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Liula [17]
3 years ago
8

One gram of pure gold would contain lots of gold

Physics
1 answer:
Agata [3.3K]3 years ago
3 0

Answer:

One gram of pure gold would contain lots of gold atoms- A.

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A machine is supplied Energy at a rate of 4000 W and does useful work at a rate of 3760 W. What is the efficiency of the machine
ICE Princess25 [194]
The efficiency of a machine is the ratio between the output power and the power in input:
\epsilon =  \frac{P_{out}}{P_{in}}
in our problem, the output power is 3760 W while the input power is 4000 W, so the efficiency is
\epsilon =  \frac{3760 W}{4000 W} =0.94
So the efficiency of the machine is 94%.
6 0
3 years ago
Read 2 more answers
Suppose you increase your walking speed from 4m/s to 12m/s in a period of 1 s. What is your acceleration?
den301095 [7]
We use the following formula to determine our acceleration:
acceleration = \frac{(final velocity) - (initial velocity)}{time} = \frac{change of velocity}{time}
We normally use letters in this formula, which gives us:
a = \frac{v}{t}

Filling in this formula gives us:
Δv (change of velocity) = 12 - 4 = 8 m/s
t = 1s
a = \frac{8}{1} = 8 m/s^2
So our answer: the acceleration is 8m/ s^{2}
3 0
3 years ago
Date
sergiy2304 [10]

Answer:

Because Kinetic Energy(KE) is not the same as Momentum(P)

Kinetic Energy is a scalar(has magnitude only). For a body of mass M, velocity V:

KE = 0.5MV^2

The units of KE: Joules.

Energy is the ability to do work.

Momentum is not a form of energy.

Momentum is a vector(has magnitude and direction).

P = MV

Units of momentum: kg m/s

If you have rifles of mass 2, 4, 8, 16 kg, using the same cartridge, with the same load, barrel length(remember momentum of projectile is proportional to velocity), they all have the same recoil momentum.

But the kinetic energy of recoil would be inversely proportional to the mass of the gun.

Thus the 2kg gun(possible even in large powerful calibers due to modern materials like titanium etc), would have 8 times the recoil ENERGY of the 16kg gun.

A lot of confusion exists in America because of retention of old units, namely Foot Pounds(force) for KE, and Pounds(mass) Feet Per Second for Momentum(P). Because of the more awkward momentum units, a lot of old books had a bad habit of calling the momentum units Pounds Feet, leaving out the rest. Naturally this created confusion with Foot Pounds. Multiplication being commutative and all that:).

Remember that the momentum of the rifles is the same. But the ones with the highest recoil energy hurt the most.

Speaking of hurt:

If momentum killed, then consider two dinosaur killer asteroids with the same masses and velocities, striking vertically at the same time antipodal points on the Earth’s surface. Total momentum delivered would be Zero. That would not make us safe at all:)

Similarly, being shot simultaneously at close range from opposite sides with a 5 round burst from each from two M4 assault rifles(by definition must be able to fire full auto) delivered in 0.3 seconds, would deliver zero momentum. But not zero harm.

Also, the recoil momentum of any firearm is equal to the mass of projectile x velocity + mass of propellant x exit velocity of propellant. This is obviously greater, often much greater, depending on range, than the striking momentum of the projectile at the target.

The recoil kinetic energy is vastly less than the kinetic energy of the bullet/projectile. Neglecting propellant contribution:

recoil Momentum = bullet momentum

BUT:

recoil KE/bullet KE = projectile mass/gun mass

This is a very small fraction.

If we consider the M4 carried by American military:

M855(SS109 equivalent) 5.56 bullet of mass 0.004kg(62 grains)is fired from M4 assault rifle of mass, with optic and full mag 4kg, a thousand times as much!

Even allowing for the 0.0015kg powder charge, and the higher velocity of the powder(approx 1400=1500 m/s vs approx 900 m/s muzzle velocity of the bullet), the recoil energy is hundreds of times less than the muzzle energy of the bullet.

That’s why you want to be behind the gun, and not in front.

Explanation:

7 0
3 years ago
A student project is required to be portable and hand held. It requires 6 V DC power at a current of 150 mA. The batteries for t
egoroff_w [7]

Answer:

a. 5 batteries b. 1050 mAh

Explanation:

Here is the complete question

A student project is required to be portable and hand held. It requires 6 V DC power at a current of 150 mA. The batteries for the power supply must last for a minimum of 7 hours of continuous operation. NiMH rechargeable batteries in AA size are to be used. A) How many batteries are needed? B) What mAh capacity should the batteries have?

Solution

A) How many batteries are needed?

Since the nominal voltage for a single NiMH battery is 1.2 V per battery and we require 6V DC power, we combine the batteries in series to obtain a total voltage of 6 V. The number of batteries required, n = total voltage/voltage per cell = 6V/1.2V per battery = 5 batteries

So, the number of batteries needed is 5.

B) What mAh capacity should the batteries have?

Since the batteries are in series, they would each deliver a current of 150 mA. Since we require a current of 150 mA for 7 hours, the number of milliampere-hours capacity mAh of batteries required is Q = It where I = current = 150 mA and t = time = 7 hours.

So, Q = It = 150 mA × 7 h = 1050 mAh.

So, the batteries should have a mAh of 1050 mAh

4 0
3 years ago
If the sun became cooler but kept the same size it would emit
jek_recluse [69]
If the sun would become cooler having constant size, it would emit less ultraviolet light and less visible light than what it currently gives to earth. Hope this answers the question. Have a nice day. Thank you for posting here.
4 0
3 years ago
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