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const2013 [10]
3 years ago
14

The electric force between electric charges is much larger than the gravitational force between the charges. Why, then, is the g

ravitational force between Earth and the Moon much larger than the electric force between Earth and the Moon?
Physics
1 answer:
Mamont248 [21]3 years ago
5 0

Answer:

Explanation:

The electric force between charges is much larger than the gravitational force but Gravitational force between earth and moon is dominant over Electric force because earth and moon are the electrically neutral body.

Electrically neutral bodies are those bodies that contain equal no of electron and proton in the body.

An example of electrically neutral bodies is a neutron.    

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suppose that you look into a photometer's eyepiece and the fluorescent disks appear to be equal in intensity. If the distance be
Y_Kistochka [10]
75 candela

Hope this helps ;)

7 0
3 years ago
Some hydrogen gas is enclosed within a chamber being held at 200^\ { C} with a volume of 0.025 \rm m^3. The chamber is fitted wi
vlada-n [284]

Answer:

The final volume is 0.039 m^3

Explanation:

<u>Data:</u>

Initial temperature: T1=200C

Final temperature: T2=200C

Initial pressure: P1=1.50 \times10^6 Pa

Final pressure: P2=0.950 \times10^6 Pa

Initial volume: V1=0.025m^{3}

Final volume: V2=?

Assuming hydrogen gas as a perfect gas it satisfies the perfect gas equation:

\frac{PV}{T}=nR (1)

With P the pressure, V the volume, T the temperature, R the perfect gas constant and n the number of moles. If no gas escapes the number of moles of the gas remain constant so the right side of equation (1) is a constant, that allows to equate:

\frac{P_{1}V_{1}}{T_{1}}=\frac{P_{2}V_{2}}{T_{2}}

Subscript 2 referring to final state and 1 to initial state.

solving for V2:

V_{2}=\frac{P_{1}V_{1}T_{2}}{T_{1}P_{2}}=\frac{(1.50 \times10^6)(0.025)(200)}{(200)(0.950 \times10^6)}

V_{2}=0.039 m^3

3 0
3 years ago
A ball is dropped from rest. What will be its speed when it hits the ground in each case. a. It is dropped from 0.5 meter above
Papessa [141]

Answer:

(a) 3.13 m/s

(b) 9.9 m/s

(c) 7.73 m/s

Explanation:

u = 0 m/s, g = 9.8 m/s^2

Let v be the velocity of ball as it hit the ground.

(a) h = 0.5 m

Use third equation of motion.

v^2 = u^2 + 2 g h

v^2 = 0 + 2 x 9.8 x 0.5

v^2 = 9.8

v = 3.13 m/s

(b) h = 5 m

Use third equation of motion.

v^2 = u^2 + 2 g h

v^2 = 0 + 2 x 9.8 x 5

v^2 = 98

v = 9.9 m/s

(c) h = 10 feet = 3.048 m

Use third equation of motion.

v^2 = u^2 + 2 g h

v^2 = 0 + 2 x 9.8 x 3.048

v^2 = 59.74

v = 7.73 m/s

5 0
3 years ago
An object acted on by three forces moves with constant velocity. One force acting on the object is in the positive xx direction
Rudik [331]

Answer with Explanation:

We are given that

F_1=6.9 N

F_2=4.5 N

We have to find the direction and magnitude of the third force acting on the object.

Resultant force,F=\sqrt{F^2_1+F^2_2}

F=\sqrt{(6.9)^2+(4.5)^2}=8.24 N

The object moves with constant velocity .Therefore, net force on object is equal to zero

So,Third force,F_3=F=8.24 N

Direction,\theta=tan^{-1}(\frac{F_2}{F_1})

\theta=tan^{-1}(\frac{4.5}{6.9})=33.02^{\circ}

Angle lies in second quadrant because the direction of third force is opposite to the direction of the resultant force of F1 and F2.

Therefore,\theta=\pi-\theta=180-33.02=146.98^{\circ}

8 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
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