A bat can detect small objects, such as an insect, whose size is approximately equal to one wavelength of the sound the bat makes. If bats emit a chirp at a frequency of 7.84 104 Hz, and if the speed of sound in air is 343 m/s, what is the smallest insect a bat can detect?

Answer:

Doubled

Explanation:

F = 1.8 x 10^-6 N, q1 = q2

The force between the two charges is directly proportional to the product of two charges and inversely proportional to the square of distance between them.

Now one of the charge is doubled but the distance remains same so the force between the two charges becomes doubled.

Answer:

[tex]x = -1.437 cm[/tex]

Explanation:

The general equation for position of Simple harmonic motion is given as:

[tex]x = A sin(\omega t)[/tex]          ........(1)

where,

x = Position of the wave

A = Amplitude of the wave

ω = Angular velocity

t = time

In this case, the amplitude is just half the range,

thus,

[tex]A =\frac{3cm}{2}=1.5cm[/tex]  (Given range = 3cm)

A = 1.5 cm  

Now, The angular velocity is given as:

[tex]\omega=\frac{2\pi}{T}[/tex]

Where, T = time period of the wave =0.27s (given)

[tex]\omega=\frac{2\pi}{0.27s}[/tex]

or

[tex]\omega=23.27s^{-1}[/tex]

so, at time t = 55 s, the equation (1) becomes as:

[tex]x = 1.5 sin(23.27\times 55)[/tex]

on solving the above equation we get,

[tex]x = -1.437 cm[/tex]

here the negative sign depicts the position in the opposite direction of +x

To determine the position of a machine part in simple harmonic motion after 55 seconds, we utilize the period and amplitude to establish the displacement function. The part completes full periods, and at t = 55 seconds, it returns to the central position. The exact position can be calculated using the cosine function for displacement in SHM.

The machine part is undergoing simple harmonic motion (SHM), which is a type of periodic oscillation. Given the period T is 0.27 seconds and the range of 3.0 cm, which is the total distance from the maximum position in one direction to the maximum in the other, we can determine the amplitude (A) is half of this range, which is 1.5 cm or 0.015 meters.

The displacement in simple harmonic motion as a function of time can be modeled using the cosine function:
x(t) = A cos(2πt/T). The part starts at its central position at t = 0 and moves in the +x direction. To find its position at t = 55 s, we can substitute these values into the displacement formula. Since 55 is not a multiple of the period, we need to account for the number of full periods that have passed. After 55 seconds, which is 55/0.27 or approximately 203.7 periods, the part will have completed 203 full periods and be halfway through the 204th cycle. The next step is to find the remainder of the division to obtain the phase of the cycle at t = 55 s, which is 55 mod 0.27. The cosine function is periodic with a period of 2π, hence we consider the fraction of the period completed to model the motion at that specific time.

The exact calculation for the phase at t = 55 leads us to determine that the machine part's displacement will be back at the central position. Since the cosine function repeats every 2π and the part was initially at the central position moving in the +x direction (which corresponds to cos(0) = 1), after an even number of complete cycles at t = 55 s, the part will again be at the central position (x = 0) and about to start moving in the +x direction.

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[tex]E_{k}[/tex] = 652500J

The easiest way to solve this problem is using the kinetic energy equation:

[tex]E_{k}[/tex] = mv²/2

A 1450kg sports car traveling down the road at a speed of 30m/s is kinectic energy will be:

[tex]E_{k}[/tex] = [(1450kg)(30m/s)²]/2 = 1305000kg-m²/s²/2

[tex]E_{k}[/tex] = 652500J

Answer:

652500 J

Explanation:

Answer:

Magnetic field = 1.41 T

Explanation:

Magnetic field of solenoid, B = μnI

μ = 4π x 10⁻⁷N/A²

Number of turns per meter, [tex]n=\frac{N}{L}=\frac{5601}{9.1\times 10^{-2}}=6.15\times 10^4turns/m[/tex]

Current, i = 18.2 A

B = μnI = 4π x 10⁻⁷ x 6.15 x 10⁴ x 18.2 = 1.41 T

Magnetic field = 1.41 T

Answer:

Explanation:

The efficiency of a refrigerator is defined in the terms of coefficient of performance (COP).

