Answer:
v = 17.5 m/s = 63 km/h
Explanation:
The general expression for the kinetic energy of one moving object is as follows:[tex]K = \frac{1}{2}*m *v^{2} (1)[/tex]
where m = mass of the object, v= speed of the object.
In order to get the value of the kinetic energy of the truck in Joules, we need to convert km/hr to m/s first, as follows:[tex]21 km/hr * \frac{1 hr}{3600s}*\frac{1000m}{1 km} = 5.83 m/seg (2)[/tex]
Now, replacing (2) and m = 18000 kg in (1), we get:[tex]K = \frac{1}{2}*18000 kg *(5.83m/s)^{2} = 306250 J (3)[/tex]
This value must be the same for the 2000 kg compact car, so we can write:[tex]K = 306250 J = \frac{1}{2}*2000 kg *v^{2} (4)[/tex]
Solving for v, we get:[tex]v = \sqrt{\frac{306250}{1000} (m/s)2} = 17. 5 m/s = 63 km/h (5)[/tex]
38. You are fishing and catch a fish with a mass of
6kg. If the fishing line can withstand a maximum
tension of 30 N, what is the maximum acceleration
you can give the fish as you reel it in?..*
(10 Points)
Enter your answer
Answer:
1.7333333m/s²
Explanation:
Tension of the line = the weight + force from pulling up the fish
30N = mg + ma
30 = (6)(9.8) + (6)a
10.4 = 6a
∴ a = 1.7333333m/s²
You are fishing and catch a fish with a mass of 6 kg. If the fishing line can withstand a maximum tension of 30 N, the maximum acceleration is 1.7333333 m/s².
What is acceleration?The rate at which an item changes its velocity is known as acceleration, a vector quantity. If an object's velocity is changing, it is acceleration.
Tension of the line = the weight + force from pulling up the fish
30 N = mg + ma
30 = (6)(9.8) + (6)a
10.4 = 6 a
a = 1.7333333 m/s²
You are fishing and catch a fish with a mass of 6 kg. If the fishing line can withstand a maximum tension of 30 N, the maximum acceleration is 1.7333333 m/s².
To learn more about acceleration refer to the link:
brainly.com/question/12550364
#SPJ2
A guitar string produces 3 beats/s when sounded with a 352 Hz tuning fork and 8 beat/s when sounded with a 357 Hz tuning fork. What is the vibrational frequency (in Hz) of the string
Answer:
349 Hz
Explanation:
We are told that the guitar string produces 3 beats/s when sounded with a 352 Hz tuning fork and 8 beat/s when sounded with a 357 Hz tuning fork.
This means that for the 352 Hz tuning fork, the vibrational frequency is;
f = 352 ± 3
f = (352 + 3) or (352 - 3)
f = 355 Hz or 349Hz
For the 357 Hz tuning fork, the vibrational frequency is;
f = 357 ± 8
f = (357 + 8) or (357 - 8)
f = 365 Hz or 349 Hz
In both cases, 349 Hz is common;
Thus, the vibrational frequency of the string = 349 Hz
Two particles are separated by 0.38 m and have charges of -6.25 x 10-9C
and 2.91 x 10-9 C. Use Coulomb's law to predict the force between the
particles if the distance is doubled. The equation for Coulomb's law is
Fe = kq92, and the constant, k, equals 9.00 x 10°N-m/c2.
A. -2.83 x 10-7N
B. 2.83 x 10-7N
C. -1.13 x 10-6N
D. 1.13 x 10-6N
Answer:A
Explanation:
Answer:
A. -2.83 x 10-7N
Explanation:
What specific changes in two climate variables are expected to lead to major decreases in soil moisture southern Africa and the Mediterranean region?
Answer:
Less precipitation, droughts9: How might agriculture in southern Europe change by the end of the century if conditions follow the RCP8.
Explanation:
Precipitation and droughts are the specific changes in two climate variables that are expected to lead to major decreases in soil moisture.
What is drought?
