When a car is on an inclined bank of angle θ and rounding a curve with no friction, what is the centripetal force equal to? a. The weight of the car b. N cos (θ) c. N sin (θ) d. Zero

Answer:

distance between both oasis ( 1 and 2) is  27.83 Km

Explanation:

let d is the distance between oasis1 and oasis 2

from figure

OC  = 25cos 30

OE = 25sin30

OE = CD

Therefore BC =  30-25sin30

distance between both oasis ( 1 and 2) is calculated by using phytogoras theorem

in[tex]\Delta BCO[/tex]

[tex]OB^2 = BC^2 + OC^2[/tex]

PUTTING ALL VALUE IN ABOVE EQUATION

[tex]d^2 = 930-25sin30)^2 + (25cos30)^2[/tex]

[tex]d^2 = 775[/tex]

d = 27.83 Km

distance between both oasis ( 1 and 2) is  27.83 Km

Using vector addition and the Pythagorean theorem, the distance separating the two oases is found to be approximately 27.95 km after the camel walks a two-leg journey with specified directions and distances.

To determine the distance that separates the two oases, we can use vector addition and Pythagorean theorem. The camel walks 25 km in a direction 30° south of west, which can be represented as a vector with components to the south and to the west.

Then, the camel walks 30 km towards the north. In terms of vectors, these two displacements will partly cancel each other out in the north-south direction.

First, let's resolve the initial 25 km walk into components. The southern component is 25 km * sin(30°) = 12.5 km, and the western component is 25 km * cos(30°) = 21.65 km. After the camel walks 30 km north, the remaining southward component is 30 km - 12.5 km = 17.5 km north. The westward component remains unchanged at 21.65km.

Now, we can use the Pythagorean theorem to find the resultant distance between the two oases, which is the hypotenuse of the right triangle formed by the northward and westward components.

The distance is √(17.5 km² + 21.65 km²), which is approximately 27.95 km. Therefore, the distance that separates the two oases is around 27.95 km.

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Answer:

The acceleration of the car is [tex]20 m/s^{2}[/tex]

Solution:

According to the question:

Initial velocity of the car, u = 12 m/s

Final velocity of the car, v = 52 m/s

Time interval for the change in velocity, [tex]\DeltaT = 2 s[/tex]

The rate at which the velocity of an object varies is referred to as its acceleration:

[tex]a = \frac{v - u}{\Delta T}[/tex]

[tex]a = \frac{52 - 12}{2} = 20 m/s^{2}[/tex]

Answer:B

Explanation:

Given

First motor has three times as much as power as another

Let the power of smaller motor be P

Therefore bigger motor has power 3 P

and we know that energy [tex]E=power\times time[/tex]

Larger motor [tex]E=3P\times t=3Pt[/tex]

Smaller motor [tex]E=P\times 3t=3Pt[/tex]

Therefore smaller motor will do same work if it takes three times the time taken by Larger motor

Answer:

The inside Pressure of the tank is [tex]4499.12 lb/ft^{2}[/tex]

Solution:

As per the question:

Volume of tank, [tex]V = 0.25 ft^{3}[/tex]

The capacity of tank, [tex]V' = 50ft^{3}[/tex]

Temperature, T' = [tex]80^{\circ}F[/tex] = 299.8 K

Temperature, T = [tex]59^{\circ}F[/tex] = 288.2 K

Now, from the eqn:

PV = nRT                      (1)

Volume of the gas in the container is constant.

V = V'

Similarly,

P'V' = n'RT'                       (2)

Also,

The amount of gas is double of the first case in the cylinder then:

n' = 2n

[tex]\]frac{n'}{n} = 2[/tex]

where

n and n' are the no. of moles

Now, from eqn (1) and (2):

[tex]\frac{PV}{P'V'} = \frac{nRT}{n'RT'}[/tex]

[tex]P' = 2P\frac{T'}{T} = 2\times 2116\times \frac{299.8}{288.2} = 4499.12 lb/ft^{2}[/tex]

Answer:

they meet from point o at distance 50.46 m and time taken is 11.6 seconds

Explanation:

given data

acceleration = 0.75 m/s²

speed B = 6 m/s

time B = 20 s

to find out

when and where the vehicles passed each other

solution

we consider here distance = x , when they meet after o point

and time = t for meet point z

we find first Bus B distance for 20 s ec

distance B = velocity × time

distance B = 6 × 20

distance B = 120 m

so

B take time to meet is calculate by distance formula

distance = velocity × time

120 - x = 6 × t

x = 120 - 6t   .................1

and

distance of A when they meet by distance formula

distance = ut + 1/2 × at²

here u is initial speed = 0 and t is time

x = 0 + 1/2 × 0.75 × t²

x = 0.375 × t²      .............2

so from equation 1 and 2

0.375 × t²  = 120 - 6t

t = 11.6

so time is 11.6 second

and

distance from point o from equation 2

x = 0.375 (11.6)²

x = 50.46

so distance from point o is 50.46 m

Automobile A and bus B pass each other approximately 11.6 seconds after automobile A started, at a position 50.5 meters from point O. This is determined by setting their displacement equations equal and solving the quadratic equation. The final position of the meeting point is found to be around 50.5 meters from point O.

