Two copper spheres are currently 0.5 meters apart. One sphere has a charge of
+3.4x10-4 C and the other has a charge of -5.6*10-4 C. What is the force between
the charged spheres? (k=8.89*10' Nm²/C2)
0 -6839.17 N
0 1200 N
0 5497.45 N
6839.17 N
Answer:
6839.17 N
Explanation:
From the question given above, the following information were obtained:
Distance apart (r) = 0.5 m
Charge 1 (q₁) = +3.4×10¯⁴ C
Charge 2 (q₂) = -5.6×10¯⁴ C
Electrical constant (K) = 8.98×10⁹ Nm²/C²
Force (F) =.?
F = Kq₁q₂ / r²
F = 8.98×10⁹ × 3.4×10¯⁴ × 5.6×10¯⁴ / 0.5²
F = 1709.792 / 0.25
F = 6839.17 N
Thus, the force of attraction between the two charged spheres is 6839.17 N
if I accelerated at a rate of 20 mi/hr/s how fast would I be going after 5 seconds?
Answer: 100
Explanation: 20 *5=100
Hope this helps!!Brainlist?plz
Question 1
3 pts
There is an apple with a mass of 14 kg that is attached to the tree 12
m above the ground (on earth). How much Gravitational Potential
Energy does the apple have? (Round to the nearest tenth)
Hint g= 9.8 (m/s^2)
*
Answer:
[tex]\boxed {\boxed {\sf 1646.4 \ J }}[/tex]
Explanation:
Gravitation potential energy can be found using this formula:
[tex]E_P=m*g*h[/tex]
where m is the mass, g is the gravtiational acceleration, and h is the height.
The mass of the apple is 14 kilograms and it is 12 meters above the ground. Since this is on Earth, the acceleration due to gravity is 9.8 m/s².
[tex]m= 14 \ kg \\g= 9.8 \ m/s^2 \\h= 12 \ m[/tex]
Substitute the values into the formula.
[tex]E_p= ( 14 \ kg )( 9.8 \ m/s^2)(12 \ m)[/tex]
Multiply.
[tex]E_p= (137.2 \ kg*m/s^2)(12 \ m )[/tex]
Multiply again.
[tex]E_P= 1646.4 \ kg*m^2/s^2[/tex]
1 kg*m²/s² is equal to 1 Joule. Our answer of 1646.4 kg*m²/s² is equal to 1646.4 Joules[tex]E_p=1646.4 \ J[/tex]
This is already rounded to the tenth place, so it's the final answer.
The apple has 1,646.4 Joules of gravitational potential energy.
Two trucks with equal mass are attracted to each other with a gravitational force of 6.7 x 10^-4 N. The trucks are separated by a distance of 3.0 m. What is the mass of one of the trucks?
Answer:
9508.1kg
Explanation:
Given parameters:
Gravitational force = 6.7 x 10⁻⁴N
Distance of separation = 3m
Unknown:
Mass of one of the truck = ?
Solution:
Let the mass of one of the truck be = m
According to the Newton's law of universal gravitation:
Fg = [tex]\frac{G m1 m2}{d^{2} }[/tex]
G is the universal gravitation constant = 6.67 x 10⁻¹¹
m1 and m2 are the masses of the two bodies, here m1 = m2 = m
d is the distance of separation between them
6.7 x 10⁻⁴ = [tex]\frac{6.67 x 10^{-11} x m^{2} }{3^{2} }[/tex]
6.67 x 10⁻¹¹m² = 9 x 6.7 x 10⁻⁴
m = 9508.1kg
The mass of one of the trucks is 9510 kg.
The problem above can be solved using the mathematical expression of newton's law of universal gravitation
Using Newton's law of universal gravitation,
F = Gm²/r²................... Equation 1
Where F = Gravitational force on the truck, m = mass of each truck, r = distance between the trucks, G = universal constant
make m the subject of the equation
m = √(Fr²/G)................ Equation 2
From the question,
Given: F = 6.7×10⁻⁴ N, r = 3.0 m.