Drought is defined as a period of protracted water scarcity, whether it is due to atmospheric surface water, or groundwater constraints.
Droughts can last months or years, although they can be proclaimed in as little as 15 days.
It has the potential to have a significant influence on the afflicted region's ecology and agriculture as well as harm the local economy.
Precipitation and droughts are the specific changes in two climate variables that are expected to lead to major decreases in soil moisture in southern Africa and the Mediterranean region.
Hence Precipitation and droughts are the specific changes in two climate variables.
To learn more about the drought refer to the link;
https://brainly.com/question/26693108
Which is the goal of technology?
to expand comprehension
to make life easier
to apply knowledge
to improve communication
pls help
Answer:
i think it should be to make life easier
Which image illustrates the interaction of a light wave with a mirror?
t J
A
с
.
A. A
B. B
C. C
D. D
0
Answer:
I'm pretty sure its A
Explanation:
because its a reflection- Hope you get a good grade!
A bullet is fired horizontally at a height of 2 meters at a velocity of 930 m/s. Assume no air resistance. How long until the bullet reaches the ground?
0.32 s
0.57 s
0.64 s
0.25 s
A box of mass 7.0 kg is accelerated from rest across a floor at a rate of 2.0 m/s2 for 9.0 s .Find the net work done on the box. Express your answer to two significant figures and include the appropriate units.
Answer:
Explanation:
Step one:
given data
mass = 7kg
acceleration =2m/s^2
time= 9seconds
acceleration = velocity/time
velocity= acceleration *time
velocity=2*9
velocity= 18m/s
distance moved= velocity* time
distance= 18*9
distance=162m
we also know that the force on impulse is given as
Ft=mv
F=mv/t
F=7*18/9
F=126/9
F=14N
work done = Force* distance
work done=14*162
work=2268Joules
work= 2.27kJ
Please help!!! I will give brainliest,
Answer:
C. a liter of salt water.
Explanation:
Defination of Solution =>
a liquid mixture in which the minor component (the solute) is uniformly distributed within the major component (the solvent).
A 0.20-kg object is attached to the end of an ideal horizontal spring that has a spring constant of 120 N/m. The simple harmonic motion that occurs has a maximum speed of 0.70 m/s. Determine the amplitude A of the motion.
Answer:
The amplitude of the motion is 0.0286 m.
Explanation:
Given;
mass of the object, m = 0.2 kg
spring constant, k = 120 N/m
maximum speed of the simple harmonic motion, [tex]V_m[/tex] = 0.70 m/s
The amplitude A of the motion is given by;
[tex]V_m = \omega A\\\\[/tex]
where;
ω is the angular velocity given as;
[tex]\omega = \sqrt{\frac{k}{m} }\\\\\omega = \sqrt{\frac{120}{0.2} }\\\\\omega =24.5 \ rad/s[/tex]
Now, substitute the value of angular velocity and solve the amplitude;
[tex]V_m = \omega A\\\\A = \frac{V_m}{\omega}\\\\A = \frac{0.7}{24.5}\\\\A = 0.0286 \ m[/tex]
Therefore, the amplitude of the motion is 0.0286 m.
A runner taking part in the 200 m dash must run around the end of a track that has a circular arc with a radius of curvature of 50 m. If he completes the 200 m dash in 29.6 s and runs at constant speed throughout the race, what is the magnitude of his centripetal acceleration (in m/s2) as he runs the curved portion of the track?
Answer:
The centripetal acceleration of the runner as he runs the curved portion of the track is 0.91 m/s²
Explanation:
Given;
distance traveled in the given time = 200 m
time to cover the distance, t = 29.6 s
speed of the runner, v = d / t
v = 200 / 29.6
v = 6.757 m/s
The centripetal acceleration of the runner is given by;
[tex]a_c = \frac{V^2}{r}[/tex]
where;
r is the radius of the circular arc, given as 50 m
Substitute the givens;
[tex]a_c = \frac{V^2}{r}\\\\a_c = \frac{(6.757)^2}{50}\\\\a_c = 0.91 \ m/s^2[/tex]
Therefore, the centripetal acceleration of the runner as he runs the curved portion of the track is 0.91 m/s².
