A ball is thrown vertically upwards with an initial velocity of 20.00 m/s. Neglecting air resistance, how long is the ball in the air? What is the greatest height reached by the ball? Also calculate the time at which the ascending ball reaches a height of 15 m above the ground.

Answer:

The statue is 1.67 miles west of both the runners

Explanation:

Wherever Amy and Alejandro meet they will have covered a total distance of 5+4 = 9 mi

They will also have run the same amount of time if they started at the same moment = t

Speed of Amy = 5 mi/h

Speed of Alejandro = 7 mi/h

Distance = Speed × Time

Distance travelled by Amy = 5t

Distance travelled by Alejandro = 7t

Total distance run by Amy and Alejandro is

5t+7t = 9

[tex]\\\Rightarrow 12t=9\\\Rightarrow t=\frac{12}{9}\\\Rightarrow t=\frac{4}{3}\ hours[/tex]

Distance travelled by Amy

[tex]5\times t=5\times \frac{4}{3}=\frac{20}{3}\\ =6.67\ miles[/tex]

The distance of Amy from the statue would be

6.67 - 5 = 1.67 miles

So, the statue is 1.67 miles west of both the runners

The electric field strengths at various points around a charged object can be calculated using Coulomb's Law. They depend on the charge of the object and the distance from it. Fields produced by multiple charges need to be added vectorially.

You're asking about the electric field strengths at varying distances from a charged glass rod, located next to a plastic rod with a charge of equal magnitude, but of opposite sign. The electric field strength E at a point in the vicinity of a charged object can be derived using Coulomb's Law. This law states that the force F between two charges q1 and q2 is proportional to the product of the charges and inversely proportional to the square of the distance r between them.

Given this, the electric field E created by a charge q at a distance r from the charge is given by E = k|q| / r², where k is Coulomb's constant, approximately 9 × 10⁹ N•m²/C². The direction of the electric field is determined by the sign of the charge. When it's positive, the field lines are going outward, and when it's negative, they are directed inward.

To calculate the net electric field at a point on the line connecting the midpoints of the rods, we consider the electric fields produced by both rods. Since they're opposite in sign, the fields will have opposite directions, and they should be added vectorially. The magnitudes and directions of the fields will also depend on the specific distances (1.0 cm, 2.0 cm, and 3.0 cm) from the glass rod.

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Answer:

a)[tex] C_p=35.42\ \rm J/mol.K[/tex]

b)[tex] C_v=27.1\ \rm J/mol.K[/tex]

Explanation:

Given:

Let R be the gas constant which has value [tex]R=8.31432\times10^3\ \rm N\ m\ kmol^{-1}K^{-1}[/tex]

Work done in the process

[tex]W=PdV\\W=0.947\times(104-56.7)\times10^{-3}\ \rm L\ atm\\W=44.8\times10^{-3}\times101.33\ \rm J\\W=4.43\ \rm J[/tex]

Now Using First Law of thermodynamics

[tex]Q=W+\Delta U\\19.3=4.43+\Delta U\\\Delta U=14.77\ \rm J[/tex]

Let[tex]C_v[/tex] be the molar specific heat of the gas at constant volume given by

[tex]\Delta U=nC_v\Delta T\\14.77=\dfrac{C_v}{R}{PdV}\\14.77=\dfrac{C_v}{R}(0.987\times10^5(104-56.7)\times10^-6})\\C_v=3.26R\\C_v=27.1\ \rm J/mol.K[/tex]

Also we know that

Let[tex]C_p[/tex] be the molar specific heat of the gas at constant pressure given by

[tex]C_p-C_v=R\\C_p=R+C_v\\C_p=4.26R\\C_p=35.42\ \rm J/mol.K[/tex]

Answer:

Approximate linear dimension is 2 light years.

Explanation:

Radius of the spiral galaxy r = 62000 LY

Thickness of the galaxy h = 700 LY

Volume of the galaxy = πr²h

                                   = (3.14)(62000)²(700)

                                   = (3.14)(62)²(7)(10)⁸

                                   = 84568×10⁸

                                   = [tex]8.45\times 10^{12}[/tex] (LY)³

Since galaxy contains number of stars = 1078 billion stars ≈ [tex]1.078\times 10^{12}[/tex]

Now volume covered by each star of the galaxy = [tex]\frac{\text{Total volume of the galaxy}}{\text{Number of stars}}[/tex]

= [tex]\frac{8.45\times 10^{12} }{1.078\times 10^{12}}[/tex]

= 7.839 Light Years

Now the linear dimension across the volume

= [tex](\text{Average volume per star})^{\frac{1}{3}}[/tex]

= [tex](7.839)^{\frac{1}{3}}[/tex]

= 1.99 LY

≈ 2 Light Years

Therefore, approximate linear dimension is 2 light years.

