Answer:
<em>e. 4.1 m/s^2, 52 degrees north of east</em>
Explanation:
Two forces act on a 16 kg object. The first force has a magnitude of 68 N and is directed 24 degrees north of east. The second force is 32N , 48 degrees of west. What is the acceleration of the object resulting from the application of these forces? a. 1.6 m/s^2, 5.5 degrees north of west b. 1.9 m/s^2, 18 degrees north of west c. 2.4 m/s^2 , 34 degrees north of east d. 3.6 m/s^2 , 5.5 degrees north of west e.4.1 m/s^2, 52 degrees north of east
There are two components of force
finding the sum of the horizontal component of the forces
Efx
68 cos 24 - 32cos48 = 40.7088 N
Resolve the sum of the vertical components of the forces
68 sin 24 + 32 Sin 48 = 51.438
now we find the resultant of the two components
R=(Fy^2+Fx^2)^0.5
resultant force = (51.438 ^ 2 + 40.7088 ^ 2)^ 1/2
= 65.597
recall that force is the product of mass and acceleration
ma = 65.597
m=16 kg
a = 65.597/16 = 4.09 m/s/s
tan=EFy/EFx
tan = 51.438/ 40.7088 =
1.263
taking the tan inverse of both sides
= 51.64 degree north of east
Answer:
c) The pitch of the two sirens sounds the same to you
Explanation:
The pitch does not depend on the distance of the object from the observer.
As per the given data
pitch = frequency
Frequency =
=
Hence, the pitch of the two sirens remains the same for the observer.
Answer:
Force = 186 N
Explanation:
Torque is the rotational equivalent of linear force. It can be easely calculated using the formula :
Where is a vector that from the origin of the coordinate system to the point at which the force is applied (the position vector),
is the applied force.
The easiest way of computing the force is by setting the origin of the coordinate system to the lowest point of the torque wrench. By doing this we have that (the magnitud of the position vector) is 35cm.
Before computing the force we need to set all our values to the international system of units (SI). The torque is already in SI. The one missing is the length of the torque wrench (it is in centimeters and we need it in meters). So :
Now using the torque formula:
Where is the smaller angle between the force and the position vector. Because the force is applied perpendiculary to the position vector
, thus :
so the force is approximately 186 N.
The potential energy will be 1.46*10^-4J.
To find the answer, we have to know about the torque acting on a current loop in a uniform magnetic field.
<h3>How to find the potential energy of the loop?</h3>
where; M is the magnetic dipole moment, B is the magnetic field , and theta is the angle between M and B.
Thus, we can conclude that, the potential energy will be 1.46*10^-4J.
Learn more about the torque here:
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