PHYSICS COLLEGE

Beginning 156 miles directly east of the city of Uniontown, a truck travels due south. If the truck is travelling at a speed of 31 miles per hour, determine the rate of change of the distance between Uniontown and the truck when the truck has been travelling for 81 miles. (Do not include units in your answer, and round to the nearest tenth.)

Answers

Answer 1

Answer:

Answer:

14.3

Explanation:

The distance s as a function of time can be written as:

$s(t) = \sqrt{156^(2) + (31t)^2}$

The rate of change is the derivative of d with respect to time:

$(ds)/(dt) =\frac{961t}{\sqrt{156^(2)+(31t)^(2)}}$

The time t when the track has been traveling for 81 miles is given by:

$81 = 31t\n t = (81)/(31)$

Using t in the previous equation gives:

$(ds)/(dt)((81)/(31) ) =\frac{2511}{\sqrt{156^(2)+81^(2)}}=14.3$

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A wire with mass 90.0 g is stretched so that its ends are tied down at points 98.0 cm apart. The wire vibrates in its fundamental mode with frequency 60.0 Hz and with an amplitude of 0.300 cm at the antinodes. Part A What is the speed of propagation of transverse waves in the wire

Answers

Answer:

118 m/s

Explanation:

Given :

We know that

$f\ =\ (1)/(2l) \sqrt{(T)/(U) }$ ......Eq(1)

Where $\sqrt{(T)/(u) }$ =v

l=length

f=frequency

l= 98.0 cm= 0.98 m

f=60.0 Hz

Now from the Eq(1)

$f\ =\ (v)/(2l)$

This equation can be written as

v=2fl.............Eq(2)

Putting the value f and l in Eq(2)

v=2*60*0.98

v=117.6 m/s ~ 118 m/s

A glass plate 2.95 mmmm thick, with an index of refraction of 1.60, is placed between a point source of light with wavelength 600 nmnm (in vacuum) and a screen. The distance from source to screen is 1.25 cm. How many wavelengths are there between the source and the screen?

Answers

Answer:

$N_T=2086285.67$

Explanation:

Given;

Thickness of the glass plate, $x=2.95* 10^(-3)\ m$

refractive index of the glass plate, $n=1.6$

wavelength of light source in vacuum, $\lambda=600* 10^(-9)\ m$

distance between the source and the screen, $d=1.25\ m$

Distance travelled by the light from source to screen in vacuum:

$d_v=d-x$

$d_v=1.25-0.00295$

$d_v=1.24705\ m$

So the no. of wavelengths in the vacuum:

$N=(d_v)/(\lambda)$

$N=(1.24705)/(6* 10^(-7))$

$N\approx2.0784* 10^(6)$ .......................(1)

Now we find the wavelength of the light wave in the glass:

$n=(\lambda)/(\lambda')$

where:

$\lambda'=$ wavelength of light in the medium of glass.

$1.6=(600* 10^(-9))/(\lambda')$

$\lambda'=375* 10^(-9)\ m=375\ nm$

Now the no. of wavelengths in the glass:

$N'=(x)/(\lambda')$

$N'=(2.95* 10^(-3))/(375* 10^(-9))$

$N'=7.8667* 10^(3)$ ............................(2)

From (1) & (2):

total no. of wavelengths are there between the source and the screen:

$N_T=N+N'$

$N_T=2086285.67$

Impulse is the______of the force and time of contact

Answers

Answer:

Product

Explanation:

Impulse is defined as the average force acting on an object times the time the force acts:

Impulse = F · Δt

In this example we will use pendulum motion to actually measure the acceleration of gravity on a different planet. An astronaut on the surface of Mars measures the frequency of oscillation of a simple pendulum consisting of a ball on the end of a string. He finds that the pendulum oscillates with a period of 1.5 s. But the acceleration due to gravity on Mars is less than that on earth, gMars=0.38gearth. Later, during a journey to another planet, the astronaut finds that his simple pendulum oscillates with a period of 0.92 s. What planet is he now on?SOLUTIONSET UP Each planet has a different value of the gravitational acceleration g near its surface. The astronaut can measure g at his location, and from this he can determine what planet he's on. First we use the information about Mars to find the length L of the string that the astronaut is swinging. Then we use that length to find the acceleration due to gravity on the unknown planet.

Answers

Answer:

Explanation:

Let length of the pendulum be l . The expression for time period of pendulum is as follows

T = 2π $\sqrt{(l)/(g) }$

For Mars planet ,

1.5 = $2\pi\sqrt{(l)/(.38*9.8) }$

For other planet

.92 = $2\pi\sqrt{(l)/(g_1) }$

Squiring and dividing the two equations

$(1.5^2)/(.92^2) = (g_1)/(3.8*9.8)$

$g_1 = 9.9$

The second planet appears to be earth.

For every increase in mass the gravitational force blank If the total mass increase by effective for the gravitational force

Answers

For every increase in mass, the gravitational force increases. Gravitational force is directly proportional to the mass of the object.

What is gravitational force?

Gravitational force is the force by which an object attracts other objects into its center of mass. Earth attracts other objects gravitationally and that keep everyone stand to the ground.

Gravitational force directly proportional to the mass and inversely proportional to the distance between the objects. The expression relating the force and mass is written as:

g = G m/r²

Where G is the universal gravitational constant.

Therefore, as the mass of the object increase, the gravitational force exerted also increases. Similarly massive object experience more gravitation force by earth.

Find more on gravitational force:

brainly.com/question/12528243

#SPJ5

Answer:

Increases by the same amount.Increases by a factor of 4.

Explanation:

i took it

A person's body is producing energy internally due to metabolic processes. If the body loses more energy than metabolic processes are generating, its temperature will drop. If the drop is severe, it can be life-threatening. Suppose that a person is unclothed and energy is being lost via radiation from a body surface area of 1.36 m2, which has a temperature of 34° C and an emissivity of 0.700. Also suppose that metabolic processes are producing energy at a rate of 122 J/s. What is the temperature of the coldest room in which this person could stand and not experience a drop in body temperature

Answers

To solve this problem it is necessary to apply the concepts related to the Stefan-Boltzman law that is responsible for calculating radioactive energy.

Mathematically this expression can be given as

$P = \sigma Ae\Delta T^4$