individual plate

A=[tex] a^2 = (16.5*10^{-2})^2 = 272.25 *10^{-4}[/tex] m2

capacitance

[tex]C = QV = 18.5 *10^{-9} *37.8 = 699.3 * 10^{-9} C[/tex]

separation between plates d  is given as[tex] = \frac{A\epsilon _{0}}{C }[/tex]

[tex]d =  \frac{272.25 *10^{-4} *8.85*10^{-12}}{699.3 *10^-9}[/tex]

d = 3.44 *10^{-7} m  

Answer:

(a-1) d₂=4.89 m: The object slides 4.89 m along the rough surface

(a-2) Work (Wf) done by the friction force while the mass is sliding down the in- clined plane:

Wf=  -20.4 J    is negative

(b) Work (Wg) done by the gravitational force while the mass is sliding down the inclined plane:

Wg= 58.8 J is positive

Explanation:

Nomenclature

vf: final velocity

v₀ :initial velocity

a: acceleleration

d: distance

Ff: Friction force

W: weight

m:mass

g: acceleration due to gravity

Graphic attached

The attached graph describes the variables related to the kinetics of the object (forces and accelerations)

Calculation de of the components of W in the inclined plane

W=m*g

Wx₁ = m*g*sin30°

Wy₁=  m*g*cos30°

Object kinematics on the inclined plane

vf₁²=v₀₁²+2*a₁*d₁

v₀₁=0

vf₁²=2*a₁*d₁

[tex]v_{f1} = \sqrt{2*a_{1}*d_{1}  }[/tex]  Equation (1)

Object kinetics on the inclined plane (μ= 0.2)

∑Fx₁=ma₁  :Newton's second law

-Ff₁+Wx₁ = ma₁   , Ff₁=μN₁

-μ₁N₁+Wx₁ = ma₁      Equation (2)

∑Fy₁=0   : Newton's first law

N₁-Wy₁= 0

N₁- m*g*cos30°=0

N₁  =  m*g*cos30°

We replace   N₁  =  m*g*cos30 and  Wx₁ = m*g*sin30° in the equation (2)

-μ₁m*g*cos30₁+m*g*sin30° = ma₁   :  We divide by m

-μ₁*g*cos30°+g*sin30° = a₁  

g*(-μ₁*cos30°+sin30°) = a₁  

a₁ =9.8(-0.2*cos30°+sin30°)=3.2 m/s²

We replace a₁ =3.2 m/s² and d₁= 3m in the equation (1)

[tex]v_{f1} = \sqrt{2*3.2*3}  }[/tex]

[tex]v_{f1} =\sqrt{2*3.2*3}[/tex]

[tex]v_{f1} = 4.38 m/s[/tex]

Rough surface  kinematics

vf₂²=v₀₂²+2*a₂*d₂   v₀₂=vf₁=4.38 m/s

0   =4.38²+2*a₂*d₂  Equation (3)

Rough surface  kinetics (μ= 0.3)

∑Fx₂=ma₂  :Newton's second law

-Ff₂=ma₂

--μ₂*N₂ = ma₂   Equation (4)

∑Fy₂= 0  :Newton's first law

N₂-W=0

N₂=W=m*g

We replace N₂=m*g inthe equation (4)

--μ₂*m*g = ma₂   We divide by m

--μ₂*g = a₂

a₂ =-0.2*9.8= -1.96m/s²

We replace a₂ = -1.96m/s² in the equation (3)

0   =4.38²+2*-1.96*d₂

3.92*d₂ = 4.38²

d₂=4.38²/3.92

d₂=4.38²/3.92

(a-1) d₂=4.89 m: The object slides 4.89 m along the rough surface

(a-2) Work (Wf) done by the friction force while the mass is sliding down the in- clined plane:

Wf = - Ff₁*d₁

Ff₁= μ₁N₁= μ₁*m*g*cos30°= -0.2*4*9.8*cos30° = 6,79 N

Wf= -  6.79*3 = 20.4 N*m

Wf=  -20.4 J    is negative

(b) Work (Wg) done by the gravitational force while the mass is sliding down the inclined plane

Wg=W₁x*d= m*g*sin30*3=4*9.8*0.5*3= 58.8 N*m

Wg= 58.8 J is positive

The work done by the friction force is negative, and the work done by the gravitational force is positive as the object slides down the inclined plane.

The work done by the friction force while the object slides down the inclined plane is negative. The work done by the gravitational force while the object slides down the inclined plane is positive.

When the object slides down the inclined plane, the friction force acts in the opposite direction to its motion. Since friction always opposes the motion, the work done by friction is negative.

The gravitational force, on the other hand, acts in the same direction as the object's motion. Therefore, the work done by the gravitational force is positive.

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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.

Answer:

total kinetic energy is 8 × [tex]10^{-19}[/tex] J

Explanation:

given data

potential difference = 5 V

e = 1.60 × [tex]10^{-19}[/tex] C

to find out

what is kinetic energy

solution

we will apply here conservation of energy that is

change in potential energy is equal to change in kinetic energy

so

change potential energy is e × potential difference

change potential energy =  1.60 × [tex]10^{-19}[/tex] × 5

change potential energy = 8 × [tex]10^{-19}[/tex] J

so change in kinetic energy  = 8 × [tex]10^{-19}[/tex] J

and we know proton start from rest that mean ( kinetic energy is 0 ) so

change in KE is total KE

total kinetic energy is 8 × [tex]10^{-19}[/tex] J

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:

[tex]-1.144\ \mu C[/tex]

Explanation:

Given:

Assume:

We know that the electric field is the electric force applied on a unit positive charge i.e.,

[tex]\vec{E}=\dfrac{\vec{F}}{Q}[/tex]

This means the electric force applied on this additional charge placed in the field is given by:

[tex]\vec{F}=Q\vec{E}\\\Rightarrow \vec{F} =  1.04\times 10^{-8}\ n C\times (148.0\ \hat{i}-110.0\ \hat{j})\ N/C\\\Rightarrow \vec{F} = (1.539\ \hat{i}-1.144\ \hat{j})\ \mu N\\[/tex]

From the above expression of force, we have the following y-component of force on this additional charge.

[tex]F_y = -1.144\ \mu N[/tex]

Hence, the y-component of the electric force on the this charge is [tex]-1.144\ \mu N[/tex].

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:

[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]

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:

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

Answer:

The fraction of time for turn on is 0.3852

Solution:

As per the question:

Temperature at which oil bath is maintained, [tex]T_{o} = 50.5^{\circ}[/tex]

Heat loss at rate, q = 4.68 kJ/min

Resistance, R = [tex]60\Omega[/tex]

Operating Voltage, [tex]V_{o} = 110 V[/tex]

Now,

Power that the resistor releases, [tex]P_{R} = \frac{V_{o}^{2}}{R}[/tex]

[tex]P_{R} = \frac{110^{2}}{60} = 201.67 W = 12.148 J/min[/tex]

The fraction of time for the current to be turned on:

[tex]P_{R} = \frac{q}{t}[/tex]

[tex]12.148 = \frac{4.68}{t}[/tex]

t = 0.3852