A hot-air balloon of diameter 12 mm rises vertically at a constant speed of 15 m/sm/s. A passenger accidentally drops his camera from the railing of the basket when it is 18 mm above the ground. If the balloon continues to rise at the same speed, how high is the railing when the camera hits the ground?

Answer:

The velocity of the pin is opposite its acceleration on the way up.

(d) option is correct.

Explanation:

when the juggler throws a bowling pin straight in the air, the acceleration working on the pin is in the downward direction due to the gravitational force of the earth.

According to Newton's Universal Law of Gravitation

''The gravitational force is a force that attracts any objects with mass''

Final answer:

The acceleration due to gravity is always downward, so when the pin is thrown up, its velocity is opposite to its acceleration. At the peak, the velocity is zero and then aligns with the direction of gravity on the descent.

Explanation:

When a juggler throws a bowling pin straight up in the air, the acceleration due to gravity is always directed towards the ground, which means it is downwards. As the pin moves upwards, its velocity is in the opposite direction of the acceleration. At the peak of its motion, the velocity of the pin is zero, after which it starts to fall back down, and its velocity is then in the same direction as acceleration. Thus, the correct statement is (d) The velocity of the pin is opposite its acceleration on the way up.

To solve this problem we will apply the concepts given by the kinematic equations of motion. For this purpose it will be necessary with the given data to obtain the deceleration. With this it will be possible again to apply one of the kinematic equations of motion that does not depend on time, but on distance, to find how far the block would slide with the quadruplicate velocity

Our values are given as,

[tex]\text{Initial speed} =V_i = 2.3 m/s[/tex]

[tex]\text{Final speed}= V_f = 0 m/s[/tex]

[tex]\text{Stopping distance = }d = 1.9 m[/tex]

[tex]a = acceleration[/tex]

[tex]\text{mass} = m = 3.1kg[/tex]

Using the kinematic equation of motion we have

[tex]V_f^2 = V_i^2 + 2 a d[/tex]

[tex]0^2 = 2.3^2 + 2 a (1.9)[/tex]

[tex]a = -1.39211 m/s^2[/tex]

Now if the initial velocity is quadrupled we have that,

[tex]\text{Initial speed} =V_i' = 2.3*4 m/s = 9.2m/s[/tex]

[tex]\text{Final speed}= V_f' = 0 m/s[/tex]

[tex]\text{Stopping distance = }d'[/tex]

[tex]V_f^2' = V_i^2' + 2 a d'[/tex]

Replacing the values

[tex]0^2 = 9.2^2+ 2 (-1.39211) d'[/tex]

[tex]d' = 30.44m[/tex]

Therefore the block would have slipped around 30.44 if its initial velocity quadrupled.

Answer

given,

position of particle

x(t)= A t + B t²

A = -3.5 m/s

B = 3.9 m/s²

t = 3 s

a)  x(t)= -3.5 t + 3.9 t²

   velocity of the particle is equal to the differentiation of position w.r.t. time.

[tex]\dfrac{dx}{dt}=\dfrac{d}{dt}(-3.5t + 3.9t^2)[/tex]

[tex]v= -3.5 + 7.8 t [/tex]------(1)

velocity of the particle at t = 3 s

  v = -3.5 + 7.8 x 3

 v = 19.9 m/s

b) velocity of the particle at origin

  time at which particle is at origin

  x(t)= -3.5 t + 3.9 t²

   0 = t (-3.5  + 3.9 t )

   t = 0, [tex]t=\dfrac{3.5}{3.9}[/tex]

   t = 0 , 0.897 s

speed of the particle at t = 0.897 s

from equation (1)

 v = -3.9 + 7.8 t

 v = -3.9 + 7.8 x 0.897

  v = 3.1 m/s

To solve the problem we should know about velocity.

Velocity is the rate of change of its position with respect to time.

