Answer:
The change in momentum is Δp= 0.028 kg m/sExplanation:
An impulse describes a change in momentum. The change in momentum of an object is its mass times the change in its velocity.
The change in moment is given by the expression below
Δp=m⋅(Δv)=m⋅(vf−vi) .
Given data
mass m= 0.140-kg
initial velocity vi= 120 m/s
final velocity vf= 1 m/s
substituting we have
Δp=m⋅(Δv)=0.14⋅(1−1.2)
Δp=m⋅(Δv)=0.14⋅(-0.2)
Δp= 0.028 kg m/s
The change in momentum was found to be Δp= 0.028 kg m/s
What is the magnitude of the electric force on the charge at the bottom left-hand vertex of the triangle due to the other two charges
Answer:
[tex]\frac{kq^{2} }{2a^2}[/tex]
Explanation:
The magnitude on the charge at the bottom-left corner due to the charge on the top vertex of the triangle will act along the +ve x-axis and the +ve y-axis.
From Coulomb's law the magnitude of the forces on the charge at the bottom-left corner, due to the charge on the top vertex of the triangle are [tex]\frac{kq^{2} }{a^2}[/tex]cos 60, and
The magnitude on the charge due to the charge at the bottom-right corner will only act in the -ve x-axis, since they repel each other (like charges repel). The magnitude is [tex]\frac{kq^{2} }{a^2}[/tex]
The angle made by the upper charge to the charge we're considering is 60° with the horizontal.
The total force on the charge along the x-axis is
[tex]F_{x}[/tex] = [tex]\frac{kq^{2} }{a^2}[/tex]cos 60 -
[tex]F_{x}[/tex] = [tex]\frac{kq^{2} }{2a^2}[/tex] - [tex]\frac{kq^{2} }{a^2}[/tex]
==> -[tex]\frac{kq^{2} }{2a^2}[/tex]
For the y-axis, we have
[tex]F_{y}[/tex] = [tex]\frac{kq^{2} }{a}[/tex]sin 60
[tex]F_{y}[/tex] = [tex]\frac{\sqrt{3}* kq^{2} }{2a^2}[/tex]
The resultant force is
[tex]|F| = \sqrt{F_{x}^{2}+ F_{y}^2 }[/tex]
The common factors between the two x-axis force, and the y-axis force is
[tex]\frac{kq^{2} }{2a^2}[/tex], we put this outside the square root (squaring this and square rooting will give us the initial value)
[tex]|F|[/tex] = [tex]\frac{kq^{2} }{2a^2}[/tex][tex]\sqrt{(\frac{1}{2})^2 + (\frac{\sqrt{3} }{2})^2 }[/tex]
[tex]|F|[/tex] = [tex]\frac{kq^{2} }{2a^2}[/tex][tex]\sqrt{\frac{1}{4} +\frac{3}{4} }[/tex]
==> [tex]\frac{kq^{2} }{2a^2}[/tex][tex]\sqrt{1}[/tex]
the magnitude of the electric force on the charge at the bottom left-hand vertex of the triangle due to the other two charges is
[tex]|F|[/tex] = [tex]\frac{kq^{2} }{2a^2}[/tex]
Which event took place during Copernican revolution, When most people started to believe in a heliocentric model of the solar system
Answer:
Copernicus rediscovered Aristarchus’s heliocentric model
Explanation:
If the radiant energy from the Sun comes in as a plane EM wave of intensity 1340 W/m2, calculate the peak values of E and B
Answer:
Given that
I = 1340 W/m²
µo = 4πx 10^-7 Tm/A
sc = 3 x 10^8 m/s
So to find the peak values of B we use
I = (c Bm²) / (2 µo)
1340 = (3 x 10^8 x Bm²) / (2 x 4πx 10^-7)
1340 = (3 x 10^8 x Bm²) / (25.12 x 10^-7)
(3 x 10^8 x Bm²) = 1340 x (25.12 x 10^-7)
(3 x 10^8 x Bm²) = 3.366 x 10^-3
Bm² = 3.366 x 10^-3 / 3 x 10^8
Bm² = 1.122 x 10^-11
Bm = 3.4 x 10^-6 T
Also to find the peak values of E we
use Em = c x Bm
Em = (3 x 10^8) (3.4 x 10^-6)
Em = 1.005 N/C
If two identical blocks are glued together and pulled with twice the original force, what will their acceleration be
Answer:
The acceleration will be a
Explanation:
Using F = ma
But we are given that F2= 2F1 and
m2= 2m1
So
a2= F2/m2= F1/m1
and
F1/m1=a
A class contains 9 boys and 11 girls. In how
many
ways
can a committee of two boys
and two
girls be formed
in the class?
