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25% part (c) assume that d is the distance the cheetah is away from the gazelle when it reaches full speed. Derive an expression in terms of the variables d, vcmax and vg for the time, tc, it takes the cheetah to catch the gazelle.

Answers

Answer 1

Answer:

maximum speed of cheetah is

$v_1 = v_(max)$

speed of gazelle is given as

$v_2 = v_(g)$

Now the relative speed of Cheetah with respect to Gazelle

$v_(12) = v_1 - v_2$

$v_(12) = v_(max) - v_g$

now the relative distance between Cheetah and Gazelle is given initially as "d"

now the time taken by Cheetah to catch the Gazelle is given as

$d = v_(12)* t$

so by rearranging the terms we can say

$t = (d)/(v_(12))$

$t = (d)/(v_(max) - v_g)$

so above is the relation between all given variable

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A car (m = 2000 kg) is going around an unbanked curve at the recommended speed of 11m/s (24.6 MPH). (a) If the radius of the curvature of the path is 25m and the coefficient of static friction between the rubber tires and the road is µs = 0.70, does the car skid as it goes around the curve? (b) What will happen if the driver ignores the highway speed limit sign and travels at 18 m/s (40.3 MPH)? (c) What speed is safe for traveling around the curve if the road surface is wet from a recent rainstorm and the coefficient of static friction between the wet roud and the rubber tires is µs = 0.50?

A 1500 kg car moving at 25 m/s hits an initially uncompressed horizontal ideal spring with spring constant (force constant) of 2.0 × 106 N/m. What is the maximum distance the spring compresses?

Answers

Answer:

x = 0.68 meters

Explanation:

It is given that,

Mass of the car, m = 1500 kg

Speed of the car, v = 25 m/s

Spring constant of the spring, $k=2* 10^6\ N/m$

When the car hits the uncompressed horizontal ideal spring the kinetic energy of the car is converted to the potential energy of the spring. Let x is the maximum distance compressed by the spring such that,

$(1)/(2)mv^2=(1)/(2)kx^2$

$x=\sqrt{(mv^2)/(k)}$

$x=\sqrt{(1500* (25)^2)/(2* 10^6)}$

x = 0.68 meters

So, the spring is compressed by a distance of 0.68 meters. Hence, this is the required solution.

Final answer:

The maximum distance the spring compresses when a 1500 kg car moving at 25 m/s hits it, given a spring constant of 2.0 × 10⁶N/m, is approximately 0.53 meters or 53 centimeters.

Explanation:

In this specific problem, we can apply the conservation of energy principle, where the initial kinetic energy of the car is converted into potential energy stored in the spring when the car comes to a stop. The formula for kinetic energy is K = 1/2 × m× v² and for potential energy stored in a spring is U = 1/2×k × x², where m = mass of the car, v = velocity of the car, k = spring constant, and x = maximum distance the spring is compressed.

By setting the kinetic energy equal to potential energy (since no energy is lost), we get 1/2 × m×v² = 1/2×k×x². Solving this equation for x (maximum compression of the spring), we obtain x = sqrt((m×v²)/k). Substituting the given values, x = sqrt((1500 kg× (25 m/s)²) / (2.0 × 10⁶ N/m)), which yields approximately 0.53 meters or 53 centimeters. Therefore, the maximum distance the spring compresses is 53 cm.

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You are using a Geiger counter to measure the activity of a radioactive substance over the course of several minutes. If the reading of 400. counts has diminished to 100. counts after 90.3 minutes, what is the half-life of this substance?

Answers

Answer : The half-life of this substance will be, 45 minutes.

Explanation :

First we have to calculate the value of rate constant.

Expression for rate law for first order kinetics is given by:

$k=(2.303)/(t)\log(a)/(a-x)$

where,

k = rate constant = ?

t = time passed by the sample = 90.3 min

a = initial amount of the reactant = 400

a - x = amount left after decay process = 100

Now put all the given values in above equation, we get

$k=(2.303)/(90.3min)\log(400)/(100)$

$k=1.54* 10^(-2)\text{ min}^(-1)$

Now we have to calculate the half-life of substance, we use the formula :

$k=(0.693)/(t_(1/2))$

$1.54* 10^(-2)\text{ min}^(-1)=(0.693)/(t_(1/2))$

$t_(1/2)=45min$

Therefore, the half-life of this substance will be, 45 minutes.