Part A: -,+

The elevator is moving downward, this is what determines the direction of the velocity, as it will follow the direction of the movement. As we are told that the positive direction is upward, then the velocity has negative direction. Also, after the button is pressed, the elevator starts to stop, in other words, its velocity starts to decreased. This means that the acceleration has an opposite direction to the velocity, therefore, its sign is +.

Part B: +, -

The ball is moving upward, and as said before, this is what determines the direction of the velocity, as it will follow the direction of the movement. Then, velocity has a + sign.

Also, after the ball is thrown, there is no other force other than gravity, which will oppose to the movement of the ball, trying to make it come back to the ground. This means that the acceleration has an opposite direction to the velocity, in other words, it's directed downward, therefore, its sign is -.

Part C: 0, -

The acceleration of the ball since it was thrown until it fell to the ground will always be the gravity, which will always go downward (-).

After being thrown, the ball's velocity will start to decrease because of gravity. When its velocity has turned to 0, the ball will have reached maximum height . At this point it will start to fall again, accelerated by gravity. But at the very top, the velocity of the ball is 0.

Answer:

4.1 eV

Explanation:

Kinetic energy, K = 0.8 eV = 0.8 x 1.6 x 10^-19 J = 1.28 x 10^-19 J

wavelength, λ = 253.5 nm = 253.5 x 10^-9 m

According to the Einstein energy equation

[tex]E = W_{o}+K[/tex]

Where, E be the energy incident, Wo is the work function and K is the kinetic energy.

h = 6.634 x 10^-34 Js

c = 3 x 10^8 m/s

[tex]E=\frac{hc}{\lambda }=\frac{6.634 \times 10^{-34} \times 3 \times 10^{8}}{253.5\times 10^{-9}}=7.85 \times 10^{-19} J[/tex]

So, the work function, Wo = E - K

Wo = 7.85 x 10^-19 - 1.28 x 10^-19

Wo = 6.57 x 10^-19 J

Wo = 4.1 eV

Thus, the work function of the metal is 4.1 eV.

Answer:

E = 1.655 x 10⁷ N/C towards the filament

Explanation:

Electric field due to a line charge is given by the expression

E = [tex][tex]\frac{\lambda}{2\pi\times\epsilon_0\times r}[/tex][/tex]

where λ is linear charge density of line charge , r is distance of given point from line charge and ε₀ is a constant called permittivity and whose value is

8.85 x 10⁻¹².

Putting the given values in the equation given above

E = [tex]\frac{92\times10^{-6}}{2\times3.14\times8.85\times10^{-12}\times10^{-1}}[/tex]

E = 1.655 x 10⁷ N/C

Answer:

11.1 s

Explanation:  

Speed of the police car as given = v = 18 m/s

Speed of the car = V = 42 m/s

Reaction time = t = 0.8 s

Distance traveled by the police car during the reaction time = d₁= 0.8 x 18 = 14.4 m

Distance traveled by speeding car = d₂ =0.8 x 42 = 33.6 m

Acceleration of the police car = a = 5 m/s/s

The police car can catch the speeding car only if it travels a distance equal to the speeding car in a time t.

Distance traveled by the police car = D = d₁ + v t +0.5 at², according to the kinematic equation.

⇒ D = 14.4 + 18 t + 0.5 (5) t²

⇒ D = 14.4 + 18 t+2.5 t²  → (1)

For the speeding car, distance traveled is D = 33.6 + 42 t, since it is constant velocity. Substitute for D from the above equation (1).

