A boat is headed due north on a river that's flowing directly east. What is the general direction of travel for this boat with r
2 answers:
Answer:
It must be moving at an angle with East towards North
Explanation:
As we know that boat headed towards North while river is flowing towards East
So here the resultant velocity of boat is vector sum of velocity of river and velocity of boat.
So we will find the vector resultant as
here we know
= velocity of boat towards North
= velocity of river towards East
now if we add two vectors along North and and east then the resultant must be between North and East so the boat will move at some angle with East towards North
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3) The work done is D. zero
4) The kinetic energy is B. 180 J
5) The potential energy is A. 120 J
6) The work done depends on B. position
7) The example of non-renewable energy is C. coal
8) The power expended is
9) The efficiency is A. 100%
10) The velocity ratio is 5
Explanation:
3)
The work done by a force acting an object is given by:
where :
F is the magnitude of the force
d is the displacement
is the angle between the direction of the force and the displacement
When the force is applied perpendicular to the direction of motion,
Therefore, the work done is:
4)
The kinetic energy of a body is given by
where
m is the mass of the body
v is its speed
For the girl in this problem, we have
m = 40 kg
v = 3 m/s
Therefore her kinetic energy is
5)
The potential energy of an object is given by
where
m is the mass
is the acceleration of gravity
h is the heigth of the object relative to the ground
For the ball in this problem,
m = 0.4 kg
h = 30 m
So, the potential energy is
6)
A conservative field is a field for which the work done by the field on an object does not depend on the path taken, but only on the initial and final position of the object.
Gravitational and electric fields are examples of conservative fields. In fact:
- When an object is pulled down by gravity (free fall), the work done by the gravitational field only depends on the change in height
between the two points, not on the path taken during the fall
- When an electric charge is pushed by the electric field, the work done by the field depends only on the initial and final position of the charge in the field
For any conservative field, it is possible to define a "potential" function, which represents the energy per unit mass/charge, and depends only on the position of the object.
7.
- Non-renewable energy sources are sources of energy whose rate of consumption is faster than the rate at which they are re-created. Examples of non-renewable sources are coal, oil, natural gas. These energy sources are consumed at a fast rate, while they take million of years to regenerate, so at the current rate they will eventually run out.
- Renewable energy sources are sources of energy that replenish at faster rate than the rate at which it is consumed. Examples of renewable sources are solar energy, wind, hydroelectric power.
Therefore, the example of non-renewable energy in this case is
C. Coal
8.
For an object pushed by a force F and moving at a constant velocity v, the power expended is given by
where F is the force and v is the velocity.
for the rocket in this problem, we have:
F = 10 N is the force propelling the rocket
v = 3000 m/s is its velocity
Substituting into the equation, we find the power expended:
9.
The efficiency of a machine is given by
where
is the energy in input to the machine
is the useful work in output from the machine
For a real machine, the useful work in output is always lower than the energy input, because part of the energy is "wasted" and converted into thermal energy due to the presence of internal frictions. However, for an ideal machine, all the input energy is converted into useful work, so
And therefore the efficiency is
which means 100%.
10.
The velocity ratio of a block and tackle system is the ratio between the distance moved by the effort and the distance moved by the load.
In a block and tackle system, the velocity ratio is also equal to the number of pulleys in the system.