Constant: G = 6.67×10⁻¹¹ Nm²/kg²
Substitute these values into equation 2
m = √[(6.7×10⁻⁴×3²)/(6.67×10⁻¹¹)]
m = √(9.04×10⁷)
m = 9.51×10³ kg.
m = 9510 kg
Hence, The mass of one of the trucks is 9510 kg
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a 30-kg child sitting in a chair applies a force of approximately 300 N downward. on the chair approximately how much force does the chair apply upward to the child
A. 10N
B. 30N
C. 300N
D .600N
Answer:
moment's equilibrium required
moment = force x perpendicular distance from force's line of action (from pivot)
counter clockwise = clockwise moments
=> 15g(1.5) = F(0.3)
=> F = 15*9.81*5 = 735.75 ~= 736 N
Explanation:
A vector
of magnitude 5 units and another vector ū of magnitude 3 units point in directions
differing by 60°. Find out (a) the scalar product of the two vectors and (b) the vector product of the two vectors
Answer:
7.5 units
13 units
Explanation:
[tex]|v|=5\ \text{units}[/tex]
[tex]|u|=3\ \text{units}[/tex]
[tex]\theta[/tex] = Angle between the vectors = [tex]60^{\circ}[/tex]
Scalar product is given by
[tex]u\cdot v=|u||v|\cos\theta\\ =3\cdot 5\cdot \cos60^{\circ}\\ =7.5\ \text{units}[/tex]
The scalar product of the vectors is 7.5 units.
Vector product is given by
[tex]u\times v=|u||v|\sin\theta\\ =3\times 5\sin60^{\circ}\\ =13\ \text{units}[/tex]
The vector product of the vectors is 13 units.
When fireworks explode, sound and light are produced. These are examples of(1 point)
macroscopic inputs.
macroscopic outputs.
microscopic inputs.
microscopic outputs.
Answer: macroscopic outputs
Explanation:
When fireworks explode, sound and light are produced. These are examples of macroscopic outputs. Because, explosion from fireworks is an exothermic process which releases massive heat energy to the surroundings.
What is exothermic reaction?Exothermic reaction are those which evolve heat energy to the surroundings. The change in enthalpy of the reaction is negative here. Whereas, in an endothermic reaction energy is absorbed by the reactants.
Exothermic reactions sometimes results in massive explosion. The heat energy released to the surroundings from the fire works is macroscopic level.
The small scale process or quantity that cannot be measured using normal scales are called microscopic units. Therefore, the sound, light, and heat from the explosion all are macroscopic outputs.
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Find the work done by a man who is pulling a box of 45kg of mass by means of rope which makes angle of 45 degrees.
Answer:
0
Explanation:
Since no distance is given, the force is not doing any work
No work is done by the man since we do not know the distance or displacement.
Work is only said to be done when the force applied on an object moves it through a particular distance.
Work done = Force x distance.
Since no distance is given in this problem, we can as well assume that the force applied is doing no work on the object.
PLEASE HELP ASAP A ball is rolling on a flat, frictionless surface. What will happen to the ball if no unbalanced force acts on the ball?
A.
The ball will roll faster and faster.
B.
The ball will begin to roll in the opposite direction.
C.
The ball will begin to roll faster for a time and then begin to roll slower.
D.
The ball will continue to roll in the same direction and at the same speed.
Rolling friction causes a ball to roll more slowly when it rolls on a horizontal surface. Since static friction is to blame for the friction force, it produces no net work and has no effect on energy dissipation. Thus, option A is correct.
What happens to the ball if no unbalanced force on the ball?On a frictionless horizontal surface, the ball could neither speed up no slow down because the horizontal surface was intermediate between downhill and uphill. Galileo concluded that in the absences of friction, a ball that once started rolling on a horizontal surface would roll forever.
Friction causes a ball rolling on a flat surface to slow down. The ball finally comes to a stop as a result of friction, which operates in opposition to the ball's motion. If not, it will continue to roll indefinitely.
Therefore, The friction force is brought on by static friction; it has no effect on releasing energy or producing net work.
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Se necesita subir una carga de 500 kg (4900 N) a una altura de 1.5 m deslizándola sobre una rampa inclinada. ¿Qué longitud debe tener la rampa si sólo se puede aplicar una fuerza de 1633.33 N?
Answer:
4.22 m
Explanation:
Una rampa es una máquina que se utiliza para levantar un objeto con una fuerza menor a la que realmente necesitarías. Cuanto mayor sea la longitud de la rampa, menor será la magnitud de la fuerza necesaria para levantar el objeto.
