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2 years ago

Use your calculator to evaluate -3.7 meter/second-13.9 meter/second 21.4 second-72 second

Physics

1 answer:

cluponka [151]2 years ago

6 0

Answer :

(-3.7 meter/second) - (13.9 meter/second) = -17.6 meter/second

(21.4 second) - (72 second) = -50.6 second

Explanation :

(1) As we are given the expression :

(-3.7 meter/second) - (13.9 meter/second)

Now we have to evaluate this expression, we get:

⇒ -17.6 meter/second

(2) As we are given the expression :

(21.4 second) - (72 second)

Now we have to evaluate this expression, we get:

⇒ -50.6 second

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How long( in hours, will it take for 500 000 C of charge to flow through a diode if it requires

Alina [70]

Answer:

277.78 hours

Explanation:

The formula for calculating the amount of charge is expressed as;

Q = It

I is the current

t is the time

Given

I =0.05A

Q = 50,000C

Required

Time t

Recall that: Q = It

t = Q/I

t = 50,000/0.05

t = 1,000,000secs

Convert to hours

1,000,000secs = 1,000,000/3600

1,000,000secs = 277.78 hours

Hence it will take 277.78 hours for the charge to flow through the diode

6 0

3 years ago

A sealed tank containing seawater to a height of 10.5 mm also contains air above the water at a gauge pressure of 2.95 atmatm. W

weqwewe [10]

Answer:

The water is flowing at the rate of 28.04 m/s.

Explanation:

Given;

Height of sea water, z₁ = 10.5 m

gauge pressure, $P_{gauge \ pressure}$ = 2.95 atm

Atmospheric pressure, $P_{atm}$ = 101325 Pa

To determine the speed of the water, apply Bernoulli's equation;

$P_1 + \rho gz_1 + \frac{1}{2}\rho v_1^2 = P_2 + \rho gz_2 + \frac{1}{2}\rho v_2^2$

where;

P₁ = $P_{gauge \ pressure} + P_{atm \ pressure}$

P₂ = $P_{atm}$

v₁ = 0

z₂ = 0

Substitute in these values and the Bernoulli's equation will reduce to;

$P_1 + \rho gz_1 + \frac{1}{2}\rho v_1^2 = P_2 + \rho gz_2 + \frac{1}{2}\rho v_2^2\\\\P_1 + \rho gz_1 + \frac{1}{2}\rho (0)^2 = P_2 + \rho g(0) + \frac{1}{2}\rho v_2^2\\\\P_1 + \rho gz_1 = P_2 + \frac{1}{2}\rho v_2^2\\\\P_{gauge} + P_{atm} + \rho gz_1 = P_{atm} + \frac{1}{2}\rho v_2^2\\\\P_{gauge} + \rho gz_1 = \frac{1}{2}\rho v_2^2\\\\v_2^2 = \frac{2(P_{gauge} + \rho gz_1)}{\rho} \\\\v_2 = \sqrt{ \frac{2(P_{gauge} + \rho gz_1)}{\rho} }$