The ratio of amount of heat in cold reservoir to the work done is termed as the COP.

COP = QL / W

COP = T2 / (T1 - T2)

Where, T1 be the temperature of hot reservoir, T2 be the temperature of cold reservoir.

Answer: Cells require a constant input of energy to maintain their high level of organization.

Explanation:

According to the second principle of thermodynamics:

"The amount of entropy in the universe tends to increase over time ".

That is, in any cyclic process, entropy will increase, or remain the same.

So, in this context, entropy is a thermodynamic quantity defined as a criterion to predict the evolution or transformation of thermodynamic systems. In addition, it is used to measure the degree of organization of a system.  

In other words: Entropy is the measure of the disorder of a system and is a function of state.

Now, in the specific case of cells, in order to maintain their high level of organization, which goes against the natural tendency to disorder, a constant input of energy is necessary to maintain that level.

The Second Law of Thermodynamics explains how energy transfers result in some energy being lost in unusable forms, such as heat energy.

The Second Law of Thermodynamics states that in every energy transfer, some amount of energy is lost in a form that is unusable, often in the form of heat energy. This principle explains why energy transfers are never completely efficient, leading to increased disorder or entropy in the universe.

Answer:

Centripetal acceleration of the car is 17.4 m/s²

Explanation:

It is given that,

Radius of the track, r = 57 m

Speed of car, v = 31.5 m/s

We need to find the centripetal acceleration of the race car. The formula for the centripetal acceleration is given by :

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

[tex]a=\dfrac{(31.5\ m/s)^2}{57\ m}[/tex]

[tex]a=17.4\ m/s^2[/tex]

So, the centripetal acceleration of the race car is 17.4 m/s². Hence, this is the required solution.

Answer:

8.5 kg

Explanation:

Angular momentum is conserved.

I₁ ω₁ = I₂ ω₂

I ω₀ = (I + mr²) ω

(49 kg m²) (40 rpm) = (49 kg m² + m (1.2 m)²) (32 rpm)

61.25 kg m² = 49 kg m² + (1.44 m²) m

m = 8.5 kg

The mass of the clay is 8.5 kg.

Answer:

4.39 x 10^-19 J

Explanation:

q1 = 1.6 x 10^-19 C

q2 = - 1.6 x 10^-19 C

r1 = 3 x 10^-10 m

r2 = 7 x 10^-10 m

The formula for the potential energy is given by

U1 = k q1 q2 / r1 = - (9 x 10^9 x 1.6 x 10^-19 x 1.6 x 10^-19) / (3 x 10^-10)

U1 = - 7.68 x 10^-19 J

U2 = k q1 q2 / r2 = - (9 x 10^9 x 1.6 x 10^-19 x 1.6 x 10^-19) / (7 x 10^-10)

U2 = - 3.29 x 10^-19 J

Change in potential energy is

U2 - U1 = - 3.29 x 10^-19 + 7.68 x 10^-19 = 4.39 x 10^-19 J

Final answer:

The change in electric potential energy is calculated using the formula ΔU = kq1q2/r, with the distances and charges plugged into the formula, resulting in a negative value, indicating a decrease in potential energy.

Explanation:

The change in electric potential energy when moving an electron from 3 × 10-10 m to 7 × 10-10 m from a proton can be calculated using the formula for the potential energy between two charges: ΔU = kq1q2/r. Plugging in the constants and distances, we have ΔU = k(-e)(+e)(1/7 × 10-10 - 1/3 × 10-10) where k is the Coulomb's constant (8.987 × 109 Nm2/C2) and e is the elementary charge (1.602 × 10-19 C).

The result, calculated in joules, will be negative, signifying that the electron is moving to a region of lower potential energy as it moves away from the proton.

Answer:

F0 / 81

Explanation:

Let the two charges by Q and q which are separated by d.

By use of coulomb's law

F0 = k Q q / d^2      ......(1)

Now the charges are decreased by factor of 9.

Q' = Q / 9

q' = q / 9                 ......(2)

Now the Force is

F' = k Q' q' / d^2

F' = k (Q /9) (q / 9) / d^2

F' = k Q q / 81d^2

F' = F0 / 81      

Answer:

a) 0.124 m

b) 0.93 ms⁻¹

c) 0.5 k A² cos ² ( ωt )  

Explanation:

1) Potential energy = U = 0.5 k A² , where A is the amplitude and k = 850 N/m is the spring constant.

