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1 year ago

2 grenades have been dropped, one in the pool, one on land, where do you dive for cover?

Physics

1 answer:

ELEN [110]1 year ago

3 0

Answer:

It is safer to endure an explosion on land rather than in water. So it is better to take cover on lanad.

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A constant voltage of 12.00 V has been observed over a certain time interval across a 1.20 H inductor. The current through the i

kiruha [24]

Answer:

The time is 0.5 sec.

Explanation:

Given that,

Voltage V= 12.00 V

Inductance L= 1.20 H

Current = 3.00 A

Increases rate = 8.00 A

We need to calculate change in current

$\Delta A = 8.00-3.00= 5.00\ A$

We need to calculate the time interval

Using formula of inductor

$V=L\dfrac{\Delta A}{\Delta t}$

$\Delta t =\dfrac{L\Delta A}{V}$

Where, $\Delta A$ = change in current

V = voltage

L = inductance

Put the value into the formula

$\Delta t=\dfrac{1.20\times5.00}{12.00}$

$\Delta t=0.5\ sec$

Hence, The time is 0.5 sec.

5 0

3 years ago

When you stop at a railroad crossing at night, instead of correctly perceiving two red lights that flash alternately you may per

Nezavi [6.7K]

I'm pretty sure that there should be an options to choose. Anyway, I've seen this question before and I know that this is an example of <span>the phi phenomenon.</span>

7 0

3 years ago

Read 2 more answers

Which is the best example of Newton's First Law of Motion? A small, lightweight ball and a large, heavy ball are dropped off the

Marizza181 [45]

I believe the best example of Newton's First Law of motion would be the example or illustration with the basketball player. An object will move in a straight line or a given direction at a constant speed unless or until another force acts upon the object, causing a change in speed and or direction.

4 0

3 years ago

Read 2 more answers

Air as an ideal gas enters a diffuser operating at steady state at 5 bar, 280 K with a velocity of 510 m/s. The exit velocity is

Nataly [62]

Answer:

Explanation:

Calculating the exit temperature for K = 1.4

The value of $c_p$ is determined via the expression:

$c_p = \frac{KR}{K_1}$

where ;

R = universal gas constant = $\frac{8.314 \ J}{28.97 \ kg.K}$

k = constant = 1.4

$c_p = \frac{1.4(\frac{8.314}{28.97} )}{1.4 -1}$

$c_p= 1.004 \ kJ/kg.K$

The derived expression from mass and energy rate balances reduce for the isothermal process of ideal gas is :

$0=(h_1-h_2)+\frac{(v_1^2-v_2^2)}{2}$ ------ equation(1)

we can rewrite the above equation as :

$0 = c_p(T_1-T_2)+ \frac{(v_1^2-v_2^2)}{2}$

$T_2 =T_1+ \frac{(v_1^2-v_2^2)}{2 c_p}$

where:

$T_1 = 280 K \\ \\ v_1 = 510 m/s \\ \\ v_2 = 120 m/s \\ \\c_p = 1.0004 \ kJ/kg.K$

$T_2= 280+\frac{((510)^2-(120)^2)}{2(1.004)} *\frac{1}{10^3}$

$T_2 = 402.36 \ K$

Thus, the exit temperature = 402.36 K

The exit pressure is determined by using the relation: $\frac{T_2}{T_1} = (\frac{P_2}{P_1})^\frac{k}{k-1}$

$P_2=P_1(\frac{T_2}{T_1})^\frac{k}{k-1}$

$P_2 = 5 (\frac{402.36}{280} )^\frac{1.4}{1.4-1}$

$P_2 = 17.79 \ bar$

Therefore, the exit pressure is 17.79 bar

7 0

3 years ago

A dart with mass md is launched toward a block of mass mb that is suspended from a string of length L. The dart is moving horizo

Yuki888 [10]

Answer:

A) Impulse is the same for both the objects

B) The higher is the speed, the greater will be the height.

Explanation:

Part a)

The time of interaction of the two bodies i.e the hanging mass and the stick is same. Thus, force caused by dart on the block = force caused by block on the dart. Hence, impulse is the same for both the objects.

Part B

The energy will be conserved in the entire reaction process

Hence, Kinetic energy = potential energy

0.5Mv^2 = gh(md+mb)

H is directly proportional to the square of speed.

Hence, the higher is the speed, the greater will be the height.

5 0

2 years ago