To solve this problem, we need to determine the position and time at which automobile A and bus B pass each other.

Determine the position of automobile A:

Determine the position of bus B:

Set the positions equal to find when they pass each other:

Multiplying through by 8 to clear the decimal:

We have two potential solutions: t = (69.62 / 6) ≈ 11.6 seconds (since a negative time is not meaningful in this context)

Determine the meeting point:

Answer:

3.83 s

Explanation:

Initially for 50 m A falls with acceleration equal to g ie 9.8 m /s².In this journey his initial velocity u = 0 , a = 9.8 , v = ? , t = ?

h = ut + 1/ 2 g t²

50 = .5 x 9.8x t²

t = 3.19 s

v = u +gt

= o + 9.8 x 3.19

= 31.26 m /s

For motion under deceleration

initial speed  u = 31.26 m/s

Final speed v = 14 m/s

deceleration a = - 27 m/s²

v = u - at

14 = 31.26 - 27 t

t = 0.64 s

So total time in the air

= 3.19 + .64 = 3.83 s

Explanation:

Given that,

Number density [tex]n= 1\ atom/cm^{3} =10^{6}\ atom/m^3[/tex]

Temperature = 2.7 K

(a). We need to calculate the pressure in interstellar space

Using ideal gas equation

[tex]PV=nRT[/tex]

[tex]P=\dfrac{nRT}{V}[/tex]

[tex]P=\dfrac{10^{6}\times8.314\times2.7}{6.023\times10^{23}}[/tex]

[tex]P=3.727\times10^{-17}\ Pa[/tex]

[tex]P=36.78\times10^{-23}\ atm[/tex]

The pressure in interstellar space is [tex]36.78\times10^{-23}\ atm[/tex]

(b). We need to calculate the root-mean square speed of the atom

Using formula of rms

[tex]v_{rms}=\sqrt{\dfrac{3RT}{Nm}}[/tex]

Put the value into the formula

[tex]v_{rms}=\sqrt{\dfrac{3\times8.314\times2.7}{1.007\times10^{-3}}}[/tex]

[tex]v_{rms}=258.6\ m/s[/tex]

The root-mean square speed of the atom is 258.6 m/s.

(c). We need to calculate the  kinetic energy

Average kinetic energy of atom

[tex]E=\dfrac{3}{2}kT[/tex]

Where, k = Boltzmann constant

Put the value into the formula

[tex]E=\dfrac{3}{2}\times1.38\times10^{-23}\times2.7[/tex]

[tex]E=5.58\times10^{-23}\ J[/tex]

The kinetic energy stored in 1 km³ of space is [tex]5.58\times10^{-23}\ J[/tex].

Hence, This is the required solution.

Answer:

[tex]8.66\times 10^{-6}\ C[/tex] or [tex]8.66\ \mu C[/tex].

Explanation:

Given:

For the moving particle to circular motion, the electrostatic force between the two must be balanced by the centripetal force on the moving particle.

The electrostatic force on the moving particle due to the charge Q at origin is given by Coulomb's law as:

[tex]\rm F_e = \dfrac{kqQ}{r^2}.[/tex]

where, [tex]\rm k[/tex] is the Coulomb's constant having value [tex]\rm 9\times 10^9\ Nm^2/C^2.[/tex]

The centripetal force on the moving particle due to particle at origin is given as:

[tex]\rm F_c = \dfrac{mv^2}{r}.[/tex]

For the two forces to be balanced,

[tex]\rm F_e = F_c\\\dfrac{kqQ}{r^2}=\dfrac{mv^2}{r}\\\Rightarrow Q = \dfrac{mv^2}{r}\times \dfrac{r^2}{kq}\\=\dfrac{mv^2r}{kq}\\=\dfrac{(0.959\times 10^{-3})\times (47.9)^2\times (0.207)}{(9\times 10^9)\times (5.84\times 10^{-6})}\\=8.66\times 10^{-6}\ C\\=8.66\ \mu C.[/tex]

For the moving particle to execute circular motion, the charge Q must be approximately 6.75 µC and have the same sign as the particle's charge (5.84 µC) to ensure attraction and centripetal motion.

To determine the value of charge Q that will allow the moving particle to execute circular motion, we must consider the forces acting on the charged particle due to the electric field created by Q.

For the charged particle to execute circular motion, the centripetal force required to keep it in circular motion must equal the electric force acting on it due to charge Q.

Centripetal Force:
The centripetal force (
F_c) is given by:
[tex]F_c = \frac{mv^2}{r}[/tex]
where r is the radius of the circular motion. Here, r = 0.207 m (distance from the origin).

Substituting the given values, we get:
[tex]F_c = \frac{0.000959 \text{ kg} \times (47.9 ext{ m/s})^2}{0.207 ext{ m}}[/tex]
[tex]F_c = \frac{0.000959 \times 2299.61}{0.207} \approx 0.008291 \text{ N}[/tex]

Electric Force:
The electric force (F_e) on the particle due to the charge Q at the origin is given by Coulomb's Law:
[tex]F_e = \frac{k |Q| |q|}{r^2}[/tex]
where k is Coulomb's constant [tex]k \approx 8.99 \times 10^9 \text{ N m}^2/ ext{C}^2[/tex].