[tex]V = \dfrac{dy}{dt}[/tex]

Given to us

x(t)= At+Bt²

[tex]V(t) = \dfrac{dy}{dt} = \dfrac{d(At+Bt^2)}{dt} = A+2Bt[/tex]

A.)  the velocity, in meters per second. of the particle at the time t1= 3.0 s,

[tex]V(t) = A +2Bt[/tex]

Substituting the values,

[tex]V(t_1=3) = (-3.5) +2(3.9)(3.0)\\\\V(t_1=3) = 19.9\ m/s[/tex]

B.)  the velocity, in meters per second: of the particle when it is at the origin (x=0) at t ≥ 0

[tex]x(t)= At+Bt^2\\\\0 = At+Bt^2\\\\[/tex]

Taking t as common,

[tex]0 = t(A+Bt)\\\\[/tex]

[tex]0 = (A+Bt)\\\\[/tex]

Substituting the values and solving or t,

[tex]0 = A+ Bt\\0 = -3.5 + (3.9)t\\3.5=3.9t\\t= \dfrac{3.5}{3.9}\\\\t = 0.8974\ s[/tex]

Substituting the value in the formula of velocity,

[tex]V(t) = A +2Bt[/tex]

Substituting the values,

[tex]V(t_1=0.8974) = (-3.5) +2(3.9)(0.8974)\\\\V(t_1=3) = 3.5 \ m/s[/tex]

Hence, the velocity of the particle is 3.5 m/s.

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The new power is: New P_avg = 2.52 W / 4 ≈ 0.63 W

Average Power Carried by a Wave

To solve this problem, we need the following information:

Mass of piano wire: 2.95 g = 0.00295 kg

Length of wire: 79.0 cm = 0.79 m

Tension: 29.0 N

Frequency: 105 Hz

Amplitude: 1.80 mm = 0.00180 m

First, calculate the linear mass density (μ) of the wire:

μ = mass / length = 0.00295 kg / 0.79 m ≈ 0.00373 kg/m

Next, find the wave speed using the formula for the speed of a wave on a string:

v = [tex]\sqrt{Tension / \mu}[/tex] =[tex]\sqrt{29.0 N / 0.00373 kg/m}[/tex]≈ 88.19 m/s

Now, we calculate the average power (P_avg) carried by the wave using the formula:

P_avg = 0.5 x μ x v x ω² x A²

Where:

ω = 2πf (angular frequency)

ω = 2 x π x 105 ≈ 659.73 rad/s

Therefore,

P_avg = 0.5 x 0.00373 kg/m x 88.19 m/s x (659.73 rad/s)² x (0.00180 m)² ≈ 2.52 W

Average Power if Amplitude is Halved

If the amplitude (A) is halved, the new amplitude is:

New A = 0.00180 m / 2 = 0.00090 m

Since power is proportional to the square of the amplitude (A²), halving the amplitude reduces the power by a factor of 4.

Thus, the new power is:

New P_avg = 2.52 W / 4 ≈ 0.63 W

Answer:

[tex]3.26198\times 10^{-12}\ kg[/tex]

Explanation:

E = Electric field = 20 MN/C

q = Charge of electron = [tex]1.6\times 10^{-19}\ C[/tex]

g = Acceleration due to gravity = 9.81 m/s²

m = Mass of drop

The electrical force will balance the weight

[tex]Eq=mg\\\Rightarrow 20\times 10^{6}\times 10\times 1.6\times 10^{-19}=m\times 9.81\\\Rightarrow m=\dfrac{20\times 10^{6}\times 10\times 1.6\times 10^{-19}}{9.81}\\\Rightarrow m=3.26198\times 10^{-12}\ kg[/tex]

The mass that can be suspended is [tex]3.26198\times 10^{-12}\ kg[/tex]

From charge to mass ratio, the mass of the charges is 3.2  × 10^-12 Kg.

