Which of the following terms describes a process for which Q = -73 J?
1 answer:
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Answer: 62.22 meters
Explanation:
Two solve this, we need to know first how the Sled interact with the snow. While the Sled runs over the snow, the snow opposes to it movement, this is call friction. Frictions creates is a force that goes against any movement and is caused by the interaction between surfaces.
Friction can be calculated using the following equation:
Friction's Force (Ff) = m*N
Where m is the Frictions coficient and N is call Normal, which is a force created by the contact between any weight and a surface. Because the sled is and the surface are totally horizontal, the Normal is equal to the Sled's Weight.
So for this case, we can calculate the Friction's Force:
Ff = 0.045 * 735 N
Ff = 33,07 N
This force will cause a negative acceleration the Sled causing it to slow down and finally stop. To know this accelaration we can use the following equation:
Force (F) = Mass (m) * Accelaration (a)
Looking for the accelaration:
Acceleration (a) = F / M
As a side note:
Mass (M) = F / a (1)
Where we know that F is equal to our Friction Force (Ff) and the Mass for the girl and the Sled can be obtained using the previous equation (1) like this
Girl and sled Mass (M) = 735 N / (9.81 m/s)
M = 74,92 Kg
This is posible, because the Girl and The Sled are aplying a Force on the ground equal to 735 N. The accelartion for their mass is the gravity
Now, with the mass we can found the Accelaration caused by friction:
Acceleration (a) = Ff / M
Accelaration (a) = 33,07 N / 74,92 Kg
a = 0,44
To know how far the Sled travel we can use the following equation:
Where Vf is the Final Velocity, which is when the sled finally stops (Vf = 0), Vo is the Initial Speed of the Sled equal to 7.4 m/s and a is equal to the negative accelaration that causes the friction.
Searching for x:
The girls will run over the snow 62.22 meter until finally stopping
Answer:
a) t = 2.0 s, b) x_f = - 24.56 m, Δx = 16.56 m
Explanation:
This is an exercise in kinematics, the relationship of position and time is indicated
x = 5 t³ - 9t² -24 t - 8
a) ask when the velocity is zero
speed is defined by
v =
let's perform the derivative
v = 15 t² - 18t - 24
0 = 15 t² - 18t - 24
let's solve the quadratic equation
t =
t1 = -0.8 s
t2 = 2.0 s
the time has to be positive therefore the correct answer is t = 2.0 s
b) the position and distance traveled for a = 0
acceleration is defined by
a = dv / dt
a = 30 t - 18
a = 0
30 t = 18
t = 18/30
t = 0.6 s
we substitute this time in the expression of the position
x = 5 0.6³ - 9 0.6² - 24 0.6 - 8
x = 1.08 - 3.24 - 14.4 - 8
x = -24.56 m
we see that all the movement is in one dimension so the distance traveled is the change in position between t = 0 and t = 0.6 s
the position for t = 0
x₀ = -8 m
the position for t = 0.6 s
x_f = - 24.56 m
the distance
ΔX = x_f - x₀
Δx = | -24.56 -(-8) |
Δx = 16.56 m