The first step that Enrique must take in order to calculate the tangential speed of the satellite is to convert the period from days to seconds.
We know that the SI unit of speed is meter per second and now, we with to obtain the tangential speed of the satellite.
Since the period is given in days, the first step is to convert the period from days to seconds.
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Answer:
Gravity. An object is moving across a surface, but it does not gain or lose speed.
Explanation:
The basic idea. Physicists see gravity as one of the four fundamental forces that govern the universe, alongside electromagnetism and the strong and weak nuclear forces.
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Answer:
The total amount of energy that would have been released if the asteroid hit earth = The kinetic energy of the asteroid = 1.29 × 10¹⁵ J = 1.29 PetaJoules = 1.29 PJ
1 PJ = 10¹⁵ J
Explanation:
Kinetic energy = mv²/2
velocity of the asteroid is given as 7.8 km/s = 7800 m/s
To obtain the mass, we get it from the specific gravity and diameter information given.
Density = specific gravity × 1000 = 3 × 1000 = 3000 kg/m³
But density = mass/volume
So, mass = density × volume.
Taking the informed assumption that the asteroid is a sphere,
Volume = 4πr³/3
Diameter = 30 m, r = D/2 = 15 m
Volume = 4π(15)³/3 = 14137.2 m³
Mass of the asteroid = density × volume = 3000 × 14137.2 = 42411501 kg = 4.24 × 10⁷ kg
Kinetic energy of the asteroid = mv²/2 = (4.24 × 10⁷)(7800²)/2 = 1.29 × 10¹⁵ J
Answer:
A) F=-20.16×10⁹N
B) if the distance doubles, force is 4 times smaller.
Explanation:
q1=-28C
q2=5mC=0.005C
d=25cm=0.25m
Electrostatic force between charges: F=k×q1×q2/d², where k is a coefficient that has the value k=9 × 10⁹ N⋅m²⋅C^(-2) for air.
Thus:
F=9×10⁹×(-28)×0.005/0.25²
F=-20.16×10⁹N
The minus sign indicates attraction.
If distance doubles, d1=2×d, then we have 4d² at the denominator and the force is 4 times smaller.
Explanation:
Work done is given by the product of force and displacement.
Case 1,
1. A boy lifts a 2-newton box 0.8 meters.
W = 2 N × 0.8 m = 1.6 J
2. A boy lifts a 5-newton box 0.8 meters.
W = 5 N × 0.8 m = 4 J
3. A boy lifts a 8-newton box 0.2 meters.
W = 8 N × 0.2 m = 1.6 J
4. A boy lifts a 10-newton box 0.2 meters.
W = 10 N × 0.2 m = 2 J
Out of the four options, in option (2) ''A boy lifts a 5-newton box 0.8 meters'', the work done is 4 J. Hence, the greatest work done is 4 J.