Dado que:
altura de la rampa = 1.5 m, carga = 4900 N, fuerza aplicada = 1633.33 N.
La fórmula de la rampa se da como:
fuerza aplicada * longitud de la rampa = peso de la carga * altura de la rampa
1633.33 * longitud de la rampa = 4900 * 1.5
longitud de la rampa = 4900 * 1.5 / 1633.33
longitud de la rampa = 4.22 m
HELPPPPPPP !!!!!! PLS A 0.050 kg Truck moving right at 0.20 m/s collides with a toy race car weighing 0.015 kg initially at rest on a frictionless track. the truck keeps moving right at 0.15 m/s after Collision what is the final speed of the race car?
Answer:
[tex]v_f=0.15\:\mathrm{m/s}[/tex]
Explanation:
From the Law of Conservation of Momentum, we can write the following equation:
[tex]m_1v_1+m_2v_2=m_fv_f[/tex]
Because the toy car was initially at rest, [tex]m_2v_2=0[/tex].
Therefore, we can plug in given values and solve for [tex]v_f[/tex]:
[tex]m_1v_1=m_fv_f,\\0.050\cdot 0.20 = (0.050+0.015)v_f,\\v_f=\fbox{$0.15\:\mathrm{m/s}$}[/tex] (two significant figures).
3 write
down
two method to increase
any
the effiencey of a simple machine.
Ang two method to increase the effienee.
of a simple machine aree
ns
y of
1)
Answer:
two method of increasing efficiency of machine are 1st is give lubricants to machine to reduce friction or heat loss and
2nd. is by increasing the input of energy
A physics student is driving his pick-up
truck down Lake Avenue. The pick-up is
equipped with a projectile launcher that
imparts a vertical velocity to a water-filled
rubber projectile. While traveling 20.0 m/s
in an eastward direction, the projectile is
launched vertically with a velocity of 58.8
m/s.
Fill in the table at the right. showing the
horizontal and vertical displacement of the
projectile every second for the first 12
seconds.
TOTAL FLIGHT TIME
Rising Projectile
vi = initial velocity = 58.8 m/s
vf = final velocity = 0
g = gravity acceleration = -9.81 m/s²
t = elapsed time = to be determined
vf = vi + gt
vf - vi = gt
(vf - vi)/g = t
t = (vf - vi)/g
t = (0 - 58.8)/-9.81
t = -58.0/-9.81
t = 5.99 s ≈ 6.00 s
Falling Projectile
vi = initial velocity = 0
vf = final velocity = 58.8 m/s
g = gravity acceleration = 9.81 m/s²
t = elapsed time = to be determined
vf = vi + gt
vf - vi = gt
(vf - vi)/g = t
t = (vf - vi)/g
t = (58.8 - 0)/9.81
t = 58.0/9.81
t = 5.99 s ≈ 6.00 s
t(total) = t(up) + t(down) = 6.00 s + 6.00 s = 12.0 s Total Flight Time
HORIZONTAL DISTANCE TRAVELED
v = velocity, horizontal = 20.0 m/s
d = to be determined
t = elapsed time = 12.0 s
v = d/t
vt = d
d = vt
d = (20.0 m/s)(12.0 s)
d = 240 m Horizontal Distance Traveled
00
in which diagram 1 or 2 would the person be experiencing winter? explain
Answer: Diagram 2
In diagram 1, the person in the northern hemisphere gets more sunlight compared to someone in the southern hemisphere. So diagram 1 has summertime in the northern hemisphere.
Conversely, diagram 2 has the southern hemisphere get more sunlight, so the northern hemisphere is experiencing winter.
A body moving in simple harmonic motion has an amplitude of 10cm and a frequency of 100Hz. Find i.mthe period of oscillation, ii. the acceleration at the maximum displacement, iii. the velocity at the centre of motion
Explanation:
= 5 cm = 0.05 m
T = 0.2 s
ω=2π/T=2π/0.2=10πrad/s
When displacement is y, then acceleration, a=−ω2y
Velocity, V=ωr2−y2
Case (a) When y=5cm=0.05m
a=−(10π)2×0.05=−5π2m/s2
V=10π×(0.05)2−(0.05)2=0
Case (b) When y=3cm=0.03m
a=−(10π)2×0.03=−3π2m/s2
V=10π×(0.05)2−(0.03)2=10π×0.04=0.4π m/s
Case (c) When y=0
a=−(10π)2×0=0
V=10π×(0.05)2−02=
hope it helps. pls follow and mark as brainliest
a train has an initial velocity of 44m/s and an accelaration of _4m/s calculate its velocity
Complete question:
A train has an initial velocity of 44m/s and an acceleration of -4m/s². calculate its velocity after 10s ?