The charge to mass ratio experiment was used by R. A. Millikan to accurately determine the charge to mass ratio of the electron. We have the following information from the question;

Field strength = 20 MN/C

Number of charges = 10

Now;

The magnitude of electric field strength is obtained from;

E = F/q

F = Eq

Where;

F = electric force

E = electric field intensity

q = magnitude of charge

F = 10 × 20  × 10^6  × 1.6 × 10^-19 = 3.2  × 10^-11 N

Where the charges fall freely under gravity;

F = mg

m = F/g

m = 3.2  × 10^-11 N/10 ms-2

m = 3.2  × 10^-12 Kg

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Answer:

D = 72.68 m

Explanation:

given,

R = 100 m

angular speed = 0.1 rad/s

distance she can be pulled before the centripetal acceleration reaches 5g = 49 m/s².

using conservation of Angular momentum

[tex]I_i\omega_i= I_f\omega_f[/tex]

[tex]mr_i^2\omega_i=m r_f^2\omega_f[/tex]

[tex]\omega_f = \dfrac{r_i^2}{r_f^2}\times \omega_i[/tex]

[tex]\omega_f = \dfrac{r_i^2}{r_f^2}\times \omega_i[/tex]

we know,

centripetal acceleration

[tex]a = \dfrac{v^2}{r}[/tex]

v = r ω

[tex]a =\omega_f^2 r_f [/tex]

[tex]a =(\dfrac{r_i^2}{r_f^2}\times \omega_i)^2 r_f [/tex]

[tex]a =\dfrac{r_i^4\times \omega_i^2}{r_f^3}[/tex]

[tex]r_f^3=\dfrac{100^4\times 0.1^2}{5\times 9.8}[/tex]

[tex]r_f^3=20408.1632[/tex]

[tex]r_f = 27.32\ m[/tex]

distance she has reached inward is equal to

D = 100 - 27.23

D = 72.68 m

To solve this problem we will apply the concepts related to the potential, defined from the Coulomb laws for which it is defined as the product between the Coulomb constant and the load, over the distance that separates the two objects. Mathematically this is

[tex]V = \frac{kq}{r}[/tex]

k = Coulomb's constant

q = Charge

r = Distance between them

[tex]q = 18 pC \rightarrow q = 1.8*10^-11 C[/tex]

[tex]d = 2.4mm \rightarrow r = 1.2 mm = 1.2*10^-3 m[/tex]

Replacing,

[tex]V = \frac{kq}{r}[/tex]

[tex]V = \frac{ (9*10^9)*(1.8*10^{-11})}{(1.2*10^{-3})}[/tex]

[tex]V = 135 V[/tex]

Therefore the potential at the surface of the raindrop is 135 V

Final answer:

The current through the wire is 1.95 A.

Explanation:

To find the current through the wire, we can use Ampere's law. Ampere's law states that the magnetic field around a long straight wire is directly proportional to the current through the wire and inversely proportional to the distance from the wire.

So, we can use the equation B = μ0 * I / (2π * r), where B is the magnetic field, μ0 is the magnetic constant, I is the current, and r is the distance from the wire.

Plugging in the given values, we have 1.30x10-3 T = (4πx10-7 T*m/A) * I / (2π * 0.03 m). Solving for I, we get I = 1.30x10-3 * (2*0.03)/(4x10-7) = 1.95 A.

Answer:

B

Explanation:

The sailor, the boat and the wrench are all moving at he same constant rate, so the wrench will appear to fall straight down. This due to that fact there is no relative motion among them and all are at rest w.r.t to one another. Hence the correct answer would be B.

The correct option is (B). The wrench will fall straight down relative to the moving boat and will land directly at the base of the mast.

To determine where the wrench will hit the deck, let's analyze the motion of the wrench.

1. Horizontal Motion:

2. Vertical Motion*: The wrench will accelerate downwards due to gravity.

Answer:

n = 7.5 times/minute

Explanation:

Given that,

Wavelength of the ocean wave, [tex]\lambda=40\ m[/tex]

The speed of the ocean wave, v = 5 m/s

To find,

Number of times a minute does a boat bob up and down on ocean waves.

Solution,

The relation between the speed of wave, wavelength and frequency is given by :

[tex]v=f\times \lambda[/tex]

[tex]f=\dfrac{v}{\lambda}[/tex]

[tex]f=\dfrac{5\ m/s}{40\ m}[/tex]

f = 0.125 Hz

The number of times per minute the bob moves up and down is given by :

[tex]n=f\times t[/tex]

[tex]n=0.125\times 60[/tex]

n = 7.5 times/minute

So, its will move up and down in 7.5 times/minute. Therefore, this is the required solution.