Setting the electric force equal to the centripetal force for circular motion:
[tex]\frac{k |Q| |q|}{r^2} = F_c[/tex]
Plugging in the numbers:
[tex]\frac{(8.99 \times 10^9) |Q| (5.84 \times 10^{-6})}{(0.207)^2} = 0.008291[/tex]

Solving for Q:
Rearranging for Q:
[tex]|Q| = \frac{0.008291 \cdot (0.207)^2}{8.99 \times 10^9 imes (5.84 \times 10^{-6})}[/tex]
Calculate the right side:
[tex]|Q| = \frac{0.008291 \cdot 0.042849}{8.99 \times 10^9 imes 5.84 \times 10^{-6}}[/tex]
[tex]|Q| \approx \frac{0.000355}{0.05256} \approx 6.75 \times 10^{-6} \text{ C}[/tex] or 6.75 µC.

Answer:

440 N

Explanation:

Force applied, F = 440 N

mass of crate, m = 170 kg

μs = 0.5, μk = 0.2

g = 9.81 m/s^2

The normal reaction acting on the crate, N = m g = 170 x 9.81 = 1667.7 N

The maximum value of static friction force acting on the crate

[tex]f_{s}=\mu _{s}N=0.5 \times 1667.7 = 833.85 N[/tex]

The maximum value of static friction force is more than the applied force so the crate does not move and teh applied force becomes friction force.

thus, the friction force acting on the crate is 440 N.

The frictional force exerted by the carpet on the crate is equal to the force of static friction.

To find the frictional force exerted by the carpet on the crate, we need to calculate the force of static friction first. The maximum force of static friction is given by fs(max) = μsN, where μs is the coefficient of static friction and N is the normal force. The normal force is equal to the weight of the crate, which is given by N = mg. In this case, the normal force N = (170 kg)(9.81 m/s^2) = 1666.7 N.

Therefore, the maximum force of static friction is fs(max) = (0.5)(1666.7 N) = 833.35 N. Since the worker is pushing with a force of 440 N, which is less than the maximum force of static friction, the crate remains at rest and the frictional force exerted by the carpet on the crate is equal to the force of static friction, which is 833.35 N.

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Answer:

The power is 0.73 kW.

(B) option is correct.

Explanation:

Given that,

Charge = 9000 C

Potential difference = 220 V

Time = 45 min

We need to calculate the energy required to transfer charge Q through V

Using formula of energy

[tex]E =QV[/tex]

Put the value into the formula

[tex]E=9000\times220[/tex]

[tex]E=1980000\ J[/tex]

We need to calculate the power

Using formula of power

[tex]P=\dfrac{1980000}{45\times60}[/tex]

[tex]P=733.33\ W[/tex]

[tex]P=0.73\ kW[/tex]

Hence, The power is 0.73 kW.

Answer:

(a): [tex]\rm 1.133\ eV\ \ \ or\ \ \ 1.8128\times 10^{-19}\ J.[/tex]

(b): [tex]\rm 1.298\ eV \ \ \ or \ \ \ 2.077\times 10^{-19}\ J.[/tex]

Explanation:

The energy of the photon that absorbed by a hydrogen atom causes a transition is equal to the difference in energy levels of the hydrogen atom corresponding to that transition.

According to Rydberg's formula, the energy corresponding to [tex]\rm n^{th}[/tex] level in hydrogen atom is given by

[tex]\rm E_n = -\dfrac{E_o}{n^2}.[/tex]

where,

[tex]\rm E_o=13.6\ eV.[/tex]

Part (a): For the electronic transition from the n = 3 state to the n = 6 state.

The energy of the photon which cause this transition is given by

[tex]\rm \Delta E=E_6-E_3.\\\\where,\\E_6=-\dfrac{E_o}{6^2}=-\dfrac{13.6}{36}=-0.378\ eV.\\E_3=-\dfrac{E_o}{3^2}=-\dfrac{13.6}{9}=-1.511\ eV.\\\\\therefore \Delta E = (-0.378)-(-1.511)=1.133\ eV\\or\ \ \ \Delta E = 1.133\times 1.6\times 10^{-19}\ J=1.8128\times 10^{-19}\ J.[/tex]

Part (b): For the electronic transition from the n = 3 state to the n = 8 state.

The energy of the photon which cause this transition is given by

[tex]\rm \Delta E=E_8-E_3.\\\\where,\\E_8=-\dfrac{E_o}{8^2}=-\dfrac{13.6}{64}=-0.2125\ eV.\\E_3=-\dfrac{E_o}{3^2}=-\dfrac{13.6}{9}=-1.511\ eV.\\\\\therefore \Delta E = (-02125)-(-1.511)=1.298\ eV\\or\ \ \ \Delta E = 1.298\times 1.6\times 10^{-19}\ J=2.077\times 10^{-19}\ J.[/tex]

Final answer:

The energy of the photon that can cause an electronic transition in a hydrogen atom from n = 3 to n = 6 or n = 8 is -3.04 x 10^-19 J. The frequency of the photon is -4.60 x 10^14 Hz.

Explanation:

To calculate the energy of a photon that can cause an electronic transition in a hydrogen atom, we can use the equation:

E = Ef - Ei

where Ef is the energy of the final state and Ei is the energy of the initial state. The energy of a photon is given by:

E = hf

where h is Planck's constant (6.626 x 10-34 J·s) and f is the frequency of the photon. By substituting these equations, we can determine the frequency and energy of the photon.

