Answer:
40 m/s
Explanation:
Three equal point charges, each with charge 1.05 μCμC , are placed at the vertices of an equilateral triangle whose sides are of length 0.500 mm . What is the electric potential energy UUU of the system? (Take as zero the potential energy of the three charges when they are infinitely far apart.)
Answer:
The value is [tex]U = 0.06 \ J [/tex]
Explanation:
From the question we are told that
The value of charge on each three point charge is
[tex]q_1 = q_2 = q_3 =q= 1.05 \mu C = 1.05 *10^{-6} \ C[/tex]
The length of the sides of the equilateral triangle is [tex]r = 0.500 \ [/tex]
Generally the total potential energy is mathematically represented as
[tex]U = k * [ \frac{q_1 * q_2}{r} + \frac{q_2 * q_3}{r} + \frac{q_3 * q_1}{r} ][/tex]
=> [tex]U = k * 3 * \frac{q^2}{r} [/tex]
Here k is coulomb constant with value [tex]k = 9*10^{9}\ kg\cdot m^3\cdot s^{-4} \cdot A^{-2}.[/tex]
=> [tex]U = 9*10^9 * 3 * \frac{(1.05 *10^{-6})^2}{0.5 } [/tex]
[tex]U = 0.06 \ J [/tex]
The relatively high resistivity of dry skin, about 1 × 106 O·m, can safely limit the flow of current into deeper tissues of the body. Suppose an electrical worker places his palm on an instrument whose metal case is accidentally connected to a high voltage. The skin of the palm is about 1.5 mm thick. Estimate the area of skin on the worker’s palm that would contact a flat panel, then calculate the approximate resistance of the skin of the palm.
Answer:
The area is [tex]A = 0.015 \ m^2 [/tex]
The resistance is [tex]R = 1.0 *10^{5} \ \Omega [/tex]
Explanation:
From the question we are told that
The relativity of dry skin is [tex]\rho = 1 * 10^6\ \Omega\cdot m[/tex]
The thickness of the skin is [tex]d = 1.5 \ mm = 1.5 *10^{-3} \ m[/tex]
Generally the area of a flat palm with length = 0.10 m and breath b = 0.15 m is mathematically represented as
[tex]A = d * L[/tex]
[tex]A = 0.10 * 0.15 [/tex]
[tex]A = 0.015 \ m^2 [/tex]
Generally the resistance is mathematically represented as
[tex]R = \frac{\rho * d}{ A }[/tex]
=> [tex]R = \frac{ 1.0 *10^{6} * 1.5*10^{-3}}{ 0.015 }[/tex]
=> [tex]R = 1.0 *10^{5} \ \Omega [/tex]
6
Following an inelastic collision, what can kinetic energy be converted into? Check all that apply.
heat
impulse
momentum
potential energy
sound
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Answer:
Heat, potential energy, and sound.
Explanation:
Edgenuity says so
The characteristics of the inelastic shocks allow to find the correct answers of in which the kinetic energy can be transformed are:
Heat Potential energy Sound.
In a collision the total momentum of the system is conserved, this is an important principle of physics, we have two types of collisions:
Elastic. In this case the kinetic energy of the system is conserved. Inelastic. In this case the initial kinetic energy is not conserved.
In inelastic collision, part of the energy of the movement is transformed into potential energy of configuration of the system.
Let's analyze the different answers:
a) Heat.
True. Heat is a form of energy that, due to the change in the relative friction motion of bodies, is one of the greatest sources of transformation of energy into configuration power energy.
b) Impulse.
False. The impulse is the change of the momentum of the system and this is conserved, therefore the momentum is also conserved.
c) Momentum
False. Momentum is preserved in all crashes.
d) Potential energy.
True. Power energy is a configuration energy, therefore kinetic energy can be transformed into potential energy.
e) The sound.
True. Sound is a wave that carries energy, therefore kinetic energy can be transformed into traveling sound waves.
In conclusion, using the characteristics of inelastic shocks, we can find the correct answers that the kinetic energy can be transformed into are:
Heat Potential energy Sound.