A vector is parallel to the y axis, what is it x component?
Answer:
Explanation:
A vector is parallel to the y axis .
Let its magnitude be A . So the vector can be represented as A j .
where i and j are unit vectors in x and y axis direction .
The x component of A j will be dot product of A j with i
The x component of A j = A j . i
= A x 0 [ Since j . i = 0 ]
= 0
Solve for x
–30 = 5(x + 1)
Answer:
-30=5(x+1) is -7
Explanation:
distribute flip subtract 5 from both sides divide both sides by 5
If the diameter of the low-power field is 1.5 mm, an object that occupies approximately a third of that field has an estimated diameter of
Answer:
d' = 0.86 mm
Explanation:
First we need to find the area of low power field:
A = πd²/4
where,
A = Area = ?
d = diameter = 1.5 mm
Therefore, using the value of diameter, we get:
A = π(1.5 mm)²/4
A = 1.767 mm²
Now, the new object occupies 1/3rd of this area:
A' = (1/3)A
πd'²/4 = (1/3)A
d'² = (4/3π) A
d'² = (4/3π)(1.767 mm²)
d' = √(0.75 mm²)
d' = 0.86 mm
An ideal gas is compressed in a well-insulated chamber using a well-insulated piston. This process is
Answer:
The options are
A) isochoric.
B) isothermal.
C) adiabatic.
D) isobaric.
The answer is C. Adiabatic
Adiabatic process involves zero loss or gain of heat in a system. This is usually depicted as Q= 0.
An ideal gas being compressed in a well-insulated chamber using a well-insulated piston involves the use of adiabatic process. The insulated chamber and piston helps to prevent heat loss or gain of heat. This is because insulators are poor conductors of heat and electricity.
The process of compressing an ideal gas in a well-insulated chamber using a well-insulated piston is known as adiabatic process.
When an ideal gas in compressed adiabatically, the heat lost is zero and the work is done it which causes increase its temperature.
Q = 0
The well-insulated chamber and piston helps to prevent heat loss by conduction and convection. The results in a zero heat lost to the surroundings.
Thus, we can conclude that the process of compressing an ideal gas in a well-insulated chamber using a well-insulated piston is known as adiabatic process.
Learn more about adiabatic process here: https://brainly.com/question/17185468
What is the wavelength of the waves you create in a swimming pool if you splash your hand at a rate of 2.00 Hz and the waves propagate at 0.800 m/s
Answer:
The wavelength of the waves created in the swimming pool is 0.4 m
Explanation:
Given;
frequency of the wave, f = 2 Hz
velocity of the wave, v = 0.8 m/s
The wavelength of the wave is given by;
λ = v / f
where;
λ is the wavelength
f is the frequency
v is the wavelength
λ = 0.8 / 2
λ = 0.4 m
Therefore, the wavelength of the waves created in the swimming pool is 0.4 m
Determine the magnitude of the current flowing through a 4.7 kilo ohms resistor if the voltage across it is (a) 1mV (b) 10 V (c) 4e^-t
Answer:
213 nA
2.13 mA
851e^-t μA
Explanation:
We have a pretty straightforward question here.
