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sergiy2304 [10]

2 years ago

11

Two particles move along an x axis. The position of particle 1 is given by x ! 6.00t 2 # 3.00t # 2.00 (in meters and seconds); t

he acceleration of particle 2 is given by a ! "8.00t (in meters per second squared and seconds) and, at t ! 0, its velocity is 20 m/s. When the velocities of the particles match, what is their velocity?

1 answer:

s344n2d4d5 [400]2 years ago

7 0

Answer:

15.6m/s

Explanation:

V1= $\frac{dx}{dt}=\frac{d}{dt}(6t^{2}+3t+2)$ because the derivate of the position is the velocity

V1=12t+3

V2=20+ $\int\limits^_ {} \,$ -8t because the integral of the acceleration is the velocity

V2= $20-4t^{2}$

V1=V2 to see when the velocities of particles match

12t+3=20-4t^2

4t^2+12t-17=0 we resolve this with $\frac{-b+-(\sqrt{b^{2} -4ac} )}{2a}$

and we take the positif root

t=1.05 sec

if we evaluate the velocity (V1 or V2) the result is 15.6m/s

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Answer: Phillipe wins the race

Explanation:

Given

Length of track=20 m

Velocity of Phillippe=2 m/s w.r.t sidewalk

Velocity of sidewalk=1.5 m/s

Rena velocity=2 m/s

Therefore absolute velocity of P is 2+1.5=3.5 m/s

Time taken by phillippe $\frac{20}{3.5}+\frac{20}{3.5-1.5}$

$t_P$ =5.714+10=15.714 s

Time taken by rena

$t_r=\frac{20}{2}+\frac{20}{2}=20 s$

as the time taken by phillippe is less than rena therefore Phillippe wins the race.

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3 years ago

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A Ferris wheel on a California pier is 27 m high and rotates once every 32 seconds in the counterclockwise direction. When the w

Leto [7]

A) 140 degrees

First of all, we need to find the angular velocity of the Ferris wheel. We know that its period is

T = 32 s

So the angular velocity is

$\omega=\frac{2\pi}{T}=\frac{2\pi}{32 s}=0.20 rad/s$

Assuming the wheel is moving at constant angular velocity, we can now calculate the angular displacement with respect to the initial position:

$\theta= \omega t$

and substituting t = 75 seconds, we find

$\theta= (0.20 rad/s)(75 s)=15 rad$

In degrees, it is

$15 rad: x = 2\pi rad : 360^{\circ}\\
x=\frac{(15 rad)(360^{\circ})}{2\pi}=860^{\circ} = 140^{\circ}$

So, the new position is 140 degrees from the initial position at the top.

B) 2.7 m/s

The tangential speed, v, of a point at the egde of the wheel is given by

$v=\omega r$

where we have

$\omega=0.20 rad/s$

r = d/2 = (27 m)/2=13.5 m is the radius of the wheel

Substituting into the equation, we find

$v=(0.20 rad/s)(13.5 m)=2.7 m/s$

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2 years ago

A small block with a mass of 0.0600 kg is attached to a cord passing through a hole in a frictionless, horizontal surface (Figur

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Answer with Explanation:

We are given that mass of block=0.0600 kg

Initial speed of block=0.63 m/s

Distance of block from the hole when the block is revolved=0.47 m

Final speed=3.29 m/s

Distance of block from the hole when the block is revolved= $9\times 10^{-2}m$

a.We have to find the tension in the cord in the original situation when the block has speed = $v_0=0.63 m/s$

$T=\frac{mv^2}{r}$

Because tension is equal to centripetal force

Substitute the values

$T=\frac{0.06\times (0.63)^2}{0.47}=0.05 N$

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$T=\frac{mv^2}{r}=\frac{0.06\times (3.29)^2}{0.09}=7.2 N$

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$W=\frac{1}{2}m(v^2-v^2_0)$

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2 years ago

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How much force is needed to keep the 750000 Newton Space Shuttle moving at a constant speed of 28000 km/h, in a straight line?

Alika [10]

The force needed to keep the space shuttle moving at constant speed is 0.

The given parameters;

<em>weight of the space shuttle, F = 750,000 N</em>
<em>constant speed of the space shuttle, v = 28,000 km/h</em>

The mass of the space shuttle is calculated as follows;

$W = mg\\\\m = \frac{W}{g} \\\\m = \frac{750,000}{9.8} \\\\m = 76,530.61 \ kg$

The force needed to keep the space shuttle moving at constant speed is calculated as follows;

$F = ma$

$F = 76,530.61 \times a$

where;

a is the acceleration of the space shuttle

At a constant speed, acceleration is zero.

F = 76,530.61 x 0

F = 0

Thus, the force needed to keep the space shuttle moving at constant speed is 0.

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2 years ago

Write a numerical expression for the emissive intensity (in W/m^2.sr) coming out of a tiny hole in an enclosure of surface tempe

stiks02 [169]

Answer:

6.0 × $10^{11}$ W/ $m^{2}$

Explanation:

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Where: Q is the quantity of heat transferred, e is the emissivity of the surface, A is the area, and T is the temperature.

The emissive intensity = $\frac{Q}{A}$ = e $T^{4}$

Given from the question that: e = 0.6 and T = 1000K, thus;

emissive intensity = 0.6 × $(1000)^{4}$

= 0.6 × 1.0 × $10^{12}$

= 6.0 × $10^{11}$ $\frac{W}{m^{2} }$

Therefore, the emissive intensity coming out of the surface is 6.0 × $10^{11}$ W/ $m^{2}$ .

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2 years ago