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3 years ago

13

A tide of increased range that occurs during the new and full moons is called

1 answer:

irina [24]3 years ago

4 0

A tide of increased range that occurs during the new and full moons is called Spring Tide.

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How do you get derived units from derived quantities

Tasya [4]

Answer:

A derived quantities is terms of the 7 base quantities via a system of quantity equations which are called SI derived units.

Explanation: there you go:)

4 0

3 years ago

What is the momentum of a 750kg car traculing at a velocity of 24m/s

m_a_m_a [10]

The answer is 18000 kgm/s

Momentum is mass times velocity so just do 750•24.

7 0

3 years ago

PLEASE HELP ME TIMED TEST!!!

malfutka [58]

Answer:

56

Explanation:

I just want the points to be completely honest with you.

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2 years ago

Read 2 more answers

Steam enters a well-insulated nozzle at 200 lbf/in.2 , 500F, with a velocity of 200 ft/s and exits at 60 lbf/in.2 with a velocit

Ede4ka [16]

Answer:

$386.2^{\circ}F$

Explanation:

We are given that

$P_1=200lbf/in^2$

$P_2=60lbf/in^2$

$v_1=200ft/s$

$v_2=1700ft/s$

$T_1=500^{\circ}F$

$Q=0$

$C_p=1BTU/lb^{\circ}F$

We have to find the exit temperature.

By steady energy flow equation

$h_1+v^2_1+Q=h_2+v^2_2$

$C_pT_1+\frac{P^2_1}{25037}+Q=C_pT_2+\frac{P^2_2}{25037}$

$1BTU/lb=25037ft^2/s^2$

Substitute the values

$1\times 500+\frac{(200)^2}{25037}+0=1\times T_2+\frac{(1700)^2}{25037}$

$500+1.598=T_2+115.4$

$T_2=500+1.598-115.4$

$T_2=386.2^{\circ}F$

7 0

3 years ago

A flywheel is a solid disk that rotates about an axis that is perpendicular to the disk at its center. Rotating flywheels provid

Sergeeva-Olga [200]

Answer:

$8.37*10^5$ rpm

Explanation:

Given that rotational kinetic energy = $4.66*10^9J$

Mass of the fly wheel (m) = 19.7 kg

Radius of the fly wheel (r) = 0.351 m

Moment of inertia (I) = $\frac{1}{2} mr ^2$

Rotational K.E is illustrated as $(K.E)_{rt} = \frac{1}{2} I \omega^2$

$\omega = \sqrt{\frac{2(K.E)_{rt}}{I} }$

$\omega = \sqrt{\frac{2(KE)_{rt}}{1/2 mr^2} }$

$\omega = \sqrt{\frac{4(K.E)_{rt}}{mr^2} }$

$\omega = \sqrt{\frac{4*4.66*10^9J}{19.7kg*(0.351)^2} }$

$\omega = 87636.04$

$\omega = 8.76*10^4 rad/s$

Since 1 rpm = $\frac{2 \pi}{60} rad/s$

$\omega = 8.76*10^4(\frac{60}{2 \pi})$

$\omega = 836518.38$

$\omega = 8.37 *10^5 rpm$

3 0

3 years ago