At a certain instant, a particle is moving in the direction with momentum 18 kg·m/s. During the next 0.5 s, a constant force <-4, 12, 0> N acts on the particle. What is the momentum of the particle at the end of this 0.5 s interval?

An additional force of 33.73 lb must be applied to the left to cancel out the initial force and create a resultant force of zero, according to Newton's first law of motion.

To ensure that the resultant force on the object is zero, another force of equal magnitude but in the opposite direction must be applied. In this case, since a force of 33.73 lb is directed toward the right, an additional force must be applied to the left with the same magnitude, i.e., 33.73 lb to the left. This balances out the forces acting upon the object, creating a state of equilibrium where the sum of the forces equals zero, thus satisfying Newton's first law of motion.

Now, considering the provided example where a man applies a force of +50 N, it's clear that the force of friction opposing the motion must also be 50 N but in the negative direction (-50 N) for the forces to cancel each other and fulfill the condition of Newton’s first law where Fnet = 0.

In the context of three forces acting on an object, discussing the necessary condition for an object to move straight down, the forces in the horizontal plane (to the right and left) must be balanced. This means that force A and force B, which act in opposite directions, must be equal in magnitude.

Answer:1.902 m

Explanation:

Given

height of apartment=1.5 m

It takes 0.21 sec to reach the bottom from apartment

So

[tex]s=u_1t+\frac{gt^2}{2}[/tex]

[tex]1.5=u_1\times 0.21+\frac{9.81\times 0.21^2}{2}[/tex]

[tex]u_1=6.11 m/s[/tex]

i.e. if ball is dropped from top its velocity at window is 6.11 m/s

So height of upper floor above window

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

where s= height of upper floor above window

here u=0

[tex]6.11^2=2\times 9.81\times s[/tex]

[tex]s=\frac{6.11^2}{2\times 9.81}[/tex]

s=1.902 m

Answer:

The initial horizontal velocity was 21 m/s

Explanation:

Please, see the figure for a better understanding of the problem.

The equation for the position of an object moving in a parabolic trajectory is as follows:

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

Where:

r = vector position at time t

x0 = initial horizontal position

v0 = initial velocity

t = time

α = launching angle

y0 = initial vertical position

g = acceleration due to gravity

Notice that at time t = 2.4 s the vector "r" is the one in the figure. We know that the x-component of that vector is 50 m. Then using the equation for the x-component of the vector "r", we can calculate the initial velocity:

x = x0 + v0 · t · cos α

Let´s place the center of the frame of reference at the point of the kick so that x0 = 0.

x =  v0 · t · cos α

x/t = v0 · cos α

Notice in the figure that v0 · cos 30° = v0x which is the initial horizontal velocity. Remember trigonometry of right triangles:

cos α = adjacent / hypotenuse = v0x / v0

Then:

50 m/ 2.4 s = v0 · cos 30° = v0x

v0x = 21 m/s

Answer:

1) -13570.3 Btu

2) 2.25

Explanation:

The formula for potential energy is:

[tex]E_p = m*g*h[/tex]

Where m is the mass in slugs, g is the gravity and h is the change in height. Soliving the equation:

[tex]E_p = 2000lb*\frac{1 slug}{32.174 lb} *32.174 \frac{ft}{s^2} *(0ft-5280ft) = -10560000 lbf*ft * \frac{1 Btu}{778.17 lbf*ft} = -13570.3 Btu[/tex]

The car would lose -13570.3 Btu in potential energy.

2) The kinetic energy is given by:

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

Then:

[tex]\frac{E_k_2}{E_k_1} = \frac{\frac{1}{2}mv_2^2}{\frac{1}{2}mv_1^2}  = \frac{\frac{1}{2}m(1.5v_1)^2}{\frac{1}{2}mv_1^2}=1.5^2\frac{v_1^2}{v_1^2} = 2.25[/tex]

Answer:

The gravitational force is significantly constant.

Explanation:

The gravitational force is expressed as:

F= G*M*m/d^2

G= Gravitational constant

M= in this case it is the mass of the planet

m= mass of the rock

d= distance of the rocks from the ground (suppose both at the same height for better comparison)

At the same time, we know that the force that each rock experiences is equal to the product of the mass due to acceleration:

F=m*a

We can match both expressions for F:

G*M*m/d^2 = m*a

we simplify m, and we obtain that the acceleration is independent of the mass of the attracted bodies:

a= G*M/d^2

Answer:

33 seconds.

Explanation:

The equation for speed with constant acceleration at time t its:

[tex]V(t) \ = \ V_0 \ + \ a \ t[/tex]

where [tex]V_0[/tex] is the initial speed, and  a its the acceleration.

Starting at rest, the initial speed will be zero, so

[tex]V_0 = 0[/tex]

the final speed is

[tex]V(t_{f1}) = 23 \frac{m}{s}[/tex]

and the acceleration is

[tex]a = 2.3 \frac{m}{s^2}[/tex].