Ohms Law states that the current in an electric circuit is directly proportional to the voltage and inversely proportional to the resistance in the circuit. It is mathematically written as
V = IR, since we need I, we can write that
I = V/R
a) at V = 1 mV
I = (1 * 10^-3) / 4.7 * 10^3
I = 2.13 * 10^-7 A or 213 nA
b) at V = 10 V
I = 10 / 4.7 * 10^3
I = 0.00213 A or 2.13 mA
c) at V = 4e^-t
I = 4e^-t / 4.7 * 10^3
I = 0.000851e^-t A or 851e^-t μA
the process of _ is when tow atoms combine to form a larger atom, releasing energy
Answer:Nuclear fusion is the process by which two or more atomic nuclei join together, or “fuse,” to form a single heavier nucleus. During this process, matter is not conserved because some of the mass of the fusing nuclei is converted to energy, which is released.
Explanation:
Atoms of different elements can combine to make new substances. A molecule is formed when two or more atoms join together chemically. If atoms combine that are of two or more different elements, we call that a compound. ... When two hydrogen atoms combine with one oxygen atom, it becomes the compound water.
Answer:
nuclear fusion
Explanation:
If 1.34 ✕ 1020 electrons move through a pocket calculator during a full day's operation, how many coulombs of charge moved through it?
Given :
Number of operations move through a pocket calculator during a full day's operation , [tex]n=1.34 \times 10^{20}[/tex] .
To Find :
How many coulombs of charge moved through it .
Solution :
We know , charge in one electron is :
[tex]e^-=-1.6\times 10^{-19}\ coulombs[/tex]
So , charge on n electron is :
[tex]C=e^-\times n\\C=-1.6\times 10^{-19}\times 1.34\times 10^{20} \ C\\C=-21.44\ C[/tex]
Therefore , -21.44 coulombs of charge is moved through it .
Hence , this is the required solution .
Name at 2 areas of physics that make video games possible
Answer:
projectiles
electromagnetic
Answer:
Explanation:
física cuántica y Quantum Moves
According to Einstein, increasing the brightness (or intensity) of a beam of light without changing its color will increase (circle all that apply):
The question is incomplete,so the complete question is as follows:
According to Einstein, increasing the brightness (or intensity) of a beam of light without changing its color will increase (circle all that apply):
A) the number of photons per second traveling in the beam.
B) the energy of each photon.
C) the speed of the photons.
D) the frequency of the light.
E) the wavelength of the photons
Answer:
A)
Explanation:
According to Einstein, an increase in the intensity or brightness of light beam will increase the number of photons over a given time interval.
It means that the energy emitted will depend on the energy of individual photon rather than the intensity of the incoming light.
Hence, the correct option is "A)"
5. Haley throws her ball downward with speed v from height 2h. Simultaneously, Joe drops his ball from rest at height h. What should v be so that the two balls hit the ground at the same time
Answer:
We know that the second equation of motion is
S= ut + 1/2a²
And S is displacement and u is initial velocity
So in the case of Haley lets take downwards as positive Y-axis
S = 2h and
initial velocity = v
a = g (acceleration due to gravity = 9.8)
Substituting
2h = vt + 1/2gt²
And for Joe we take ownwards as positive Y-axis
S = h and
initial velocity = 0 (since the ball is dropped from rest)
a = g
h = 0x t + 1/2gt2²
t2= √ 2h/g
Now since both balls reach ground at same time: t1=t2
So
putting value of t2 in Hayley's equation:
2h= v(√2h/g) + 1/2 g( √2h/g)²
So v= √gh/2
What is the rotational inertia (moment of inertia) of a 12-kg uniform rod, 0.30 m long, rotating about an axis perpendicular to the rod and passing through the center of the rod
Answer:
[tex]I=0.54\ kg-m^2[/tex]
Explanation:
Given that,
Mass of a uniform rod, m = 12 kg
Length of the rod, l = 0.3 m
The moment of inertia rotating about an axis perpendicular to the rod and passing through the center of the rod is given by :
[tex]I=\dfrac{ml^2}{2}\\\\I=\dfrac{12\times (0.3)^2}{2}\\\\I=0.54\ kg-m^2[/tex]
So, the moment of inertia of the rod is [tex]0.54\ kg-m^2[/tex]
A steel tank of weight 600 lb is to be accelerated straight upward at a rate of 1.5 ft/sec2. Knowing the magnitude of the force P is 525 lb, determine the values of the angle a and the required magnitude of the vertical force, F.