Taking into account the definition of wavelength, frecuency and propagation speed, the number of times per minute the bob moves up and down is 7.5 times per minute.

Wavelength is the minimum distance between two successive points on the wave that are in the same state of vibration. It is expressed in units of length (m).

Frequency is the number of vibrations that occur in a unit of time. Its unit is s⁻¹ or hertz (Hz).

Finally, the propagation speed is the speed with which the wave propagates in the medium, that is, it is the magnitude that measures the speed at which the wave disturbance propagates along its displacement.

The propagation speed relate the wavelength (λ) and the frequency (f) inversely proportional using the following equation:

v = f× λ

In this case, you know:

Replacing in the definition of propagation speed:

5 [tex]\frac{m}{s^{2} }[/tex]= f× 40 m

Solving:

f= 5 [tex]\frac{m}{s^{2} }[/tex]÷ 40 m

f= 0.125 Hz= 0.125 [tex]\frac{1}{seconds}[/tex]

Then, a boat bob up and down on ocean waves 0.125 times in a second.

So, the number of times per minute the bob moves up and down is given by:

n= f× time

n= 0.125 Hz× 60 minutes in 1 second

n=7.5 times per minute

Finally, the number of times per minute the bob moves up and down is 7.5 times per minute.

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Answer

given,

x = 4 t³ - 6 t² + 12

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

[tex]\dfrac{dx}{dt}=\dfrac{d}{dt}(4t^3-6t^2+12)[/tex]

[tex]v =12t^2-12t[/tex]

For minimum velocity calculation we have differentiate it and put it equal to zero.

[tex]\dfrac{dv}{dt} =\dfrac{d}{dt}12t^2-12t[/tex]

[tex]\dfrac{dv}{dt} =24t-12[/tex].........(1)

putting it equal to zero

24 t - 12 =0

t = 0.5 s

At t = 0.5 s velocity will be minimum.

b) minimum velocity

    v = 12t² -12 t

    v = 12 x 0.5² -12 x 0.5

    v = -3 m/s

c) derivative of velocity w.r.t. time is acceleration

from equation 1

     a = 24 t - 12

time at which acceleration will be zero

     0 = 24 t - 12

     t = 0.5 s

At t = 0.5 s acceleration will be zero.

Part A. The particle reaches its minimum velocity at 0.5 seconds.

Part B. The minimum velocity of the particle is -3 m/s.

Part C. The acceleration of the particle will be zero at the time t = 0.5 seconds.

Given that the position of a particle is given by the function x.

[tex]f(x) = 4t^2 -6t^2 +12[/tex]

The function of the velocity of a particle can be obtained by the time function.

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

[tex]v = \dfrac {d}{dt} ( 4t^3-6t^2 +12)[/tex]

[tex]v = 12t^2 -12 t[/tex]

The velocity function of the particle is [tex]v = 12t^2 - 12t[/tex].

The minimum velocity of the particle is obtained by the differentiation of velocity function with respect to the time and put it equal to zero.

[tex]\dfrac {dv}{dt} = \dfrac {d}{dt} (12t^2 - 12t) = 0[/tex]

[tex]\dfrac {dv}{dt} = 24 t-12 = 0[/tex]

[tex]t = 0.5\;\rm s[/tex]

Hence we can conclude that the particle reaches its minimum velocity at 0.5 seconds.

The velocity function is [tex]v = 12t^2 - 12t[/tex]. Substituting the value of t = 0.5 s to calculate the minimum velocity.

[tex]v = 12(0.5)^2 - 12(0.5)[/tex]

[tex]v = 3 - 6[/tex]

[tex]v = -3 \;\rm m/s[/tex]

The minimum velocity of the particle is -3 m/s.

The acceleration is defined as the change in the velocity with respect to time. Hence,

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

[tex]a = 24 t-12[/tex]

Substituting the value of a = 0, we get the time.

[tex]0 = 24t - 12[/tex]

[tex]t = 0.5 \;\rm s[/tex]

The acceleration of the particle will be zero at the time t = 0.5 seconds.