(a) For an electronic transition from the n = 3 state to the n = 6 state, we have:

E = E6 - E3

E = (-3.4 eV) - (-1.5 eV)

E = -1.9 eV

Using the conversion factor 1 eV = 1.6 x 10-19 J, we can convert the energy to joules:

E = -1.9 eV x (1.6 x 10-19 J/eV)

E = -3.04 x 10-19 J

Now, we can find the frequency of the photon by rearranging the equation:

f = E/h

f = (-3.04 x 10-19 J) / (6.626 x 10-34 J·s)

f = -4.60 x 1014 Hz

(b) For an electronic transition from the n = 3 state to the n = 8 state, we follow the same steps:

E = E8 - E3

E = (-3.4 eV) - (-1.5 eV)

E = -1.9 eV

E = -3.04 x 10-19 J

f = E/h

f = (-3.04 x 10-19 J) / (6.626 x 10-34 J·s)

f = -4.60 x 1014 Hz

Therefore, the energy of the photon that can cause the electronic transition from the n = 3 state to the n = 6 state or the n = 8 state is -3.04 x 10-19 J and the frequency is -4.60 x 1014 Hz.

Answer:

Distance from the net when she meets the return: 14,3 m

Explanation:

First we need to know how much time it takes the ball to reach the net, the kinematycs general equation for position:

(1) [tex]x = x_{0}  + V_{s}  t + \frac{1}{2}  a t^{2}[/tex]

Taking the net as the origin (x = 0), [tex]x_{0} = 25m[/tex], velocity will be nagative [tex]V_{s} = - 50m/s[/tex] and assuming there is no friction wit air acceleration would be 0, so:

(2) [tex]x = x_{0}  + V_{s}  t [/tex]

we want to know the time when it reaches the net so when x=0, replacing de values:

(3) [tex]0 = 25m  - 50 m/s t_{net} [/tex]

So:  [tex]t_{net}  = 0,5 s [/tex]

The opponent will return the ball at [tex]V_{ret} = 25m/s[/tex], the equation for the return of the ball will be:

(4) [tex]y = y_{0}  + V_{ret}  t + \frac{1}{2}  a t^{2}[/tex]

Note that here it start from the origin, [tex]y_{0} = 0[/tex], as in the other case acceleration equals 0, and here we have to consider that the time starts when the ball reaches the net ([tex]t_{net} [/tex]) so the time for this equiation will be [tex]t - t_{net} [/tex], this is only valid for [tex]t >= t_{net} [/tex]:

(5) [tex]y = V_{ret}  (t - t_{net}) [/tex]

Coco starts running as soon as he serves so his equiation for position will be:

(6) [tex]z = z_{0}  + V_{c}  t + \frac{1}{2}  a t^{2}[/tex]

As in the first case it starts from 25m, [tex]z_{0} = 25m[/tex], acceleration equals 0 and velocity is negative [tex]V_{c} = - 10m/s[/tex]:

(7) [tex]z = z_{0} + V_{c}  t[/tex]

To get the time when they meet we have that [tex]z = y[/tex], so from equiations (5) and (7):

[tex]V_{ret}  (t - t_{net}) =  z_{0} + V_{c}  t[/tex]

[tex]t = \frac{z_{0} + V_{ret}* t_{net}}{V_{ret}-V_{c}}[/tex]

Replacing the values:

[tex]t = 1,071 s[/tex]

Replacing t in either (5) or (7):

[tex]z = y = 14,3 m[/tex]

This is the distance to the net when she meets the return

Answer:

[tex]\theta = 58.98[/tex]°

Explanation:

given data:

h = 18 ft = 5.48 m

from figure

[tex]h_{max} = 5.48 - 0.50 = 4.98 m[/tex]

[tex]h_{max} = \frac{ v_y^2}{2g}[/tex]

[tex]v_y =\sqrt{2gh_{max}[/tex]

[tex]v_y = \sqrt{2*9.8* 4.98} m/s[/tex]

[tex]v_y = 9.88 m/s[/tex]

[tex]t = \frac{v_y}{g} =\frac{9.88}{9.8} = 1.01 s[/tex]

[tex]v_x = \frac{d}{t} = \frac{6}{1.01}  = 5.49 m/s

[tex]v = \sqrt{v_x^2+v_y^2} = \sqrt{5.95^2+9.88^2}[/tex]

v = 11.53 m/s

[tex]tan\theta = \frac[v_x}{v_y}[/tex]

[tex]\theta = tan^{-1} \frac{9.88}{5.94}[/tex]

[tex]\theta = 58.98[/tex]°

Answer:

a) 49500 Kg.m/s

b)  1.65 m/s

Explanation:

Given:

Mass of the car, m₁ = 11000 kg

Mass of the second car, m₂ = 19000 kg

Initial Speed of the first car, u₁ = 4.5 m/s

Initial velocity of the second car , u₂ = 0

Now,

Momentum = Mass × Velocity

Initial momentum = m₁u₁

Thus

Initial momentum P₁ = 11000 × 4.5 = 49500 kg-m/sec

b)By using the concept of momentum conservation

Initial momentum = Final momentum

m₁u₁ + m₂u₂ = ( m₁ + m₂ )v

Where, v is the velocity after collision

thus,

49500 + 19000 × 0 = ( 11000 + 19000 ) × v

or

V = [tex]\frac{\textup{49500}}{\textup{30000}}[/tex]

or

v = 1.65 m/s

a) The initial momentum of the first car is 49,500 kg·m/s. b) The final velocity of the coupled cars is 1.65 m/s.