A 40-cm-diameter, 300 g beach ball is dropped with a 4.0 mg ant riding on the top. The ball experiences air resistance, but the ant does not. What is the magnitude of the normal force exerted on the ant when the ball's speed is 4.0 m/s?
Answer:
The normal force exerted on the ant is 0.75 N.
Explanation:
Given;
diameter of the ball, D = 40 cm = 0.4m
radius of the ball, r = 0.2m
mass of the beach ball, m₁ = 300 g = 0.3 kg
mass of the ant, m₂ = 4 x 10⁻⁶ kg
speed of the ball, v = 4 m/s
The area of the ball, assuming spherical ball is given by;
A = 4πr²
A = 4π(0.2)² = 0.5027 m²
The drag force (resistance) experienced by the spherical ball is given as;
[tex]F_D = \frac{1}{2}C\rho Av^2[/tex]
where;
C is the drag coefficient of the spherical ball = 0.45
ρ is density of air = 1.21 kg/m³
[tex]F_D = \frac{1}{2}C\rho Av^2\\\\F_D = \frac{1}{2}(0.45)(1.21) (0.5027)(4)^2\\\\F_D = 2.19 \ N[/tex]
The downward force of the ball due to its weight and that of the ant is given by;
[tex]F_g = mg\\\\F_g =g(m_{ant} + m_{ball})\\\\F_g = g(4*10^{-6} \ kg\ + \ 0.3\ kg)\\\\F_g = g(0.300004 \ kg) \ \ \ (mass \ of \ the \ ant \ is \ insignificant)\\\\F_g = 9.8(0.3)\\\\F_g = 2.94 \ N[/tex]
The net downward force experienced by the ball is given by;
[tex]F_{net} = F_g - F_D\\\\F_{net} = 2.94 \ N - 2.19 \ N\\\\F_{net} = 0.75 \ N[/tex]
This downward force experienced by the ball is equal to the normal reaction it exerts on the ant.
Thus, the normal force exerted on the ant is 0.75 N.
A rope is wrapped around the rim of a large uniform solid disk of mass 325 kg and radius 3.00 m. The horizontal disk is made to rotate by pulling on the rope with a constant force of 195 N. If the disk starts from rest, what is its angular speed in rev/s at the end of 2.05 s?
Answer:
The angular speed is 0.13 rev/s
Explanation:
From the formula
[tex]\tau = I\alpha[/tex]
Where [tex]\tau[/tex] is the torque
[tex]I[/tex] is the moment of inertia
[tex]\alpha[/tex] is the angular acceleration
But, the angular acceleration is given by
[tex]\alpha = \frac{\omega}{t}[/tex]
Where [tex]\omega[/tex] is the angular speed
and [tex]t[/tex] is time
Then, we can write that
[tex]\tau = \frac{I\omega}{t}[/tex]
Hence,
[tex]\omega = \frac{\tau t}{I}[/tex]
Now, to determine the angular speed, we first determine the Torque [tex]\tau[/tex] and the moment of inertia [tex]I[/tex].
Here, The torque is given by,
[tex]\tau = rF[/tex]
Where r is the radius
and F is the force
From the question
r = 3.00 m
F = 195 N
∴ [tex]\tau = 3.00 \times 195[/tex]
[tex]\tau = 585[/tex] Nm
For the moment of inertia,
The moment of inertia of the solid disk is given by
[tex]I = \frac{1}{2}MR^{2}[/tex]
Where M is the mass and
R is the radius
∴[tex]I = \frac{1}{2} \times 325 \times (3.00)^{2}[/tex]
[tex]I = 1462.5[/tex] kgm²
From the question, time t = 2.05 s.
