A piece of purple plastic is charged with 8.45 × 10^6 extra electrons compared to its neutral state. What is its net electric charge (including its sign) in coulombs.

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:

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:

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:

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.

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:

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:

(a) 34.47 cm

(b) [tex]24.09^\circ[/tex] south of west

Explanation:

Let us draw a figure representing the individual displacement vectors in the four jumps as shown in the figure attached with this solution.

Now, let us try to write the four displacement vectors in in terms of unit vectors along the horizontal and the vertical axis.

[tex]\vec{d}_1= 31\ cm\ west = -31\ cm\ \hat{i}\\\vec{d}_2= 26\ cm\ south\ of\ west = -26\cos 44^\circ\ \hat{i} -26 \sin 44^\circ\ \hat{j}=(-18.72\ \hat{i}-18.06\ \hat{i})\ cm\\\vec{d}_3= 22\ cm\ south\ of\ east = 22\cos 56^\circ\ \hat{i} -22 \sin 56^\circ\ \hat{j}=(12.30\ \hat{i}-18.23\ \hat{i})\ cm\\\vec{d}_4= 23\ cm\ north\ of\ east = 23\cos 75^\circ\ \hat{i} +23\sin \sin 75^\circ\ \hat{j}=(5.95\ \hat{i}+22.22\ \hat{i})\ cm\\[/tex]

Now, the vector sum of all these vector will give the resultant displacement vector.

[tex]\vec{D} = \vec{d}_1+\vec{d}_2+\vec{d}_3+\vec{d}_4\\\Rightarrow \vec{D} = -31\ cm\ \hat{i}+(-18.72\ \hat{i}-18.06\ \hat{i})\ cm+(12.30\ \hat{i}-18.23\ \hat{i})\ cm+(5.95\ \hat{i}+22.22\ \hat{i})\ cm\\\Rightarrow \vec{D} =(-31.47\ \hat{i}-14.07\ \hat{i})\ cm[/tex]

Part (a):

The magnitude of the resultant displacement vector is given by:

[tex]D=\sqrt{(-31.47)^2+(-14.07)^2}\ m = 34.47\ m[/tex]

Part (b):

Since the resultant displacement vector indicates that the final position of the vector lies in the third quadrant, the vector will make some positive angle in the direction south of west given by:

[tex]\theta = \tan^{-1}(\dfrac{14.07}{31.47})= 24.09^\circ[/tex]

To find the resultant displacement of the grasshopper, we can break down the vectors into their x and y components, and then sum up the components separately. After performing the calculations, we find that the magnitude of the resultant displacement is approximately 39.4 cm and the direction is approximately 38.3° south of west.

To find the resultant displacement of the grasshopper, we need to add the individual displacement vectors. We can do this by breaking down each vector into its x and y components.

For vector (1) with a magnitude of 31.0 cm due west, the x component is -31.0 cm and the y component is 0.

Similarly, for the other vectors, the x and y components are:

Now, we can sum up the x components and y components separately to find the resultant displacement.

The magnitude of the resultant displacement can be found using the formula:

resultant magnitude = sqrt((sum of x components)^2 + (sum of y components)^2)

The direction of the resultant displacement can be found using the formula:

resultant direction = atan2((sum of y components), (sum of x components))

Plugging in the values and performing the calculations, we find that the magnitude of the resultant displacement is approximately 39.4 cm and the direction of the resultant displacement is approximately 38.3° south of west.

Answer:

v = 0.85 m/s

Explanation:

Given that,

Mass of the ball, m = 0.01 kg

Centripetal force on the ball, F = 0.025 N

Radius of the circular path, r = 0.29 m

Let v is the speed of the ball. The centripetal force of the ball is given by :

[tex]F=\dfrac{mv^2}{r}[/tex]

[tex]v=\sqrt{\dfrac{Fr}{m}}[/tex]

[tex]v=\sqrt{\dfrac{0.025\times 0.29}{0.01}}[/tex]

v = 0.85 m/s

So, the speed of the ball is 0.85 m/s. Hence, this is the required solution.