Learn more here: brainly.com/question/24616147
Earth's atmosphere is constantly bombarded by cosmic ray protons that originate somewhere in space. If the protons all passed through the atmosphere, each square meter of Earth's surface would intercept protons at the average rate of 1800 protons per second. What would be the electric current in amperes intercepted by a 41 × 107 km2 area on the planet
Answer:
Electric current in amperes = 1.1808 A
Explanation:
Given:
Intercept protons rate = 1800 protons per second
Area = 41 × 10⁷ km²
Find:
Electric current in amperes
Computation:
Current density = Intercept protons rate × 1.6 × 10⁻¹⁹
Current density = 1800 × 1.6 × 10⁻¹⁹
Current density = 2.88 × 10⁻¹⁶
1 km² = 10⁶m²
So,
Electric current in amperes = 2.88 × 10⁻¹⁶ × 41 × 10⁷ × 10⁶
Electric current in amperes = 1.1808 A
Question 3
A box is being pulled by a rope that makes a 25 degree angle with the ground. The
pulling force is 100.0 N
along the rope. Find the horizontal and vertical components of the force vector.
A. 90.63 N, 42.26 N
B. 86.87 N, 32.17 N
C. 60.87 N, 75,63 N
D. 80.9 N, 45.5 N
What is the answer A,B,C or D?
Answer:
A
Explanation:
Enunciado: Una bola se lanza verticalmente de la parte superior de un edificio con una velocidad inicial de 25 m/s. La bola impacta al suelo en la base del edificio 7 segundos después de ser lanzada. (Marque la respuesta correcta)
¿Qué altura subió la bola (medida desde la parte superior del edificio)? a) 19.6 m b) 12.75 m c) 31.88 m d) 40 m e) 20 m
Responder:
Explicación:
Usaremos la ecuación de movimiento para determinar la altura de la bola medida desde la parte superior del edificio.
Usando la ecuación para obtener la altura de caída
S = ut + 1 / 2gt²
u es la velocidad inicial = 25 m / s
g es la aceleración debida a la gravedad = 9,81 m / s²
t es el tiempo = 7 segundos
S es la altura de la caída
S = 25 (7) +1/2 (9,81) × 7²
S = 175 + 4,905 (49)
S = 175 + 240,345
S = 415,35 m
Esto significa que la pelota se elevó a 415,35 m de altura
La altura vertical máxima alcanzada es de 31,88 m.
Tenemos la siguiente información de la pregunta;
Velocidad inicial = 25 m/s
Velocidad final = 0 m/s (a la altura máxima)
tiempo empleado = 3,5 minutos (el tiempo empleado para subir y bajar es igual).
Usando la ecuación;
v^2 = u^2 - 2gh
Dado que v = 0
u^2 = 2gh
h = tu^2/2g
h = (25)^2/2 *9.8
h = 31,88 m
Obtenga más información sobre las ecuaciones de movimiento: https://brainly.com/question/8898885
Jesse is swinging Miguel in a circle at a tangential speed of 3.50 m/s. If the radius of the circle is 0.600 m and Miguel has a mass of 11.0 kg, what is the centripetal force on Miguel? Round to the nearest whole number. N
Answer:
224.6 N
Explanation:
We can use first the formula to calculate the centripetal acceleration, given by:
[tex]a_c=\frac{v_t^2}{R}[/tex]
where the Vt is the tangential velocity, and R is the radius of the circular motion.
Then, for our case we have:
[tex]a_c=\frac{v_t^2}{R}=\frac{3.5^2}{0.6}\approx 20.42\,\,\frac{m}{s^2}[/tex]
And now we multiply this acceleration by Miguel's mass (11 kg) to obtain the centripetal force acting on him:
[tex]F_c=11 \,*\,20.42\,N = 224.6\,\,N[/tex]
Answer:
225 N
Explanation:
Just answered the question
How does an electrometer differ from a voltmeter? What is meant by an electrical ground? What must you do before each measurement in this experiment and why?
Answer:
n an electrometer, it is built in such a way that its resistance in parallel is extremely high
Ground in a circuit is a reference point from which voltages are measured
all the instruments must be grounded and we must ground ourselves
Explanation:
When you build a voltmeter you have a resistance in parallel with the galvanometer, therefore when measuring the voltage of a circuit, so that there is no effect (load effect) by the voltmeter, a resistance must be much greater than the resistance where it is is measuring.
