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The acceleration of a particle moving along a straight line is given by a = −kt2 m/s2 where k is a constant and time t is in seconds. The initial velocity of the particle at t = 0 is v0 = 12 m/s and the particle reverses it direction of motion at t = 6 s. Determine the constant k and the displacement of the particle over the same 6-second interval of motion. Ans: k = 1/6 m/s4, Δs = 54 m

Answers

Answer 1

Answer:

Answer:

$X - Xo = 54m$

k = 1/18

Explanation:

Data:

a = -k $t^(2)$ $(m)/(s^(2) )$

to = 0s Vo = 12m/s

t = 6s the particle chage it's moviment, so v = 0 m/s

We know that acceleration is the derivative of velocity related to time:

$a = (dV)/(dT)$

rearranging...

$a*dT = dV$

Then, we must integrate both sides:

$\int\limits^f_i {dV} \, dV =-k \int\limits^f_i {t^(2) } \, dT$

$V - Vo = -k(t^(3) )/(3)$

V = 0 because the exercise says that the car change it's direction:

$0 - 12 = -k(6^(3) )/(3)$

k = 1/6

In order to find X - Xo we must integer v*dT = dX

$V - Vo = -k(t^(3) )/(3)$

so...

$(Vo -k(t^(3) )/(3))dT = dX$

$\int\limits^f_i {dX} \, dX = \int\limits^f_i {Vo -k(t^(3) )/(3) } \, dT$

integrating...

$X - Xo = Vot -k(t^(4) )/(12)$

$X - Xo = 12*6 -(1)/(6)* (6^(4) )/(12)$

X - Xo = 54m

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Two children, Ferdinand and Isabella, are playing with a waterhose on a sunny summer day. Isabella is holding the hose in herhand 1.0 meters above the ground and is trying to spray Ferdinand,who is standing 10.0 meters away. I know so far that she cannotspray Ferdinand at the current position and with the curreentspeed of spray. I got stuck inthe following question:To increase the range of the water, Isabellaplaces her thumb on the hose hole and partially covers it. Assuming that the flow remains steady, what fraction f of the cross-sectional area of the hose hole does shehave to cover to be able to spray her friend?

Assume that the cross section of the hoseopening is circular with a radius of 1.5 centimeters.

Answers

Answer:

Explanation:

According to the formula below, with constant flow rate, the less cross-sectional area there is, the faster water would flow, and vice-versa

$\dot{V} = A*v$

where $\dot{V} m^3/s$ is the constant flow rate,

A m2 is the cross-sectional area

v m/s is the water speed.

So if the flow rate is constant, when A decreases, v must increase proportionally.

Since this problem is missing the water speed, here are the steps to solve it

Step 1: find the new spray speed that could reach Ferdinand

Step 2: find the ratio of this new spray speed to the old one, this will also be the ratio of the old cross-sectional area to the new one.

Step 3: find the fraction f of the cross-sectional area of the hose hole

An artificial satellite circling the Earth completes each orbit in 144 minutes. (a) Find the altitude of the satellite.

Answers

Answer:

Explanation:

given,

time taken to complete the each orbit = 144 minutes

t = 144 x 60 = 8640 s

mass of the earth = 5.98 x 10²⁴ Kg

radius of earth,R = 6.38 x 10⁶ m

Using Kepler's 3rd law

$T^2 = ((4\pi^2)/(GM_e))r^3$

$r = ((T^2GM_e)/(4\pi^2))^(1/3)$

$r = ((8640^2* 6.67 * 10^(-11)* 5.98* 10^(24))/(4\pi^2))^(1/3)$

r = 9.1 x 10⁶ m

the altitude of the satellite

H = r - R

H = 9.1 x 10⁶ - 6.38 x 10⁶

H = 2.72 x 10⁶ m

A 81.0 kg diver falls from rest into a swimming pool from a height of 4.70 m. It takes 1.84 s for the diver to stop after entering the water. Find the magnitude of the average force exerted on the diver during that time.

Answers

Explanation:

The given data is as follows.

height (h) = 4.70 m, mass = 81.0 kg

t = 1.84 s

As formula to calculate the velocity is as follows.

$\nu$ = 2gh

= $2 * 9.8 m/s^(2) * 4.70 m$

= 92.12 $s^(2)$

As relation between force, time and velocity is as follows.

F = $(m * \nu)/(t)$

Hence, putting the given values into the above formula as follows.

F = $(m * \nu)/(t)$

= $(81.0 kg * 92.12 s^(2))/(1.84 s)$

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Thus, we can conclude that the magnitude of the average force exerted on the diver during that time is 4055.28 N.

In a super-heater (A) pressure rises, temperature drops (B) pressure rises, temperature remains constant (C) pressure remains constant and temperature rises (D) both pressure and temperature remains constant

Answers

Answer:

i believe that it is d

Explanation:

Final answer:

In a super heater, the temperature of the steam rises while the pressure remains constant. This process helps to remove the last traces of moisture from the saturated steam.

Explanation:

In a super heater, the conclusion is that option (C) pressure remains constant and temperature rises is the correct choice. A super heater is a device used in a steam power plant to increase the temperature of the steam, above its saturation temperature. The function of the super heater is to remove the last traces of moisture (1 to 2%) from the saturated steam and to increase its temperature above the saturation temperature. The pressure, however, remains constant during this process because the super heater operates at the same pressure as the boiler.

Learn more about Super heater here:

brainly.com/question/32665042

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