For the system in the problem, there are 5 pulleys: therefore, this means that when the effort moves 5 metres, the load moves 1 metres, therefore the velocity ratio is
Learn more about kinetic and potential energy:
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Answer:
The resultant tension of the two ropes is approximately 42.4 N
The length of the line representing the resultant tension is approximately 8.48 cm
Please find included with the answer the scale drawing created with Microsoft Word
Explanation:
The given parameters are;
The tension in rope P, = 30 N
The tension in rope Q, = 30 N
The angle the rope, 'P', makes with the horizontal = 45°
The angle the rope, 'Q', makes with the horizontal = 45°
The scale factor of the scale diagram, S.F. = 5.0 N/cm
By the resolution of forces at equilibrium, we have;
The sum of the vertical forces, =
+
+ W = 0
∴ W = -( +
)
W = -(30 × sin(45°) + 30 × sin(45°)) = -42.4264068712
The weight of the heavy ball, W ≈ 42.4 N acting downwards
The sum of the horizontal forces, =
+
= 0
The length of the resultant force, W = W/(S.F.) ≈ 42.4 N/(5.0 N/cm) = 8.48 cm
The drawing of the vectors using the scale factor of 5.0 N/cm is created using Microsoft Word is included
Answer:
a) ( 0.0015139 i^ + 0.0020185 j^ + 0.00060556 k^ ) N.m
b) ΔU = -0.000747871 J
c) w = 47.97 rad / s
Explanation:
Given:-
- The area of the circular ring, A = 4.45 cm^2
- The current carried by circular ring, I = 13.5 Amps
- The magnetic field strength, vec ( B ) = (1.05×10−2T).(12i^+3j^−4k^)
- The magnetic moment initial orientation, vec ( μi ) = μ.(−0.8i^+0.6j^)
- The magnetic moment final orientation, vec ( μf ) = -μ k^
- The inertia of ring, T = 6.50×10^−7 kg⋅m2
Solution:-
- First we will determine the magnitude of magnetic moment ( μ ) from the following relation:
μ = N*I*A
Where,
N: The number of turns
I : Current in coil
A: the cross sectional area of coil
- Use the given values and determine the magnitude ( μ ) for a single coil i.e ( N = 1 ):
μ = 1*( 13.5 ) * ( 4.45 / 100^2 )
μ = 0.0060075 A-m^2
- From definition the torque on the ring is the determined from cross product of the magnetic moment vec ( μ ) and magnetic field strength vec ( B ). The torque on the ring in initial position:
vec ( τi ) = vec ( μi ) x vec ( B )
= 0.0060075*( -0.8 i^ + 0.6 j^ ) x 0.0105*( 12 i^ + 3 j^ -4 k^ )
= ( -0.004806 i^ + 0.0036045 j^ ) x ( 0.126 i^ + 0.0315 j^ -0.042 k^ )
- Perform cross product:
- The initial torque ( τi ) is written as follows:
vec ( τi ) = ( 0.0015139 i^ + 0.0020185 j^ + 0.00060556 k^ )
- The magnetic potential energy ( U ) is the dot product of magnetic moment vec ( μ ) and magnetic field strength vec ( B ):
- The initial potential energy stored in the circular ring ( Ui ) is:
Ui = - vec ( μi ) . vec ( B )
Ui =- ( -0.004806 i^ + 0.0036045 j^ ) . ( 0.126 i^ + 0.0315 j^ -0.042 k^ )
Ui = -[( -0.004806*0.126 ) + ( 0.0036045*0.0315 ) + ( 0*-0.042 )]
Ui = - [(-0.000605556 + 0.00011)]
Ui = 0.000495556 J
- The final potential energy stored in the circular ring ( Uf ) is determined in the similar manner after the ring is rotated by 90 degrees with a new magnetic moment orientation ( μf ) :
Uf = - vec ( μf ) . vec ( B )
Uf = - ( -0.0060075 k^ ) . ( 0.126 i^ + 0.0315 j^ -0.042 k^ )
Uf = - [( 0*0.126 ) + ( 0*0.0315 ) + ( -0.0060075*-0.042 ) ]
Uf = -0.000252315 J
- The decrease in magnetic potential energy of the ring is arithmetically determined:
ΔU = Uf - Ui
ΔU = -0.000252315 - 0.000495556
ΔU = -0.000747871 J
Answer: There was a decrease of ΔU = -0.000747871 J of potential energy stored in the ring.
- We will consider the system to be isolated from any fictitious forces and gravitational effects are negligible on the current carrying ring.
- The conservation of magnetic potential ( U ) energy in the form of Kinetic energy ( Ek ) is valid for the given application:
Ui + Eki = Uf + Ekf
Where,
Eki : The initial kinetic energy ( initially at rest ) = 0
Ekf : The final kinetic energy at second position
- The loss in potential energy stored is due to the conversion of potential energy into rotational kinetic energy of current carrying ring.
-ΔU = Ekf
0.5*T*w^2 = -ΔU
w^2 = -ΔU*2 / T
Where,
w: The angular speed at second position
w = √(0.000747871*2 / 6.50×10^−7)
w = 47.97 rad / s