Answer:
As the "plates" on each side of ridges in the seafloor are pulled away, lava comes up from the middle, hardens and "records" the current magnetic field.
Explanation:
Explanation:
The question says that "Does a 60 kg person running at 4 m/s have more kinetic energy than a 10 gram projectile at 300 m/s ?
"
Speed of a person is 4 m/s
Mass of a person is 60 kg
Kinetic energy of a person is : 
So,
Mass of a projectile is 10 grams or 0.01 kg
Speed of a projectile is 300 m/s
Kinetic energy of a projectile is :
So, it is clear that the kinetic energy of a person is more than that of the kinetic energy of a projectile.
Answer:
94 kg
Explanation:
The mass registered by the scale is based on the assumption that the force applied is due entirely to gravity. If the force is greater, then the indicated mass will be greater.
__
<h3>how many g's</h3>
As a fraction of the acceleration of gravity, the elevator's acceleration is ...
(1.2 m/s²)/(9.8 m/s²) ≈ 6/49
<h3>net force</h3>
The force required to produce a given acceleration is found by the formula ...
F = ma . . . . . . . force on mass m to produce acceleration 'a'
When the man is stationary on the scale, the upward force it supplies is balanced by the downward force on the man due to gravity. The force and the mass are proportional, and the constant of proportionality (the acceleration due to gravity) is used to calibrate the scale. More force is thus translated to a higher mass reading.
Since the man's net acceleration is upward at the rate of 6/49×g, the total force applied by the scale is (1 +6/49) = 55/49 times as great as when the man is stationary. This greater force gets translated to a greater mass reading.
The force is equivalent to what would be required to support a stationary man with a mass of ...
(84 kg)(55/49) = 94 2/7 kg
The scale would read about 94 kg during the upward acceleration period.
Use the equation q=mc/\T, where q is the heat lost, m is mass, and /\T is the change in temperature, and c is the specific heat.
q=50kg(3470J/kg K)(2K)
q=<span>347000 J
Any other questions, just ask.
</span>
Answer:
Part 1
20 N
Part 2
0.4 m/s²
Part 3
4 m/s
Explanation:
The force which pulls the sled right = 50 N
The friction force exterted towards left by the snow = -30 N
The mass of the sled = 50 kg
Part 1
The sum of the forces on the sled, F = 50 N + (-30) N = 20 N
Part 2
The acceleration of the sled is given as follows;
F = m·a
Where;
m = The mass of the sled
a = The accelertion
a = F/m
∴ a = (20 N)/(50 kg) = 0.4 m/s²
The acceleration of the sled, a = 0.4 m/s²
Part 3
The initial velocity of the sled, u = 2 m/s
The kinematic equation of motion to determine the speed of the sled is v = u + a·t
The speed, <em>v</em>, of the sled after t = 5 seconds is therefore;
v = 2 m/s + 0.4 m/s² × 5 s = 4 m/s.
