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finlep [7]
3 years ago
8

A bullet fired horizontally hits the ground in 0.5 sec. If it had been fired with a much higher speed in the same direction, and

neglecting air resistance and the earth’s curvature, it would have hit the ground in1. There is no way to tell from the information given.
2. less than 0.5 sec.
3. 0.5 sec.
4. more than 0.5 sec.
Physics
1 answer:
stira [4]3 years ago
4 0

Answer:

3. 0.5 sec.

Explanation:

A bullet fired horizontally follows a projectile motion, which consists of two independent motions:

- A horizontal motion with constant speed

- A vertical motion with  constant acceleration, g = 9.8 m/s^2, towards the ground

The time taken for the bullet to reach the ground can be calculated just by considering the vertical motion:

y(t) = h + v_{0y} t - \frac{1}{2}gt^2

where y is the vertical position at time t, h is the initial height, and v_{0y} is the initial vertical velocity of the bullet.

Since the bullet is fired horizontally, v_{0y}=0. So the equation becomes

y(t) = h - \frac{1}{2}gt^2

And the time that the bullet takes to reach the ground can be found by requiring y=0 and solving for t:

t=\sqrt{\frac{2h}{g}}

As we can see, in this equation there is no dependance on the initial speed of the bullet: therefore, if the bullet is fired still horizontally but with a different speed, it will still take the same time (0.5 s) to reach the ground.

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Answer:

C

Explanation:

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In a year it is very hard to determine because of the night and different light levels.

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A mole of ideal gas expands at T=27 °C. The pressure changes from 20 atm to 1 atm. What’s the work that the gas has done and wha
Airida [17]

Answer:

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  • The heat absorbed is 7475.69 joules

Explanation:

<h3>Work</h3>

We know that the differential work made by the gas  its defined as:

dW =  P \ dv

We can solve this by integration:

\Delta W = \int\limits_{s_1}^{s_2}\,dW = \int\limits_{v_1}^{v_2} P \ dv

but, first, we need to find the dependence of Pressure with Volume. For this, we can use the ideal gas law

P \ V = \ n \ R \ T

P = \frac{\ n \ R \ T}{V}

This give us

\int\limits_{v_1}^{v_2} P \ dv = \int\limits_{v_1}^{v_2} \frac{\ n \ R \ T}{V} \ dv

As n, R and T are constants

\int\limits_{v_1}^{v_2} P \ dv = \ n \ R \ T \int\limits_{v_1}^{v_2} \frac{1}{V} \ dv

\Delta W= \ n \ R \ T  \left [ ln (V) \right ]^{v_2}_{v_1}

\Delta W = \ n \ R \ T  ( ln (v_2) - ln (v_1 )

\Delta W = \ n \ R \ T  ( ln (v_2) - ln (v_1 )

\Delta W = \ n \ R \ T  ln (\frac{v_2}{v_1})

But the volume is:

V = \frac{\ n \ R \ T}{P}

\Delta W = \ n \ R \ T  ln(\frac{\frac{\ n \ R \ T}{P_2}}{\frac{\ n \ R \ T}{P_1}} )

\Delta W = \ n \ R \ T  ln(\frac{P_1}{P_2})

Now, lets use the value from the problem.

The temperature its:

T = 27 \° C = 300.15 \ K

The ideal gas constant:

R = 8.314 \frac{m^3 \ Pa}{K \ mol}

So:

\Delta W = \ 1 mol \ 8.314 \frac{m^3 \ Pa}{K \ mol} \ 300.15 \ K  ln (\frac{20 atm}{1 atm})

\Delta W = 7475.69 joules

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We know that, for an ideal gas, the energy is:

E= c_v n R T

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By the first law of thermodynamics, we know

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where \Delta W is the Work made by the gas (please, be careful with this sign convention, its not always the same.)

So:

\Delta E = 0

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Recall that power is given by P= V^2/R where;

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