Putting the values into the equation,
[tex]\omega = \frac{\tau t}{I}[/tex]
[tex]\omega = \frac{585 \times 2.05}{1462.5}[/tex]
[tex]\omega = 0.82[/tex] rad/s
Now, we will convert from rad/s to rev/s. To do that, we will divide our answer by 2π
0.82 rad/s = 0.82/2π rev/s
= 0.13 rev/s
Hence, the angular speed is 0.13 rev/s,
How does cycling of matter occur in Earth’s mantle?
A. Hot, soft rock rises from the bottom of the mantle toward the top, cools, and sinks back through the mantle.
B. Hot, soft rock sinks to the mantle, cools, and rises to Earth’s crust.
C. Solid rock rises from the bottom of the mantle, cools, and sinks back through the mantle.
D. Solid rock sinks to Earth’s core and then rises to form lava.
i need help fast plz and thank you
11/24/2020
Answer:
Correct answer to this is: A
Hopefully, this helped out a bit :)
Zinc has a work function of 4.3 eV. a. What is the longest wavelength of light that will release an electron from a zinc surface? b. A 4.7 eV photon strikes the surface and an electron is emitted. What is the maximum possible speed of the electron?
Answer:
a
[tex]\lambda_{long} = 288.5 \ nm[/tex]
b
The velocity is [tex]v = 3.7 *0^{5} \ m/s[/tex]
Explanation:
From the question we are told that
The work function of Zinc is [tex]W = 4.3 eV[/tex]
Generally the work function can be mathematically represented as
[tex]E_o = \frac{hc}{\lambda_{long}}[/tex]
=> [tex]\lambda_{long} = \frac{hc}{E_o}[/tex]
Here h is the Planck constant with the value [tex]h = 4.1357 * 10^{-15} eV s[/tex]
and c is the speed of light with value [tex]c = 3.0 *10^{8} \ m/s[/tex]
So
[tex]\lambda_{long} = \frac{4.1357 * 10^{-15} * 3.0 *10^{8}}{4.3}[/tex]
=> [tex]\lambda_{long} = 2.885 *10^{-7} \ m[/tex]
=> [tex]\lambda_{long} = 288.5 \ nm[/tex]
Generally the kinetic energy of the emitted electron is mathematically represented as
[tex]K = E -E_o[/tex]
Here E is the energy of the photon that strikes the surface
So
[tex]E- E_o = \frac{1}{2} m * v^2[/tex]
Here m is the mass of electron with value [tex]m = 9.11*10^{-31 } \ kg[/tex]
Generally [tex]1 ev = 1.60 *10^{-19} \ J[/tex]
=> [tex]v = \sqrt{ \frac{2 (E - E_o ) }{ m } }[/tex]
=> [tex]v = \sqrt{ \frac{2 (4.7 - 4.3 )* 1.60 *10^{-19} }{ 9.11 *10^{-31} } }[/tex]
=> [tex]v = 3.7 *0^{5} \ m/s[/tex]
Two parallel 3.0-meter long wires conduct current. The current in the top wire is 12.5 A and flows to the right. The top wire feels a repulsive force of 2.4 x 10^-4 N created by the interaction of the 12.5 A current and the magnetic field created by the bottom current (I). Find the magnitude and direction of the bottom current.
Complete question:
Two parallel 3.0-meter long wires conduct current. The current in the top wire is 12.5 A and flows to the right. The top wire feels a repulsive force of 2.4 x 10^-4 N created by the interaction of the 12.5 A current and the magnetic field created by the bottom current (I). Find the magnitude and direction of the bottom current, if the distance between the two wires is 40cm.
Answer:
The bottom current is 12.8 A to the right.
Explanation:
Given;
length of the wires, L = 3.0 m
current in the top wire, I₁ = 12.5 A
repulsive force between the two wires, F = 2.4 x 10⁻⁴ N
distance between the two wires, r = 40 cm = 0.4 m
The repulsive force between the two wires is given by;
[tex]F = \frac{\mu_oI_1I_2L}{2\pi r}\\\\I_{2} = \frac{2F\pi r}{\mu_oI_1L}[/tex]
Where;
I₂ is the bottom current
The direction of the bottom current must be in the same direction as the top current since the force between the two wires is repulsive.