In an electrometer, it is built in such a way that its resistance in parallel is extremely high in the order of 10¹²Ω, so its load effect is very small and can be measured with high resistance mu
Electric ground in home and industrial installations is a protection system consisting of a metal piece connected to a buried ground electrode.
Ground in a circuit is a reference point from which voltages are measured and is common to all parts of the circuit
In an experiment where an electrometer is used, all the instruments must be grounded and we must ground ourselves, since it must be an instrument where very small voltages are measured at high impedances.
what is the mass of an object that is experiencing a net force of 200N and an acceleration of 500m/s2
Answer:
F=200N
a=500m/s2
Mass=?
Explanation:
F=ma
200=m*500
200/500=m
Mass=0.4kg
5.
An object with zero acceleration is not changing its velocity.
TRUE
FALSE
What is an example of intellectual development?
Answer:
learning new activities like riding a bike and playing sports. and also having the ability to collect information and remember them (memory). <3
Which are best examples of projectile motion
Answer:
Please add the multiple choice awensers too then I'll get back to you when I can
Answer:
Answer is D
Explanation:
Which atom is most likely to accept electrons to form an ionic bond? a mercury ion with a negative 2 charge a potassium ion with a negative 1 charge radon, a noble gas with 8 valence electrons sulfur, a nonmetal with 6 valence electrons
Sulfur, a nonmetal with 6 valence electrons atom is most likely to accept electrons to form an ionic bond option (D) correct.
What is an ionic bond?Ionic bonds, also known as electrovalent bonds, are a type of connection created in a chemical molecule by the electrostatic attraction of ions with opposing charges.
As we know,
It has been shown that the top of the electronegativity scale is defined by the fundamental atomic particles F>O>N.
Ionic bonds need an electron, often a nonmetal, and an electron, typically a metal. Metals display ionic bonding because there aren't many-electron in outer orbitals.
Thus, sulfur, a nonmetal with 6 valence electrons atom is most likely to accept electrons to form an ionic bond option (D) correct.
Learn more about the ionic bond here:
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If you go 45 meters at a speed of 16 m/s, how long does the movement take? Exact answer, do NOT round.
Answer:
2 seconds
Explanation:
The time taken for the movement is 2.81 seconds.
To find the time, the given values are,
Distance = 45 meters
speed = 16 m/s.
What is distance?Distance is the total movement of an object without any regard to direction.Distance can be evaluated how much an object moves from starting point to the end point.The distance completely depends upon the speed and time, i,e., the object covering some area with some particular time interval with the particular speed.Distance will be measured in meter, kilometer, etc..Distance is a Scalar quantity.Formula of distance,
Distance = Speed × Time.
As we have to find the time, Formula can be rearranged as,
Time = Distance / Speed
Substituting the given datas in the formula,
Time = 45 / 16
= 2.81 seconds.
So, the time taken for the movement is 2.81 seconds.
Learn more about the distance,
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What is the velocity of a plane to travel 3000 miles from New York to California in 5.0 hours
Answer:
10 miles per minute.
a steel ball bearing will sink in water but a large ship floats because
Answer:
water displacement.
the bigger the item means more water displacement meaning bouyance
A satellite in outer space is moving at a constant velocity of 21.4 m/s in the y direction when one of its onboard thruster turns on, causing an acceleration of 0.250 m/s2 in the x direction. The acceleration lasts for 45.0 s, at which point the thruster turns off. (a) What is the magnitude of the satellite's velocity when the thruster turns off
Answer:
a) The magnitude of the satellite's velocity when the thruster turns off is approximately 24.177 meters per second.
b) The direction of the satellite's velocity when the thruster turns off is approximately 62.266º.
Explanation:
Statement is incomplete. The complete description is now described below:
A satellite in outer space is moving at a constant velocity of 21.4 m/s in the y direction when one of its onboard thruster turns on, causing an acceleration of 0.250 m/s2 in the x direction. The acceleration lasts for 45.0 s, at which point the thruster turns off.
