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elixir [45]
3 years ago
10

A hammer of mass M is moving at speed v0 when it strikes a nail of negligible mass that is stuck in a wooden block. The hammer i

s observed to drive the nail a distance L deeper into the block.
(a) Find the work W done on the hammer by the nail. Express your answer as a function of F and L, where F is the force the nail exerts on the hammer. Don't forget to consider the sign of your answer.
(b) Find the change in kinetic energy of the hammer. Express your answer in terms of M and v0.
(c) Find the magnitude F of the force that the wooden block exerts on the nail, assuming that this force is independent of the depth of penetration of the nail into the wood.
(d) Now evaluate the magnitude of the holding force of the wooden block on the nail by assuming that the force necessary to pull the nail out is the same as that needed to drive it in, which we just derived.
Assume a relatively heavy M = 0.5 kg hammer (about 18 ounces), moving with speed v0 = 10 m/s. (If such a hammer were swung this hard upward and released, it would rise 5 m). Take the penetration depth L to be 2 cm, which is appropriate for one hit on a relatively heavy construction nail. Express your answer to the nearest pound. (Note: 1 lb = 4:45 N.)
Physics
1 answer:
OleMash [197]3 years ago
8 0

Answer:

i think it would be D

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The diver uses compressed air to breathe under water. 1700 litres of air from the atmosphere is compressed into a 12-litre gas c
ivann1987 [24]

Answer:

The answer is below

Explanation:

A diver works in the sea on a day when the atmospheric pressure is 101 kPa. The diver uses compressed air to breathe under water. 1700 litres of air from the atmosphere is compressed into a 12-litre gas cylinder. The compressed air quickly cools to its original temperature. Calculate the pressure of the air in the cylinder.

Solution:

Boyles law states that the volume of a given gas is inversely proportional to the pressure exerted by the gas, provided that the temperature is constant.

That is:

P ∝ 1/V;   PV = constant

P₁V₁ = P₂V₂

Given that P₁ = initial pressure = 101 kPa, V₁ = initial volume = 1700 L, P₂ = cylinder pressure, V₂ = cylinder volume = 12 L. Hence:

P₁V₁ = P₂V₂

100 kPa * 1700 L = P₂ * 12 L

P₂ = (100 kPa * 1700 L) / 12 L

P₂ = 14308 kPa

6 0
2 years ago
I need to lift a 2000kg car, 1.798m and the joules required is 35240.8. Converted to watt (W = 35240.8/5 (s)) I got 7048.16 W. I
marusya05 [52]
This is a very interesting problem ... mainly because it's different from
the usual questions in the Physics neighborhood.

I can discuss it with you, but maybe not quite give you a final answer
with the information you've given in the question.

I agree with all of your calculations so far ... the total energy required,
and the power implied if the lift has to happen in 5 seconds.

First of all, let's talk about power.  I'm assuming that your battery is
a "car" battery, and I'm guessing you measured the battery voltage
while the car was running.  Turn off the car, and you're likely to read
something more like 13 to 13.8 volts.
But that's not important right now.  What I'm looking for is the CURRENT
that your application would require, and then to look around and see whether
a car battery would be capable of delivering it.

   Power = (volts) x (current)

   7,050 W  =  (14 volts) x (current)

   Current = (7,050 watts / 14 volts) =  503 Amperes. 

That kind of current knocks the wind out of me.  I've never seen
that kind of number outside of a power distribution yard.
BUT ... I also know that the current demand from a car battery during
starting is enormous, so I'd better look around online and try to find out
what a car battery is actually capable of.

I picked a manufacturer's name that I'd heard of, then picked their
recommended battery for a monster 2003-model car, and looked at
the specs for the battery.

The spec I looked at was the 'CCA' ... cold cranking Amps.
That's the current the battery is guaranteed to deliver for 30 seconds,
at a temperature of 0°F, without dropping below 12 volts.

This battery that I saw is rated  803 Amps  CCA !

OK.  Let's back up a little bit.  I'm pretty sure the battery you have
is a nominal "12-volt" battery.  Let's say you use to start lifting the lift. 
As the lift lifts, the battery voltage sags.  What is the required current
if the battery immediately droops to 12V and stays there, while delivering
7,050 watts continuously ?

          Power = (volts) x (current)

          7,050 W = (12 V) x (current)

            Current = (7,050 W / 12 V)  =  588 Amps . 

Amazingly, we may be in the ball park.
If the battery you have is rated by the manufacturer for 600 Amps
CCA (0°F) or CA (32°F), then the battery can deliver the current
you need.
BUT ... you can't conduct that kind of current through ear-bud wire,
or house wiring wire.  I'm not even so sure of jumper-cables. 
You need thick, no-nonsense cable, AND connections with a lot of
area ... No alligator clips.  Shiny nuts and bolts with no crud on them.

Now ... I still want to check the matter of the total energy.
I'm sure you're OK, because the CCA and CA specifications talk about
30 seconds of cranking, and you're only talking about 5 seconds of lifting.
But I still want to see the total energy requirement compared to the typical
battery specification ... 'AH' ... ampere-hours.

You're talking about 35,000 joules

                          = 35,000 watt-seconds

                         =  35,000 volt-amp-seconds.

               (35,000 volt-amp-sec) x (1 hour/3600 sec) / (12 volt)               

           =  (35,000 x 1) / (3600 x 12)  volt-amp-sec-hour / sec-volt

           =    0.81 Amp-Hour  .

That's an absurdly small depletion from your car battery.
But just because it's only  810 mAh, don't get the idea that you can
do it with a few rechargeable AA batteries out of your camera.
You still need those 600 cranking amps.  That would be a dead short
for a stack of camera batteries, and they would shrivel up and die.

Have I helped you at all ?
5 0
3 years ago
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hoa [83]
It expands and pushes the crack further aprt
3 0
2 years ago
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