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4 years ago

15

ASAP!!! A cell phone charger supplies 2.0A of current to your phone. you plug in your phone overnight (8 hours). how much charge

does the charger move through the phone?

1 answer:

Ksivusya [100]4 years ago

6 0

current = charges / times (sec)

if a current of 2A flows for 8 hours(28800sec), then 2 * 28800 = 57600 coulombs of electrical charge move through the phone

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A centrifugal pump discharges 300 gpm against a head of 55 ft when the speed is 1500 rpm. The diameter of the impeller is 15.5 i

Ksivusya [100]

Answer: Question 1: Efficiency is 0.6944

Question 2: speed of similar pump is 2067rpm

Explanation:

Question 1:

Flow rate of pump 1 (Q1) = 300gpm

Flow rate of pump 2 (Q2) = 400gpm

Head of pump (H)= 55ft

Speed of pump1 (v1)= 1500rpm

Speed of pump2(v2) = ?

Diameter of impeller in pump 1= 15.5in = 0.3937m

Diameter of impeller in pump 2= 15in = 0.381

B.H.P= 6.0

Assuming cold water, S.G = 1.0

eff= (H x Q x S.G)/ 3960 x B.H.P

= (55x 300x 1)/3960x 6

= 0.6944

Question 2:

Q = A x V. (1)

A1 x v1 = A2 x V2. (2)

Since A1 = A2 = A ( since they are geometrically similar

A = Q1/V1 = Q2/V2. (3)

V1(m/s) = r x 2π x N(rpm)/60

= (0.3937x 2 x π x 1500)/2x 60

= 30.925m/s

Using equation (3)

V2 = (400 x 30.925)/300

= 41.2335m/s

To rpm:

N(rpm) = (60 x V(m/s))/2 x π x r

= (60 x 41.2335)/ 2× π × 0.1905

= 2067rpm.

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3 years ago

How much heat must be removed from 200 pounds of blanched vegetables at 189 degrees F to freeze them to a temperature of 22 degr

Virty [35]

To solve this problem it is necessary to apply the concepts related to heat exchange in the vegetable and water.

By definition the exchange of heat is given by

$Q =mc\Delta T$

where,

m = mass

c = specific heat

$\Delta T$ = Change in temperature

Therefore the total heat exchange is given as

$\Delta Q = Q_w+Q_v$

$\Delta Q = m_wc_w(T_i-T_f)+m_vc_v(T_i-T_f)$

Our values are given as,

Total mass is $M_T$ = 200lb ,however the mass of solid vegetable and water is given as,

$m_v= 0.4*200lb = 80lb$

$m_w=0.6*200lb=120lb$

$T_i = 183\°F\\T_f = 22\°F\\c_w = 1Btu/lbF\\c_v = 0.24Btu/lbF$

Replacing at our equation we have,

$\Delta Q = m_wc_w(T_i-T_f)+m_vc_v(T_i-T_f)$

$\Delta Q = (120)(1)(183-22)+(80)(0.24)(183-22)$

$\Delta Q = 22411.2Btu$

Therefore the heat removed is 22411.2 Btu

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3 years ago

J-s. If your 1400-kg car is parked in an 8.54-m-long garage, what is the uncertainty in its velocity? cm/s the tolerance is +/-2

IgorC [24]

Answer:

$\Delta v = 4.41 \times 10^{-37} cm/s$

Explanation:

As per Heisenberg's uncertainty principle we know that

$\Delta P \times \Delta x = \frac{h}{4\pi}$

so here we have

$\Delta P = m\Delta v$

$\Delta x = 8.54 m$

now from above equation we have

$m\Delta v \times (8.54) = \frac{h}{4\pi}$

$1400(\Delta v) \times (8.54) = \frac{6.626 \times 10^{-34}}{4\pi}$

$\Delta v = 4.41 \times 10^{-39} m/s$

$\Delta v = 4.41 \times 10^{-37} cm/s$

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3 years ago

What is the average speed of a car that travels 60 meters in 2<br> seconds?

blondinia [14]

Answer:

30 m/s

Explanation:

Speed is distance over time. 60 meters / 2 seconds, = 30 m/s.

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3 years ago

A bar magnet whose magnetic dipole moment is 21 A·m2 is aligned with an applied magnetic field of 3.6 T. How much work must you

iren [92.7K]

Answer:

The amount of work must be do to rotate the bar magnet is 151.2 J

Explanation:

Given:

Magnetic moment $\mu = 21$ $A. m^{2}$

Magnetic field $B = 3.6$ T

To find work do to rotate the bar magnet,

From the formula of work done in case of magnetic field,

$U = \mu .B \cos 0 -\mu .B \cos 180$

Here $\theta$ changes 0 to 180

But $\cos 180 = -1$

$U = 2\mu B$

$U = 2 \times 21 \times 3.6$

$U = 151.2$ J

Therefore, the amount of work must be do to rotate the bar magnet is 151.2 J

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3 years ago