The question involves a physics concept called the conservation of momentum, which is particularly applicable to collisions, such as the one described between two railroad cars.

a) The initial momentum (p) of the first car can be calculated using the formula p = mv, where m is the mass and v is the velocity of the car. For the first car with a mass m of 11,000 kg moving with a velocity v of 4.5 m/s, the initial momentum is: p = 11,000 kg × 4.5 m/s = 49,500 kg·m/s.

b) For the final velocity, we apply the principle of conservation of linear momentum. As external forces are ignored, the total momentum before the collision is equal to the total momentum after the collision. The combined mass of the coupled cars is 11,000 kg + 19,000 kg = 30,000 kg. The final velocity (v_f) is calculated by setting the initial total momentum equal to the final total momentum: (11,000 kg × 4.5 m/s) = 30,000 kg × v_f, hence v_f = 49,500 kg·m/s / 30,000 kg = 1.65 m/s.

Answer:

magnitude of the average acceleration is 4.16 m/s²

Explanation:

given data

initial velocity u = 5 m/s

distance s = 3 m

final velocity v = 0

to find out

magnitude of the average acceleration

solution

we will apply here linear motion equation that is

v²-u² = 2×a×s   ..............1

put here all these value

v is final velocity and u is initial velocity and s is distance

0²-5² = 2×a×3

a = -4.16

so magnitude of the average acceleration is 4.16 m/s²

Answer: The magnitude of the average acceleration is 4.16m/s^2

Answer:

16 m

Explanation:

given,

power of hydraulic plant = 770 kW

volume of water pass through the turbine = 300 m³

density of water = 1000 kg/m³

m  =ρ × V

mass of water pass each minute  = 300 × 1000 = 3 × 10⁵

assume height of the fall be h

potential head of the water = mgh

[tex]\dfrac{mgh}{60}= 770 \times 10^3[/tex]

[tex]3\times 10^5 \times 9.81\times h= 770 \times 10^3\times 60[/tex]

[tex]h = \dfrac{770 \times 10^3\times 60}{3\times 10^5 \times 9.81}[/tex]

h = 15.69 m ≈ 16 m

the distance of the water fall is equal to 16 m.

Answer:

a) [tex]F=-3.1465 N[/tex]

b) Greater attractive force

Explanation:

Assuming they are point charges, we can use coulomb's law to know the magnitude of the force between these objects:

[tex]F=\frac{kq_1q_2}{d^2}[/tex]

Here, K is the Coulomb constant, [tex]q_1[/tex] and [tex]q_2[/tex] are the point charges and d is the distance bewteen the charges.

a) So, we have:

[tex]F=\frac{(8.99*10^9\frac{Nm^2}{C^2})(7*10^{-6}C)(-6*10^{-6}C)}{12*10^{-2}m}\\F=-3.1465 N[/tex]

b) There will be a greater amount of charge on the side closest to the object with opposite charge, therefore this will increase the charge at that point, increasing the attractive force.

The attractive force between a 0.700 µC charged glass rod and a -0.600 µC charged silk cloth 12.0 cm apart is 0.2625 N, as calculated by Coulomb's law. If charges are distributed over an area instead of acting as point charges, the force calculation might differ, depending on the geometry and distribution of the charges.

To calculate the attractive force between a glass rod with a 0.700 µC charge and a silk cloth with a -0.600 µC charge that are 12.0 cm apart, we can use Coulomb's law. Coulomb's law is given by the formula F = k * |q1*q2| / r^2, where F is the force between the charges, q1 and q2 are the values of the charges, r is the distance between the charges, and k is Coulomb's constant (approximately 8.988 × 10^9 Nm^2/C^2). Plugging in the values, we get F = (8.988 × 10^9 Nm^2/C^2) * |0.700 × 10^-6 C * -0.600 × 10^-6 C| / (0.12 m)^2, which results in a force of approximately 0.2625 N.

Discussing how the answer to this problem might be affected if the charges are distributed over some area rather than acting like point charges: The resulting force might differ from the point charge approximation. This is because the distribution of charge affects the electric field produced by the charges, potentially reducing the force if the charges are spread out over a larger area compared to being concentrated at a point. The effect typically depends on the actual distribution and geometry of the charges.