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Answer:

6.08 cm

Explanation:

We are given that

Ratio =1:125000

Let x be the distance on a map between two features .

The distance between two features on ground=y=7.6 km

According to question

[tex]\frac{x}{y}=\frac{1}{125000}[/tex]

Substitute the values then we get

[tex]\frac{x}{7.6}=\frac{1}{125000}[/tex]

[tex]x=\frac{7.6}{125000}=0.0000608 km[/tex]

We know that

[tex]1km=100000 cm[/tex]

0.0000608 km=[tex]0.0000608\times 100000=6.08cm[/tex]

Hence, the distance between two features on the map=6.08 cm

The distance on the map between the two features is 6.08 centimeters.

To find the distance on a map in centimeters between two features if they are 7.6 km apart on the ground and the map has a scale of 1:125000, you can use the scale factor formula: Distance on Map = Distance on Ground / Scale

Plugging in the values: Distance on Map = 7.6 km / 125000 = 0.0000608 km

Since 1 km = 100000 centimeters, multiply by 100000 to convert km to cm:

Distance on Map = 0.0000608 km × 100000 cm/km = 6.08 cm

Therefore, the distance on the map between the two features is 6.08 centimeters.

Answer:

The detect frequency is 1622.72 Hz.

Explanation:

Given that,

Frequency = 1500 Hz

Suppose you move with a speed of 27.0 m/s toward the fire engine. what frequency do you detect ?

We need to calculate the frequency

Using formula of frequency

[tex]f=(\dfrac{v+v_{0}}{v})f_{0}[/tex]

Where, v = speed of sound

v₀ = speed of source

f₀ = frequency of siren

Put the value into the formula

[tex]f=\dfrac{330+27.0}{330}\times1500[/tex]

[tex]f=1622.72\ Hz[/tex]

Hence, The detect frequency is 1622.72 Hz.

Answer:

Explanation:

The "traditional" form of Coulomb's law, explicitly the force between two point charges. To establish a similar relationship, you can use the integral form for a continuous charge distribution and calculate the field strength at a given point.

In the case of moving charges, we are in presence of a current, which generates magnetic effects that in turn exert force on moving charges, therefore, no longer can consider only the electrostatic force.

Answer:

n/(FG) = 3.

Explanation:

At the top of the loop-the-loop, the normal force is directed downwards as well as the weight of the car. So, the total net force of the car is

[tex]F_{net} = N + mg[/tex]

By Newton's Second Law, this force is equal to the centripetal force, because the car is making circular motion in the loop.

[tex]F_{net} = ma = \frac{mv^2}{R}\\N + mg = \frac{mv^2}{R}[/tex]

The critical speed is the minimum speed at which the car does not fall. So, at the critical speed the normal force is zero.

[tex]0 + mg = \frac{mv_c^2}{R}\\v_c = \sqrt{gR}[/tex]

If the car is moving twice the critical speed, then

[tex]N + mg = \frac{m(2v_c)^2}{R} = \frac{m4gR}{R} = 4mg\\N = 3mg[/tex]

Finally, the ratio of the normal force to the gravitational force is

[tex]\frac{3mg}{mg} = 3[/tex]

Answer:

34.9 kg

Explanation:

Since there are no net external forces on the system, the center of gravity does not move.

Let's say that m is the mass of the man, M is the mass of the boat, and L is the length of the boat.

When the man is at the stern, the distance between the center of gravity and the pier is:

CG = (m L + M (L/2)) / (m + M)

When the man reaches the prow, the distance between the center of gravity and the pier is:

CG = (m x + M (x + L/2)) / (m + M)

Since these are equal:

(m L + M (L/2)) / (m + M) = (m x + M (x + L/2)) / (m + M)

m L + M (L/2) = m x + M (x + L/2)

m L + M (L/2) = m x + M x + M (L/2)

m L = m x + M x

m L − m x = M x

m (L − x) = M x

M = m (L − x) / x

Plugging in values:

M = 89.3 kg (5.8 m − 4.17 m) / 4.17 m

M = 34.9 kg

The required mass of boat is 34.9 kg.