[tex]I_{2} = \frac{2F\pi r}{\mu_oI_1L}\\\\I_{2} = \frac{2(2.4*10^{-4})(\pi)(0.4)}{(4\pi*10^{-7})(12.5)(3)}\\\\I_{2} = 12.8 \ A[/tex]
Therefore, the bottom current is 12.8 A to the right.
How much would a spring scale with k = 120 N/m stretch, if it had a 3.75 J of work done
on it?
Answer:
0.25m
Explanation:
Given parameters:
Spring constant , K = 120N/m
Work done = 3.75J
Unknown:
magnitude of extension = ?
Solution:
To solve this problem;
Work done = [tex]\frac{1}{2}[/tex]kx²
K is the spring constant
x is the extension
3.75 = [tex]\frac{1}{2}[/tex] x 120x²
3.75 = 60x²
x² = 0.06
x = √0.06 = 0.25m
Resistor is made of a very thin metal wire that is 3.2 mm long, with a diameter of 0.4 mm. What is the electric field inside this metal resistor?
Complete question:
Resistor is made of a very thin metal wire that is 3.2 mm long, with a diameter of 0.4 mm. What is the electric field inside this metal resistor? If the potential difference due to electric field between the two ends of the resistor is 10 V.
Answer:
The electric field inside this metal resistor is 3125 V/m
Explanation:
Given;
length of the wire, L = 3.2 mm = 3.2 x 10⁻³ m
diameter of the wire, d = 0.4 mm = 0.4 x 10⁻³ m
the potential difference due to electric field between the two ends of the resistor, V = 10 V
The electric field inside this metal resistor is given by;
ΔV = EL
where;
ΔV is change in electric potential
E = ΔV / L
E = 10 / (3.2 x 10⁻³ )
E = 3125 V/m
Therefore, the electric field inside this metal resistor is 3125 V/m
In the video your blood is compared to a __________________ that delivers oxygen to your body and picks up CO2 to be released out when you breath.
Answer:
delivery truck
Explanation:
because i got it right
what can i yeet baby or toddler
Answer:
both
Explanation:
baby for fun, toddler for vengence
A train is traveling at 55m/s begins to slow down as it approaches a bend in the tracks. If it travels around the bend at a speed of9 m/s and it takes 49 s to properly slow down, what distance does the train travel while slowing down?
Answer:
x = 1127 [m]
Explanation:
In order to solve this problem, we must use the equations of kinematics. With the first equation, we must find the acceleration and with the second equation we must find the distance.
[tex]v_{f} =v_{o} -a*t[/tex]
where:
Vf = final velocity = 9 [m/s]
Vo = initial velocity = 55 [m/s]
a = acceleration o desacceleration [m/s²]
t = time = 49 [s]
Now replacing:
9 = 55 - a*49
a*49 = 55 + 9
a = 1.306 [m/s²]
Note: The negative sign in the above equation means that the speed decreases.
Now using the second equation.
[tex]v_{f}^{2} =v_{o}^{2} -2*a*x[/tex]
(9)² = (55)² - 2*(1.306)*x
2944 = 2.612*x
x = 1127 [m]
Any conclusion reached by analogy is worth accepting
True
False
Answered: A 4 kg mass is attached to a horizontal spring with the spring constant of 600 N/m and rests on a frictionless surface on the ground. The spring is compressed 0.5 m past its equilibrium. What is the initial energy of the system.
Answer: 75 joules
Suppose a certain object has a mass of 5.00 kilograms on the earth. On the
Moon, where g is 1.6 m/s/s what would its mass be?*
Answer:
it would be 49.03325 Newton.
What two methods are the best choices to factor this expression 18x2-8
Answer:
Please check the explanation
Explanation:
The best two methods will be:
Factor by groupingFactor out the GCFFactor by grouping
Factor by grouping deals with establishing a smaller groups from each term.