Answer:

[tex]r_{cm}\ =\ \dfrac{m_2d\ +\ m_1v_0 (t_1\ -\ t_0)}{m_1\ +\ m_2}[/tex]

Explanation:

Given,

Let [tex]v_{cm}[/tex] be the initial velocity of the center of mass of the system of particle at time [tex]t_o[/tex]

[tex]\therefore v_{cm}\ =\ \dfrac{m_1v_1\ +\ m_2v_2}{m_1\ +\ m_2}\\\Rightarrow v_{cm}\ =\ \dfrac{m_1v_0}{m_1\ +\ m_2}[/tex]

Assuming that the first particle is at origin, distance of the second particle from the origin is 'd'

Center of mass of the system of particles

[tex]x_{cm}\ =\ \dfrac{m_1x_1\ +\ m_2x_2}{m_1\ +\ m_2}\\\Rightarrow x_{cm}\ =\ \dfrac{m_2d}{m_1\ +\ m_2}\\[/tex]

Hence, at time [tex]t_0[/tex], the center of mass of the system is at [tex]x_0\ =\ \dfrac{m_2d}{m_1\ +\ m_2}[/tex] at an initial speed of [tex]v_{cm}[/tex]

Both the particles are assumed to be the point masses, therefore at the time [tex]t_1[/tex] the center of mass is at the position of the second particle which should be equal to the total distance traveled by the first particle because the second particle is at rest.

Let [tex]r_{cm}[/tex] be the distance traveled by the center of mass of the system of particles in the time interval [tex](t_1\ +\ t_0)[/tex]

From the kinematics,

[tex]s\ =\ x_0\ +\ vt\\\Rightarrow r_{cm}\ =\ x_{cm}\ +\ v_{cm}{t_1\ -\ t_0}\\\Rightarrow r_{cm}\ =\ \dfrac{m_2d}{m_1\ +\ m_2}\ +\ \left ( \dfrac{m_1v_0}{m_1\ +\ m_2}\ \right )\times (t_1\ -\ t_0)\\\Rightarow r_{cm}\ =\ \dfrac{m_2d\ +\ m_1v_0 (t_1\ -\ t_0)}{m_1\ +\ m_2}[/tex]

Hence, this is the required distance traveled by the first mass to collide with the second mass which is at rest.

Answer:

(1) 42.94 m

(2) [tex]16.02^\circ[/tex]

Explanation:

Let us first draw a figure, for the given question as below:

In the figure, we assume that the person starts walking from point A to travel 11 m exactly [tex]24^\circ[/tex] south of west to point B and from there, it walks 21 m exactly [tex]39^\circ[/tex] west of north to reach point C.

Let us first write the two displacements in the vector form:

[tex]\vec{AB} = (-11\cos 24^\circ\ \hat{i}-11\sin 24^\circ\ \hat{j})\ m =(-10.05\ \hat{i}-4.47\ \hat{j})\ m\\\vec{BC} = (-21\sin 39^\circ\ \hat{i}+21\cos 39^\circ\ \hat{j})\ m =(-31.22\ \hat{i}+16.32\ \hat{j})\ m[/tex]

Now, the vector sum of both these vectors will give us displacement vector from point A to point C.

[tex]\vec{AC}=\vec{AB}+\vec{BC}\\\Rightarrow \vec{AC}=(-10.05\ \hat{i}-4.47\ \hat{j})\ m+(-31.22\ \hat{i}+16.32\ \hat{j})\ m\\\Rightarrow \vec{AC}=(-41.25\ \hat{i}+11.85\ \hat{j})\ m[/tex]

Part (1):

the magnitude of the shortest displacement from the starting point A to point the final position C is given by:

[tex]AC=\sqrt{(-41.25)^2+(11.85)^2}\ m= 42.94\ m[/tex]

Part (2):

As the vector AC is coordinates lie in the third quadrant of the cartesian vector plane whose angle with the west will be positive in the north direction.

The angle of the shortest line connecting the starting point and the final position measured north of west is given by:

[tex]\theta = \tan^{-1}(\dfrac{11.85}{41.27})\\\Rightarrow \theta = 16.02^\circ[/tex]

Answer:

0.6 m

Explanation:

When a spring is compressed it stores potential energy. This energy is:

Ep = 1/2 * k * x^2

Being x the distance it compressed/stretched.

When the spring bounces the ice cube back it will transfer that energy to the cube, it will raise up the slope, reaching a high point where it will have a speed of zero and a potential energy equal to what the spring gave it.

The potential energy of the ice cube is:

Ep = m * g * h

This is vertical height and is related to the distance up the slope by:

sin(a) = h/d

h = sin(a) * d

Replacing:

Ep = m * g * sin(a) * d

Equating both potential energies:

1/2 * k * x^2 = m * g * sin(a) * d

d = (1/2 * k * x^2) / (m * g * sin(a))

d= (1/2 * 25 * 0.1^2) / (0.05 * 9.81 * sin(25)) = 0.6 m

Using the conservation of mechanical energy principle, we relate the initial potential energy in the compressed spring to the final potential energy due to the height gained by the ice cube sliding up a slope. In doing so, we calculate how far the ice cube travels up the slope before reversing direction to be approximately 0.607 meters.

The question pertains to the concept of energy conservation involving the energies of spring compression (potential energy) and gravity (also potential energy). Here the energy initially stored in the compressed spring is used to propel the ice cube up the slope until its potential energy (due to height gained) fully consumes the initial energy from the spring.

The conservation of mechanical energy principle states the initial total energy is equal to the final total energy. Initially we have spring potential energy and finally it's all converted to gravitational potential energy. So, 1/2*k*x^2=m*g*h, where k is the spring constant, x is the spring compression, m is mass of the ice block, g is acceleration due to gravity, and h is height gained by the ice cube.