The given problem is based on the concept of the center of mass. The point of analysis where the entire mass of the system is known to be concentrated is known as the center of mass.

Given data:

The mass of man is, m = 89.3 kg.

The length of boat is, L = 5.8 m.

The distance away from the pier is, d = 4.17 m.

Since there are no net external forces on the system, the center of gravity does not move. Let's say that m is the mass of the man, M is the mass of the boat

When the man is at the stern, the distance between the center of gravity and the pier is:

CG = (m L + M (L/2)) / (m + M)

When the man reaches the prow, the distance between the center of gravity and the pier is:

CG = (m d + M (d + L/2)) / (m + M)

Since these are equal:

(m L + M (L/2)) / (m + M) = (m d + M (d + L/2)) / (m + M)

m L + M (L/2) = m d + M (d + L/2)

m L + M (L/2) = m d + M d + M (L/2)

Further solving as,

m L = m d + M d

m L − m d = M d

m (L − x) = M x

M = m (L − x) / x

M = 89.3 kg (5.8 m − 4.17 m) / 4.17 m

M = 34.9 kg

Thus, we can conclude that the required mass of boat is 34.9 kg.

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Answer:

No

Explanation:

Although the direction of position, velocity or acceleration of an object in marry-go-round problem changes continuously,however the magnitude of the position, velocity and acceleration do remains the same. Marry-go-round is nothing but a machine found in fairs that turn round in circular motion. So, the laws of circular motion are applicable in it.

Answer:

The details of bands is given in explanation.

Explanation:

The electromagnetic waves are differentiated into different bands based upon their wavelengths and frequencies. The names of different bands are as follows:

1. Radio Waves

2. Micro Waves

3. Infra-red

4. Visible light

5. Ultra Violet

6. X-rays

7. Gamma Rays

The frequency of every region or rays increases from 1 through 7. The energy of rays also increase from 1 through 7. Since, the wave length is inversely related to energy and frequency, thus the wavelength of rays decrease from 1 through 7.

A detailed information of the bands is provided in the picture attached.

Answer:

Option A is correct (0) ( The electric field at the center of circular charged ring is zero)

Explanation:

Option A is correct (0) ( The electric field at the center of circular charged ring is zero)

The reason why electric field at the center of circular charged ring is zero because the fields at the center of the circular ring due to any point are cancelled by electric fields of another point which is at 180 degree to that point i.e opposite to that charge.

Answer: a) 0

The electric field from opposite directions at the centre of the circular ring cancel each other out and give a resultant of zero

Explanation:

Given that the ring is perfectly circular with radius b which has a uniform distributed charge q around it.

The electric field experienced at the centre of the ring from opposite directions are given as

E1 = kₑq/b²

E2 = -kₑq/b²

It experience the two electric field E1 and E2 from opposite directions at the centre. So the resultant electric field is given by:

E = E1 + E2

E = kₑq/b² - kₑq/b²

E = 0

The electric field from opposite directions at the centre of the circular ring cancel each other out and give a resultant of zero

Answer:

(a). The magnitude of the force is 0.38416 N.

(b). The original length is 0.0869 m.

Explanation:

Given that,

Tensile strength = 196 MPa

Maximum strain = 0.380

Diameter = 50.0 μm

Length = 12.0 cm

We need to calculate the area

Using formula of area

[tex]A=\dfrac{\pi}{4}\times d^2[/tex]

Put the value into the formula

[tex]A=\dfrac{\pi}{4}\times(50.0\times10^{-6})^2[/tex]

[tex]A=1.96\times10^{-9}\ m^2[/tex]

We need to calculate the magnitude of the force

Using formula of force

[tex]F=\sigma A[/tex]

Put the value into the formula

[tex]F=196\times10^{6}\times1.96\times10^{-9}[/tex]

[tex]F=0.38416\ N[/tex]

(b). If the length of a strand of the hair is 12.0 cm at its breaking point

We need to calculate the unstressed length

Using formula of strain

[tex]strain=\dfrac{\Delta l}{l_{0}}[/tex]