Taking all this together, we got

[tex]V(t_{f1}) = 23 \frac{m}{s} = 0 + 2.3 \ \frac{m}{s^2} t_{f1}[/tex]

[tex]23 \frac{m}{s} = 2.3 \ \frac{m}{s^2} t_{f1}[/tex]

[tex]\frac{23 \frac{m}{s}}{2.3 \ \frac{m}{s^2}} =  t_{f1}[/tex]

[tex]10 s =  t_{f1}[/tex]

So, for the first half of the problem we got a time of 10 seconds.

Now, the initial speed will be

[tex]V_0 = 23 \frac{m}{s}[/tex],

the acceleration

[tex]a=-1.0 \frac{m}{s^2}[/tex],

with a minus sign cause its slowing down, the final speed will be

[tex]V(t_{f2}) = 0[/tex]

Taking all together:

[tex]V(t_{f2}) = 0 = 23 \frac{m}{s} -  1.0 \frac{m}{s^2} t_{f2}[/tex]

[tex] 23 \frac{m}{s} =  1.0 \frac{m}{s^2} t_{f2}[/tex]

[tex] \frac{23 \frac{m}{s}}{1.0 \frac{m}{s^2}} = t_{f2}[/tex]

[tex] 23 s = t_{f2}[/tex]

So, for the first half of the problem we got a time of 23 seconds.

[tex]t_total = t_{f1} + t_{f2} = 33  \ s[/tex]

Final answer:

To estimate the electric field near a uniformly charged wire, one must determine the linear charge density and integrate the contributions from infinitesimal charge elements along the wire, considering the symmetry that leads to the cancellation of horizontal components.

Explanation:

To estimate the electric field at a point 2.40 cm perpendicular to the midpoint of a uniformly charged 2.00-m-long thin wire, we use the principle of superposition and integrate the contributions of the infinitesimally small charge elements along the wire.

First, we determine the linear charge density λ by dividing the total charge Q by the length L of the wire:

λ = Q / L

Then, we apply the formula for the electric field due to an infinitesimal charge element dQ at a distance r:

dE = (k * dQ) / r²

Where k is the Coulomb's constant (approximately 8.99 × 10⁹ Nm²/C²). Since we need to integrate over the length of the wire to find the total electric field, we use the linear charge density λ and a differential element of the wire's length dl:

dQ = λ * dl

The resulting integral, when evaluated, will give us the magnitude of the electric field E at the specified distance from the wire.

It is important to note that due to the symmetry of the problem, the horizontal components of the electric field due to charge elements on the wire will cancel out, leaving only the vertical components to contribute to the total electric field at the point of interest

Answer:

Size of image: 1.8 cm inverted

Magnification of camera: -0.1

Explanation:

Given:

Assume:

Using sign convention, we have

[tex]u = -200\ cm\\v = 20\ cm\\h_o = 18\ cm[/tex]

The hole in the pinhole camera behaves as a convex lens. The other face opposite to the hole face behaves like a film where the image is formed to be seen. So, using the formula of magnification, we have

[tex]m = \dfrac{h_i}{h_o}=\dfrac{v}{u}\\\Rightarrow  \dfrac{h_i}{h_o}=\dfrac{v}{u}\\\Rightarrow  h_i=\dfrac{v}{u}h_o\\\Rightarrow  h_i=\dfrac{20}{-200}\times 18\\\Rightarrow  h_i=-1.8 cm[/tex]

This means the image of the bird measures 1.8 cm in length where negative sign in the calculation represents that the image formed is inverted.

Hence, the image of the bird on the film is 1.8 cm large.

Now, again using the formula of magnification, we have

[tex]m = \dfrac{h_i}{h_o}\\\Rightarrow m = \dfrac{-1.8}{18}\\\Rightarrow m =-0.1[/tex]

Hence, the magnification of the camera is -0.1.

In a pinhole camera, image size = object size * (camera depth / object distance). For a chicken 18 cm tall standing 2 meters away, the image size will be 0.198 cm. The magnification, or size of the image compared to the object, will be 0.011 or 1.1%

The size of the image formed by a pinhole camera is given by the formula: image size = object size * (camera depth / object distance). Substituting the values in this formula, image size = 18 cm * (22 cm / 2 m) = 0.198 cm. This tells us that the fierce chicken's image will be about 0.198 cm high on the film.

Next, the magnification of the camera is the ratio of the image size to the object size. So the magnification is 0.198 cm / 18 cm = 0.011. In other words, the image on the film is about 1.1% the size of the actual object.

Note:

These calculations are based on the simple model of a pinhole camera where only rays of light traveling in straight lines are considered. The actual picture may differ due to factors like the scattering of light.

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Answer:

Ff= 8.9N:Force of friction acting on the block.

