Answer:
The resultant velocity is 86.1 mi/h.
Explanation:
The law of cosines is given by:
Where:
c: is the resultant velocity =?
a: is the velocity of the plane = 75.0 mi/h
b: is the velocity of the wind = 15.0 mi/h
θ: is the angle between "a" and "b"
The angle between "a" and "b" can be found as follows:
Now, by using the law of cosines we have:
Therefore, the resultant velocity is 86.1 mi/h.
The law of sines is:
Where:
γ: is the angle between "b" and "c"
α: is the angle between "a" and "c"
So, if we want to find "c" by using the law of sines, we need to know another angle besides θ (γ or α), and the statement does not give us.
I hope it helps you!
Answer:
<em>2.78m/s²</em>
Explanation:
Complete question:
<em>A box is placed on a 30° frictionless incline. What is the acceleration of the box as it slides down the incline when the co-efficient of friction is 0.25?</em>
According to Newton's second law of motion:
Where:
is the coefficient of friction
g is the acceleration due to gravity
Fm is the moving force acting on the body
Ff is the frictional force
m is the mass of the box
a is the acceleration'
Given
Required
acceleration of the box
Substitute the given parameters into the resulting expression above:
Recall that:
9.8sin30 - 0.25(9.8)cos30 = ax
9.8(0.5) - 0.25(9.8)(0.866) = ax
4.9 - 2.1217 = ax
ax = 2.78m/s²
<em>Hence the acceleration of the box as it slides down the incline is 2.78m/s²</em>
1) According to the law of conservation of momentum ..
<span>Horiz recoil mom of gun (M x v) = horiz. mon acquired by shell (m x Vh) </span>
<span>1.22^6kg x 5.0 m/s = 7502kg x Vh </span>
<span>Vh = 1.22^6 x 5 / 7502 .. .. Vh = 813 m/s </span>
<span>Barrel velocity V .. .. cos20 = Vh / V .. ..V = 813 /cos20 .. .. ►V = 865 m/s </span>
<span>2) Using the standard range equation .. R = u² sin2θ /g </span>
<span>R = 865² x sin40 / 9.80 .. .. ►R = 49077 m .. (49 km)</span>
Answer:
0.3405V
Explanation:
#Given a magnetic field of 
, diameter= 18.5cm(r=9.25cm or 0.0925m), we find the magnetic flux of the loop as:
we can now calculate the induced emf, 
:
Hence, the induced emf of the loop is 0.3405V
The angular momentum of the person is given by:
where
is the mass of the person,
is the tangential velocity, and
is the distance of the person from the center of rotation. Using these data, we find