⇒ 14.4 + 18 t+2.5 t²=  33.6 + 42 t

⇒ 2.5 t² -24 t - 19.2 = 0

⇒ t = 10.3 s

Total time = t +0.8 s

⇒ Time taken for the police car to reach the speeding car = 10.3+0.8= 11.1 s

Answer:

v =4.36 m/s

Explanation:

given,

mass of object A = 18.5 Kg

initial velocity of object A = 8.15 m/s in east

mass of object B = 30.5 kg

initial velocity of object B = 5 m/s

[tex]P = P_A+P_B[/tex]

[tex]P = m_Av_A\widehat{i} + m_B v_B\widehat{j}[/tex]

[tex]P = 18.5\times 8.15 \widehat{i} + 30.5\times 5\widehat{j}[/tex]

[tex]P = 150.775 \widehat{i} + 152.5 \widehat{j}[/tex]

[tex]P = \sqrt{150.775^2+152.5^2}[/tex]

P = 214. 45 N s

velocity after collision is equal to

[tex]v =\dfrac{214.45}{18.5+30.5}[/tex]

v =4.36 m/s

hence, velocity after collision is equal to 4.36 m/s

Answer:

The magnitude of the final velocity of the two-object system is [tex]v=4.37\frac{m}{s}[/tex]

Explanation:

As the Momentum is conserved, we can compare the instant before the collision, and the instant after. Also, we have to take in account the two components of the problem (x-direction and y-direction).

To do that, we put our 0 of coordinates where the collision takes place.

So, for the initial momentum we have that

[tex]p_{ix}=m_{a}v_{0a}+0[/tex]

[tex]p_{iy}=0+m_{b}v_{0b}[/tex]

Now, this is equal to the final momentum (in each coordinate)

[tex]p_{fx}=(m_{a}+m_{b}) v_{fx}[/tex]

[tex]p_{fy}=(m_{a}+m_{b}) v_{fy}[/tex]

So, we equalize each coordinate and get each final velocity

[tex]m_{a}v_{0a}=(m_{a}+m_{b}) v_{fx} \Leftrightarrow v_{fx}=\frac{m_{a}v_{0a}}{(m_{a}+m_{b})}[/tex]

[tex]m_{b}v_{0b}=(m_{a}+m_{b}) v_{fy} \Leftrightarrow v_{fy}=\frac{m_{b}v_{0b}}{(m_{a}+m_{b})}[/tex]

Finally, to calculate the magnitude of the final velocity, we need to calculate

[tex]v_{f}=\sqrt{(v_{fx})^{2}+(v_{fy})^{2}}[/tex]

which, replacing with the previous results, is

[tex]v_{f}=\sqrt{(v_{fx})^{2}+(v_{fy})^{2}}=(\sqrt{(\frac{18.5*8.15}{49})^{2}+(\frac{30.5*5.00}{49})^{2}})\frac{m}{s}[/tex]

Therefore, the outcome is

[tex]v_{f}=4.37\frac{m}{s}[/tex]

Answer:displacement =787.71 m

Explanation:

Given

Driver driver the car 690 ft to the east

then turn 380 ft to the north

(a)magnitude of acceleration is [tex]=\sqrt{380^2+690^2}=\sqrt{620500}[/tex]

displacement=787.71 m

(b)direction of displacement

[tex]tan\theta =\frac{380}{690}=0.5507[/tex]

[tex]\theta =28.84^{\circ}[/tex] with east direction

The magnitude of the displacement is approximately 775 ft and its direction is roughly 29 degrees north of east.

Your total displacement after driving a car 690 ft to the east and then 380 ft to the north is calculated using the Pythagorean theorem, which states that the magnitude of the hypotenuse (displacement) of a right triangle (formed by the eastwards and northwards journeys of the car) can be found by sqrt((eastwards travel)^2 + (northwards travel)^2). Applying the theorem, we obtain sqrt((690 ft)^2 + (380 ft)^2) = 775 ft approximately.

The direction of the displacement can be found using the tangent of the angle, which is the ratio of the opposite (northwards) to the adjacent side (eastwards). Applying the inverse tangent function, we get tan^-1(380/690), which gives us approximately 29 degrees. Therefore, the direction of the displacement is 29 degrees north of east.

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Answer:

a) 0.76 m

b) 1.53 s

c) 0.39 s

d) 3.8 m/s

Explanation:

This is a problem in which we need to use the equations pertaining Uniformly Accelerated Motion, as the acceleration during all this process is constant: The gravitational pull on the ball, 9.8 m/s².