Answer:
a) the values of the angle α is 45.5°
b) the required magnitude of the vertical force, F is 41 lb
Explanation:
Applying the free equilibrium equation along x-direction
from the diagram
we say
∑Fₓ = 0
Pcosα - 425cos30° = 0
525cosα - 368.06 = 0
cosα = 368.06/525
cosα = 0.701
α = cos⁻¹ (0.701)
α = 45.5°
Also Applying the force equation of motion along y-direction
∑Fₓ = ma
Psinα + F + 425sin30° - 600 = (600/32.2)(1.5)
525sin45.5° + F + 212.5 - 600 = 27.95
374.46 + F + 212.5 - 600 = 27.95
F - 13.04 = 27.95
F = 27.95 + 13.04
F = 40.99 ≈ 41 lb
In a thermal circulation what is happening at the surface of the warm column of air?
Answer:
Explanation: Thermals are created by the irregular heating of Earth's surface from solar radiation, and are an example of convection, specifically atmospheric convection.
At this point, air is converging on low pressure and rising air forms clouds.
answers If the motor is to accelerate the elevator car upward at 1.8 m/s2, how much torque must it generate
Answer:
Hello your question is incomplete attached is the missing diagram and solution
One elevator arrangement includes the passenger car, a counterweight, and two large pulleys, as shown in (Figure 1). Each pulley has a radius of 1.2 m and a moment of inertia of 410 kg⋅m2. The top pulley is driven by a motor. The elevator car plus passengers has a mass of 3000 kg, and the counterweight has a mass of 2600 kg.
a) If the motor is to accelerate the elevator car upward at 1.8 m/s2, how much torque must it generate? Express your answer to two significant figures and include appropriate units.
Answer : 6030 N.m ≈ 6000 N.m
Explanation:
To determine how much torque the motor accelerating at 1.8 m/s^s will generate we will have to determine T1 ( tension generated for upward acceleration in the rope ) and Tg ( tension generated in the rope when there is a counterweight accelerating downwards)
Torque in the pulley (T2) = (T1 - Tg ) r
torque generated by motor = T1 + T2 = 1230 + 4800 = 6030 N.m to two significant figures = 6000 N.m
r = 1.2
a = 1.8
attached below is the remaining part of the solution
A 0.250-kg rubber ball is dropped from a height of 2.00 m. It hits the floor and rebounds to a height of 1.80 m. What is the magnitude of impulse the floor applies to the ball
Answer:
The magnitude of impulse the floor applies to the ball is 3.05 kg.m/s.
Explanation:
Given;
mass of the rubber ball, m = 0.25 kg
height of drop, h₁ = 2 m
height of rebounds, h₂ = 1.8 m
Determine the initial velocity of the ball as it moves downwards;
[tex]v_i = \sqrt{2gh}\\\\ v_i = \sqrt{2*9.8*2}\\\\v_i = 6.26 \ m/s[/tex]this initial velocity is acting downwards = - 6.26 m/s
Determine the final velocity of the ball as it rebounds
[tex]v_f = \sqrt{2gh} \\\\v_f = \sqrt{2*9.8*1.8}\\\\ v_f = 5.94 \ m/s[/tex]this final velocity is acting upwards = 5.94 m/s
Impulse is given by;
J = mΔv
[tex]J = m(v_f-v_i)[/tex]
J = 0.25(5.94 - (6.26))
J = 0.25(5.94 + 6.26)
J = 3.05 kgm/s
Therefore, the magnitude of impulse the floor applies to the ball is 3.05 kg.m/s.
When an x-ray photon passes through matter it undergoes a process called:_________.
Answer:
when an x-ray passes trough matter, it undergoes a process called attenuation
What is the cell emf for the concentrations given? Express your answer using two significant figures.