(a) What is the magnitude of the satellite's velocity when the thruster turns off
(b) What is the direction of the satellite's velocity when the thruster turns off? Give your answer as an angle measured counterclockwise from the +x-axis. ° counterclockwise from the +x-axis
Let be x and y-directions orthogonal to each other and the satellite is accelerated uniformly from rest in the +x direction and moves at constant velocity in the +y direction. The velocity vector of the satellite ([tex]\vec{v}_{S}[/tex]), measured in meters per second, is:
[tex]\vec{v}_{S} = (v_{o,x}+a_{x}\cdot t)\,\hat{i}+v_{y}\,\hat{j}[/tex]
Where:
[tex]v_{o,x}[/tex] - Initial velocity in +x direction, measured in meters per second.
[tex]a_{x}[/tex] - Acceleration in +x direction, measured in meter per square second.
[tex]t[/tex] - Time, measured in seconds.
[tex]v_{y}[/tex] - Velocity in +y direction, measured in meters per second.
If we know that [tex]v_{o,x} = 0\,\frac{m}{s}[/tex], [tex]a_{x} = 0.250\,\frac{m}{s^{2}}[/tex], [tex]t = 45\,s[/tex] and [tex]v_{y} = 21.4\,\frac{m}{s}[/tex], the final velocity of the satellite is:
[tex]\vec{v}_{S} = \left[0\,\frac{m}{s}+\left(0.250\,\frac{m}{s^{2}} \right)\cdot (45\,s) \right]\,\hat{i}+\left(21.4\,\frac{m}{s} \right)\,\hat{j}[/tex]
[tex]\vec{v_{S}} = 11.25\,\hat{i}+21.4\,\hat{j}\,\,\left[\frac{m}{s} \right][/tex]
a) The magnitud of the satellite's velocity can be found by the resource of the Pythagorean Theorem:
[tex]\|\vec {v}_{S}\| = \sqrt{\left(11.25\,\frac{m}{s} \right)^{2}+\left(21.4\,\frac{m}{s} \right)^{2}}[/tex]
[tex]\|\vec{v}_{S}\| \approx 24.177\,\frac{m}{s}[/tex]
The magnitude of the satellite's velocity when the thruster turns off is approximately 24.177 meters per second.
b) The direction of the satellite's velocity when the thruster turns off is determined with the help of trigonometric functions:
[tex]\tan \alpha = \frac{v_{y}}{v_{x}} = \frac{21.4\,\frac{m}{s} }{11.25\,\frac{m}{s} }[/tex]
[tex]\tan \alpha = 1.902[/tex]
[tex]\alpha = \tan^{-1}1.902[/tex]
[tex]\alpha \approx 62.266^{\circ}[/tex]
The direction of the satellite's velocity when the thruster turns off is approximately 62.266º.
What is the kinetic energy when the pendulum is at the halfway point
(height=1 m) when the total mechanical energy is 4 J
What must be the acceleration of a train in order for it to stop from 12m/s in a distance of 541m?
Please help!!
Answer:
The acceleration of the train must be - 0.133 m/s²
Explanation:
A train in order for it to stop 12 m/s in a distance if 541 m
That means the initial velocity of the train is 12 m/s
Its final velocity is zero (stop)
The distance it covers is 541 m
P.S. This is part of the answer of another brainly user this is not my knowledge.
But Hope I helped you
Answer:
-0.133 m/s²
Explanation:
Given:
Δx = 541 m
v₀ = 12 m/s
v = 0 m/s
Find: a
v² = v₀² + 2aΔx
(0 m/s)² = (12 m/s)² + 2a (541 m)
a = -0.133 m/s²
lol
idk what this is
Answer:
I know
Explanation:
Physics= hard
If two tug boats are towing a ship with force of 5 tons each and the angle between the two ropes is 60 degrees, what is the resultant force on the ship? Explain how to use a force table to verify answer.
Answer:
8.6602 tons
Explanation:
We first draw the known vector forces.
2fcos30⁰
We have f to be equal to 5tons
Inserting into formula
Σfx = 2(5)cos30⁰
= 8.6602 tons
Σfy is equal to 0, this is because in the y direction, the forces cancel themselves out.