To make things easier, we can divide this process in two parts: The first one (A) is from the moment the ball is thrown, until the moment it reaches it highest point and momentarily stops. The second one (B) from the moment it starts descending until it hits the ground.

a) During part A, we use the formula Vf²=Vi² +2*a*x . Where Vf is the final velocity (0 m/s, as the ball stopped in midair), Vi is the initial velocity (15 m/s), a is the acceleration (-9.8 m/s², it has a minus sign, as it goes against the direction of the movement) and x is the distance; thus, we're left with:

0 m/s=(15.0 m/s)²+2*(-9.8 m/s²)*x

We solve for x

x = 0.76 m

b) The formula is Vf=Vi +a*t, where t is the time. We're left with:

0 m/s=15.0 m/s + (-9.8 m/s²)*t

We solve for t

t= 1.53 s

c) Now we focus on part B, and use the formula x=Vi * t + [tex]\frac{at^{2}}{2}[/tex] , with the difference that the Vi is 0 m/s. We already know the value of x in exercise a). Note that a does not have a negative sign, as the direction of movement is opposite to the direction of part A

0.76 m=0 m/s * t +[tex]\frac{9.8\frac{m}{s^{2} } *t^{2} }{2}[/tex]

Solve for t

0.76=4.9t²

t=0.39 s

d) Once again we use the formula Vf=Vi +a*t, using the value of t previously calculated in exercise c).

Vf=0 m/s + 9.8 m/s² * 0,39 s

Vf=3.8 m/s

The ball rises to a height of 11.48 m, takes 1.53 s to reach its highest point, the same amount of time to fall back down, and has a velocity of -15.0 m/s when it returns to the starting level.

The question involves concepts from physics, specifically the kinematics of one-dimensional motion under constant acceleration due to gravity. Here is how we can solve each part of the given problem:

(a) How high does it rise?

To find the maximum height reached by the ball, we can use the kinematic equation that relates initial velocity, final velocity, acceleration, and displacement:

v² = u² + 2as

At the highest point, the final velocity (v) is 0 m/s, the initial velocity (u) is 15.0 m/s (upward), the acceleration due to gravity (a) is -9.8 m/s² (downward), and s represents the height. Solving for s:

0 = (15.0)² + 2(-9.8)s

s = (15.0)² / (2 * 9.8)

s = 11.48 m

(b) How long does it take to reach its highest point?

To find the time (t) it takes for the ball to reach its highest point, we can use the equation:

v = u + at

Since the final velocity at the highest point is 0 m/s:

0 = 15.0 + (-9.8)t

t = 15.0 / 9.8

t = 1.53 s

(c) How long does the ball take to hit the ground after it reaches its highest point?

The time for the ball to fall back down is the same as the time taken to reach the highest point, so this is also 1.53 s.

(d) What is its velocity when it returns to the level from which it started?

The velocity on returning to the starting level will be the same magnitude as the initial velocity but in the opposite direction, so it will be -15.0 m/s, with the negative sign indicating the downward direction.

Answer:

R = 3170.36m   or  R = 186.5m

Explanation:

For this problem, we have either trajectory (a), assuming that the car was going south-east, or trajectory (b), assuming the car was going north-east.

In both cases, we know that S = 830m = θ * R. Finding θ, will lead us to the value of R.

For option a:

θ = 15° = 0.2618 rad

[tex]R = \frac{S}{\theta} = 3170.36m[/tex]

For option b:

θ = 270° - 15° = 4.45 rad

[tex]R = \frac{S}{\theta} = 186.5m[/tex]

Answer:

The wavelength required is 102.9 nm.

Explanation:

The energy levels for the hydrogen atom are

[tex]E(n) = \frac{-13.6 \ eV}{n^2}[/tex]

So, for a transition from the first level to the third level we got

[tex]\Delta E = E(3) - E (1)[/tex]

[tex]\Delta E = \frac{-13.6 \ eV}{ 3 ^2} - \frac{-13.6 \ eV}{1^2}[/tex]

[tex]\Delta E = \frac{-13.6 \ eV}{ 9} - \frac{-13.6 \ eV}{1}[/tex]

[tex]\Delta E = \frac{8}{9} 13.6 \ eV[/tex]

[tex]\Delta E = 12.09 \ eV[/tex]

[tex]\Delta E = 12.09 \ eV * \frac{1.6 \ 10^-19 \ Joules}{1 \ eV}[/tex]

[tex]\Delta E = 1.93 \ 10^-18 \ Joules[/tex]

So we need a photon with this energy.