Complete Question
A voltaic cell is constructed with two [tex]Zn^{2+}[/tex][tex]- Zn[/tex] electrodes. The two cell compartment have [tex][Zn^{2+}] = 1.6 \ M[/tex] and [tex][Zn^{2+}] = 2.00*10^{-2} \ M[/tex] respectively.
What is the cell emf for the concentrations given? Express your answer using two significant figures
Answer:
The value is [tex]E = 0.06 V[/tex]
Explanation:
Generally from the question we are told that
The concentration of [tex][Zn^{2+}][/tex] at the cathode is [tex][Zn^{2+}]_a = 1.6 \ M[/tex]
The concentration of [tex][Zn^{2+}][/tex] at the anode is [tex][Zn^{2+}]_c = 2.00*10^{-2} \ M[/tex]
Generally the the cell emf for the concentration is mathematically represented as
[tex]E = E^o - \frac{0.0591}{2} log\frac{[Zn^{2+}]a}{ [Zn^{2+}]c}[/tex]
Generally the [tex]E^o[/tex]is the standard emf of a cell, the value is 0 V
So
[tex]E = 0 - \frac{0.0591}{2} * log[\frac{ 2.00*10^{-2}}{1.6} ][/tex]
=> [tex]E = 0.06 V[/tex]
A hollow sphere has a mass of 15 kg, an inner radius of 12 cm and an outer radius of 18 cm. What is the rotational inertia (moment of inertia) of the sphere about an axis passing through its center
Answer:
0.2396
Explanation:
Given that
Mass of the hollow sphere, m = 15 kg
Inner radius, r = 12 cm = 0.12 m
Outer radius, R = 18 cm = 0.18 m
Volume of a sphere is expressed as
V = 4/3.π.R³
Density ρ = mass / volume, therefore
Mass = ρ * v
Mass of the hollow sphere is given as
Mass of the outer sphere - mass of the inner sphere
M = ρV(o) - ρV(i)
V(o) = 4/3 * 3.142 * 0.18³
V(o) = 0.0244
V(i) = 4/3 * 3.142 * 0.12³
V(i) = 0.00724
15 = ρ (0.0244 - 0.00724)
15 = ρ (0.01716)
ρ = 15 / 0.01716
ρ = 874 kg/m³
moment of inertia about its centroidal axis is
I = 2/5 ρVR²
I(h) = I(o) - I(i)
I(h) = (2/5 * 874 * 0.0244 * 0.18²) - (2/5 * 874 * 0.00724 * 0.12²)
I(h) = 0.276 - 0.0364
I(h) = 0.2396
About how many times greater is the density of a neutron star compared to a white dwarf?
Answer:
Explanation:
over a million times
To calculate an answer to the correct number of significant figures, you must complete all calculations first and then round your final answer as the very last step. In Part C, you are asked to calculate the volume of a rectangular prism that has a length of 5.6 cm, a width of 2.1 cm, and a height of 6.6 cm. You can calculate the volume of a prism by multiplying the area of the base times the height. You already calculated the area of the base as 2.1cm×5.6cm in Part A. Part B What value should you use as the area of the base when calculating the answer to Part C? What value should you use as the area of the base when calculating the answer to Part C?
Answer:
B) We should use the complete value of 11.76 cm², because the rounding off is done at the end of all calculations. Until then exact and complete values should be used.
C) Volume of Prism = 78 cm³
Explanation:
PART C:
Since, the volume of prism s given by the following formula:
Volume of Prism = (Area of Base of Prism)(Height of Prism)
Therefore,
Height of Prism = 6.6 cm
Area of Base of Prism = Length x Width
Area of Base of Prism = 5.6 cm x 2.1 cm
Area of Base of Prism = 11.76 cm²
Hence, the equation gives:
Volume of Prism = (11.76 cm²)(6.6 cm)
Volume of Prism = 77.616 cm³
Since, the initial values had 2 significant figures, so rounding off to two significant figures:
Volume of Prism = 78 cm²
PART B:
We should use the complete value of 11.76 cm², because the rounding off is done at the end of all calculations. Until then exact and complete values should be used.
