Answer:11.602 KW
Explanation:
mass of vehicle
speed=40Km/h
Resistance=600 N
Gear ratio
Net force to overcome by engine is
F=Resistance + sin component of weight
F=600+
Where
F=600+235.38=835.38 N
power=
Engine Power==11.602 KW
Read through the solution below. Then, answer the questions that follow.
2 answers:
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Answer:
a) 2.18 m/s^2
b) 9.83 m/s
Explanation:
The flywheel has a moment of inertia
J = m * k^2
Where
J: moment of inertia
k: radius of gyration
In this case:
J = 144 * 0.45^2 = 29.2 kg*m^2
The block is attached through a wire that is wrapped around the wheel. The weight of the block causes a torque.
T = p * r
r is the radius of the wheel.
T = m1 * g * r
T = 18 * 9.81 * 0.6 = 106 N*m
The torque will cause an acceleration on the flywheel:
T = J * γ
γ = T/J
γ = 106/29.2 = 3.63 rad/s^2
SInce the block is attached to the wheel the acceleration of the block is the same as the tangential acceleration at the eddge of the wheel:
at = γ * r
at = 3.63 * 0.6 = 2.81 m/s^2
Now that we know the acceleration of the block we can forget about the flywheel.
The equation for uniformly accelerated movement is:
X(t) = X0 + V0*t + 1/2*a*t^2
We can set a frame of reference that has X0 = 0, V0 = 0 and the X axis points in the direction the block will move. Then:
X(t) = 1/2*a*t^2
Rearranging
t^2 = 2*X(t)/a
It will reach the 1.8 m in 3.6 s.
Now we use the equation for speed under constant acceleration:
V(t) = V0 + a*t
V(3.6) = 2.81 * 3.6 = 9.83 m/s
Answer:
a. 0.11
b. 110 students
c. 50 students
d. 0.46
e. 460 students
f. 540 students
g. 0.96
Explanation:
(See attachment below)
a. Probability that a student got an A
To get an A, the student needs to get 9 or 10 questions right.
That means we want P(X≥9);
P(X>9) = P(9)+P(10)
= 0.06+0.05=0.11
b. How many students got an A on the quiz
Total students = 1000
Probability of getting A = 0.11 ---- Calculated from (a)
Number of students = 0.11 * 1000
Number of students = 110 students
So,the number of students that got A is 110
c. How many students did not miss a single question
For a student not to miss a single question, then that student scores a total of 10 out of possible 10
P(10) = 0.05
Total Students = 1000
Number of Students = 0.05 * 1000
Number of Students = 50 students
We see that 5
d. Probability that a student pass the quiz
To pass, a student needed to get at least 6 questions right.
So we want P(X>=6);
P(X>=) =P(6)+P(7)+P(8)+P(9)+P(10)
=0.08+0.12+0.15+0.06+0.05=0.46
So, the probability of a student passing the quiz is 0.46
e. Number of students that pass the quiz
Total students = 1000
Probability of passing the quiz = 0.46 ----- Calculated from (d)
Number of students = 0.46 * 1000
Number of students = 460 students
So,the number of students that passed the test is 460
f. Number of students that failed the quiz
Total students = 1000
Total students that passed = 460 ----- Calculated from (e)
Number of students that failed = 1000 - 460
Number of students that failed = 540
So,the number of students that failed is 540
g. Probability that a student got at least one question right
This means that we want to solve for P(X>=1)
Using the complement rule,
P(X>=1) = 1 - P(X<1)
P(X>=1) = 1 - P(X=0)
P(X>=1) = 1 - 0.04
P(X>=1) = 0.96