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

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

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:

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.

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:

The ball never passes 2m high, Hmax=1.27m

Explanation:

we assume the ball doesn't bounce when it hits the ground.

We calculate the maximum height, Vf = 0.

[tex]v_{o}^{2}=2gH_{max}\\H_{max}=v_{o}^{2}/(2g)=5^{2}/(2*9.81)=1.27m[/tex]

So, the ball never passes 2m high.

Kinematics equations:

[tex]x(t)=v_{o}t-1/2*g*t^{2}\\v(t)=v_{o}-gt[/tex]

Find annexed the graphics of x(t) and v(t)

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: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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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:

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: 0.95 ft/year

Explanation: In order to explain this question we have to convert the units so

if we have a rate equal to 0.033 mm/hr then 0.033 mm *24*365 hr/year

then 1 m=3.28 feet

8760*0.033 *10^-3m* 3.28 feet/m=1.08*10^-4 feet*8760=0.95 feet/year

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:

r=2.743

Explanation:

The energy stored on a capacitor is of type potencial, therfore depends on the capacity to "store" energy. Inthe case of the capacitor, it stores charge (Q), and the equations you use to calculate it are:

[tex]E_p=\frac{Q^2}{2C}=\frac{QV}{2}=\frac{CV^2}{2}[/tex]

In this case we know V and C, therefore we use the last expression:

[tex]E_{p1}=\frac{CV_1^2}{2}[/tex]

[tex]E_{p2}=\frac{CV_2^2}{2}[/tex]

[tex]\frac{E_{p1}}{E_{p2}}=r=\frac{\frac{CV_1^2}{2}}{\frac{CV_2^2}{2}}  \\r=(\frac{V_1}{V_2})^2\\r=(\frac{21.2}{12.8})^2[/tex]

r=2.743

Final answer:

The mass of the object is 1.53 kg.

Explanation:

To find the mass of the object, we need to use Hooke's Law which states that the force exerted by a spring is directly proportional to its extension. In this case, the spring stretches 0.2 cm per newton of applied force. The deflection of 3 cm corresponds to an applied force of 15 newtons (0.2 cm per newton * 3 cm).

Using the equation F = mg, where F is the force, m is the mass, and g is the acceleration due to gravity (9.81 m/s^2), we can find the mass:

15 newtons = m * 9.81 m/s^2

m = 15 newtons / 9.81 m/s^2 = 1.53 kg

Therefore, the mass of the object is approximately 1.53 kg.

The mass of the object is approximately 1.53 kg.

Given that the spring stretches 0.2 cm per newton of applied force. Thus, we can say that

for F = - kx.

1 N = - k (0.2 cm)

or, k = spring constant of the given spring = [tex]\frac{F}{x}[/tex] = 5 N/cm

Now, for an object producing deflection of 3 cm, we can say that:

F = - k x = 5 N/cm × 3 cm

or, F = 15 N

This concludes that the weight of the object is 15 N.

Now, W = F = mg

hence, [tex]m = \frac{F}{g}[/tex]

or, m = [tex]\frac{15 \hspace{0.6mm} N}{9.8 \hspace{0.5mm} m/s^2}[/tex]

or, m ≈ 1.53 kg

Answer:

Second ball

Explanation:

When a ball is thrown up with a certain velocity when the object reaches the same point from where it was thrown the velocity of the object becomes equal to the velocity with which the ball was thrown.

First ball

[tex]v_g_1^2-u^2=2as\\\Rightarrow v_g_1=\sqrt{2as+u^2}\\\Rightarrow v_g_1=\sqrt{2as+v^2}[/tex]

Second ball

[tex]v_g_2^2-u^2=2as\\\Rightarrow v_g_2=\sqrt{2as+u^2}\\\Rightarrow v_g_2=\sqrt{2as+4v^2}[/tex]

Third ball

[tex]v_g_3^2-u^2=2as\\\Rightarrow v_g_3=\sqrt{2as+0^2}\\\Rightarrow v_g_3=\sqrt{2as}[/tex]

From the equations above it can be seen that the second ball will have the highest velocity when it hits the ground.