What is the maximum power that can be delivered by a 1.4-cm-diameter laser beam propagating through air
Complete Question
The maximum electric field strength in air is 4.0 MV/m. Stronger electric fields ionize the air and create a spark. What is the maximum power that can be delivered by a 1.4-cm-diameter laser beam propagating through air
Answer:
The value is [tex]P = 3.270960 *10^{6} \ W[/tex]
Explanation:
From the question we are told that
The electric field strength is [tex]E = 4.0 \ M \ V/m = 4.0 *10^6 \ V/m[/tex]
The diameter is [tex]d = 1.4 \ cm = 0.014 \ m[/tex]
Generally the radius is mathematically represented as
[tex]r = \frac{d}{2}[/tex]
=> [tex]r = \frac{0.014}{2}[/tex]
=> [tex]r = 0.007 \ m[/tex]
Generally the cross-sectional area is mathematically represented as
[tex]A = \pi r^2[/tex]
[tex]A = 3.142 * (0.007)^2[/tex]
[tex]A = 0.000154 \ m^2[/tex]
Generally the maximum power that can be delivered is mathematically represented as
[tex]P = \frac{c * \epsilon_o * E^2 * A }{2}[/tex]
Here c is the speed of light with value [tex]c = 3.0*10^{8} \ m/s[/tex]
[tex]\epsilon_o[/tex] is the permittivity of free space with value [tex]\epsilon_o = 8.85 *10^{-12} \ m^{-3} \cdot kg^{-1}\cdot s^4 \cdot A^2[/tex]
[tex]P = \frac{3,0*10^8 * 8.85*10^{-12} * (4 *10^6)^2 * 0.00154}{ 2}[/tex]
[tex]P = 3.270960 *10^{6} \ W[/tex]
A uniform disk, a thin hoop, and a uniform sphere, all with the same mass and same outer radius, are each free to rotate about a fixed axis through its center. Assume the hoop is connected to the rotational axis by light spokes. With the objects starting from rest, identical forces are simultaneously applied to the rims, as shown. Rank the objects according to their angular acceleration, least to greatest.1. disk, hoop, sphere2. disk, sphere, hoop3. hoop, sphere, disk4. hoop, disk, shpere5. sphere, disk, hoop6. sphere, hoop, disk
Answer:
4 hoop, disk, sphere
Explanation:
Because
We are given data that
Hoop, disk, sphere have Same mass and radius
So let
And Initial angular velocity, = 0
The Force on each be F
And Time = t
Also let
Radius of each = r
So let's find the inertia shall we!!
I1 = m r² /2
= 0.5 mr² the his is for dis
I2 = m r² for hoop
And
Moment of inertia of sphere wiil be
I3 = (2/5) mr²
= 0.4 mr²
So
ωf = ωi + α t
= 0 + ( τ / I ) t
= ( F r / I ) t
So we can see that
ωf is inversely proportional to moment of inertia.
And so we take the
Order of I ( least to greatest ) :
I3 (sphere) , I1 (disk) , I2 (hoop) , ,
Order of ωf: ( least to greatest)
That of omega xf is the reverse of inertial so
hoop, disk, sphere
Option - 4
The ranking of the objects according to their angular acceleration is option 4 hoop, disk, sphere.
Ranking of the objects:Since
Hoop, disk, sphere contain Same mass and radius
So here
Initial angular velocity, = 0
The Force on each be F
And Time = t
Radius of each = r
Now
I1 = m r² /2
= 0.5 mr² his is for dis
I2 = m r² for hoop
And
Moment of inertia of sphere should be
I3 = (2/5) mr²
= 0.4 mr²
Now
ωf = ωi + α t
= 0 + ( τ / I ) t
= ( F r / I ) t
here,
ωf is inversely proportional to moment of inertia.