The energy of a photon its given by

[tex]E = h \nu = h \frac{c}{\lambda}[/tex]

So, the wavelength will be

[tex]\lambda = \frac{h c}{E}[/tex]

[tex]\lambda = \frac{6.62 \ 10^{-34} \ \frac{m^2 kg}{s} \ *  3.00 \ 10^8 \ \frac{m}{s}}{1.93 \ 10^-18 \ Joules}[/tex]

[tex]\lambda = 10.29 \ 10^{-8} m[/tex]

[tex]\lambda = 1.029 \ 10^{-7} m[/tex]

[tex]\lambda = 102.9 \ nm[/tex]

Answer:

V1=5ft3

V2=2ft3

n=1.377

Explanation:

PART A:

the volume of each state is obtained by multiplying the mass by the specific volume in each state

V=volume

v=especific volume

m=mass

V=mv

state 1

V1=m.v1

V1=4lb*1.25ft3/lb=5ft3

state 2

V2=m.v2

V2=4lb*0.5ft3/lb=   2ft3

PART B:

since the PV ^ n is constant we can equal the equations of state 1 and state 2

P1V1^n=P2V2^n

P1/P2=(V2/V1)^n

ln(P1/P2)=n . ln (V2/V1)

n=ln(P1/P2)/ ln (V2/V1)

n=ln(15/53)/ ln (2/5)

n=1.377

Explanation:

The position of a particle along x - axis is given by :

[tex]x=2.08+3.06t-1t^2[/tex]

(a) Position at t = 3.3 s

[tex]x=2.08+3.06(3.3)-1(3.3)^2[/tex]

x = 1.288 m

(b) Velocity at t = 3.3 s

[tex]v=\dfrac{dx}{dt}[/tex]

[tex]v=\dfrac{d(2.08+3.06t-1t^2)}{dt}[/tex]  

[tex]v=3.06-2t[/tex]

at t = 3.3 s

[tex]v=3.06-2(3.3)[/tex]

v = -3.54 m/s

(c) Acceleration,

[tex]a=\dfrac{dv}{dt}[/tex]

[tex]a=\dfrac{d(3.06-2t)}{dt}[/tex]  

[tex]a=-2\ m/s^2[/tex]

Hence, this is the required solution.

Final answer:

At t = 3.30 s, the particle is at the position of 1.288 meters, moves with a velocity of -3.54 m/s, and is subjected to a constant acceleration of -2 m/s^2.

Explanation:

The student's question about a particle's motion along the x-axis can be answered using kinematic equations from physics, which relate to the position, velocity, and acceleration of the particle.

Position at t = 3.30 s

To find the position of the particle at t = 3.30 s, we substitute the time into the position function x = 2.08 + 3.06t - 1.00t2.

2)>

Calculating this gives x(3.30 s) = 2.08 + 10.098 - 10.89 = 1.288 meters. The particle is at 1.288 meters from the origin at 3.30 seconds.

Velocity at t = 3.30 s

To find the velocity, we differentiate the position function with respect to time: v(t) = 3.06 - 2.00t.

The particle's velocity at 3.30 seconds is -3.54 meters per second.

Acceleration at t = 3.30 s

The acceleration is the second derivative of the position, which in this case is a constant because the t2 term has a constant coefficient: a(t) = -2.00 m/s2.

Therefore, the acceleration of the particle at any time, including at t = 3.30 s, is -2 meters per second squared.

Answer:

The arrow will hit 112.07 m from the point of release.

Explanation:

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

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

Where:

x0 = initial horizontal position

v0 = initial velocity

α = launching angle

y0 = initial vertical position

t = time

g = acceleration due to gravity

We know that at the final time the y-component of the vector "r" (see figure") is -1.88 m. The x-component of that vector will be the horizontal distance traveled by the arrow. Using the equation of the y-component of "r", we can obtain the final time and with that time we can calculate the value of the x-component (horizontal distance).

Then:

y = y0 + v0 · t · sin α + 1/2 · g · t²

Since the origin of the frame of reference is located at the point where the arrow is released, y0 = 0. Notice that the angle α = 26° + 15° = 41° ( see figure)

-1.88 m = 33 m/s · sin 41° · t - 1/2 · 9.8 m/s² · t²    (g is downward)

0 = -4.9 m/s² · t² + 33 m/s · sin 41° · t + 1.88 m

Solving the quadratic equation:

t = 4.5 s   ( the negative value is discarded)

Now, with this time we can calculate the horizontal distance:

x = x0 + v0 · t · cos α    (x0 = 0, the same as y0)

x = 33 m/s · 4.5 s · cos 41° = 112.07 m

The object comes to 112.07 m below its original point of release the arrow hit if it is shot with a speed of 33 m/s from a height of 1.88 m above the ground.