Therefore the resultant force on the ship is equal to 8.6602 tons
I hope this helps!
Please check attachment for diagram.
What is the mass of the cube
Density is 40 g/cm3
Length of the side= 2 cm
Answer:
80
Explanation:
because of the denstiy formula
Please help me slow this I don't understand how this works
Answer:
[tex]A=90^0-B\\B+C=180^0-90^0=90^0\\C=90^0-B\\A=C[/tex]
Explanation:
Part A
What is the magnitude of the momentum of a 0.0073-kg marble whose speed is 0.65 m/s?
Express your answer to two significant figures and include appropriate units.
Part B
What is the speed of a 0.136-kg baseball whose momentum has a magnitude of 3.14 kg⋅m/s?
Express your answer to two significant figures and include appropriate units.
Answer:
(A). The magnitude of the momentum of the marble is 0.004745 kg m/s.
(B). The speed of baseball is 23.0 m/s
Explanation:
Given that,
Mass of marble = 0.0073 kg
Speed = 0.65 m/s
(A). We need to calculate the magnitude of the momentum of the marble
Using formula of momentum
[tex]p = mv[/tex]
Where, m = mass
v = velocity
Put the value into the formula
[tex]p=0.0073\times0.65[/tex]
[tex]p=0.004745\ kg m/s[/tex]
(B). Mass of baseball = 0.136 kg
Momentum of baseball = 3.14 kg m/s
We need to calculate the speed of baseball
Using formula of momentum
[tex]p=mv[/tex]
[tex]v=\dfrac{p}{m}[/tex]
Put the value into the formula
[tex]v=\dfrac{3.14}{0.136}[/tex]
[tex]v=23.0\ m/s[/tex]
Hence, (A). The magnitude of the momentum of the marble is 0.004745 kg m/s.
(B). The speed of baseball is 23.0 m/s
the force which slow down a paper aeroplane moving through the air is called
The paper plane slows down when flying in the air due to air resistance or the friction due to the air around it. As the paper plane have flat surfaces and more surface area, the air resists it's motion more than before. If the paper plane is moving towards left, then friction acts towards left.
The friction always act opposite to the direction of paper plane and resists its motion. Hence, we can say that paper plane slows due to Frictional Force [tex]\red{\bullet}[/tex]
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Compare the de Broglie wavelength of a golf ball moving at 70.0 miles per hour (31.3 m/s) to that of an alpha particle moving at 3.40E+7 miles per hour (1.52E+7 m/s) and a bullet with a speed of 700 miles per hour (313 m/s). Which region of the electromagnetic spectrum are each of these wavelengths near
Answer:
1. Golf ball is in the range of only particulate detectable range
2, alpha partcle is in ultraviolet range of wavelength
3.bullet is in xray range of wavelength
All in the EMW spectrum
Give an example of a situation in which you would describe an
object's position in
a. one dimension.
b. two dimensions.
three dimensions.
An excited squirrel starts from rest and accelerates towards a nut at 1.0 m/s2 for 2.0 seconds. What is the squirrel’s displacement?
Answer:
[tex]x=8m[/tex]
Explanation:
Hello,
In this case, the most suitable kinematic equation for the calculation of the squirrel's displacement is:
[tex]x=v_ot+\frac{1}{2}at^2[/tex]
Whereas the initial velocity is zero as it starts from rest, thus, we obtain:
[tex]x=\frac{1}{2}*1.0\frac{m}{s^2}*(2.0s)^2\\ \\x=8m[/tex]
Best regards.
If 13 is added to a number, the result is 43 less than twice the number. Find the number,
Explanation:
13+x=43>2x
13+x=43>2x+43>2x
13+x=86>4x
x-4x=86-13
3x=73
x=73/3
x=24.333
x=24.4
A basketball leaves a player's hands at a height of 2.20 m above the floor. The basket is 2.70 m above the floor. The player likes to shoot the ball at a 36.0 ∘ angle. Of the shot is made from a horizontal distance of 9.10 m and must be accurate to ±0.23m (horizontally), what is the range of initial speeds allowed to make the basket
Answer:
The range of initial speeds allowed to make the basket is: [tex]9.954\,\frac{m}{s}\leq v \leq 10.185\,\frac{m}{s}[/tex].