[tex]18x^2=\:\left(2\cdot 3\cdot 3\cdot x^2\right)[/tex]
[tex]8\:=\:\:2\cdot 2\cdot 2[/tex]
Therefore, the expression becomes
[tex]18x^2=\:\left(2\cdot 3\cdot 3\cdot x^2\right)-\left(2\cdot \:2\cdot \:2\right)[/tex]
Now factor out the greatest common factor (GCF) which is 2
[tex]=\:2\left(3\cdot \:\:3x^2-\left(2\right)\left(2\right)\right)[/tex]
[tex]=2\left(9x^2-2\cdot \:2\right)[/tex]
[tex]=2\left(9x^2-4\right)[/tex]
Factor out the GCF
Given the expression
[tex]18x^2-8\:\:\:[/tex]
factor out common term 2
[tex]=2\left(9x^2-4\right)[/tex]
[tex]=2\left(3x+2\right)\left(3x-2\right)[/tex] ∵ [tex]Factors\:\:\left(9x^2-4\right)=\left(3x+2\right)\left(3x-2\right)[/tex]
What becomes V if we use 2 resistors of 4W in parallel?
A. 2.66 V
B. 6 V
C. 12 V
D. 24 V
Answer:
This question is incomplete.
Explanation:
This question is incomplete. However, it should be noted that the voltage, V, across resistors in parallel is the same (although there currents are not the same). Thus, if a voltage has been provided, it remains the same but if not provided, you can solve for it using the formulas below
V = IR
where V is the voltage. I is the current and R is the resistance
R in parallel can be calculated as R = 1/R₁ + 1/R₂ + 1/R₃ + ......
A crate of books rests on a level floor. To move it along the floor at a constant velocity, why do you exert less force if you pull it at an angle Ï above the horizontal than if you push it at the same angle below the horizontal?
Answer:should be a matter of vector analysis.
Pulling above the horizontal has less surface area for the opposing friction
Explanation:
15 points.
An object of mass 100 kg is observed to accelerate at a rate of 15
m/s/s. Calculate the force required to produce this acceleration.
Answer:
its 0.5 for all i beleive
Explanation:
There are two identical, positively charged conducting spheres fixed in space. The spheres are 40.4 cm apart (center to center) and repel each other with an electrostatic force of F1=0.0720 N . A thin conducting wire connects the spheres, redistributing the charge on each sphere. When the wire is removed, the spheres still repel, but with a force of F2=0.115 N . The Coulomb force constant is k=1/(4π????0)=8.99×109 N⋅m2/C2 . Using this information, find the initial charge on each sphere, q1 and q2 , if q1 is initially less than q2 .
Answer:
[tex]q_1=5.64\times 10^{-7}\ \text{C}[/tex] and [tex]q_2=2.32\times 10^{-6}\ \text{C}[/tex]
Explanation:
[tex]F_1=0.072\ \text{N}[/tex]
[tex]F_2=0.115\ \text{N}[/tex]
r = Distance between shells = 40.4 cm
[tex]q_1[/tex] and [tex]q_2[/tex] are the charges
[tex]k[/tex] = Coulomb constant = [tex]8.99\times10^{9}\ \text{Nm}^2/\text{C}^2[/tex]
Force is given by
[tex]F_1=\dfrac{kq_1q_2}{r^2}\\\Rightarrow q_1q_2=\dfrac{F_1r^2}{k}\\\Rightarrow q_1q_2=\dfrac{0.072\times 0.404^2}{8.99\times 10^{9}}\\\Rightarrow q_1q_2=1.307\times 10^{-12}\\\Rightarrow q_1=\dfrac{1.307\times 10^{-12}}{q_2}[/tex]
