Answer:
[tex]14400\ \text{N}[/tex], Attractive
[tex]3240\ \text{N}[/tex], Repulsive
Explanation:
[tex]q_1[/tex] = -20 μC
[tex]q_2[/tex] = 50 μC
r = Distance between the charges = 2.5 cm
k = Coulomb constant = [tex]9\times 10^9\ \text{Nm}^2/\text{C}^2[/tex]
Electrical force is given by
[tex]F=\dfrac{kq_1q_2}{r^2}\\\Rightarrow F=\dfrac{9\times 10^9\times (-20\times 10^{-6})\times (50\times 10^{-6})}{(2.5\times10^{-2})^2}\\\Rightarrow F=-14400\ \text{N}[/tex]
The magnitude of force each sphere will experience is [tex]14400\ \text{N}[/tex]
Since the charges have opposite charges they will attract each other.
Now the charges are brought into contact with each other so the resultant charge will be
[tex]q=\dfrac{q_1+q_2}{2}\\\Rightarrow q=\dfrac{-20+50}{2}\\\Rightarrow q=15\ \mu\text{C}[/tex]
[tex]F=\dfrac{kq^2}{r^2}\\\Rightarrow F=\dfrac{9\times 10^9\times (15\times 10^{-6})^2}{(2.5\times 10^{-2})^2}\\\Rightarrow F=3240\ \text{N}[/tex]
The magntude of the force the spheres experience will be [tex]3240\ \text{N}[/tex]
The spheres have the same charge now so they will repel each other.
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.
If a penny is dropped from rest from a tower takes 2 seconds to hit the ground, how far did it travel?
a 29.4 m
b 19.6 m
с 6.8 m
d 9.8 m
Answer:
B
Explanation:
t = 2s
u = 0m/s (released from rest)
a = +g = 9.8m/s²
s = H = ?
using,
s = ut + 1/2at²
H = 0(2) + 1/2(9.8)(2²)
H = 0 + 9.8(2)
s = H = 19.6m
''The Evolution Of Sakura And Hinata''
Who looks better?! Who do you think has improved a lot in the Naruto Series?
Answer:
hinata
Explanation:
unlike sakura she has not change at all
In my opinion, Sakura looks prettier and she developed way more than Hinata. Personally, I dislike the both of them because of their Tsundere side, but there's no doubt Sakura developed way more than Hinata. In the original Naruto series, Sakura was honestly a burden. There were some missions where Team 7 had to rescue Sakura. However, after the training with Tsunade, she gained monster strength and healing abilities. Additionally, throughout Naruto Shippuden she learned new jutsu like Yin Seal: Release.
However with Hinata, it's a bit more different. In the original Naruto series, I feel believe she was stronger than Sakura. She relied on her infamous visual prowess--the Byakugan. However in Naruto Shippuden, she improved slightly compared to Sakura. In my opinion, she would have been the most useless shinobi out of the eleven characters (YES, more useless than Tenten) if it weren't for her visual prowess. She's fortunate that she was born into the Hyuga clan. Every jutsu that's in her arsenal relies heavily on the Byakugan. And if Sakura and Hinata were to battle each other, it's most likely that Sakura would win. It's true that Hinata can block Sakura's chakra points, but I think she could dodge her attacks. And even if her chakra points are blocked, Sakura can rely on Taijutsu and raw strength to win.
Though both kunoichi are honorable, I believe Sakura is more prettier and stronger than Hinata.
show all work and round to nearest 100th thank you and you will get braineist
Answer: 16.78 miles
Explanation:
distance = rate * time
we're given the rate ( 5 m/s) and we're given the time it takes to get home (1.5 hrs). But notice how the units of hours don't match the seconds of the rate so we need to convert the hours into seconds.
1 hr = 3600 seconds so 1.5 hours = 5400 seconds
now we can plug it in and solve for the distance
distance = (5 m/s) * 5400 seconds
distance = 27000 m
now we have to convert meters to miles, so we divide our answer by 1609 and get 16.78
is 2/2 1 or 0? please help lol
Answer:
1.
Explanation:
Hello!
In this case, for such mathematical operations, we can wee that the slash represents a fraction or a division, say 8 ÷ 4 = 2, 6 ÷ 3 = 2, 20 ÷ 4 = 5, etc. In such a way, since the operation 2/2, represents 2 ÷ 2, it is clear that two is once in 2, therefore, the result is:
2 ÷ 2 = 1.
Best regards!
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.
Why do stretching exercises increase flexibility more than cardio exercises?
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
Discuss two ways to determine your muscular strength explain the advantages
Answer:
Two ways to determine ones muscular strength is to either lift the heaviest weight possible or by doing one repetition. The advantages these two methods is that it is easy to determine and does not require calculations or estimating.
Explanation:
Answer:
Two ways to determine ones muscular strength is to either lift the heaviest weight possible or by doing one repetition. The advantages these two methods is that it is easy to determine and does not require calculations or estimating.
Explanation:
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 + ......
