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RUDIKE [14]
2 years ago
10

2. A 3.45-kg centrifuge takes 100 s to spin up from rest to its final angular speed with constant angular acceleration. A point

located 8.00 cm from the axis of rotation of the centrifuge moves with a speed of 150 m/s when the centrifuge is at full speed. (a) What is the angular acceleration (in rad/s2) of the centrifuge as it spins up
Physics
2 answers:
irina1246 [14]2 years ago
4 0

Answer:

18.75 rad/s²

Explanation:

Given that

Mass of the centrifuge, m = 3.45 kg

Time taken to spin, t = 100 s

Distance from axis of rotation, r = 8 cm = 0.08 m

Speed of the centrifuge, v = 150 m/s

First, we find the angular velocity

Angular velocity, w = v / r

w = 150 / 0.08

w = 1875 rad/s

And from the angular velocity, we get our angular acceleration.

Angular acceleration = angular velocity / time taken

Angular acceleration = 1875 rad/s / 100 s

Angular acceleration = 18.75 rad/s²

Therefore, the angular acceleration is 18.75 rad/s²

Yuki888 [10]2 years ago
3 0

Answer:

The angular acceleration of the centrifuge as it spins up is 18.75 rad/s²

Explanation:

Given;

mass of centrifuge, m = 3.45 kg

time taken to spin, t = 100 s

distance from the axis of rotation, r = 8.00 cm = 0.08 m

final velocity of the centrifuge, v = 150 m/s

initial velocity of the centrifuge, u = 0

Determine the linear acceleration of the centrifuge at the given time;

a = \frac{v-u}{t} \\\\a = \frac{150-0}{100} \\\\a = 1.5 \ m/s^2

Finally, determine the angular acceleration of the centrifuge as it spins up;

α = a/r

where;

α is the angular acceleration

α = 1.5 / 0.08

α = 18.75 rad/s²

Therefore, the angular acceleration (in rad/s2) of the centrifuge as it spins up is 18.75 rad/s²

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Maslowich

Answer:

695800 N/m^2 or Pa

Explanation:

Height of the water from the ground H  =  71 m

Acceleration due to gravity g =9.8 m/s^2

density of water ρ= 1000 kg/m^3

The minimum output gauge pressure to make water reach height H

P= ρgH

= 1000×9.8×71= 695800 N/m^2 or Pa

5 0
3 years ago
A hawk is flying horizontally at 18.0 m/s in a straight line, 230 m above the ground. A mouse it has been carrying struggles fre
Lisa [10]

Answer:

a) vd = 47.88 m/s

b) θ = 80.9°

c) t = 6.8 s

Explanation:

In the situation of the problem, you can assume that the trajectory of the hawk and the trajectory of the mouse form a rectangle triangle.

One side of the triangle is the horizontal trajectory of the hawk after 2.00s of flight, the other side of the triangle is the distance traveled by the mouse when it is falling down. And the hypotenuse is the trajectory of the hawk when it is trying to recover the mouse.

(a) In order to calculate the diving speed of the hawk, you first calculate the hypotenuse of the triangle.

One side of the triangle is c1 = (18.0m/s)(2.0s) = 36m

The other side of the triangle is c2 = 230m - 3m = 227 m

Then, the hypotenuse is:

h=\sqrt{(36m)^2+(227m)^2}=229.83m    (1)

Next, it is necessary to calculate the falling down time of the mouse, this can be done by using the following formula:

y=y_o+v_ot+\frac{1}{2}gt^2    (2)

yo: initial height = 230m

vo: initial vertical speed of the mouse = 0m/s

g: gravitational acceleration = -9.8m/s^2

y: final height of the mouse = 3 m

You replace the values of the parameters in (2) and solve for t:

3=230-4.9t^2\\\\t=\sqrt{\frac{227}{4.9}}=6.8s

The hawk traveled during 2.00 second in the horizontal trajectory, hence, the hawk needed 6.8s - 2.0s = 4.8 s to travel the distance equivalent to the hypotenuse to catch the mouse.

You use the value of h and 4.8s to find the diving speed of the hawk:

v_d=\frac{229.83m}{4.8s}=47.88\frac{m}{s}

The diving speed of the Hawk is 47.88m/s

(b) The angle is given by:

\theta=cos^{-1}(\frac{c_1}{h})=cos^{-1}(\frac{36m}{229.83m})=80.9 \°

Then angle between the horizontal and the trajectory of the Hawk when it is descending is 80.9°

(c) The mouse is falling down during 6.8 s

4 0
3 years ago
IF YOUR GOOD AT SCIENCE THEN PLEASE ANSWER THIS ASAP I WILL MARK YOU THE BRAINLIEST
Digiron [165]

Answer:

This is an open circuit

Explanation:

An open circuit I believe

It needs to be closed for the bulb to be turned on

8 0
2 years ago
Describe how an electromagnet can be switched on and off.​
Phoenix [80]

Answer:The magnetic field around an electromagnet is just the same as the one around a bar magnet. It can, however, be reversed by turning the battery around. Unlike bar magnets, which are permanent magnets, the magnetism of electromagnets can be turned on and off just by closing or opening the switch.

8 0
3 years ago
A silver wire 2.6 mm in diameter transfers a charge of 420 C in 80 min. Silver contains 5.8 x 10- free electrons per cubic meter
kifflom [539]

Answer:

a). 87.5 mA or 87.5 x10^{-3}A

b). 1.78 \frac{m}{s}

Explanation:

d=2.6 mm \\Q=420C\\t=80min\\n=5.8x10^{28} \\q=1.6x10^{-19}

n the number of free electrons is 28 in text reference and if they don't give q is take as the charge of electron.

a).

I=\frac{Q}{t}\\ I= \frac{420 C}{80 min}*\frac{1min}{60 s} =\frac{420 C}{4800s}\\  I=87.5 x10^{-3}A

b).

I=n*abs (q)*V_{d}*A

A= \pi * (\frac{d}{2})^{2} \\A=\pi (*\frac{2.6x10^{-3} m}{2})^{2}  \\A=5.309x10^{-6}

V_{d} =\frac{I}{n*abs(q)*A} \\V_{d}=\frac{87.5 x10^{-2} }{5.8x^10{28} *1.6x^{-19} *5.3x^{6} }\\V_{d}=1.78 \frac{m}{s}

8 0
3 years ago
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