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Sholpan [36]
3 years ago
15

In another solar system is planet Driff, which

Physics
1 answer:
Fed [463]3 years ago
5 0

Answer:

It is (1/5)th as much.

Explanation:

If we apply the equation

F = G*m*M / r²

where

m = mass of a man

M₀ = mass of the planet Driff

M = mass of the Earth

r₀ = radius of the planet Driff

r = radius of the Earth

G = The gravitational constant

F = The gravitational force on the Earth

F₀ = The gravitational force on the planet Driff

g = the gravitational acceleration on the surface of the earth

g₀ = the gravitational acceleration on the surface of the planet Driff

we have

F₀ = G*m*M₀ / r₀² = G*m*(5*M) / (5*r)²    

⇒  F₀ = G*m*M / (5*r²) = (1/5)*F

If

F₀ = (1/5)*F

then

W₀ = (1/5)*W   ⇒  m*g₀ = (1/5)*m*g   ⇒   g₀ = (1/5)*g

It is (1/5)th as much.

You might be interested in
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 truck is traveling east at 80 km/h. At an intersection 32 km ahead, a car is traveling north at 50 km/h. How long after this m
nirvana33 [79]

The time elapsed when the vehicles are closest to each other is 20 min.

The given parameters:

  • Speed of the truck, u = 80 km/h
  • Distance, d = 32 km
  • Speed of the car, v = 50 km/h

<h3>Principles of relative speed</h3>

The time elapsed when the cars are close to each other is calculated by applying the principles of relative speed.

(V_r) t = d\\\\V_r^2 = 50^2 + 80^2\\\\V_r =\sqrt{50^2 + 80^2} \\\\V_r = 94.34 \ km/h

94.34 t = 32\\\\t = \frac{32}{94.34} \\\\t = 0.34 \ hr\\\\t \approx 20 \min

Thus, the time elapsed when the vehicles are closest to each other is 20 min.

Learn more about relative velocity here: brainly.com/question/24430414

3 0
3 years ago
Answer and I will give you brainiliest <br><br><br>Please heeeelp​
Lostsunrise [7]

Answer:

T = 4.905[N]

Explanation:

In order to solve this problem we must perform a sum of forces on the vertical axis.

∑Fy = 0

We have two forces acting only, the weight of the body down and the tension force T up, as the body does not move we can say that it is system is in static equilibrium, therefore the sum of forces is equal to zero.

T-m*g=0\\T=0.5*9.81\\T=4.905[N]

5 0
3 years ago
A projectile is launched at an angle above the
gtnhenbr [62]
The first rule of vectors is that the horizontal and vertical components are separate. Disregarding air resistance, the only thing we have to worry about is gravity.

The appropriate suvat to use for the vertical component is v = u +at
I will take a to be -9.81, you may have to change it to be 10 if your qualification likes g to be 10.

v = 30 + (-9.81x2)
v = 30 - 19.62
=10.38m/s

Therefore we know that after 2.0 s the vertical component will be 10.38ms^-1, ie 10m/s as the answers given are all to 2sf.

The horizontal component is completely separate to the vertical component and since there is no air resistance, it will remain constant throughout the projectiles trajectory. Therefore it will remain at 40ms^-1.

Combining this together we get:
(1) vx=40m/s and vy=10m/s

7 0
3 years ago
A train 471 m long is moving on a straight track with a speed of 75.1 km/h. The engineer applies the brakes at a crossing, and l
hram777 [196]

Answer:

t = 37.6 s

Explanation:

As we know that train is initially moving with the speed

v_i = 75.1 km/h

now we know that

v_i = 20.86 m/s

now the final speed of the train when it crossed the crossing

v_f = 15 km/h

v_f = 4.17 m/s

now we can use kinematics here

v_f^2 - v_i^2 = 2 a d

4.17^2 - 20.86^2 = 2 a(471)

a = -0.44 m/s^2

Now the time to cross that junction is given as

v_f - v_i = at

4.17 - 20.86 = a(-0.44)

t = 37.6 s

4 0
3 years ago
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