We solved for h to get h = (1/2*25*(0.100)^2) / (50*9.8) = 0.255 m. This is the vertical height gained, not the distance along the slope. To get the slope distance traveled by the cube (d), we divide the height by sin(25 degrees) to get: d = 0.255/sin(25) = 0.607 m.

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Answer:

Explanation:

The steam was earlier at 300 ° C. and pressure of 1 MPa. When the gas is allowed to expand against vacuum , work done by the gas is nil because there is no external pressure against which it has to work . Therefore there will not be any change in its internal energy. Since the tank is insulated therefore there is no possibility of external heat to increase its internal energy.

Hence the temperature of gas will remain unaffected.  It will remain stagnant at 300

Answer:

a) [tex]y=y_{0}+v_{oy} t+\frac{1}{2} gt^{2}  =640 ft+(65ft/s)(2.018s)+(\frac{1}{2})(-32.2ft/s^{2} )(2.018s)^{2}[/tex]

[tex]y=705.6ft[/tex]

b) [tex]t=8.63 s[/tex]

Explanation:

We start the exercise knowing that a ball is thrown up with an initial velocity of 65 ft/s with an initial height of 680 ft.

To calculate the maximum heigh, we know that at the top of the motion the ball stop going up and start going down because of the gravity

First of all, we need to calculate the time that takes the ball to reach the maximum point.

a) [tex]v_{y}=v_{oy}+gt[/tex]

[tex]t=\frac{-v_{oy} }{g}=\frac{-65ft/s}{-32.2ft/s^{2} } =2.018s[/tex]

Knowing that time, we can calculate the height to which the ball rises:

[tex]y=y_{0}+v_{oy} t+\frac{1}{2} gt^{2}  =640 ft+(65ft/s)(2.018s)+(\frac{1}{2})(-32.2ft/s^{2} )(2.018s)^{2}[/tex]

[tex]y=705.6ft[/tex]

b) Now, to know the time that the ball reach the bottom of the cliff, we know that the final height is y=0ft

[tex]y=y_{0}+v_{oy} t+\frac{1}{2} gt^{2}[/tex]

[tex]0=640ft+(65ft/s)t-\frac{1}{2}(32.2ft/s^{2})t^{2}[/tex]

This is a classic quadratic equation, that can be solve using the quadratic formula

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

a=-16.1

b=65

c=640

Solving for t, we have that

[tex]t=-4.6015 s[/tex] or [tex]t=8.6387s[/tex]

Since the time can not be negative:

[tex]t=8.6387s[/tex]

Answer:

a) 0.067 seconds

b) 112.5 m/s²

Explanation:

t = Time taken

u = Initial velocity = 7.5 m/s

v = Final velocity = 0

s = Displacement = 0.25 m

a = Acceleration

Eqaution of motion

[tex]v^2-u^2=2as\\\Rightarrow a=\frac{v^2-u^2}{2s}\\\Rightarrow a=\frac{0^2-7.5^2}{2\times 0.25}\\\Rightarrow a=-112.5\ m/s^2[/tex]

b) The deceleration of the football player is 112.5 m/s²

[tex]v=u+at\\\Rightarrow t=\frac{v-u}{a}\\\Rightarrow t=\frac{0-7.5}{-112.5}\\\Rightarrow t=0.067\ s[/tex]

a) The collision lasts for 0.067 seconds

Answer:

Force, F = 550000 N

Explanation:

Given that,

Pressure at the bottom of the Marianas Trench, [tex]P=1100\ kPa=11\times 10^5\ Pa[/tex]

Surface area, [tex]A=0.5\ m^2[/tex]

We need to find the force with which the water pressure at the bottom of the Marianas Trench push on a fish. Mathematically, the pressure is given by :

[tex]P=\dfrac{F}{A}[/tex]

[tex]F=P\times A[/tex]

[tex]F=11\times 10^5\ Pa\times 0.5\ m^2[/tex]

F = 550000 N

So, the force with which the water pressure at the bottom of the Marianas Trench push on a fish is 550000 N. Hence, this is the required solution.

The water pressure at the bottom of the Marianas Trench would exert a force of 550,000 Newtons on a fish with a surface area of 0.50 m^2.

The water pressure exerted on a fish at the bottom of the Marianas Trench can be determined by using the formula for pressure (P), which is force (F) divided by area (A). Given the pressure at the bottom of the trench is approximately 1,100 kPa and the surface area of the fish is 0.50 m2, the force can be calculated using the equation F = P x A. By inserting the given values, we get F = 1,100,000 Pa x 0.50 m2 = 550,000 N. Therefore, the water pressure would push on the fish with a force of 550,000 Newton.

Final answer:

The speed of the box as it leaves the spring can be calculated using conservation of energy, accounting for the initial potential energy in the spring and work done by friction. The maximum speed of the box is achieved when all the spring's potential energy is converted into kinetic energy, before friction does any work.

Explanation:

To solve for the speed of the box at the instant it leaves the spring and its maximum speed during the motion, we can apply the principle of conservation of energy and the work-energy theorem.

