A battery supplies 0.71 A to three resistors connected in parallel. The current through the first resistor is 0.27 A and the cur
1 answer:
You might be interested in
1) He has both potential and kinetic energy
2) Before the parachute opens, the potential energy decreases and the kinetic energy increases
Explanation:
1)
The gravitational potential energy of a body is the energy possessed by the object due to its position in a gravitational field, and it is given by:
where
m is the mass of the body
g is the acceleration of gravity
h is the height of the body above the ground
On the other hand, the kinetic energy of a body is the energy possessed by the body due to its motion; it is given by
where
v is the speed of the object
Here Mr. Leppold has both potential and kinetic energy before opening the parachute, because:
- It is moving at a certain speed, so , therefore he has kinetic energy
- He is at a certain height above the ground, , therefore he has potential energy
2)
The total mechanical energy of Mr.Leppold is the sum of the potential and the kinetic energy:
According to the law of conservation of energy, in absence of air resistance, this quantity remains constant.
During the fall, the height of Leppold decreases: this means that as decreases, the potential energy decreases too.
However, the total energy E must remain constant: therefore, this means that the kinetic energy KE must increase, and this occurs because the speed of Mr. Leppold increases as he falls.
Learn more about kinetic and potential energy:
#LearnwithBrainly
Answer:
a) The speed of the car when it reaches the edge of the cliff is 19.4 m/s
b) The time it takes the car to reach the edge is 4.79 s
c) The velocity of the car when it lands in the ocean is 31.0 m/s at 60.2º below the horizontal
d) The total time interval the car is in motion is 6.34 s
e) The car lands 24 m from the base of the cliff.
Explanation:
Please, see the figure for a description of the situation.
a) The equation for the position of an accelerated object moving in a straight line is as follows:
x =x0 + v0 * t + 1/2 a * t²
where:
x = position of the car at time t
x0 = initial position
v0 = initial velocity
t = time
a = acceleration
Since the car starts from rest and the origin of the reference system is located where the car starts moving, v0 and x0 = 0. Then, the position of the car will be:
x = 1/2 a * t²
With the data we have, we can calculate the time it takes the car to reach the edge and with that time we can calculate the velocity at that point.
46.5 m = 1/2 * 4.05 m/s² * t²
2* 46.5 m / 4.05 m/s² = t²
<u>t = 4.79 s </u>
The equation for velocity is as follows:
v = v0 + a* t
Where:
v = velocity
v0 = initial velocity
a = acceleration
t = time
For the car, the velocity will be
v = a * t
at the edge, the velocity will be:
v = 4.05 m/s² * 4.79 s = <u>19.4 m/s</u>
b) The time interval was calculated above, using the equation of the position:
x = 1/2 a * t²
46.5 m = 1/2 * 4.05 m/s² * t²
2* 46.5 m / 4.05 m/s² = t²
t = 4.79 s
c) When the car falls, the position and velocity of the car are given by the following vectors:
r = (x0 + v0x * t, y0 + v0y * t + 1/2 * g * t²)
v =(v0x, v0y + g * t)
Where:
r = position vector
x0 = initial horizontal position
v0x = initial horizontal velocity
t = time
y0 = initial vertical position
v0y = initial vertical velocity
g = acceleration due to gravity
v = velocity vector
First, let´s calculate the initial vertical and horizontal velocities (v0x and v0y). For this part of the problem let´s place the center of the reference system where the car starts falling.
Seeing the figure, notice that the vectors v0x and v0y form a right triangle with the vector v0. Then, using trigonometry, we can calculate the magnitude of each velocity:
cos -37.0º = v0x / v0
(the angle is negative because it was measured clockwise and is below the horizontal)
(Note that now v0 is the velocity the car has when it reaches the edge. it was calculated in a) and is 19,4 m/s)
v0x = v0 * cos -37.0 = 19.4 m/s * cos -37.0º = 15.5 m/s
sin 37.0º = v0y/v0
v0y = v0 * sin -37.0 = 19.4 m/s * sin -37.0 = - 11. 7 m/s
Now that we have v0y, we can calculate the time it takes the car to land in the ocean, using the y-component of the vector "r final" (see figure):
y = y0 + v0y * t + 1/2 * g * t²
Notice in the figure that the y-component of the vector "r final" is -30 m, then:
-30 m = y0 + v0y * t + 1/2 * g * t²
According to our reference system, y0 = 0:
-30 m = v0y * t + 1/2 g * t²
-30 m = -11.7 m/s * t - 1/2 * 9.8 m/s² * t²
0 = 30 m - 11.7 m/s * t - 4.9 m/s² * t²
Solving this quadratic equation:
<u>t = 1.55 s</u> ( the other value was discarded because it was negative).
Now that we have the time, we can calculate the value of the y-component of the velocity vector when the car lands:
vy = v0y + g * t
vy = - 11. 7 m/s - 9.8 m/s² * 1.55s = -26.9 m/s
The x-component of the velocity vector is constant, then, vx = v0x = 15.5 m/s (calculated above).
The velocity vector when the car lands is:
v = (15.5 m/s, -26.9 m/s)
We have to express it in magnitude and direction, so let´s find the magnitude:
To find the direction, let´s use trigonometry again:
sin α = vy / v
sin α = 26.9 m/s / 31.0 m/s
α = 60.2º
(notice that the angle is measured below the horizontal, then it has to be negative).
Then, the vector velocity expressed in terms of its magnitude and direction is:
vy = v * sin -60.2º
vx = v * cos -60.2º
v = (31.0 m/s cos -60.2º, 31.0 m/s sin -60.2º)
<u>The velocity is 31.0 m/s at 60.2º below the horizontal</u>
d) The total time the car is in motion is the sum of the falling and rolling time. This times where calculated above.
total time = falling time + rolling time
total time = 1,55 s + 4.79 s = <u>6.34 s</u>
e) Using the equation for the position vector, we have to find "r final 1" (see figure):
r = (x0 + v0x * t, y0 + v0y * t + 1/2 * g * t²)
Notice that the y-component is 0 ( figure)
we have already calculated the falling time and the v0x. The initial position x0 is 0. Then.
r final 1 = ( v0x * t, 0)
r final 1 = (15.5 m/s * 1.55 s, 0)
r final 1 = (24.0 m, 0)
<u>The car lands 24 m from the base of the cliff.</u>
PHEW!, it was a very complete problem :)