0.5 ( 850) (A² ) = 6.5

⇒ A = 0.124 m = Amplitude.

b) From energy conservation,  0.5 m v² =  6.5

⇒ speed = v = 0.93 ms⁻¹

c) If x = A cos ωt ,

Potential energy = 0.5 k A² = 0.5 k A² cos ² ( ωt )  

Answer:

option (c) - 10 j

Explanation:

F = (6 i + 4 j - 2 k) N

r1 = (1.5, 3, -4.5) m = (1.5 i + 3j - 4.5 k) m

r2 = (4, -2.5, - 3) m = (4 i - 2.5 j - 3 k) m

displacement, r = r2 - r1 = ( 2.5 i - 5.5 j + 1.5 k) m

Work done is defined as the dot product of force vector and teh displacement vector.

[tex]W = \overrightarrow{F}.\overrightarrow{r}[/tex]

W = (6 i + 4 j - 2 k) . ( 2.5 i - 5.5 j + 1.5 k)

W = 15 - 22 - 3 = - 10 J

Explanation:

When a tennis ball is thrown against a wall it appears to bounce back with exactly the same speed as it struck the wall. The momentum will remain conserved in this case. The law of conservation of momentum states that when no external force is acting on a system, the initial momentum is equal to the final momentum.

Here, this is a case of inelastic collision. The kinetic energy is not conserved in this case. Some of the energy is lost in the form of heat, sound etc.

Momentum conservation in a tennis ball-wall collision implies that the slight loss of the ball's speed is balanced by the wall slightly flexing or moving, ensuring total system momentum is conserved. The ball's kinetic energy, however, is partly lost in inelastic forms such as heat, sound, or deformation.

When a tennis ball is thrown against a wall and bounces back with a slight reduction in speed, from 15 m/s to 14 m/s, the momentum is not completely conserved in terms of the ball alone because some of it is transferred to the wall. Despite the wall being much more massive and thus not visibly moving, the wall does exert a force on the ball and so it must move very slightly or flex in response, showing that the momentum of the entire system, including the wall, is conserved.

(a) Even though the tennis ball changes its momentum upon collision with the wall, the conservation of momentum principle tells us that the rest of the system must account for this change. Since the ball loses a bit of speed, resulting in a decrease in momentum, the wall gains this tiny amount of momentum. Because the wall has a vast mass compared to the tennis ball, its movement is negligible and often imperceptible. (b) The slight decrease in the rebound speed indicates that not all of the ball's kinetic energy is recovered post-collision. Kinetic energy is dissipated due to factors such as sound generation, deformation of the ball, or heat production, leading to an inelastic collision.

Answer:

b.  [tex]\Delta KE = 390 eV[/tex]

Explanation:

As we know that the electric field due to infinite line charge is given as

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

here we can find potential difference between two points using the relation

[tex]\Delta V = \int E.dr[/tex]

now we have

[tex]\Delta V = \int(\frac{\lambda}{2\pi \epsilon_0 r}).dr[/tex]

now we have

[tex]\Delta V = \frac{\lambda}{2\pi \epsilon_0}ln(\frac{r_2}{r_1})[/tex]

now plug in all values in it

[tex]\Delta V = \frac{12\times 10^{-9}}{2\pi \epsilon_0}ln(\frac{1+5}{1})[/tex]

[tex]\Delta V = 216ln6 = 387 V[/tex]

now we know by energy conservation

[tex]\Delta KE = q\Delta V[/tex]

[tex]\Delta KE = (e)(387V) = 387 eV[/tex]

The distance at which the puck of mass 0.13 kg, traveling across the ice at speed 17.4 m/sec will stop is 131.2 meters.

Given to us

Mass of the puck, m = 0.13kg

The velocity of the ice, u = 17.4 m/sec

Friction force, f = 0.15 N

We know we want to stop the puck, therefore, the final velocity of the puck will be 0.

v = 0

We know that according to the first law of motion,

Force = mass x acceleration

F = m x a

Substitute the value,

[tex]0.15 = 0.13 \times a[/tex]

[tex]a = 1.1538\rm\ m/s^2[/tex]

As we know that the final velocity of the puck will be 0, therefore, there will be a deceleration in the puck.

a = -1.1538 m/s².