[tex]\Delta l=strain\times l_{0}[/tex]

Put the value into the formula

[tex]\Delta l=0.380\times l_{0}[/tex]

Length after expansion is 12 cm

We need to calculate the original length

Using formula of length

[tex]l=l_{0}+\Delta l[/tex]

Put the value into the formula

[tex]I=l_{0}+0.380\times l_{0}[/tex]

[tex]l=1.38l_{0}[/tex]

[tex]l_{0}=\dfrac{l}{1.38}[/tex]

[tex]l_{0}=\dfrac{12\times10^{-2}}{1.38}[/tex]

[tex]l_{0}=0.0869\ m[/tex]

Hence, (a). The magnitude of the force is 0.38416 N.

(b). The original length is 0.0869 m.

As additional resistors are connected in series to a constant voltage source, the power supplied by the source decreases.

As additional resistors are connected in series to a constant voltage source, the power supplied by the source decreases.

This can be explained by Ohm's Law, which states that power is equal to the voltage squared divided by the resistance. When resistors are connected in series, the total resistance increases, which leads to a decrease in the power supplied by the source.

For example, if you connect two resistors in series to a constant voltage source, the total resistance would be the sum of the resistances of the two resistors. As a result, the power supplied by the source would decrease.

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Adding more resistors in series to a constant voltage source results in an increase in total resistance, which leads to a decrease in current. Because power is proportional to the square of the current, the power supplied by the voltage source decreases.

When additional resistors are connected in series to a constant voltage source, the power supplied by the source is affected in a specific way. Given that power (P) is calculated by P = I^2R for a given resistance (R) and current (I), and by P = V^2/R for a given voltage (V) and resistance, we can understand the impact of adding resistors in series.

In a series circuit, the current remains constant throughout the resistors, but the total resistance (sum of all individual resistances) increases as more resistors are added. Since the voltage is constant, an increase in the total resistance will result in a decrease in current according to Ohm's law (I = V/R). Consequently, if the current decreases, the power supplied by the source also decreases (P = I^2R), because the power is proportional to the square of the current flowing through the circuit. Therefore, the correct answer is (D) The power supplied by the source decreases.

The assumptions we can reasonably make in this scenario include the highway being straight and level, corresponding to options C and D. Assuming constant acceleration or velocity for each stage of the car's motion, options A and B, is not necessarily accurate.

In analyzing the motion of a car in different stages, we can make reasonable assumptions to simplify the problem. For each stage, assuming the car is moving with constant acceleration (option a) or velocity (option b) is not necessarily accurate because acceleration and velocity may change due to various factors like interaction with driver, road conditions, or appearance of a police car. The assumptions more likely to hold are that the highway is straight (option c), meaning there are no curves, and the highway is level (option d), indicating no hills or valleys. Thus, the correct answers would be C and D.

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#SPJ3

Answer:

(A)  [tex]a=2.0.37m/sec^2[/tex]

(B) s = 146.664 m

Explanation:

We have given car starts from the rest so initial velocity u = 0 m /sec

Final velocity v = 88 km/hr

We know that 1 km = 1000 m

And 1 hour = 3600 sec

So [tex]88km/hr=88\times \frac{1000}{3600}=24.444m/sec[/tex]

Time is given t = 12 sec

(A) From first equation of motion v = u+at

So [tex]24.444=0+a\times 12[/tex]

[tex]a=2.0.37m/sec^2[/tex]

So acceleration of the car will be [tex]a=2.0.37m/sec^2[/tex]

(b) From third equation of motion [tex]v^2=u^2+2as[/tex]

So [tex]24.444^2=0^2+2\times 2.037\times s[/tex]

s = 146.664 m

Distance traveled by the car in this interval will be 146.664 m

To solve this problem we will apply Ohm's law. The law establishes that the potential difference V that we apply between the ends of a given conductor is proportional to the intensity of the current I flowing through the said conductor. Ohm completed the law by introducing the notion of electrical resistance R. Mathematically it can be described as

[tex]V = IR \rightarrow R = \frac{V}{I}[/tex]

