A Plane crashes into the side of a mountain. If the crash took place over 9
1 answer:
Given parameters:
Duration of crash = 9min
Velocity before crash = 2000 yards per second
Unknown:
Acceleration of the plane = ?
Solution:
Acceleration is the change of velocity with time.
Acceleration =
where v is the final velocity
u is the initial velocity
t is the time taken
We need to convert the time into seconds from min;
60s = 1 min
So, 9min = 60 x 9 = 540s
So input the parameters and solve;
Acceleration = = 3.7yards/s²
The acceleration of the plane is 3.7yards/s²
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Answer:
Explanation:
= Permittivity of free space =
= Distance charge traveled = 2 m
v = Velocity of electron = 420 m/s
E = Electric field
= Mass of electron =
= Charge of electron =
As the energy of the system is conserved we have
For an infinite non conducting sheet electric field is given by
The surface charge density is
<h2>v = 2.5J/c</h2>
</h2><h2>Solution:</h2><h3>_________</h3><h2>V = 
</h2><h3>V = 2.5V</h3>
</h2>
Explanation:
<h2>Given:</h2><h3>Energy = E = 20J</h3><h3>Charge = Q = 8C</h3><h2>Require :</h2>___________
<h3>Potential Difference = V = ?</h3><h2>Formula :</h2>____________
<h2>V =.
.
.
<h2>a) Vector product: |d1 × d2| = (34.2 m) k
b) Scalar product: d1 · d2 = -0.874 m
c) (d1 + d2) · d2 = 16.1 m
d) Component of d1 along direction of d2: -0.21 m
Explanation:
a)
In this part, we want to calculate
|d1 × d2|
Which is the vector product between the two displacements d1 and d2.
The two vectors are:
d1 = (6.00 m)i + (5.74 m)j
d2 = (-2.92 m)i + (2.9 m)j
The vector product of two vectors and
is also a vector which has components:
We notice immediatly that in this problem ,the two vectors d1 and d2 lie in the x-y plane, so they do not have components in zero. Therefore, the vector product has only one component, which is the one in z, and it is:
Therefore, the vector product of d1 and d2 is:
|d1 × d2| = (34.2 m) k
b)
In this case, we want to calculate
d1 · d2
Which is the scalar product between the two displacements.
The scalar product of two vectors and
is a scalar given by:
In this problem,
d1 = (6.00 m)i + (5.74 m)j
d2 = (-2.92 m)i + (2.9 m)j
Therefore, the scalar product between the two vectors is:
c)
In this case, we want to calculate
(d1 + d2) · d2
Which means that first we have to calculate the resultant displacement d1 + d2, and then calculate the scalar product of the resultant vector with d2.
Given two vectors and
, the resultant vector is also a vector given by
In this case,
d1 = (6.00 m)i + (5.74 m)j
d2 = (-2.92 m)i + (2.9 m)j
So the resultant vector is
So
And calculating the scalar product with d2, we find:
d)
The component of a vector a along another vector b is given by
where is the scalar product between and b
is the magnitude of vector b
In this problem, we have the two vectors
d1 = (6.00 m)i + (5.74 m)j
d2 = (-2.92 m)i + (2.9 m)j
We want to find the component of d1 along the direction of d2.
We already calculated the scalar product of the two vectors in part b):
d1 · d2 = -0.874 m
The magnitude of a vector b is given by
So, for vector d2,
Now we can calculate the component of d1 along d2:
Learn more about operations with vectors:
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