Thus, the acceleration of the ice puck is -1.1538 m/s².

We know that according to the third equation of the motion,

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

substitute the values,

[tex]0^2-(17.4)^2 = 2(-1.1538)s[/tex]

s = 131.2012 = 131.2 meters

Hence, the distance at which the puck of mass 0.13 kg, traveling across the ice at speed 17.4 m/sec will stop is 131.2 meters.

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Final answer:

The stopping distance of a hockey puck  is 131.1 m after rounding to the nearest tenth of a meter.

Explanation:

To calculate the stopping distance of a hockey puck, we first need to determine the deceleration caused by the frictional force.

The formula for deceleration due to friction is a = f/m, where a is the acceleration (deceleration, in this case, as it's negative), f is the frictional force, and m is the mass of the puck. Given that the frictional force f is 0.15 N and the mass m of the puck is 0.13 kg, we can calculate the deceleration as follows:

a = f/m = 0.15 N / 0.13 kg ≈ 1.15 m/s2

Now, to find the stopping distance we can use the formula d = v2 / (2×a), where d is the stopping distance, v is the initial speed, and a is the deceleration. Plugging in the initial speed v = 17.4 m/s and the deceleration a = 1.15 m/s2, we get:

d = (17.4 m/s)2 / (2 × 1.15 m/s2) ≈ 131.06 m

To round to the nearest 0.1 m, the stopping distance of the puck is 131.1 m.

Answer:

i think 7

Explanation:

Answer:

2440.24 J

Explanation:

Moment of inertia, I1 = 5 kg m^2

frequency, f1 = 3 rps

ω1 = 2 x π x f1 = 2 x π x 3 = 6 π rad/s

Moment of inertia, I2 = 2 kg m^2

Let the new frequency is f2.

ω2 = 2 x π x f2

here no external torque is applied, so the angular momentum remains constant.

I1 x ω1 = I2 x ω2

5 x 6 π = 2 x 2 x π x f2

f2 = 7.5 rps

ω2 = 2 x π x 7.5 = 15 π

Initial kinetic energy, K1 = 1/2 x I1 x ω1^2 = 0.5 x 5 x (6 π)² = 887.36 J

Final kinetic energy, K2 = 1/2 x I2 x ω2^2 = 0.5 x 3 x (15 π)² = 3327.6 J

Work done, W = Change in kinetic energy = 3327.6 - 887.36 = 2440.24 J

The cyclist is behind the car, and the speed of the car is approximately 13.3 m/s.

The question you have asked is related to the Doppler effect. When a car's horn sounds the note A (440 Hz) at rest, the frequency heard by a stationary observer is 440 Hz. However, when the car is moving and approaching a bicyclist, the frequency heard by the bicyclist is lower (415 Hz). From this information, we can determine the answers to the two parts of your question:

(v - vc)/(vs - vc) = fs/fr

Where v is the speed of sound, vc is the speed of the car, vs is the speed of the cyclist, fs is the frequency heard by the cyclist, and fr is the frequency produced by the car. By substituting the known values (v = 343 m/s, fs = 415 Hz, and fr = 440 Hz) and solving for vc, we can find the speed of the car.

By plugging in the values, we get:
(343 - vc)/(0 - vc) = 415/440

Simplifying the equation further, we get:
vc = 343(415 - 440) / (415)

After substituting the values and solving the equation, we find that:
vc ≈ 13.3 m/s

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The sound heard by the bicyclist is lower than the frequency of the car's horn due to the Doppler effect. The bicyclist is behind the car and the speed of the car can be calculated using the formula for the Doppler effect.

The phenomenon described in the question is an example of the Doppler effect. The Doppler effect is the apparent change in frequency of a sound wave due to the relative motion between the source of the sound and the observer. In this case, the frequency of the horn sounds lower to the bicyclist due to their relative motion with the car.

(a) The frequency heard by the bicyclist is lower than the frequency of the car's horn, indicating that the bicyclist is behind the car.

