We need to be careful here.
The calculation of the gravitational force between two objects
refers to the distance between their centers.
The minimum possible distance between the Earth's and moon's
centers is the sum of their radii (radiuses).
Earth's radius . . . . . 6,360 km = 6.36 x 10⁶ meters
Moon's radius . . . . . 1,738 km = 1.738 x 10⁶ meters
Sum of their radii = 8.098 x 10⁶ meters
Also:
Earth's mass . . . . . 5.972 x 10²⁴ kg
Moon's mass . . . . . 7.348 x 10²² kg
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and now we're ready to go !
Gravitational force =
G M₁ M₂ / R²
= (6.67 x 10⁻¹¹ N-m²/kg²)(</span><span>5.972 x 10²⁴ kg)(7.348 x 10²² kg)/</span>(8.098 x 10⁶ m)²
= (6.67 · 5.972 · 7.348 / 8.098²) · (10²³) Newtons
= (I get ...) 4.463 x 10²³ Newtons
That's almost exactly 10²³ pounds
= 50,153,000,000,000,000,000 tons.
Those are big numbers.
All I can say is: I wouldn't exactly call that "resting" on the surface".
Answer:
v₃ = 5 [m/s]
Explanation:
To solve this problem we must use the definition of linear momentum, which tells us that momentum is equal to the product of mass by Velocity.
P = m*v
where:
P = linear momentum [kg*m/s]
m = mass [kg]
v = velocity [m/s]
We must also clarify that the momentum is preserved i.e. it is equal before the collision and after the collision
Pbeforecollision = Paftercollision
(m₁*v₁) + (m₂*v₂) = (m₁*v₃) + (m₂*v₄)
where:
m₁ = mass of the truck = 3000 [kg]
v₁ = velocity of the truck = 10 [m/s]
m₂ = mass of the car = 1000 [kg]
v₂ = velocity of the car before the collision = 0 (the car is parked)
v₃ = velocity of the truck after the collision [m/s]
v₄ = velocity of the car after the collision = 15 [m/s]
(3000*10) + (1000*0) = (3000*v₃) + (1000*15)
30000 = 3000*v₃ + 15000
3000*v₃ = 30000 - 15000
3000*v₃ = 15000
v₃ = 5 [m/s]
Because it is releasing heat from your body sweating and prespirating rlmelease heat so that you body will cool
Answer:
W₂= 10000 N
Explanation:
Pascal´s Principle can be applied in the hydraulic press:
If we apply a small force (F1) on a small area piston A1, then, a pressure (P) is generated that is transmitted equally to all the particles of the liquid until it reaches a larger area piston and therefore a force (F2) can be exerted that is proportional to the area (A2) of the piston:
Pressure is defined as the force (F) applied per unit area (A)
P=F/A (N/m²)
P1=P2

Equation (1)
Data
W₁ = weight sits on the small piston
F₁ = W₁= 500 N
A₁ = 2.0 cm²
A₂ = 40 cm²
Calculation of the weight (W₂) can the large piston support
We replace data in the equation (1)
F₂ = 10000 N
W₂= F₂= 10000 N