Explanation

Box kinetics:  in x₁-y₁ :We apply  the second law of Newton

∑Fx₁=ma : second law of Newton  Formula (1)

Where:

∑Fx₁:algebraic sum of forces in the direction of x1, positive in the direction of movement of the block, which is down and negative in the direction opposite to the movement of the block

m: is the mass of the block

The x₁ axis coincides with the plane of sliding of the block, that is, the x₁ axis forms 45 degrees with the horizontal

The total weight (W) of the block is in the vertical direction and the tip of the vector down and its magnitude is calculated as follows:

W=m*g

Where:

m: is the mass of the block

g: is the acceleration due to gravity

Calculation of the weight in the direction x₁

Wx₁=Wcos45°= m*g*cos 45 Equation( 1)

Data

m=1.8kg

g=9.8 m/s²

a=2m/s

Wx₁=m*g*cos 45°= 1.8*9.8 *[tex]\frac{\sqrt{2} }{2}[/tex]= 12.47 N

Friction force (Ff ) calculation

We apply formula (1)

∑Fx₁=m*a

Wx₁ - Ff= m*a

Ff=Wx₁ - m*a

We replace data

Ff= 12.47 - 1.8*2 =8.87

Ff= 8.9N

The rate at which the area of the triangle formed by the wall, the ground, and the ladder is changing, at the instant the bottom of the ladder is 36 feet from the wall, is -216 square feet per second.

This question revolves around the concept of related rates in calculus. The area of the triangle formed by the ground, the wall, and the ladder is given by the formula ½ * base * height. In this situation, the base is x, which represents the distance from the wall to the base of the ladder, and the height is the vertical reach of the ladder onto the wall, represented by y (which can be found using the Pythagorean theorem). Differentiating this equation with respect to time gives us dA/dt = ½ (x dy/dt + y dx/dt).

Given that dx/dt is 8 feet per second, and x=36 feet, to find dy/dt, we use the Pythagorean relationship (39² = 36² + y²), and solve for y to get y = 15 feet. Therefore, substituting all these values into the equation we find:** dA/dt = -216 sq.ft/sec**.

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The area of the triangle is changing at a rate of ( -285.6 ) square feet per second at the instant the bottom of the ladder is 36 feet from the wall. The negative sign indicates that the area is decreasing.

Let ( x ) be the distance of the bottom of the ladder from the wall, and ( y) be the height of the ladder on the wall. We are given that [tex]\( \frac{dx}{dt} = 8 \)[/tex] feet per second, and we want to find [tex]\( \frac{dA}{dt} \) when \( x = 36 \) feet.[/tex]

Using the Pythagorean theorem, we have [tex]\( x^2 + y^2 = 39^2 \).[/tex]Differentiating both sides with respect to time ( t ), we get:

[tex]\[ 2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0 \][/tex]

We can solve for [tex]\( \frac{dy}{dt} \):[/tex]

[tex]\[ 2y\frac{dy}{dt} = -2x\frac{dx}{dt} \][/tex]

[tex]\[ \frac{dy}{dt} = -\frac{x}{y}\frac{dx}{dt} \][/tex]

Now, we can express the area ( A ) in terms of ( x ) and ( y ):

[tex]\[ A = \frac{1}{2}xy \][/tex]

Differentiating ( A ) with respect to time ( t ), we get:

[tex]\[ \frac{dA}{dt} = \frac{1}{2}\left(x\frac{dy}{dt} + y\frac{dx}{dt}\right) \][/tex]

Substituting [tex]\( \frac{dy}{dt} \)[/tex] from above, we have:

[tex]\[ \frac{dA}{dt} = \frac{1}{2}\left(x\left(-\frac{x}{y}\frac{dx}{dt}\right) + y\frac{dx}{dt}\right) \][/tex]

[tex]\[ \frac{dA}{dt} = \frac{1}{2}\left(-\frac{x^2}{y}\frac{dx}{dt} + y\frac{dx}{dt}\right) \][/tex]

[tex]\[ \frac{dA}{dt} = \frac{1}{2}\left(y - \frac{x^2}{y}\right)\frac{dx}{dt} \][/tex]

We know [tex]\( \frac{dx}{dt} = 8 \)[/tex] feet per second, and we need to find [tex]\( \frac{dA}{dt} \)[/tex]when ( x = 36 ) feet. To find ( y ) at this instant, we use the Pythagorean theorem:

[tex]\[ 36^2 + y^2 = 39^2 \][/tex]

[tex]\[ y^2 = 39^2 - 36^2 \][/tex]

[tex]\[ y^2 = 1521 - 1296 \][/tex]

[tex]\[ y^2 = 225 \][/tex]

[tex]\[ y = 15 \][/tex]

Now we can find [tex]\( \frac{dA}{dt} \):[/tex]

[tex]\[ \frac{dA}{dt} = \frac{1}{2}\left(15 - \frac{36^2}{15}\right) \times 8 \][/tex]

[tex]\[ \frac{dA}{dt} = 4\left(15 - \frac{1296}{15}\right) \][/tex]

[tex]\[ \frac{dA}{dt} = 4\left(15 - 86.4\right) \][/tex]

[tex]\[ \frac{dA}{dt} = 4\left(-71.4\right) \][/tex]

[tex]\[ \frac{dA}{dt} = -285.6 \][/tex]

So, the area of the triangle is changing at a rate of ( -285.6 ) square feet per second at the instant the bottom of the ladder is 36 feet from the wall. The negative sign indicates that the area is decreasing.

