An ice cube of mass 50.0 g can slide without friction up and down a 25.0 degree slope. The ice cube is pressed against a spring
1 answer:
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Answer:
The value is
Explanation:
From the question we are told that
The mass of the car is
The acceleration is
Generally the net force applied on the rope is mathematically represented as
Here W is the weight of the car which is evaluated as
=>
=>
Generally the net force can also be mathematically represented as
So
=>
=>
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 :)
Answer:
(a) 43.2 kC
(b) 0.012V kWh
(c) 0.108V cents
Explanation:
<u>Given:</u>
- i = current flow = 3 A
- t = time interval for which the current flow =
- V = terminal voltage of the battery
- R = rate of energy = 9 cents/kWh
<u>Assume:</u>
- Q = charge transported as a result of charging
- E = energy expended
- C = cost of charging
Part (a):
We know that the charge flow rate is the electric current flow through a wire.
Hence, 43.2 kC of charge is transported as a result of charging.
Part (b):
We know the electrical energy dissipated due to current flow across a voltage drop for a time interval is given by:
Hence, 0.012V kWh is expended in charging the battery.
Part (c):
We know that the energy cost is equal to the product of energy expended and the rate of energy.
Hence, 0.108V cents is the charging cost of the battery.