[tex]F_2=\dfrac{kq^2}{r^2}\\\Rightarrow q=\sqrt{\dfrac{F_2r^2}{k}}\\\Rightarrow q=\sqrt{\dfrac{0.115\times 0.404^2}{8.99\times 10^{9}}}\\\Rightarrow q=1.44\times 10^{-6}\ \text{C}[/tex]
[tex]q=\dfrac{q_1+q_2}{2}\\\Rightarrow q_1+q_2=2q\\\Rightarrow q_1+q_2=2\times1.44\times 10^{-6}\\\Rightarrow q_1+q_2=2.88\times 10^{-6}[/tex]
Substituting the above value of [tex]q_1[/tex] we get
[tex]\dfrac{1.307\times 10^{-12}}{q_2}+q_2=2.88\times 10^{-6}\\\Rightarrow q_2^2-2.88\times 10^{-6}q_2+1.307\times 10^{-12}=0\\\Rightarrow \frac{-\left(-0.00000288\right)\pm \sqrt{\left(-0.00000288\right)^2-4\times \:1\times \:1.307\times 10^{-12}}}{2\times \:1}\\\Rightarrow q_2=2.32\times 10^{-6}, 5.64\times 10^{-7}[/tex]
[tex]q_1=\dfrac{1.307\times 10^{-12}}{q_2}=\dfrac{1.307\times 10^{-12}}{2.32\times 10^{-6}}\\\Rightarrow q_1=5.63\times 10^{-7}[/tex]
[tex]q_1=\dfrac{1.307\times 10^{-12}}{q_2}=\dfrac{1.307\times 10^{-12}}{5.64\times 10^{-7}}\\\Rightarrow q_1=2.32\times 10^{-6}[/tex]
Since we know [tex]q_1<q_2[/tex]
[tex]q_1=5.64\times 10^{-7}\ \text{C}[/tex] and [tex]q_2=2.32\times 10^{-6}\ \text{C}[/tex].
A particle with charge q1 C is moving in the positive z-direction at 5 m/s. The magnetic field at its position is B-3 4j1T What is the magnetic force on the particle? A. (20i+15j) N B. (207-15j) N C. (-20i+15j) N D. (-20/-15) N E. none of these
Answer:
D. [tex]\vec F_{B} = -20\,\hat{i}-15\,\hat{j}\,\,\,[N][/tex]
Explanation:
The statement is not correctly written, the correct form is now described:
A particle with charge [tex]q = -1\,C[/tex] is moving in the positive z-direction at 5 meters per second. The magnetic field at its position is [tex]\vec B = 3\,\hat{i}-4\,\hat{j}\,\,\,[T][/tex]. What is the magnetic force on the particle?
From classic theory on Magnetism, we remember that the magnetic force exerted on a particle ([tex]\vec F_{B}[/tex]), measured in newtons, is determined by the following vectorial formula:
[tex]\vec F_{B} = q\cdot \vec v \,\times \,\vec B[/tex] (1)
Where:
[tex]q[/tex] - Electric charge, measured in coulombs.
[tex]\vec v[/tex] - Velocity of the particle, measured in meters per second.
[tex]\vec B[/tex] - Magnetic field, measured in teslas.
If we know that [tex]q = -1\,C[/tex], [tex]\vec v = 5\,\hat{k}\,\,\,\left[\frac{m}{s} \right][/tex] and [tex]\vec B = 3\,\hat{i}-4\,\hat{j}\,\,\,[T][/tex], then the magnetic force on the particle is:
[tex]\vec F_{B} = \left|\begin{array}{ccc}\hat{i}&\hat{j}&\hat{k}\\0\,\frac{C\cdot m}{s}&0\,\frac{C\cdot m}{s} &(-1\,C)\cdot (5\,\frac{m}{s} ) \\3\,T&-4\,T&0\,T\end{array}\right|[/tex]
[tex]\vec F_{B} = -(-4\,T)\cdot (-1\,C)\cdot \left(5\,\frac{m}{s} \right)\,\hat{i}+(-1\,C)\cdot\left(5\,\frac{m}{s} \right)\cdot (3\,T)\,\hat{j}[/tex]
[tex]\vec F_{B} = -20\,\hat{i}-15\,\hat{j}\,\,\,[N][/tex]
Which corresponds to option D.