Answer:

The work done on the jet is [tex]W_{jet} = 2.49\times 10^{7} J[/tex]

Given:

Force, F = [tex]2.3\times 10^{5} N[/tex]

Distance moved by the jet, x = 87 m

Kinetic Energy, KE = [tex]4.5\times 10^{7} J[/tex]

Solution:

Now, to calculate the work done on the fire fighter jet by the catapult, we find the difference between the KE of the jet at lift off and the work done by the engines.

This difference provides the amount of work done on the jet and is given by:

[tex]W_{jet} = KE - W_{engines}[/tex]              (1)

Now, the work done by the engines is given by:

[tex]W_{engines} = Fx = 2.3\times 10^{5}\times 87 = 2.001\times 10^{7} J[/tex]

Now, using eqn (1):

[tex]W_{jet} = 4.5\times 10^{7} - 2.001\times 10^{7} = 2.49\times 10^{7} J[/tex]

Answer:

The hydro static force on the back of the dam is [tex]1.96\times10^{11}\ N[/tex]

Explanation:

Given that,

Width b= 1000 m

Depth d= 200 m

We need to calculate the average pressure

Using formula of  average pressure

[tex]P_{avg}=\rho\times g\times d_{avg}[/tex]

Put the value into the formula

[tex]P_{avg}=1000\times9.8\times100[/tex]

[tex]P_{avg}=980000\ Pa[/tex]

We need to calculate  the hydro static force on the back of the dam

Using formula of force

[tex]F = P_{avg}\times A[/tex]

Put the value into the formula

[tex]F = 980000\times1000\times200[/tex]

[tex]F=1.96\times10^{11}\ N[/tex]

Hence, The hydro static force on the back of the dam is [tex]1.96\times10^{11}\ N[/tex]

Answer:

The the decibel level be when three sleep deprived students are typing furiously in the same room is 64.77 dB.

Explanation:

Given that,

Decibel level = 60.0 dB

Using formula of decibel level

[tex]\beta= 10 log\dfrac{I}{I_{0}}[/tex]

For one student,

[tex]60.0=10 log\dfrac{I}{I_{0}}[/tex]

For 3 students,

[tex]\beta'=10 log\dfrac{3I}{I_{0}}[/tex]

[tex]\beta'=10log 3+10 log\dfrac{I}{I_{0}}[/tex]

[tex]\beta'=4.77+60.0[/tex]

[tex]beta'=64.77\ dB[/tex]

Hence, The the decibel level be when three sleep deprived students are typing furiously in the same room is 64.77 dB.

Answer:

Along North,  1.816 km

Along east , 3.564 km

Explanation:

d = 4 km 27° North of east

As we need to find the distance traveled due north and the distance traveled due east, they are the components along north and along east. The component along north is the vertical component and the distance traveled along east is the horizontal component.

Distance traveled due North

y = d Sin 27 = 4 Sin 27 = 1.816 km

Distance traveled due east

x = d cos 27 = 4 cos 27 = 3.564 km

Answer:

Height, H = 25.04 meters

Explanation:

Initially the ball is at rest, u = 0

Time taken to fall to the ground, t = 2.261 s

Let H is the height from which the ball is released. It can be calculated using the second equation of motion as :

[tex]H=ut+\dfrac{1}{2}at^2[/tex]

Here, a = g

[tex]H=\dfrac{1}{2}gt^2[/tex]            

[tex]H=\dfrac{1}{2}\times 9.8\times (2.261)^2[/tex]

H = 25.04 meters

So, the ball is released form a height of 25.04 meters. Hence, this is the required solution.

Final answer:

Using the equation of motion under gravity, it was calculated that a ball dropped and taking 2.261 s to hit the ground was released from a height of approximately 25.07 meters.

Explanation:

To find out from what height H a ball was released if it takes 2.261 s to fall to the ground, we use the equation of motion under gravity, which is H = 0.5 * g * t^2, where g is the acceleration due to gravity (approximately 9.8 m/s^2 on the surface of the Earth) and t is the time in seconds.

Substituting the given values in the equation, we get: H = 0.5 * 9.8 * (2.261)^2. After calculating, we find that the height H approximately equals 25.07 meters.