Part (a): Speed of Box When It Leaves the Spring

The initial potential energy stored in the compressed spring is equal to the kinetic energy of the box plus the work done by friction as the box slides the distance where the spring is compressed:

PE_spring = KE_box + Work_friction

Using the formula for potential energy PE_spring = (1/2)[tex]kx^2[/tex], where k is the spring constant and x is the compression distance, and the work done by friction Work_friction = μ_kmgx, where μ_k is the coefficient of kinetic friction, m is the mass, and g is the acceleration due to gravity, we can find the kinetic energy which will then be used to calculate the speed using KE = (1/2)[tex]mv^2[/tex].

To find the actual speed, we rearrange to v = √(2(PE_spring - Work_friction)/m).

Part (b): Maximum Speed of the Box

The maximum speed of the box occurs at the point where all the potential energy in the spring has been converted to kinetic energy and before any work has been done by friction. Therefore, v_max = √(2PE_spring/m).

Answer:

36 rev

Explanation:

See it in the pic

Answer:

[tex][F]=[MLT^{-2}][/tex]

Explanation:

Newton’s second law states that the acceleration a of an object is proportional to the force F acting on it is inversely proportional to its mass m. The mathematical expression for the second law of motion is given by :

F = m × a

F is the applied force

m is the mass of the object

a is the acceleration due to gravity

We need to find the dimensions of force. The dimension of force m and a are as follows :

[tex][m]=[M][/tex]

[tex][a]=[LT^{-2}][/tex]

So, the dimension of force F is, [tex][F]=[MLT^{-2}][/tex]. Hence, this is the required solution.

Final answer:

In physics, the dimensions of force are derived from Newton's second law of motion, which can be represented as F=ma. Force has dimensions of mass (M), length (L), and time (T), leading to the dimensional formula [F] = [M][L][T]^-2. The SI unit for force is the Newton (N).

Explanation:

According to Newton's second law of motion, the acceleration a of an object is directly proportional to the net external force Fnet acting upon it and inversely proportional to its mass m. This law is mathematically represented by the equation Fnet = m × a, where Fnet is the net force, m is the mass, and a is the acceleration of an object.

The weight w of an object is another type of force, which is defined as the gravitational force acting on an object with mass m. The object experiences an acceleration due to gravity g, and this is represented by the equation w = m × g. In the International System of Units (SI), force is measured in Newtons (N), and one Newton is defined as the force required to accelerate a one-kilogram mass at a rate of one meter per second squared (1 N = 1 kg × m/s2).

Therefore, the dimensions of force in terms of the base physical quantities are mass (M), length (L), and time (T), and the dimensional formula for force is [F] = [M][L][T]-2.

Answer:

DMM should be placed in the series combination with the circuit.

Explanation:

DMM is the digital multi meter. It can measure the voltage, current and resistance at a time.

Answer:

0.3067 m

Explanation:

Resolving power of a lens is given by the expression

R.P = 1.22λ / D

where λ is wavelength of light used, D is diameter of the lens

Substituting the data given

R.P in radian =

[tex]\frac{1.22\times550\times10^{-9}}{35\times10^{-2}}[/tex]

R P = 19.17 X 10⁻⁷ radian

If d be the minimum spacing between two objects on the ground that is resolvable by lens

[tex]\frac{d}{160\times10^3} =19.17\times10^{-7}[/tex]

d = 0.3067 m

Answer:

[tex]276.74\times 10^8Mg/m^3[/tex]

31.29 m/sec

Explanation:

We have given density of substance [tex]0.14lb/in^3[/tex]

We have convert this into [tex]Mg/m^3[/tex]

We know that 1 lb = 0.4535 kg. so 0.14 lb = 0.14×0.4535 = 0.06349 kg

We know that 1 kg = 1000 g ( 1000 gram )

So 0.06349 kg = 63.49 gram

And we know that 1 gram = 1000 milligram

So 63.49 gram [tex]=63.49\times 10^3\ Mg[/tex]

We know that [tex]1 in^3=1.6387\times 10^{-5}m^3[/tex]

So [tex]0.14in^3=0.14\times 1.6387\times 10^{-5}=0.2294\times 10^{-5}m^3[/tex]

So [tex]0.14lb/in^3[/tex] =\frac{63.49\times 10^3}{0.2249\times 10^{-5}}=276.74\times 10^8lb/m^3[/tex]

In second part we have to convert 70 mi/hr to m/sec

We know that 1 mi = 1609.34 meter

So 70 mi = 70×1609.34 = 112653.8 meter

1 hour = 3600 sec

So 70 mi/hr [tex]=\frac{70\times 1609.34meter}{3600sec}=31.29m/sec[/tex]

Final answer:

The density of the substance is 3.87118626 Mg/m³. The velocity of the vehicle is 31.29222 m/s.

Explanation:

The density of a substance is the ratio of its mass to its volume. In order to convert the density from lb/in³ to Mg/m³, we need to use conversion factors. 1 lb/in³ is equal to 27679.9 Mg/m³. Therefore, the density of the substance is 0.14 lb/in³ * 27679.9 Mg/m³/lb/in³ = 3.87118626 Mg/m³.

For the second question, to convert mi/hr to m/s, we can use the conversion factor of 1 mi = 1609.34 m and 1 hr = 3600 s. Therefore, the velocity of the vehicle is 70 mi/hr * 1609.34 m/mi * 1 hr/3600 s = 31.29222 m/s.