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LuckyWell [14K]
3 years ago
8

Each of the space shuttle's main engines is fed liquid hydrogen bya high-pressure pump. Turbine blades inside the pump rotateat

617 rev/s. A point on one of the blades traces out acircle with a radius of 0.020m as the blade rotates. a) What is themagnitude of the centripetal acceleration that the blade mustsustain at this point? b) Express this acceleration as amultiple of g = 9.80m/s2 .
answer: 3.0*105m/s2
Physics
1 answer:
kvasek [131]3 years ago
8 0

Answer:

a) a= \frac{4 \pi^2 (0.02 m)^2}{0.00162 s}=9.74 \frac{m}{s^2}

b) k = \frac{9.8}{9.74}=1.006

Explanation:

Part a

For this case we can begin finding the period like this:

T= \frac{1}{w} =\frac{1}{617 rad/s}=0.00162 s

Then we know that the centripetal acceleration is given by:

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

And the velocity is given by:

v=\frac{2\pi r}{T}

If we replace this into the acceleration we got:

a = \frac{(\frac{2\pi r}{T})^2}{r}= \frac{4 \pi^2 r}{T^2}

And we can replace the values and we got:

a= \frac{4 \pi^2 (0.02 m)^2}{0.00162 s}=9.74 \frac{m}{s^2}

Part b

For this case we want to find a value of k such that:

a= k 9.8

Where a = 9.74, so then we can solve for k like this:

k = \frac{9.8}{9.74}=1.006

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In your experiment, you measure a total deflection of 4.12 cm when an electric field of 1.10×103V/m is established between the p
Bond [772]

Answer:

B_0 = 1.69 \times 10^{-4}\ T

Explanation:

given,

total deflection = 4.12 cm

Electric field = 1.1 ×10³ V/m

plate length = 6 cm

distance between them = 12 cm

using formula

v_0 = \sqrt{\dfrac{q\epsilon_0d}{ym}(\dfrac{d}{2}+L)}

q = 1.6 × 10⁻¹⁹ C

m = 9.11 x 10⁻³¹ kg

d = 0.06 m

L = 0.12 m

v_0 = \sqrt{\dfrac{1.6 \times 10^{-19}\times 1.1 \times 10^{3}\times 0.06}{0.0412\times 9.11 \times 10^{-31} }(\dfrac{0.06}{2}+0.12)}

v_0 = 6496355.63 m/s

v_0 = \dfrac{E}{B_0}

B_0 = \dfrac{E}{v_0}

B_0 = \dfrac{1.1\times 10^{3}}{6496355.63}

B_0 = 1.69 \times 10^{-4}\ T

5 0
3 years ago
A man jogs at a speed of 1.6 m/s. His dog waits 1.8 s and then takes off running at a speed of 3 m/s to catch the man. How far w
inessss [21]

Answer:

The dog catches up with the man 6.1714m later.

Explanation:

The first thing to take into account is the speed formula. It is v=\frac{d}{t}, where v is speed, d is distance and t is time. From this formula, we can get the distance formula by finding d, it is d=v\cdot t

Now, the distance equation for the man would be:

d_{man}=v_{man}\cdot t=1.6\cdot t

The distance equation for the dog would be obtained by the same way with just a little detail. The dog takes off running 1.8s after the man did. So, in the equation we must subtract 1.8 from t.

d_{dog}=v_{dog}\cdot (t-1.8)=3\cdot (t-1.8)

For a better understanding, at t=1.8 the dog must be in d=0. Let's verify:

d_{dog}=v_{dog}\cdot (1.8-1.8)=3\cdot (0)=0

Now, for finding how far they have each traveled when the dog catches up with the man we must match the equations of each one.

d_{man}=d_{dog}

1.6\cdot t=3\cdot (t-1.8)

1.6\cdot t=3\cdot t-5.4

1.4\cdot t=5.4

t=\frac{5.4}{1.4}

t=3.8571s

The result obtained previously means that the dog catches up with the man 3.8571s after the man started running.

That value is used in the man's distance equation.

d_{man}=1.6\cdot t=1.6\cdot (3.8571)

d_{man}=6.1714m

Finally, the dog catches up with the man 6.1714m later.

6 0
3 years ago
In a velocity selector having electric field E and magnetic field B, the velocity selected for positively charged particles is v
maw [93]

Answer:

True or False

Explanation:

Because.....

easy 50% chance you are right

6 0
3 years ago
Course hero N4M.6 A board has one end wedged under a rock having a mass of 380 kg and is supported by another rock that touches
aleksandrvk [35]

Answer:

Therefore it is save to carry a 62kg adult

Explanation:

From the question we are told that:

Mass m=380kg

Height of supporting Rock X=85cm

Length of BoardL_r=4.5m

Mass of board M_b=22kg

Mass of adult M_a=62  

Generally the moment of balance about wedge part about  is mathematically given by

N -Q + R = Mg + mg

0.85*N - Mg*2.25 - mg*(2.25 + x) = 0

0.85*N  = + Mg*2.25 + mg*(2.25 + x)

where

N+R=4547

therefore

N = 570.70588 + 1608.3529 + 714.823 x

if N=0 at fallen person

x=3.04m

Therefore it is save to carry a 62kg adult

8 0
2 years ago
A stereo speaker produces a pure \"E\" tone, with a frequency of 329.6 Hz. What is the period of the sound wave produced by the
olya-2409 [2.1K]

Answer:

0.003034 s

1.035 m

4.5 m

Explanation:

f = frequency of the tone = 329.6 Hz

T = Time period of the sound wave

we know that, Time period and frequency are related as

T =\frac{1}{f}\\T =\frac{1}{329.6}\\T = 0.003034 s

v = speed of the sound in the air = 341 ms⁻¹

wavelength of the sound is given as

\lambda =\frac{v}{f} \\\lambda =\frac{341}{329.6}\\\lambda = 1.035 m

v = speed of the sound in the water = 1480 ms⁻¹

wavelength of the sound in water is given as

\lambda =\frac{v}{f} \\\lambda =\frac{1480}{329.6}\\\lambda = 4.5 m

8 0
3 years ago
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