Therefore, the ball was released from a height of approximately 25.07 meters.

Answer:

equation of  motion for Bill is

[tex]y(t) = 4.9t^2[/tex]

equation of  motion for Ted is

[tex]y(t) = 2 + (-4.2)(t) + 4.9t^2[/tex]

Explanation:

Taking downward position positive and upward position negative

g = 9.8 m/s^2

equation of  motion for Bill is

[tex]y(t) = y_0 +v_0 t +\frac{1}{2}gt^2[/tex]

[tex]y(t) = 0 + 0(t) +\frac{1}{2}gt^2[/tex]

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

[tex]y(t) = 4.9t^2[/tex]

equation of  motion for Ted is

[tex]y_0 = 2m -1m = 2m[/tex]

[tex]y_0 = -4.2 m/s[/tex]

[tex]y(t) = y_0 +v_0 t +\frac{1}{2}gt^2[/tex]

[tex]y(t) = 2 + (-4.2)(t) +\frac{1}{2}gt^2[/tex]

[tex]y(t) = 2 + (-4.2)(t) +\frac{1}{2}\times (9.8t)^2[/tex]

[tex]y(t) = 2 + (-4.2)(t) + 4.9t^2[/tex]

Answer:

Answer:

For Bill:

[tex]x(t)=3-(4.9*t^{2})[/tex]

For Ted:

[tex]x(t)=1+(4.2*t)+(-4.9*t^{2} )[/tex]

Explanation:

For Bill:

[tex]Initial position=x_{0}=3[/tex]

[tex]Initial velocity=v_{0}=0[/tex]

Now using [tex]2^{nd}[/tex] equation of motion,we have

[tex]x-x_{0}=(v_{0} *t)+(1/2*g*t^{2})[/tex]

[tex]x_{0} =3[/tex]  ,[tex]v_{0}=0[/tex]

Thus,equation becomes

[tex]x-3=1/2*g*t^{2}[/tex]

[tex]x=3+(0.5*g*t^{2})[/tex]

Taking acceleration upward positive and downward negative.

[tex]g=-10[/tex] [tex]m/s^{2}[/tex]

[tex]x(t)=3-4.9*t^{2}[/tex]    for bill

For Ted

[tex]x_{0} =1[/tex]

[tex]v_{0}=4.2[/tex] [tex]m/s[/tex]

Using the same equation

[tex]x-x_{0}=(v_{0} *t)+(1/2*g*t^{2})[/tex]

[tex]x_{0}=1[/tex] [tex]m[/tex]

[tex]v_{0}=4.2[/tex] [tex]m/s[/tex]

Substitute values

[tex]x-1=(4.2*t)+(1/2*g*t^{2})[/tex]

[tex]g=-10[/tex] [tex]m/s^{2}[/tex]

Thus equation becomes

[tex]x(t)=1+(4.2*t)+(-4.9*t^{2})[/tex]   for Ted

Answer:

upward acceleration is 211.04 m/s²

Explanation:

given data

velocity = 14.6 m/s

distance = 0.505 m

to find out

upward acceleration

solution

we have given distance and velocity and ball is going upward

so acceleration will be calculated by  velocity formula that is

v² - u² = 2×a×s   ............1

here v is velocity and u is initial velocity and s is distance and a is acceleration

and u = o because starting velocity zero

put here all there value in equation 1

v² - u² = 2×a×s

14.6² - 0 = 2×a×0.505

solve it we get a

a = 211.04

so upward acceleration is 211.04 m/s²

The time for projectile motion is determined completely by the vertical motion. A ball thrown vertically upwards with an initial velocity of 20.00 m/s spends approximately 4 seconds in the air and reaches its greatest height after approximately 2 seconds. The time at which the ascending ball reaches a height of 15 m above the ground can be found by setting y = 15 m and solving for t.

The time for projectile motion is determined completely by the vertical motion. Thus, any projectile that has an initial vertical velocity of 20.00 m/s and lands 10.0 m above its starting altitude spends approximately 4 seconds in the air. The greatest height reached by the ball can be determined by using the equation for vertical motion: y = yo + voyt - 1/2gt^2. Plugging in the given values and solving for t, we find that the ball reaches its greatest height after approximately 2 seconds, and the time at which the ascending ball reaches a height of 15 m above the ground can be found by setting y = 15 m and solving for t.

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Planets like Mars exhibit direct and retrograde motion, resulting in complex paths across the night sky, which differ from the Sun's steady eastward movement along the ecliptic through the zodiac signs. Earth's faster orbit leads to the phenomenon of retrograde motion when observing other planets.