(b) To find the speed of the car, we can use the formula for the Doppler effect:

fo = fs (V + Vo) / (V - Vs)

Where fo is the observed frequency, fs is the source frequency, V is the speed of sound, Vo is the speed of the observer, and Vs is the speed of the source.

In this case, fo is 415 Hz, fs is 440 Hz, and Vo is one-third the speed of the car. We can rearrange the formula and solve for V (the speed of the car):

V = Vs (fo / fs) - Vo

Plugging in the values, we find that the speed of the car is approximately 40 m/s.

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Here, given that,

Horizontal velocity V_x = 4.5 x 10⁵ m /s

vertical electric field E_y = 9.6 x 10³ N/C

so, we get,

acceleration in vertical direction a_y = force on proton / mass

= 9.6 x 10³ x 1.6 x 10⁻¹⁹ / 1.67 x 10⁻²⁷

= 9.2 x 10¹¹ m /s²

a ) In horizontal direction it will move with uniform velocity

time required = distance / velocity

= 5 x 10⁻² / 4.5 x 10⁵

= 1.11 x 10⁻⁷ s

b ) vertical displacement in time 1.11 x 10⁻⁷ s

h = 1/2 x at² , initial vertical velocity is zero.

= .5 x 9.2 x 10¹¹ x ( 1.11 x 10⁻⁷ )²

= 5.66 x 10⁻³ m

=5.66 mm

c )Horizontal velocity will be unchanged ie 4.5 x 10⁵ m /s

vertical velocity will change due to acceleration

= u + at

0 + 9.2 x 10¹¹ x1.11 x 10⁻⁷

= 10.21 x 10⁴ m / s

= 1.021 x 10⁵ m /s

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The time interval required for the proton to travel 5.00 cm horizontally is 1.11 × 10^-7 s. The vertical displacement of the proton during this time interval can be calculated using the formula displacement = 0.5 * (qE/m) * (time^2). The horizontal component of the proton's velocity remains unchanged, while the vertical component changes due to the electric field force.

(a) To find the time interval required for the proton to travel 5.00 cm horizontally, we need to use the formula: time = distance/speed. In this case, the distance is 5.00 cm, which is 0.05 m, and the speed is 4.50 × 10^5 m/s. So, the time interval is 0.05 m / (4.50 × 10^5 m/s) = 1.11 × 10^-7 s.

(b) The vertical displacement of the proton during the time interval can be found using the formula: displacement = 0.5 * acceleration * time^2. Since the proton is not affected by gravity, the only force acting on it is the electric field force, which is given by: F = qE, where q is the charge of the proton and E is the electric field magnitude. The acceleration is equal to F/m, where m is the mass of the proton. The vertical displacement is then given by: displacement = 0.5 * (qE/m) * (time^2). Substituting q = +e (the charge of a proton) and m = mass of a proton, we can calculate the vertical displacement.
(c) After traveling 5.00 cm horizontally, the horizontal component of the proton's velocity remains unchanged since there are no forces acting on it in the horizontal direction. The vertical component of the velocity, on the other hand, changes due to the electric field force. To find the new vertical velocity, we can use the formula: final velocity = initial velocity + acceleration * time, where the acceleration is given by F/m and the time is the time interval we calculated in part (a).

Answer:

0.7

Explanation:

f = 50 hz, R = 40 ohm, L = 0.3 h, C = 60 uC = 60 x 10^-6 c

XL = 2 x 3.14 x f x L = 2 x 3.14 x 50 x 0.3 = 94.2 ohm

Xc = 1 / ( 2 x 3.14 x 50 x 60 x 10^-6) = 53.078 ohm

Impedance is Z.

Z^2 = R^2 + ( XL - Xc)^2

Z^2 = 40^2 + (94.2 - 53.078)^2

Z^2 = 1600 + 1691.019

Z = 57.37 ohm

The power factor is given by

CosФ = r / Z = 40 / 57.37 = 0.697 = 0.7

The power factor of a circuit is the ratio of real power to apparent power and is calculated by the cosine of the phase angle between voltage and current. With the information given, we cannot calculate the power factor of the RLC series circuit because the rms voltage of the source is required but not provided.