Our values are

[tex]I = 500mA = 0.5A[/tex]

[tex]V = 37V[/tex]

Replacing,

[tex]R = \frac{V}{I}[/tex]

[tex]R = \frac{37}{0.5}[/tex]

[tex]R = 74 \Omega[/tex]

Therefore the smallest resistance you can measure is [tex]74 \Omega[/tex]

Answer:

a, b) part a and b on diagram attached

c) sf = -35 i + 20 j

35 km West and 20 km North

Explanation:

For part a and b refer to the attached co-ordinate system:

Note: unit vector i is in West/East direction and unit vector j is in North/South direction.

si = -15 i + 25 j

sf-si  = -20 i - 5 j

Hence,

Mark relative position from habitat sf = si + sf/i

sf = ( -15 i + 25 j ) + ( -20 i - 5 j )

sf = -35 i + 20 j

35 km West and 20 km North

Answer:

b. E (about 329 Hz)

Explanation:

Given data:

Initial length of the string l1= 24 in

initial frequency f1= 247 Hz

changed length l2= 18 in

Then we have to find the changed frequency f2= ?

We already now that

frequency f ∝ 1/length of the string l

therefore,

[tex]\frac{f_1}{f_2} =\frac{l_1}{l_2}[/tex]

⇒[tex]{f_2}=\frac{l_1}{l_2}\times{f_1}[/tex]

⇒[tex]{f_2}=\frac{24}{18}\times{247}[/tex]

⇒[tex]{f_2}=329.33 Hz[/tex]

Answer:

277 s

Explanation:

Given data

We can find the elapsed time using the following expression.

I = C/t

t = C/I

t = 56.0 C/(0.2020 C/s)

t = 277 s

It took 277 seconds.

To solve this problem we will apply the concepts related to the kinematic equations of linear motion. From there we will define the distance as the circumference of the earth (approximate as a sphere). With the speed given in the statement we will simply clear the equations below and find the time.

[tex]R= 6370*10^3 m[/tex]

[tex]v = 239m/s[/tex]

[tex]a = 16.5m/s^2[/tex]

The circumference of the earth would be

[tex]\phi = 2\pi R[/tex]

Velocity is defined as,

[tex]v = \frac{x}{t}[/tex]

[tex]t = \frac{x}{v}[/tex]

Here [tex]x = \phi[/tex], then

[tex]t = \frac{\phi}{v} = \frac{2\pi (6370*10^3)}{239}[/tex]

[tex]t = 167463.97s[/tex]

Therefore will take 167463.97 s or 1 day 22 hours 31 minutes and 3.97seconds

Answer:

The center mass (Xcm) of the two mass = (M₁X₁ + M₂X₂)/(M₁ +M₂)

Explanation:

let the first mass = M

let the position of second  = M

The formula for the trajectory of the center of mass of a two mass oscillator connected by a spring is x(t) = A * cos(ωt + φ), where A represents the amplitude, ω is the angular frequency, and φ is the phase constant.

The formula for the trajectory of the center of mass of a two mass oscillator connected by a spring can be derived using the principles of simple harmonic motion (SHM). Let's assume that the masses are m1 and m2, and the spring constant is k. The equation for the trajectory of the center of mass can be written as:

x(t) = A * cos(ωt + φ)

where:

Explanation:

(a) Velocity is given by :

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

s is the length of the distance

t is the time

The dimension of v will be, [tex][v]=[LT^-1][/tex]      

(b) The acceleration is given by :

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

v is the velocity

t is the time

The dimension of a will be, [tex][a]=[LT^{-2}][/tex]

(c) Since, [tex]d=\int\limits{v{\cdot}dt} =[LT^{-1}][T]=[L][/tex]

(d) Since, [tex]v=\int\limits{a{\cdot}dt} =[LT^{-2}][T]=[LT^{-1}][/tex]

(e)

[tex]\dfrac{da}{dt}=\dfrac{[LT^{-2}]}{[T]}[/tex]

[tex]\dfrac{da}{dt}=[LT^{-3}]}[/tex]

Hence, this is the required solution.