Two stars in a faraway part of the Milky Way are orbiting each other as a binary star system. By
careful measurement we find out that they are separated by 1.7 AU. We also determine that
their orbital period is 594 Earth-days. The total mass of the two-object system is
Answer: [tex]M_{total}=[/tex] 1.85
Explanation: Estimate the total mass of a binary system is done by a reformulation of Kepler's Third Law, which states that the square of the period of a planet's orbit is proportional to the cube of its semimajor axis, i.e.:
[tex]a^{3}=(M_{1}+M_{2})P^{2}[/tex]
where
a is semimajor axis in astronomical units (AU);
P is period measured in years;
[tex]M_{1}+M_{2}[/tex] is total mass of the two-stars system;
For the two stars faraway in the Milky Way:
1 year is equivalent of 365 days, so period in years:
[tex]P=\frac{594}{365}[/tex]
P = 1.63 years
Calculating total mass:
[tex]a^{3}=(M_{total})P^{2}[/tex]
[tex]M_{total}=\frac{a^{3}}{P^{2}}[/tex]
[tex]M_{total}=\frac{1.7^{3}}{1.63^{2}}[/tex]
[tex]M_{total}=[/tex] 1.85
The total mass of the two-object system is 1.85 mass units.
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 rack of seven spherical bolwing balls (each 7.00 kg, radius of 9.50 cm) is positioned along a line located a distance =0.850 m from a point ,
Calculate the gravitational force
the bowling balls exert on a ping-pong ball of mass 2.70 g, centered at point .
Answer:
8.72*[tex]10^{-12}[/tex] N
Explanation:
force of attraction f = G m1m2/ r^2 = [tex]\frac{6.67*10^{-11}*7*5*2.70 }{.85*.85*1000}[/tex] = 8.72*[tex]10^{-12}[/tex] N
The gravitational force the bowling balls exert on a ping-pong is given as 8.72*[tex]10^{-12}[/tex] N.
What is force?A force is an effect that can alter an object's motion according to physics. An object with mass can change its velocity, or accelerate, as a result of a force. An obvious way to describe force is as a push or a pull. A force is a vector quantity since it has both magnitude and direction.
force of attraction
f = G [tex]m_1m_2[/tex]/r²
= [tex]6.67*10^{-11}[/tex]*7*5*2.70/.85²*1000
= 8.72*[tex]10^{-12}[/tex] N
The gravitational force the bowling balls exert on a ping-pong is given as 8.72*[tex]10^{-12}[/tex] N.
To learn more about force refer to the link:
brainly.com/question/13191643
#SPJ3
A car is moving at an average speed of 20 meters per second. This is equivalent to
Answer:
44.73 MP/H or 71.98 KM/H
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
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.
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.
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:
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.
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]
A potential difference of 1.20 V will be applied to a 33.0 m length of 18-gauge copper wire (diameter = 0.0400 in.). Calculate (a) the current, (b) the magnitude of the current density, (c) the magnitude of the electric field within the wire, and (d) the rate at which thermal energy will appear in the wire.
Answer:A) Current = 1.739A, B)current density, J = 2.147x10^6 A/m2
magnitude of electric field , E = 0.036 N/C
)rate of thermal energy, P =2.086W
Explanation:
Resistance = R = ρL/A
But the cross-section area of the wire. is given as
Diameter / 2 = 0.04/2 =0.02in to m = 0.02 / 39.37= 0.000508
A = πr^2 = π x 0.000508^2 = 8.10 x 10^-7
since resistivity of copper,ρ= 17x10-9 ohm.m
so resistance is R = ρL/A
17x10-9 x 33 / 8.1x10-7
= 0.69 ohm.
A) Current = I = Voltage /Resistance =1.20/0.69 =1.739A
B)current density, J = Current /Area
= 1.739/8.1x10-7
= 2.147x10^6 A/m2
c)magnitude of electric field , E = Current density x resistivity =J ρ
E = 2.147 x 10^6 x 17 x 10^-9
E = 0.036 N/C
D)rate of thermal energy, P = I² R =1.739² X 0.69
=2.086W
Any conclusion reached by analogy is worth accepting
True
False
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
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:
9. In the graph below, what is the force being exerted on
the 16-kg cart?
A. 4N
C. 16N
B. 8N
D. 32 N
Answer:
D
Explanation:
Can someone help me with this question
Answer:
Net force: 20 N to the right
mass of the bag: 20.489 kg
acceleration: 0.976 m/s^2
Explanation:
Since the normal force and the weight are equal in magnitude but opposite in direction, they add up to zero in the vertical direction. In the horizontal direction, the 195 N tension to the right minus the 175 force of friction to the left render a net force towards the right of magnitude:
195 N - 175 N = 20 N
So net force on the bag is 20 N to the right.
The mass of the bag can be found using the value of the weight force: 201 N:
mass = Weight/g = 201 / 9.81 = 20.489 kg
and the acceleration of the bag can be found as the net force divided by the mass we just found:
acceleration = 20 N / 20.489 kg = 0.976 m/s^2
A 7.50 kg bowling ball has 70.4
kg•m/s of momentum. What is its
velocity?
Answer:
9.39 m/sExplanation:
The velocity of the bowling ball can be found by using the formula
[tex]v = \frac{p}{m} \\ [/tex]
p is the momentum
m is the mass
From the question we have
[tex]v = \frac{70.4}{7.5} \\ = 9.38666666..[/tex]
We have the final answer as
9.39 m/sHope this helps you
what can i yeet baby or toddler
Answer:
both
Explanation:
baby for fun, toddler for vengence
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 :)
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.