The power factor of a circuit represents the cosine of the phase angle between the voltage and the current in an AC circuit. In an RLC circuit, like the one described, the power factor can be calculated by dividing the real power (measured in watts) by the apparent power (volt-amps). To find the power factor, we need to first calculate the impedance (Z) of the circuit using the formula Z = √(R² + (XL - XC)²), where XL is the inductive reactance and XC is the capacitive reactance.

Inductive reactance (XL) is given by XL = 2πfL, and capacitive reactance (XC) by XC = 1/(2πfC). Since we know that f = 50 Hz, L = 0.30 H,  and C = 60 μF, we can calculate XL and XC. Substituting R for the resistance in the circuit, we can find the impedance. After finding the impedance, we can calculate the real power (P) using P = I²R, where I is the rms current.

From the real power and the apparent power, which is IZ, we can find the power factor by calculating P/(IZ). However, we need the actual values of all elements, including the voltage, to complete these calculations. With the information provided in the question, we can't calculate the power factor because the rms voltage of the source isn't given.

Answer:

248

Explanation:

L = Inductance of the slinky = 130 μH = 130 x 10⁻⁶ H

[tex]l[/tex] = length of the slinky = 3 m

N = number of turns in the slinky

r = radius of slinky = 4 cm = 0.04 m

Area of slinky is given as

A = πr²

A = (3.14) (0.04)²

A = 0.005024 m²

Inductance is given as

[tex]L = \frac{\mu _{o}N^{2}A}{l}[/tex]

[tex]130\times 10^{-6} = \frac{(12.56\times 10^{-7})N^{2}(0.005024)}{3}[/tex]

N = 248

Answer:

Option D is the correct answer.

Explanation:

We equation for elongation

   [tex]\Delta L=\frac{PL}{AE}[/tex]      

Here we need to find weight required,

We need to stretch a steel road by 2 mm, that is ΔL = 2mm = 0.002 m

[tex]A=\frac{\pi d^2}{4}=\frac{\pi ({4\times 10^{-3}})^2}{4}=1.26\times 10^{-5}m^2[/tex]

E = 200GPa = 2 x 10¹¹ N/m²

L=2m

Substituting

[tex]0.002=\frac{P\times 2}{1.26\times 10^{-5}\times 2\times 10^{11}}\\\\P=2520N=256.4kg[/tex]

Option D is the correct answer.

Answer:

Option C is the correct answer.

Explanation:

Heat required to melt solid in to liquid is calculated using the formula

            H = mL, where m is the mass and L is the latent heat of fusion.

Latent heat of fusion for water = 333.55 J/g

Mass of ice = 0.3 kg = 300 g

Heat required to convert 0.3 kilogram of ice at 0°C to water at the same temperature

          H = mL = 300 x 333.55 = 100,375 J

Option C is the correct answer.

Answer:

Energy, [tex]E=2.65\times 10^{-25}\ J[/tex]

Explanation:

It is given that,

The MRI (Magnetic Resonance Imaging) body scanners used in hospitals operate at a frequency of 400 MHz,

[tex]\nu=400\ MHz=400\times 10^6\ Hz=4\times 10^8\ Hz[/tex]

We need to find the energy for a photon having this frequency. The energy of a photon is given by :

[tex]E=h\nu[/tex]

[tex]E=6.63\times 10^{-34}\ J-s\times 4\times 10^8\ Hz[/tex]

[tex]E=2.65\times 10^{-25}\ J[/tex]

So, the energy of the photon is [tex]2.65\times 10^{-25}\ J[/tex]. Hence, this is the required solution.

Answer:

Option (c)

Explanation:

The rate of change of linear momentum is called force.

As the linear motion terms are analogous to the terms in rotational motion.

So, the rate of change of angular momentum is called torque.

Answer:

1.4719 m per sec

Explanation:

Hello

Kinetic energy is the energy associated with the movement of objects. Although there are many forms of kinetic energy  

the formula to use is

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

where m is the mass of the object and v the velocity

lets see the kinetic energy of the sprinter running

[tex]E=\frac{65 Kg*10(\frac{m}{s} ^)){2} }{2} \\\\E=\frac{65 *100 }{2} \\E=3250 Joules\\\\[/tex]

Now, the elephant must have the same kinetic energy

[tex]E=\frac{m*v_{2} ^{2} }{2} \\\\E*2=m*v_{2} ^{2}\\ \frac{2E}{m} =v_{2} ^{2} \\\sqrt{\frac{2E}{m} } =v_{2}  \\\\\\v_{2} =\sqrt{\frac{2*3250}{3000} }\\ \\v_{2} =1.4719 \frac{m}{s} \\\\[/tex]

it works only the positive root, so the elephant must to  walk  to 1.4719 m/s to have the same kinetic energy.

