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aleksklad [387]

3 years ago

10

how fast in revolutions per minute must a centrifuge rotate in order to subject the contents of a test tube (30cm) from the axis

to an acceleration equivalent to 15,000gs

1 answer:

trapecia [35]3 years ago

4 0

Answer:

ω=6684.51 rpm

Explanation:

r= 30cm= 0.3m

a= 15000gs (convert to m/s^{2}

1g = 9.8 m/s^{2}

a= 15000 *9.8 = 147000 m/s^{2}

a=\frac{v^{2} }{r}

147000 = \frac{v^{2} }{0.3}

147000*0.3 = v^{2}

44100 = v^{2}

√44100 = v

210m/s = v

v=210m/s (linear velocity)

we will convert this to angular velocity

ω=\frac{v}{r}

ω= 210/0.3

ω= 700 rads^{-1}

we will convert this to rev per minute

1rad per second = 9.5493 rev per minute

ω= 700*9.5493

ω=6684.51 rpm

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A bus leaves at 9 am with a group of tourists. They travel 350 km before they stop for lunch. Then they travel an additional 250

Anettt [7]

Average speed = Distance traveled / time taken

In this case Time taken = Difference in hours between 3 PM and 9 AM

= 6 hours

Total distance traveled = 350 km + 250 km

= 600 kilometers

So average speed = 600/6 = 100 km/hr

Average speed of bus = 27.78 m/s

So the bus's average speed = 27.78 m/s or 100 km/hr.

8 0

3 years ago

A proton that has a mass m and is moving at 270 m/s in the i hat direction undergoes a head-on elastic collision with a stationa

Nataly_w [17]

Answer:

$V_p = 267.258 m/s$

$V_n = 38.375 m/s$

Explanation:

using the law of the conservation of the linear momentum:

$P_i = P_f$

where $P_i$ is the inicial momemtum and $P_f$ is the final momentum

the linear momentum is calculated by the next equation

P = MV

where M is the mass and V is the velocity.

so:

$P_i = m(270 m/s)$

$P_f = mV_P + M_nV_n$

where m is the mass of the proton and $V_p$ is the velocity of the proton after the collision, $M_n$ is the mass of the nucleus and $V_n$ is the velocity of the nucleus after the collision.

therefore, we can formulate the following equation:

m(270 m/s) = m $V_p$ + 14m $V_n$

then, m is cancelated and we have:

270 = $V_p$ + $14V_n$

This is a elastic collision, so the kinetic energy K is conservated. Then:

$K_i = \frac{1}{2}MV^2 = \frac{1}{2}m(270)^2$

and

Kf = $\frac{1}{2}mV_p^2 +\frac{1}{2}(14m)V_n^2$

then,

$\frac{1}{2}m(270)^2$ = $\frac{1}{2}mV_p^2 +\frac{1}{2}(14m)V_n^2$

here we can cancel the m and get:

$\frac{1}{2}(270)^2$ = $\frac{1}{2}V_p^2 +\frac{1}{2}(14)V_n^2$

now, we have two equations and two incognites:

270 = $V_p$ + $14V_n$ (eq. 1)

$\frac{1}{2}(270)^2$ = $\frac{1}{2}V_p^2 +\frac{1}{2}(14)V_n^2$

in the second equation, we have:

36450 = $\frac{1}{2}V_p^2 +\frac{1}{2}(14)V_n^2$ (eq. 2)

from this last equation we solve for $V_n$ as:

$V_n = \sqrt{\frac{36450-\frac{1}{2}V_p^2 }{\frac{1}{2} } }$

and replace in the other equation as:

270 = $V_p +$ 14 $\sqrt{\frac{36450-\frac{1}{2}V_p^2 }{\frac{1}{2} } }$

so,

$V_p = -267.258 m/s$

Vp is negative because the proton go in the -i hat direction.

Finally, replacing this value on eq. 1 we get:

$V_n = \frac{270+267.258}{14}$

$V_n = 38.375 m/s$

3 0

3 years ago

In football we see unbalanced forces. When 1 player exerts an unbalanced force on another player and causes a player to

zaharov [31]

Answer:

Fall

Explanation:

3 0

3 years ago

A constant unbalanced force is applied to an object for a period of time. Which graph best represents the acceleration of the ob

balu736 [363]

Answer:

Here is an image attached with similar questions.

The correct answer is D where acceleration is function of time.

Explanation:

The force acting on the object is constant.

There is no change in the application of forces.

And we know that Force is the product of acceleration and masses.

Newtons second law: $F=m\times a$

Regarding mass we know that it can neither be created nor destroyed.

So in $F=m \times a$ we have two constant terms, constant divided by constant will give the same result as $a=\frac{Force}{mass}$

There is no change in acceleration $(a)$ with respect to time $(t)$ .

So the most appropriate graph where time (t) is changing on $x-axis$ but acceleration doesn't changes is D.

The graph will be similar to $y=any\ constant$ and will be horizontal to the $x-axis$ .

Option D depicts the same.

5 0

3 years ago

A billiard ball moving at 5 m/s strikes a stationary ball of the same mass. After the collision, the original ball moves at a ve

SIZIF [17.4K]

90 degrees - 30 = 60 degrees

Velocity = (5m/s - 4.35m/s x cos(30)) / cos(60)

Velocity = 2.47 m/s

The answer is D) 2.47 m/s at 61.9 degrees

8 0

3 years ago