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

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

y = y₀ + v₀ t  sin α + 1/2 g  t²

Since the origin of the frame of reference is located at the point where the arrow is released, y₀ = 0. Notice that the angle α = 26° + 15° = 41° ( see figure)

-1.88 m = 33 m/s  sin 41°  t - 1/2  9.8 m/s²  t²  

0 = -4.9 m/s² · t² + 33 m/s · sin 41° · t + 1.88 m

Solving the quadratic equation:

t = 4.5 s  

Now, with this time we can calculate the horizontal distance:

x = x₀ + v₀  t cos α

x = 33 m/s · 4.5 s · cos 41° = 112.07 m

The object comes to 112.07 m below its original point of release the arrow hit if it is shot with a speed of 33 m/s from a height of 1.88 m above the ground.

To know more about the horizontal distance:

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Answer:

- 2.7 x 10^-6 J

Explanation:

q1 = 1 nC  at x = 0 cm

q2 = - 1 nC at x = 1 cm

q3 = 4 nC at x = 2 cm

The formula for the potential energy between the two charges is given by

[tex]U=\frac{Kq_{1}q_{2}}{r}[/tex]

where r be the distance between the two charges

By use of superposition principle, the total energy of the system is given by

[tex]U = U_{1,2}+U_{2,3}+U_{3,1}[/tex]

[tex]U=\frac{Kq_{1}q_{2}}{0.01}+\frac{Kq_{2}q_{3}}{0.01}+\frac{Kq_{3}q_{1}}{0.02}[/tex]

[tex]U=-\frac{9\times10^{9}\times 1\times10^{-9}\times 1\times10^{-9}}}{0.01}-\frac{9\times10^{9}\times 1\times10^{-9}\times 4\times10^{-9}}}{0.01}+-\frac{9\times10^{9}\times 1\times10^{-9}\times 4\times10^{-9}}}{0.02}[/tex]

U = - 2.7 x 10^-6 J

Answer:

a) 1511 MW

b) 44%

Explanation:

The thermal power will be the electric power plus the heat taken away by the cooling water.

Qt = P + Qc

The heat taken away by the water will be:

Qc = G * Cp * (t1 - t0)

The Cp of water is 4180 J/(kg K)

The density of water is 1 kg/L

Then

G = 1.17 * 10^8 L/h * 1 kg/L * 1/3600 h/s = 32500 kg/s

Now we calculate Qc

Qc = 32500 * 4180 * (29.8 - 23.6) = 842*10^6 W = 842 MW

The total thermal power then is

Qt = 669 + 842 = 1511 MW

The efficiency is

η = P / Qt

η = 669 / 1511 = 44%

Answer:

accelerate in the direction in which the electric field is pointing.

Explanation:

The positive charge feels a force in the same direction as the electric field

F=Eq  

F and E are vectors, q is a scalar

(if it were a negative charge the force would be in the opposite direction)

that force will produce an acceleration in the same direction, that acceleration will cause the particle to move in the same direction, ie the direction of the electric field.

A free positive charge in an electric field will accelerate in the direction in which the electric field is pointing due to electrostatic forces.

In the context of physics and electric fields, a free positive charge released in an electric field will accelerate in the direction in which the electric field is pointing. This is because electric field lines are drawn from positive to negative, indicating the direction that a positive test charge would move. Therefore, when a positive charge is placed in an electric field, it is repelled from the positive source charge and attracted to the negative source charge, causing it to accelerate along the electric field lines in the direction they are pointing.

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Answer:

Zero

Explanation:

Electric flux is defined as the number of electric field lines which passes through any area in the direction of area vector.

The formula for the electric flux is given by

[tex]\phi =\overrightarrow{E}.\overrightarrow{dS}[/tex]

Here, E is the strength of electric field and dS be the area vector.

It is a scalar quantity.

According to the Gauss's theorem, the electric flux passing through any surface is equal to the [tex]\frac{1}{\epsilon _{0}}[/tex] times the total charge enclosed through the surface.

Here, the charge enclose is zero, so the total flux is also zero.  