So, [tex]v_g_3<v_g_1<v_g_2[/tex]

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:

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:

946.92 kJ

Explanation:

This process has 3 parts:

1. The first part, where the temperature of Ethyl alcohol remains constant and it changes from gas to liquid.

2. The second part, where the temperature drops from 78°C to -114°C

3. The third parts, where the temperature remains constant and it changes from liquid to solid.

The energy lost in a phase change is:

Q = m*cl

The energy lost because of the drop in temperature is:

[tex]Q = m c(T_2-T_1)[/tex]

cl is the heat of vaporization or heat of fusion, depending on the type of phase change. c is the specific heat.

So, the energy lost in each part is:

1. [tex]Q_1 = 0.651kg*879 kJ/kg =  572.23 kJ[/tex]

2. [tex]Q_2 = 0.651kg*2.43 kJ/kgK(78.0^oC - (-114^oC)) = 303.73 kJ[/tex]

3. [tex]Q_3 = 0.651kg*109kJ/kg = 70.96 kJ[/tex]

Then, the total energy removed should be:

Q = Q1 + Q2 + Q3 = 572.23 kJ + 303.73kJ + 70.96kJ = 946.92 kJ

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

Answer:

Force between two spheres will be 14400 N

And as the both charges of opposite nature so force will be attractive

Explanation:

We have given two conducting spheres of charges [tex]q_1=-20\mu C=-20\times 10^{-6}C\ and\ q_2=50\mu C=50\times 10^{-6}C[/tex]

Distance between the spheres = 2.5 cm =0.025 m

According to coulombs law we know that force between two charges is given by [tex]F=\frac{1}{4\pi \varepsilon _0}\frac{q_1q_2}{r^2}=\frac{Kq_1q_2}{r^2}=\frac{9\times 10^9\times 20\times 10^{-6}\times 50\times 10^{-6}}{0.025^2}=14400N[/tex]  

As the both charges of opposite nature so force will be attractive

Answer:

The relative density of the second liquid is 7.

Explanation:

From archimede's principle we know that the force that a liquid exerts on a object equals to the weight of the liquid that the object displaces.

Let us assume that the volume of the object is 'V'

Thus for the liquid in which the block is completely submerged

The buoyant force should be equal to weight of liquid

Mathematically

[tex]F_{buoyant}=Weight\\\\\rho _{1}\times V\times g=m\times g\\\\\therefore \rho _{1}=\frac{m}{V}...............(i)[/tex]

Thus for the liquid in which the block is 1/7 submerged

The buoyant force should be equal to weight of liquid

Mathematically

[tex]F'_{buoyant}=Weight\\\\\rho _{2}\times \frac{V}{7}\times g=m\times g\\\\\therefore \rho _{2}=\frac{7m}{V}.................(ii)[/tex]

Comparing equation 'i' and 'ii' we see that

[tex]\rho_{2}=7\times \rho _{1}[/tex]

Since the first liquid is water thus [tex]\rho _{1}=1gm/cm^3[/tex]

Thus the relative density of the second liquid is 7.

Answer:

7

Explanation:

Let the density of second liquid is d.

Density of water = 1 g/cm^3

In case of equilibrium, according to the principle of flotation, the weight of the body is balanced by the buoyant force acting on the body.

Let A be the area of cross section of block and D be the density of material of block and h be the height.

For first liquid (water):

Weight of block = m x g  = A x h x D x g .... (1)

Buoyant force in water = A x h x 1 x g  ..... (2)

Equating (1) and (2) we get

A x h x D x g = A x h x 1 x g

D = 1 g/cm^3

For second liquid:

Weight of block = m x g  = A x h x D x g .... (1)

Buoyant force in second liquid = A x h/7 x d x g  ..... (2)

Equating (1) and (2) we get

A x h x D x g = A x h/7 x d x g

D = d/7

d = 7 D = 7 x 1 = 7 g/cm^3

Thus, the relative density of second liquid is 7.

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]