The motion of planets like Mars among the stars of the zodiac is different from the motion of the Sun because planets exhibit both direct motion and retrograde motion in their paths across the sky, as observed from Earth. Planets periodically appear to move in the opposite direction in the sky, known as retrograde motion, while the Sun moves steadily along the ecliptic through the zodiac signs. This distinct planetary motion is due to the orbital speed and path of Earth relative to other planets, leading to the phenomenon where Earth, moving faster along its orbit, occasionally overtakes other planets like Mars, changing how these planets appear to move against the backdrop of stars.

Additionally, planets can have stationary points where they appear to pause before they switch from direct to retrograde motion or vice versa. These motions are perceived from Earth because of the combination of the planets' own orbital movement around the Sun and Earth's simultaneous revolution. By contrast, the Sun follows a consistent eastward path among the fixed stars, circling the zodiac once per year, spending about a month in each zodiac sign.

Answer:

(1) [tex]46.86^\circ[/tex]

(2) Diagram has been attached in the solution.

Explanation:

This question is from projectile motion.

From the given question, we will discuss the motion of the basket ball only in the vertical direction from which we will be able to find out the angle of the initial velocity with the horizontal with which it should be shoot to enter the hoop.

Part (1):

Let us assume:

Using the equation of motion for constant acceleration, we have

[tex]y_f-y_i=u_yt+\dfrac{1}{2}a_yt^2\\\Rightarrow 3.05-2.1=u\sin \theta (0.8) +\dfrac{1}{2}\times (-9.8)(0.8)^2\\\Rightarrow 0.95=7\times \sin \theta (0.8) -3.136\\\Rightarrow 0.95=5.6\sin \theta -3.136\\\Rightarrow 5.6\sin \theta= 0.95+3.136\\\Rightarrow 5.6\sin \theta= 4.086\\\Rightarrow \sin \theta= \dfrac{4.086}{5.6}\\\Rightarrow \sin \theta=0.729\\\Rightarrow \theta=\sin^{-1}(0.729)\\\Rightarrow \theta=46.86^\circ[/tex]

Hence, the angle of the shoot of the basket ball with the horizontal is [tex]46.86^\circ[/tex] such that it reaches the hoop on time.

Part (2):

For this part, a diagram has been attached.

Answer:

speed of sound = 321 m/s

Rule of thumb for kilometers: 1 km every 3 seconds

Explanation:

Hi!

If a thunder originates at a place a distance D from you, and it takes time T for its sound to reach you, then:

[tex]V = \frac{D}{T}[/tex]

Whre V is the speed of sound. Time T is the time elapsed between the moment you see the flash (becuase of the assumption of tha it takes no time for the light to reach you) and the moment you hear the thunder. Then

[tex]V = \frac{1mile}{5\;s} =\frac{1609.34 \;m}{5\;s} = 321\frac{m}{s}[/tex]

To calculate the rule in kilometers:

[tex]T = \frac{1\;km}{321*10^{-3} \frac{km}{s}} \approx 3\; s[/tex]

The estimated speed of sound derived from the rule of thumb is approximately 322 m/s. For the rule in kilometers, each five seconds would represent about 1.609 kilometers.

The rule-of-thumb that every five seconds between a lightning flash and the accompanying thunder equates to one mile of distance is based on the speed of sound. To convert this to meters per second we must know that 1 mile approximately equals 1609.34 meters. Therefore, if light travels nearly instantaneously, and one mile is covered in five seconds, the speed of sound can be estimated as about 1609.34 m/5 s or approximately 322 m/s.

For the rule in kilometers, we need to convert miles into kilometers. As 1 mile is approximately 1.609 kilometers, the rule of thumb in kilometers would be that each five seconds would represent 1.609 kilometers.

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Answer:

The correct option is 'D': Equal and opposite force.

Explanation:

We know from Newton's third law of motion

"For every action there exists an equal and opposite reaction".

The statement is further explained as

If there exists a force from Body 1 to Body 2 then Body 2 must also exert an equal and opposite force on the Body 1.

In simple terms we can say that the action and the reaction forces exist in pairs and are always present if any one among them is present.

Answer:

[tex]r=\frac{1}{75}[/tex] Ω≈0.01333Ω

Explanation:

You can check the internal resistance model in the image I attached you.

Now using Kirchhoff's voltage law:

ε=[tex]Ir+IR[/tex]  (1)

Where:

[tex]IR=V_R=LoadVoltage=10V[/tex]

ε=12V

[tex]I=150A[/tex]

Evaluating the data provide in (1)

[tex]r=\frac{12-10}{150} =\frac{1}{75}[/tex]≈0.01333Ω

Answer:

mean = 54

standard deviation = 6.24

Explanation:

given data

mean a = 72

standard deviation σx = 12

average b = 95

standard deviation σy = 4

final grade Z = 70%  = 0.70

project y = 30%  = 0.30

to find out

mean and standard deviation

solution

final grade equation will be here

final grade = 0.70x + 0.30 y   .....a

here x is statics grade and y is class project

so

mean will be 0.70x + 0.30 y

mean = 0.70 (a) + 0.30 (b)

mean = 0.70 (12) + 0.30 (95)

mean = 54

and

standard deviation = 0.70x + 0.30 y

standard deviation = 0.70(σx) + 0.30 (σy)

standard deviation = 0.70(12) + 0.30 (4)

standard deviation = 6.24

The forces acting on you are:

That is. There is not need for any other force, cause the car its going at constant speed, so the acceleration its zero, as the net force its mass multiplied by acceleration, the net force is zero.

The forces acting on you are:

That is. Again. The car its slowing down. But, it cant apply a force in you against the direction of movement to slow you down.

As you can't be slowed down, you will go forward at the same speed you had when the car went steady, and will crash against whatever its in front of you.

D)

The friction force, will pull you against the direction of movement, and slow you down will the car slows down.

When a car moves at a steady pace, the forces acting on a passenger include gravity and the normal force from the bench. When the car begins to slow down, the passenger experiences inertia, which might give the effect of being 'pushed' forward. If the bench is not slippery, friction will be the force that aids in decelerating the passenger along with the car.

(a) If you are sitting on a frictionless bench in a car that is moving at a perfectly steady speed, the forces acting on you are: the gravitational force directed down and the normal force from the bench acting upward. Since the car moves at a constant speed, there is no horizontal force.

(b) If the car starts slowing down, then in addition to the forces mentioned before, there is one more force in play, which is inertia. It's not a force like gravity or the normal force, but it is a consequence of Newton's first law of motion. You experience this as a force pushing you forward, even though the car is slowing down or stopping.

(c) Because of inertia, as the car slows down, you tend to keep moving. Without friction or anything to hold onto, you would slide off the bench and continue moving forward at the original speed of the car.

(d) If the bench is not slippery and you don't slide off as the car slows down, the force responsible for your deceleration is friction. Even though you can't feel it, friction between you and the bench affects you when the car decelerates.

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Answer:

option (d) 7.1 kN

Explanation:

Given:

Mass of the car, m = 1600 kg

Acceleration of the car, a = 1.5 m/s²

Coefficient of kinetic friction = 0.3

let the tension be 'T'

Now,

ma = T - f .................(1)

where f is the frictional force

also,

f = 0.3 × mg

where g is the acceleration due to the gravity

thus,

f = 0.3 × 1600 × 9.81 =

therefore,

equation 1 becomes

1600 × 1.5 = T - 4708.8

or

T = 2400 + 4708.8

or

T = 7108.8 N

or

T = 7.108 kN

Hence,

The correct answer is option (d) 7.1 kN

Answer:

the radius of bigger loop = 6 cm

Explanation:

given,

two concentric current loops

smaller loop radius = 3.6 cm

]current in smaller loop = 12 A

current in the bigger loop = 20 A

magnetic field at the center of loop = 0

Radius of the bigger loop = ?

[tex]B_t = B_1 + B_2[/tex]

[tex]0 = \dfrac{\mu_0I_1}{2R_1} +\dfrac{\mu_0I_2}{2R_2}[/tex]

now, on solving

[tex]\dfrac{I_1}{R_1} = \dfrac{I_2}{R_2}[/tex]

[tex]R_2 = I_2\dfrac{R_1}{I_1}[/tex]

       = [tex]20\times \dfrac{3.6}{12}[/tex]

       = 6 cm

hence, the radius of bigger loop = 6 cm

Answer:

Explanation:

First, we will find the Cartesian Representation of the [tex]\vec{X}[/tex] and [tex]\vec{Y}[/tex] vectors. We can do this, using the formula

[tex]\vec{A}= | \vec{A} | \ ( \ cos(\theta) \ , \ sin (\theta) \ )[/tex]

where [tex]| \vec{A} |[/tex] its the magnitude of the vector and θ the angle. For  [tex]\vec{X}[/tex] we have:

[tex]\vec{X}= 1430 m \ ( \ cos( 42 \°) \ , \ sin (42 \°) \ )[/tex]

[tex]\vec{X}= ( \ 1062.70 m \ , \ 956.86 m \ )[/tex]

where the unit vector [tex]\hat{i}[/tex] points east, and [tex]\hat{j}[/tex] points north. Now, the [tex]\vec{Y}[/tex] will be:

[tex]\vec{Y}= - 2200 m \hat{j} = ( \ 0 \ , \ - 2200 m \ )[/tex]

Now, taking the sum:

[tex]\vec{X} + \vec{Y} + \vec{Z} = 0[/tex]

This is

[tex]\vec{Z} = - \vec{X} - \vec{Y}[/tex]

[tex](Z_x , Z_y) = - ( \ 1062.70 m \ , \ 956.86 m \ ) - ( \ 0 \ , \ - 2200 m \ )[/tex]

[tex](Z_x , Z_y) = ( \ - 1062.70 m \ ,  \ 2200 m \ - \ 956.86 m \ )[/tex]

[tex](Z_x , Z_y) = ( \ - 1062.70 m \ ,  \ 1243.14 m\ )[/tex]

Now, for the magnitude, we just have to take its length:

[tex]|\vec{Z}| = \sqrt{Z_x^2 + Z_y^2}[/tex]

[tex]|\vec{Z}| = \sqrt{(- 1062.70 m)^2 + (1243.14 m)^2}[/tex]

[tex]|\vec{Z}| = 1635.43 m[/tex]

For its angle, as the vector lays in the second quadrant, we can use:

[tex]\theta = 180\° - arctan(\frac{1243.14 m}{ - 1062.70 m})  [/tex]

[tex]\theta = 180\° - arctan( -1.1720)  [/tex]

[tex]\theta = 180\° - 45\°31'40'' [/tex]

[tex]\theta = 130\°28'20''  [/tex]

Answer:

[tex]V=23.3kV[/tex]

Explanation:

Definition of the capacitance C, where a voltage V is applied and a charge Q is stored:

[tex]Q=C*V[/tex]

We solve to find V:

[tex]V=Q/C=0.14*10^{-3}C/6*10^{-9}F)=2.33*10^{4}V=23.3kV[/tex]

Answer:

(a): [tex]247.77^\circ.[/tex]

(b): [tex]\vec F = k\ \dfrac{q_1q_2}{|\vec r|^2}\ \dfrac{\hat i-12\hat j}{\sqrt{145}}.[/tex]

(c): Magnitude = [tex]0.604\ C.[/tex]

     Sign = negative.

Explanation:

(a):

Given that first charge is located at (3,-5) and second charge is located at (2,7).

The electrostatic force between two charges acts long the line joining the two charges, therefore, the direction of electrostatic force of attraction between these two charges is along the position vector of charge at (2,7) with respect to the position of charge at (3,-5).

Assuming, [tex]\hat i,\ \hat j[/tex] are the unit vectors along positive x and y axes respectively.

Position vector of charge at (3,-5) with respect to origin is

[tex]\vec r_1 = 3\hat i+(-5)\hat j[/tex]

Position vector of charge at (2,7) with respect to origin is

[tex]\vec r_2 = 2\hat i+7\hat j[/tex]

The position vector of at (2,7) with respect to the position of charge at (3,-5) is

[tex]\vec r = \vec r_2-\vec r_1\\=(3\hat i-5\hat j)-(2\hat i+7\hat j)\\=1\hat i-12\hat j.[/tex]

If [tex]\theta[/tex] is the angle this position vector is making with the positive x axis then,

[tex]\rm \tan\theta =\dfrac{y\ component\ of\ (\hat i-12\hat j)}{x\ component\ of\ (\hat i-12\hat j}=\dfrac{-12}{1} = -12.\\\theta =\tan^{-1}(-12)=-85.23^\circ.[/tex]

The negative sign indicates that this position vector is [tex]85.23^\circ.[/tex] below the positive x axis, therefore, in counterclockwise direction from positive x axis, its direction = [tex]360^\circ-85.23^\circ = 247.77^\circ.[/tex]

The force is also along the same direction.

(b):

According to Coulomb's law, the expression of this force is given by

[tex]\vec F = k\ \dfrac{q_1q_2}{|\vec r|^2}\ \hat r[/tex]

where,

We have,

[tex]\vec r = \hat i-12\hat j\\\therefore |\vec r| = \sqrt{1^2+(-12)^2}=\sqrt{145}.\\\Rightarrow \hat r = \dfrac{\vec r}{|\vec r|} = \dfrac{\hat i-12\hat j}{\sqrt{145}}[/tex]

Putting this value, we get,

[tex]\vec F = k\ \dfrac{q_1q_2}{|\vec r|^2}\ \dfrac{\hat i-12\hat j}{\sqrt{145}}[/tex]

(c):

If charge at (3,-5) is [tex]+2\ pC[/tex], then the magnitude of the force between the two charges is given by

[tex]F=k\ \dfrac{q_1q_2}{|\vec r|^2}[/tex]

where, we have,

[tex]F=0.75\ N\\|\vec r| = \sqrt{145}\ cm = \sqrt{145}\times 10^{-2}\ m.\\q_1 = +2\ pC = +2\times 10^{-12}\ C.[/tex]

Putting all these values,

[tex]0.75=9\times 10^9\times \dfrac{(+2\times 10^{-12})\times q_2}{(\sqrt{145}\times 10^{-2})^2}\\\Rightarrow q_2 = \dfrac{0.75\times(\sqrt{145}\times 10^{-2})^2}{9\times 10^9\times 2\times 10^{-12}} =0.604\ C.[/tex]

Since, the electric force between the two spheres is given to be attractive therefore this charge must be negative as the other charge is positive.