Now
Order of I ( least to greatest ) :
I3 (sphere) , I1 (disk) , I2 (hoop) , ,
Order of ωf: ( least to greatest)
That of omega xf is the reverse of inertial so
Therefore, the fourth option is correct.
learn more about radius here: https://brainly.com/question/17257279
The force on an object is F⃗ =−17j⃗ . For the vector v⃗ =2i⃗ +3j⃗ , find: (a) The component of F⃗ parallel to v⃗
Answer:
(a) [tex]\vec F_{\parallel} = -\frac{102}{13}\,i-\frac{103}{13}\,j[/tex] , (b) [tex]\vec F_{\perp} = \frac{102}{13}\,i -\frac{68}{13}\,j[/tex], (c) [tex]W = -51[/tex]
Explanation:
The statement is incomplete:
The force on an object is [tex]\vec F = -17\,j[/tex]. For the vector [tex]\vec v = 2\,i +3\,j[/tex]. Find: (a) The component of [tex]\vec F[/tex] parallel to [tex]\vec v[/tex], (b) The component of [tex]\vec F[/tex] perpendicular to [tex]\vec v[/tex], and (c) The work [tex]W[/tex], done by force [tex]\vec F[/tex] through displacement [tex]\vec v[/tex].
(a) The component of [tex]\vec F[/tex] parallel to [tex]\vec v[/tex] is determined by the following expression:
[tex]\vec F_{\parallel} = (\vec F \bullet \hat {v} )\cdot \hat{v}[/tex]
Where [tex]\hat{v}[/tex] is the unit vector of [tex]\vec v[/tex], which is determined by the following expression:
[tex]\hat{v} = \frac{\vec v}{\|\vec v \|}[/tex]
Where [tex]\|\vec v\|[/tex] is the norm of [tex]\vec v[/tex], whose value can be found by Pythagorean Theorem.
Then, if [tex]\vec F = -17\,j[/tex] and [tex]\vec v = 2\,i +3\,j[/tex], then:
[tex]\|\vec v\| =\sqrt{2^{2}+3^{3}}[/tex]
[tex]\|\vec v\|=\sqrt{13}[/tex]
[tex]\hat{v} = \frac{1}{\sqrt{13}} \cdot(2\,i + 3\,j)[/tex]
[tex]\hat{v} = \frac{2}{\sqrt{13}}\,i+ \frac{3}{\sqrt{13}}\,j[/tex]
[tex]\vec F \bullet \hat{v} = (0)\cdot \left(\frac{2}{\sqrt{13}} \right)+(-17)\cdot \left(\frac{3}{\sqrt{13}} \right)[/tex]
[tex]\vec F \bullet \hat{v} = -\frac{51}{\sqrt{13}}[/tex]
[tex]\vec F_{\parallel} = \left(-\frac{51}{\sqrt{13}} \right)\cdot \left(\frac{2}{\sqrt{13}}\,i+\frac{3}{\sqrt{13}}\,j \right)[/tex]
[tex]\vec F_{\parallel} = -\frac{102}{13}\,i-\frac{153}{13}\,j[/tex]
(b) Parallel and perpendicular components are orthogonal to each other and the perpendicular component can be found by using the following vectorial subtraction:
[tex]\vec F_{\perp} = \vec F - \vec F_{\parallel}[/tex]
Given that [tex]\vec F = -17\,j[/tex] and [tex]\vec F_{\parallel} = -\frac{102}{13}\,i-\frac{153}{13}\,j[/tex], the component of [tex]\vec F[/tex] perpendicular to [tex]\vec v[/tex] is:
[tex]\vec F_{\perp} = -17\,j -\left(-\frac{102}{13}\,i-\frac{153}{13}\,j \right)[/tex]
[tex]\vec F_{\perp} = \frac{102}{13}\,i + \left(\frac{153}{13}-17 \right)\,j[/tex]
[tex]\vec F_{\perp} = \frac{102}{13}\,i -\frac{68}{13}\,j[/tex]
(c) The work done by [tex]\vec F[/tex] through displacement [tex]\vec v[/tex] is:
[tex]W = \vec F \bullet \vec v[/tex]
[tex]W = (0)\cdot (2)+(-17)\cdot (3)[/tex]
[tex]W = -51[/tex]
A jogger runs 4 km in 0.4 hr, then 8 km in 0.8 hr. What is the average speed of the jogger?
Answer:
2km
Explanation: