An electron is moving northward in a magnetic field. The magnetic force on the electron is toward the northeast. What is the dir
1 answer:
To solve this problem we will use the vector concept given by the cross product between two perpendicular vectors and which results in a vector perpendicular to these two. From the definition of the Magnetic Force we have to
From the property of cross product the magnetic force should point in the direction perpendicular to the plane containing the vectors v and B.
The direction of velocity is north, and the direction of the magnetic force is northeast.
This cannot be the case, as the direction of magnetic force is not perpendicular to the direction of velocity of the charge.
Therefore the correct option for the direction of the magnetic field is <em>"This situation cannot exist because of the relative orientations of the velocity and force vectors" </em>
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The mass of an electron, me, 9.11 × 10⁻³¹ kilograms shows there are 3 important numbers in them, namely numbers 9, 1, and 1
<h3><em>Further explanation</em></h3>Significant numbers are obtained from measurement results of exact numbers and the last number estimated
This is called significant numbers
the scientific notation can be described as:
a, ... x 10ⁿ
a, ... called a significant number
10ⁿ is called a big order
Rules for significant numbers in general:
- 1. All non-zero numbers are significant numbers
- 2. a zero which is located between two non-zero numbers including a significant number
- 3. all zeros are located in the final row written behind the decimal point of the include significant number
- 4. zero decimal point is the not significant number
From the numbers known in the question amounting to 9.11 × 10⁻³¹ kilograms shows there are 3 important numbers in them,(symbol a, before a big order ,called a significant number) namely numbers 9, 1, and 1
<h3><em>Learn more</em></h3>significant figures
significant figures in the following number: 5.67 x 106
the number of significant figures is 0.025
0.080 significant figures
the fewest number of significant figures
Keywords : significant number, the scientific notation, decimal point, a big order
Answer:
h = v₀² / 2g , h = k/4g x²
Explanation:
In this exercise we can use the law of conservation of energy at two points, the lowest, before the shot and the highest point that the mouse reaches
Starting point. Lower compressed spring
Em₀ = K = ½ m v²
Final point. Highest on the path
= U = mg h
As or no friction the energy is conserved
Em₀ = Em_{f}
½ m v₀²² = m g h
h = v₀² / 2g
We can also use as initial energy the energy stored in the spring that will later be transferred to the mouse
½ k x² = 2 g h
h = k/4g x²
So her speed in still water will be 6mph :)
Answer:
v₀ =3.8 m/s
Explanation:
Newton's second law of the box:
∑F = m*a Formula (1)
∑F : algebraic sum of the forces in Newton (N)
m : mass in kilograms (kg)
a : acceleration in meters over second square (m/s²)
Known data
m=2.1 kg mass of the box
d= 5.4m length of the roof
θ = 20° angle θ of the roof with respect to the horizontal direction
μk= 0.51 : coefficient of kinetic friction between the box and the roof
g = 9.8 m/s² : acceleration due to gravity
Forces acting on the box
We define the x-axis in the direction parallel to the movement of the box on the roof and the y-axis in the direction perpendicular to it.
W: Weight of the box : In vertical direction
N : Normal force : perpendicular to the direction the roof
fk : Friction force: parallel to the direction to the roof
Calculated of the weight of the box
W= m*g = (2.1 kg)*(9.8 m/s²)= 20.58 N
x-y weight components
Wx= Wsin θ= (20.58)*sin(20)° =7.039 N
Wy= Wcos θ =(20.58)*cos(20)°= 19.34 N
Calculated of the Normal force
∑Fy = m*ay ay = 0
N-Wy= 0
N=Wy =19.34 N
Calculated of the Friction force:
fk=μk*N= 0.51* 19.34 N = 9.86 N
We apply the formula (1) to calculated acceleration of the block:
∑Fx = m*ax , ax= a : acceleration of the block
Wx-f = ( 2.1)*a
7.039 - 9.86 = ( 2.1)*a
-2.821 = ( 2.1)*a
a=(-2.821) /( 2.1)
a= -1.34 m/s²
Kinematics of the box
Because the box moves with uniformly accelerated movement we apply the following formula to calculate the final speed of the block :
vf²=v₀²+2*a*d Formula (2)
Where:
d:displacement = 5.4 m
v₀: initial speed
vf: final speed = 0
a : acceleration of the box = -1.34 m/s²
We replace data in the formula (2)
0²=v₀²+2*(-1.34)*(5.4)
2*(1.34)*(5.4)= v₀²
v₀ = 3.8 m/s