Answer:

a) 3.6 m/s

b) 1.53 m/s

c) 0.12 m

Explanation:

t = Time taken = 0.216 s

u = Initial velocity

v = Final velocity

s = Displacement = 0.54 m

a = Acceleration due to gravity = 9.81 m/s² (negative upward)

[tex]s=ut+\frac{1}{2}at^2\\\Rightarrow 0.54=u\times 0.216+\frac{1}{2}\times -9.81\times 0.216^2\\\Rightarrow u=\frac{0.54+\frac{1}{2}\times 9.81\times 0.216^2}{0.216}\\\Rightarrow u=3.6\ m/s[/tex]

Initial speed as it leaves the ground is 3.6 m/s

[tex]v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times -9.81\times 0.54+3.6^2}\\\Rightarrow v=1.53\ m/s[/tex]

Speed at the height of 0.540 m is 1.53 m/s

[tex]v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-3.6^2}{2\times -9.81}\\\Rightarrow s=0.66\ m[/tex]

The total height the armadillo leaps is 0.66 m

So, the additional height is 0.66-0.54 = 0.12 m

Answer:

[tex]t=6.96s[/tex]

Explanation:

From this exercise, our knowable variables are hight and initial velocity

[tex]v_{oy}=96ft/s[/tex]

[tex]y_{o}=112ft[/tex]

To find how much time does the ball strike the ground, we need to know that the final position of the ball is y=0ft

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

[tex]0=112ft+(96ft/s)t-\frac{1}{2}(32.2ft/s^{2})t^{2}[/tex]

Solving for t using quadratic formula

[tex]t=\frac{-b±\sqrt{b^{2}-4ac } }{2a}[/tex]

[tex]a=-\frac{1}{2} (32.2)\\b=96\\c=112[/tex]

[tex]t=-0.999s[/tex] or [tex]t=6.96s[/tex]

Since time can't be negative the answer is t=6.96s

Answer:

They're going to increase the total resistance as [tex]R_{T} = \sum\limits_{i=1}^N \left(\frac{1}{R_i} \right)^{-1}[/tex]

Explanation:

If the resistors are in parallel, the potential difference is the same for each resistor. But the total current is the sum of the currents that pass through each of the resistors. Then

[tex]I = I_1 + I_2 + ... + I_N[/tex]

where

[tex]I_i = \frac{V_i}{R_i}[/tex]

but

[tex]V_i = V_j = V[/tex] for [tex]i,j= 1, 2,..., N[/tex]

so

[tex]I = \frac{V}{R_1}+ \frac{V}{R_2} + ... + \frac{V}{R_N} = \left(\frac{1}{R_1} +\frac{1}{R_2} + ... + \frac{1}{R_N}\right)V = \frac{V}{R_T}[/tex]

where

[tex]R_T = \left(\frac{1}{R_1} +\frac{1}{R_2} + ... + \frac{1}{R_N}\right)^{-1} =\sum\limits_{i=1}^N \left(\frac{1}{R_i} \right)^{-1} [/tex]

Answer:

option B

Explanation:

the correct answer is option B

when negative charge is placed in electric field then direction will be accelerated in opposite direction of the field.

charged electric particle produce electric field and they exert force on other charged particle.

when a charged particle is positive it will accelerate in the direction of the electric field if the charge is negative then particle will accelerate in opposite direction of electric field.

Answer:

F=94.32*10⁻⁹N  , The force F is repusilve because both charges have the same sign (+)

Explanation:

Two point charges (q₁, q₂) separated by a distance (d) exert a mutual force (F) whose magnitude is determined by the following formula:  

F=K*q₁*q₂/d² Formula (1)  

F: Electric force in Newtons (N)

K : Coulomb constant in N*m²/C²

q₁,q₂:Charges in Coulombs (C)  

d: distance between the charges in meters(m)

Equivalence  

1nC= 10⁻⁹C

Data

K=8.99x10⁹N*m²/C²

q₁ = 7.94-nC= 7.94*10⁻⁹C

q₂= 4.14-nC=  4.14 *10⁻⁹C

d= 1.77 m

Magnitude of the electrostatic force that one charge exerts on the other

We apply formula (1):

[tex]F=8.99x10^{9} *\frac{7.94*10^{-9} *4.14 *10^{-9} }{1.77^{2} }[/tex]

F=94.32*10⁻⁹N  , The force F is repusilve because both charges have the same sign (+)

Answer:

-65 degC

Explanation:

