Low blood pressure. The person could faint and have an irregular heartbeat.
In this problem we have the electric field intensity E:
E = 6.5 × newtons/coulomb
We have the magnitude of the load:
q = 6.4 × coulombs
We also have the distance d that the load moved in a direction parallel to the field 1.2 × meters.
We know that the electric potential energy (PE) is:
PE = qEd
So:
PE = (6.4 × )(6.5 × )(1.2 × )
PE = 5.0 x joules
None of the options shown is correct.
The strength of the friction doesn't matter. Neither does the distance or the time the asteroid takes to stop. All that matters is that the asteroid has
1/2 (mass) (speed squared)
of kinetic energy when it lands, and zero when it stops.
So
1/2 (mass) (original speed squared)
is the energy it loses to friction in order to come to rest.
Answer:
Approximately , assuming friction between the vehicle and the ground is negligible.
Explanation:
Let denote the mass of the vehicle. Let denote the initial velocity of the vehicle. Let denote the spring constant (needs to be found.) Let denote the maximum displacement of the spring.
Convert velocity of the vehicle to standard units (meters per second):
.
Initial kinetic energy () of the vehicle:
.
When the vehicle is brought to a rest, the elastic potential energy () stored in the spring would be:
.
By the conservation of energy, if the friction between the vehicle and the ground is negligible, the initial of the vehicle should be equal to the of the vehicle. In other words:
.
Rearrange this equation to find an expression for , the spring constant:
.
Substitute in the given values , , and :
Answer:
<h2>workdone = force × distance </h2><h2>236J = 18.9cos(o) × 24.4</h2><h2>236/24.4 = 18.9cos(o)</h2><h2>(0.5117)cos^-1 = (o)</h2><h2><u>59.21°</u></h2>