Explanation:
We must notice that basketball depicts a parabolic motion, which consists of combining a constant speed motion in x-direction and free fall motion in the y-direction. The motion is described by the following kinematic formulas:
x-Direction
[tex]x = x_{o} + v_{o}\cdot t \cdot \cos \alpha[/tex]
y-Direction
[tex]y = y_{o} + v_{o}\cdot t\cdot \sin \alpha +\frac{1}{2}\cdot g\cdot t^{2}[/tex]
Where:
[tex]x_{o}[/tex], [tex]y_{o}[/tex] - Initial position of the basketball, measured in meters.
[tex]x[/tex], [tex]y[/tex] - Final position of the basketball, measured in meters.
[tex]v_{o}[/tex] - Initial speed of the basketball, measured in meters per second.
[tex]t[/tex] - Time, measured in seconds.
[tex]\alpha[/tex] - Tilt angle, measured in sexagesimal degrees.
[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.
If we know that [tex]x_{o} = 0\,m[/tex], [tex]y_{o} = 2.20\,m[/tex], [tex]\alpha = 36^{\circ}[/tex], [tex]g = -9.807\,\frac{m}{s^{2}}[/tex], [tex]x = (9.10\pm0.23)\,m[/tex] and [tex]y = 2.70\,m[/tex], the system of equation is reduce to this:
[tex](9.10\pm 0.23)\,m = 0\,m + v_{o}\cdot t \cdot \cos 36^{\circ}[/tex]
[tex]9.10\pm 0.23 = 0.809\cdot v_{o}\cdot t[/tex] (Ec. 1)
[tex]2.70\,m = 2.20\,m + v_{o}\cdot t \cdot \sin 36^{\circ} + \frac{1}{2}\cdot \left(-9.807\,\frac{m}{s^{2}} \right) \cdot t^{2}[/tex]
[tex]0.50 = 0.588\cdot v_{o}\cdot t-4.904\cdot t^{2}[/tex] (Ec. 2)
At first we clear [tex]v_{o}\cdot t[/tex] in (Ec. 1):
[tex]v_{o}\cdot t = \frac{9.10\pm 0.23}{0.809}[/tex]
[tex]v_{o}\cdot t = 11.248\pm 0.284[/tex]
(Ec. 1) in (Ec. 2):
[tex]0.5 = 0.588\cdot (11.248\pm 0.284)-4.904\cdot t^{2}[/tex]
Now we clear the time in the resulting expression:
[tex]4.904\cdot t^{2} = 0.588\cdot (11.248\pm 0.284)-0.5[/tex]
[tex]t = \sqrt{\frac{0.588\cdot (11.248\pm 0.284)-0.5}{4.904} }[/tex]
There are two solutions:
[tex]t_{1} = \sqrt{\frac{0.588\cdot (11.248- 0.284)-0.5}{4.904} }[/tex]
[tex]t_{1} \approx 1.101\,s[/tex]
[tex]t_{2} = \sqrt{\frac{0.588\cdot (11.248+ 0.284)-0.5}{4.904} }[/tex]
[tex]t_{2}\approx 1.131\,s[/tex]
The initial velocity is cleared within (Ec. 2):
[tex]v_{o}=\frac{0.50+4.904\cdot t^{2}}{0.588\cdot t}[/tex]
The bounds of the range of initial speed is determined hereafter:
[tex]t_{1} \approx 1.101\,s[/tex]
[tex]v_{o} = \frac{0.50+4.904\cdot (1.101)^{2}}{0.588\cdot (1.101)}[/tex]
[tex]v_{o} = 9.954\,\frac{m}{s}[/tex]
[tex]t_{2}\approx 1.131\,s[/tex]
[tex]v_{o} = \frac{0.50+4.904\cdot (1.131)^{2}}{0.588\cdot (1.131)}[/tex]
[tex]v_{o} = 10.185\,\frac{m}{s}[/tex]
The range of initial speeds allowed to make the basket is: [tex]9.954\,\frac{m}{s}\leq v \leq 10.185\,\frac{m}{s}[/tex].