Have a great day

A 3000-kg elephant needs to move at a speed of approximately 1.183 m/s to have the same kinetic energy as a 65.0-kg sprinter running at 10.0 m/s.

The subject of the question falls into the category of physics, specifically dealing with the concept of kinetic energy and motion. Kinetic energy is given by the equation KE = 1/2 m v², where m is the mass and v is the velocity of the object. In this case, we want the elephant and the sprinter to have the same kinetic energy.

To find the velocity at which the elephant must move, we set the kinetic energy of the elephant (1/2 * 3000 kg * v²) equal to the kinetic energy of the sprinter (1/2 * 65.0 kg * (10.0 m/s)²) and solve for v (velocity of the elephant).

This gives us v = sqrt((1/2 * 65 kg * (10 m/s)²) / (1/2 * 3000 kg)) = 1.183 m/s. Therefore, a 3000-kg elephant must move at a speed of approximately 1.183 m/s to have the same kinetic energy as a 65.0-kg sprinter running at 10.0 m/s.

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Answer:

Part a)

charge through the battery is

[tex]Q = 9.72 \times 10^5 C[/tex]

Part b)

Maximum current is

i = 386 A

Explanation:

Part a)

As we know that the rating of battery is given as

[tex]R = 270 A h[/tex]

here we also know that the charge given by the battery is same as the capacity of the battery

so we will have

[tex]Q = i t[/tex]

[tex]Q = (270 A)(3600 s)[/tex]

[tex]Q = 9.72 \times 10^5 C[/tex]

Part b)

Now we know that current in the wire is given by

[tex]i = \frac{Q}{t}[/tex]

now plug in all values in it

[tex]i = \frac{9.72 \times 10^5}{42\times 60}[/tex]

[tex]i = 386 A[/tex]

Answer:

Mass of the jet, m = 28511.5 kg

Explanation:

It is given that,

Initial velocity if the jet, u = 0

Final velocity of the jet, v = 140 mi/h = 62.58 m/s

Time, t = 1.95 s

Average force, [tex]F=9.15\times 10^5\ N[/tex]

We need to find the mass of the jet. According to second law of motion as :

F = m × a

[tex]F=m\times \dfrac{v-u}{t}[/tex]

[tex]m=\dfrac{F.t}{v-u}[/tex]

[tex]m=\dfrac{9.15\times 10^5\ N\times 1.95\ s}{62.58\ m/s-0}[/tex]

[tex]m=28511.5\ kg[/tex]

So, the mass of the jet is 28511.5 kg. Hence, this is the required solution.

Answer:

[tex]v_f = 182.3 m/s[/tex]

Explanation:

As we know that total impulse is given as the product of mass and change in velocity

so here we will have

[tex]I = m(\Delta v)[/tex]

here we will have

[tex]I = 35.0 N s[/tex]

also we know that mass of the rocket is

m = 0.192 kg

now we will have

[tex]35 = 0.192(v_f - 0)[/tex]

[tex]v_f = 182.3 m/s[/tex]

Final answer:

The half-life of radon-222 is 3.82 days.

It take approximately 9.34 days for the sample to decay to 15% of its original amount.

Explanation:

To answer the student's question regarding the half-life of radon-222, we turn to the information provided which states that radon-222 (Rn-222) has a half-life of 3.823 days. Considering this data:

(a) The half-life of radon-222 is 3.82 days.

(b) To determine how long it will take for the sample to decay to 15% of its original amount, we can utilize the radioactive decay formula:

N(t)=N_0(1/2)^(t/T)

Where:

N(t) = remaining quantity after time t

N_0 = initial quantity

T = half-life of the substance

t = time elapsed

For radon-222, substituting N(t)/N_0 = 0.15 and T = 3.823 days into the formula and solving for t, we can generate an accurate answer. The calculation would reveal that it takes approximately 9.34 days for the sample to decay to 15% of its original amount.