A UHF television loop antenna has a diameter of 11 cm. The magnetic field of a TV signal is normal to the plane of the loop and, at one instant of time, its magnitude is changing at the rate 0.16 T/s. The magnetic field is uniform. What emf is induced in the antenna

Answers

Answer:

-0.00152 V

Explanation:

Parameters given:

Diameter of the loop = 11 cm = 0.11m

Rate of change of magnetic field, dB/dt = 0.16 T/s

Radius of the loop = 0.055m

The area of the loop will be:

A = pi * r²

A = 3.142 * 0.055²

A = 0.0095 m²

The EMF induced in a loop of wire due to the presence of a changing magnetic field, dB, in a time interval, dt, is given as:

EMF = - N * A * dB/dt

In this case, there's only one loop, so N = 1.

Therefore:

EMF = -1 * 0.0095 * 0.16

EMF = -0.00152 V

The negative sign indicates that the current flowing through the loop acts opposite to the change in the magnetic field.

Answer:

The induced emf is 0.00152 V

Explanation:

Given data:

d = 11 cm = 0.11 m

$r=(d)/(2) =(0.11)/(2) =0.055m$

The area is:

$A=\pi r^(2) =\pi *(0.055^(2) )=0.0095m^(2)$

The induced emf is:

$E=-A(dB)/(dt) =-(0.0095)*(0.16)=-0.00152V$

The negative indicates the direction of E.

A motorcycle is following a car that is traveling at constant speed on a straight highway. Initially, the car and the motorcycle are both traveling at the same speed of 19.0m/s , and the distance between them is 52.0m . After t1 = 5.00s , the motorcycle starts to accelerate at a rate of 5.00m/s^2. a. How long does it take from the moment when the motorcycle starts to accelerate until it catches up with the car? In other words. b. Find t2−t1

Answers

Answer:

$t\approx4.561\ s$

Explanation:

Given:

initial speed of car and motorcycle, $v=19\ m.s^(-1)$

initial distance between the car and motorcycle, $s=52\ m$

time after which the motorcycle starts to accelerate, $t_1=5\ s$

rate of acceleration of motorcycle, $a=5\ m.s^(-2)$

The initially the relative velocity of the motorcycle is zero with respect to car.

Now using the equation of motion in the relative quantities:

$s=u.t+(1)/(2) .a.t^2$

here:

$s =$ relative distance of motorcycle with respect to the car

$u=$ initial relative velocity of the motorcycle with respect to the car

$t=$ time taken to cover the distance gap from the car.

$52=0+0.5* 5* t^2$

$t\approx4.561\ s$

Consider a space shuttle which has a mass of about 1.0 x 105 kg and circles the Earth at an altitude of about 200.0 km. Calculate the force of gravity that the space shuttle experiences

Answers

Final answer:

The force of gravity that the space shuttle experiences is 9.8 x 10^5 Newtons.

Explanation:

To calculate the force of gravity that the space shuttle experiences, we can use the equation F = mg, where F represents the force of gravity, m is the mass of the object, and g is the acceleration due to gravity (approximately 9.8 m/s² on Earth). In this case, the mass of the space shuttle is given as 1.0 x 10^5 kg. However, we need to convert the altitude of the shuttle into meters, so 200.0 km becomes 200,000 meters.

Now we can calculate the force of gravity:

F = (1.0 x 10^5 kg)(9.8 m/s²)

F = 9.8 x 10^5 N

Therefore, the space shuttle experiences a force of gravity of 9.8 x 10^5 Newtons.

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The intensity of a sound wave at a fixed distance from a speaker vibrating at 1.00 kHz is 0.750 W/m2. (a) Determine the intensity if the frequency is increased to 2.20 kHz while a constant displacement amplitude is maintained.(b) Calculate the intensity if the frequency is reduced to 0.250 kHz and the displacement amplitude is quadrupled.

Answers

Final answer:

The intensity of sounds is dependent on the square of the amplitude, not the frequency. Therefore, the intensity of sound remains the same when frequency is altered but the amplitude is constant. When the amplitude is quadrupled, the intensity of the sound becomes sixteen times greater.

Explanation:

In the field of physics, the intensity of a sound wave is defined as the power per unit area carried by the wave. This question involves calculating the change in sound wave intensity when the frequency and displacement amplitude of the source are altered.

(a) When the frequency is increased to 2.20 kHz while keeping the displacement amplitude constant, the intensity does not change, as the intensity in this case is not dependent on the frequency but on the square of the amplitude. Therefore, the intensity remains 0.750 W/m2.

(b) When the frequency is reduced to 0.250 kHz and the displacement amplitude is quadrupled, the intensity changes. Since the intensity of a sound wave is proportional to the square of the amplitude, by quadrupling the amplitude, the intensity will become 16 times greater (since 4 squared is 16). Hence, the new intensity will be 16 * 0.750 = 12 W/m2.

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