ΔD=[tex]D_{0}[/tex]αΔT

1.869-1.871=1.871(1.2E-5)ΔT

ΔT≈-89 degC

24-89=-65 degC

Answer:

given,

speed of red train = 72 km/h = 72× 0.278 = 20 m/s

speed of green train = 144 km/h = 144 × 0.278 = 40 m/s

deceleration of both the train = 1 m/s²

distance between the train when they start decelerating = 950 m

using equation of motion

v²  = u² + 2 a s

distance taken by the  red train to stop

v²  = u² + 2 a s

0 = 20² - 2×1×s

s = 200 m

distance taken by the  blue train to stop

v²  = u² + 2 a s

0 = 40² - 2×1×s

s = 800 m

so, both train will be cover 200 + 800 = 1000 m

hence, there will be collision between both the trains.

distance traveled by the green train will be 750 m

distance traveled by the red train will be 200 m

so, velocity of the red train will be zero

velocity of the green train will be

v²  = u² + 2 a s

v²  = 40² - 2 × 1× 750

v =  10 m/s

hence, velocity of the green train will be 10 m/s.

Answer:

Explanation:

Given

ambient Pressure =98.10 kPa

(a)gauge pressure 152 kPa

we know

Absolute pressure=gauge pressure+Vacuum  Pressure

[tex]P_{abs}[/tex]=152+98.10=250.1 kPa or 36.27 psi

(b)[tex]P_{gauge}[/tex]=67.5 Torr or 8.99 kpa

as 1 Torr is 0.133 kPa

[tex]P_{abs}[/tex]=8.99+98.10=107.09 kPa or 15.53 psi

(c)[tex]P_{vaccum}[/tex]=0.1 bar or 10 kPa

Thus absolute pressure=98.10-10=88.10 kPa or 12.77 psi

as 1 kPa is equal to 0.145 psi

(d)[tex]P_{vaccum}[/tex]=0.84 atm  or 85.113 kPa

as 1 atm is equal to 101.325 kPa

[tex]P_{abs}[/tex]=98.10-85.11=12.99 kPa or 1.88 psi

Answer:

(a). The average speed is 97.23 km/hr.

(b). The average velocity is 81.70 km/min.

Explanation:

Given that,

Distance in east = 14.5 km

Distance in south = 66.5 km

Time = 50 min = 0.833 hr

(a). We need to calculate the average speed

Using formula of average speed

[tex]v_{avg}=\dfrac{D}{T}[/tex]

Where, D = total distance

T = total time

Put the value into the formula

[tex]v_{avg}=\dfrac{14.5+66.5}{0.833}[/tex]

[tex]v_{avg}=97.23\ Km/hr[/tex]

(b). We need to calculate the displacement

Using Pythagorean theorem

[tex]d=\sqrt{(d_{e})^2+(d_{s})^2}[/tex]

Put the value into the formula

[tex]d=\sqrt{(14.5)^2+(66.5)^2}[/tex]

[tex]d=68.06\ km[/tex]

(b). We need to calculate the average velocity

Using formula of average velocity

[tex]v_{avg}=\dfrac{d}{t}[/tex]

Where, d = displacement

t = time

Put the value into the formula

[tex]v_{avg}=\dfrac{68.06}{0.833}[/tex]

[tex]v_{avg}=81.70\ km/min[/tex]

Here, We can not necessary to find the unit vectors for displacement because we need to the displacement for find the average velocity.

Hence, (a). The average speed is 97.23 km/hr.

(b). The average velocity is 81.70 km/hr.

Answer:

4.44 kN in the opposite direction of acceleration.

Explanation:

Given that, the initial speed of the car is, [tex]u=20m/s[/tex]

And the mass of the car is, [tex]m=1000 kg[/tex]

The total distance covered by the car before stop, [tex]s=45m[/tex]

And the final speed of the car is, [tex]u=0m/s[/tex]

Now initial kinetic energy is,

[tex]KE_{i}=\frac{1}{2}mu^{2}[/tex]

Substitute the value of u and m in the above equation, we get

[tex]KE_{i}=\frac{1}{2}(1000kg)\times (20)^{2}\\KE_{i}=20000J[/tex]

Now final kinetic energy is,

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

Substitute the value of v and m in the above equation, we get

[tex]KE_{f}=\frac{1}{2}(1000kg)\times (0)^{2}\\KE_{i}=0J[/tex]

Now applying work energy theorem.

Work done= change in kinetic energy

Therefore,

[tex]F.S=KE_{f}-KE_{i}\\F\times 45=(0-200000)J\\F=\frac{-200000J}{45}\\ F=-4444.44N\\F=-4.44kN[/tex]

Here, the force is negative because the force and acceleration in the opposite direction.