Answer:
the final velocity of the train is 4 m/s.
Explanation:
Given;
initial velocity of the train, u = 44 m/s
acceleration of the train, a = -4m/s² (the negative sign shows that the train is decelerating)
time of motion, t = 10 s
let the final velocity of the train = v
The final velocity of the train is calculated using the following kinematic equation;
v = u + at
v = 44 + (-4 x 10)
v = 44 - 40
v = 4 m/s
Therefore, the final velocity of the train is 4 m/s.
Please what is the work done by a man who is pulling a box of 45kg of mass by means of rope which makes angle of 45 degrees ?
Answer:
No work is done since no distance is given
Explanation:
Since no distance is given, the force is not doing any work
No work is done by the man since we do not know the distance or displacement.
Work is only said to be done when the force applied on an object moves it through a particular distance.
Work done = Force x distance.
Since no distance is given in this problem, we can as well assume that the force applied is doing no work on the object.
Positive feedback caused his pancreas to stop secreting insulin. Negative feedback caused his pancreas to stop secreting insulin. Negative feedback caused his pancreas to produce more insulin. Positive feedback caused his pancreas to produce more insulin.
Answer:
Negative feedback caused his pancreas to produce more insulin.
Explanation:
Positive feedback is a feedback that causes an increase due to change, that is it results in more product while negative feedback is a feedback that causes a decrease due to change, that is it result in lesser product.
Blood sugar control in the body is a negative feedback, because as the blood sugar increases, the nerves sense a change which causes the pancreas to secrete insulin until homeostasis and then the secretion stops.
HELLLPPPPPPPPPPPPPPPPPP
Answer:
If you going to ask a question ask it accurately.
Explanation:
A construction worker pushes a wheelbarrow with a total mass of 50.0 kg. What is the acceleration of the wheelbarrow if the net force on it is 75N? For physical science
Answer:
1.5m/s²
Explanation:
Given parameters:
Mass of the wheelbarrow = 50kg
Net force applied = 75N
Unknown:
Acceleration of the barrow = ?
Solution:
From Newton's second law of motion, we know that:
Force = mass x acceleration
Now insert the parameters and find the acceleration
75 = 50 x acceleration
Acceleration = 1.5m/s²
What word does this define..
The distance traveled per unit of time
Answer:..........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................Volosity???????
Explanation: THINK THATS IT PROBABLY MISPELLED IT SORRY IN ADVANCE IF WRONG
Which two life functions of animals help maintain the water cycle by recycling water back into the environment?
Answer:
Respiration and excretion
Explanation:
Respiration. When animals breathe, their lungs release water vapour to the atmosphere.
Animals excrete water by respiration and by passing urine.
In a normal Earth atmosphere, a feather will free fall through the air more slowly than a brick because of what force?
A. Terminal velocity
B. Momentum
C. Weightlessness
D. Air Resistance
Answer:
Air Resistance
Explanation:
Because air gives a much greater resistance to the falling motion of the feather than it does to the brick.
When blue light (450 nm) is shone on a particular metal surface, electrons are ejected with a kinetic energy of 2.5 x 10-20 J. Calculate the work function for this metal.
Answer:
the work function of the metal is 4.167 x 10⁻¹⁹ J .
Explanation:
Given;
wavelength of the incident light, λ = 450 nm = 450 x 10⁻⁹ m
kinetic energy, K.E = 2.5 x 10⁻²⁰ J
The energy of the incident light is calculated as;
[tex]E = hf = \frac{h c}{\lambda} \\\\where;\\\\c \ is \ speed \ of \ light = 3 \times 10^8 \ m/s\\\\ h \ is \ Planck's constant = 6.626 \times 10^{-34} Js \\\\E = \frac{(6.626 \times 10^{-34})(3\times 10^8)}{450 \times 10^{-9}} \\\\E = 4.417 \times 10^{-19} \ J[/tex]
Apply Einstein photoelectric equation to determine the work function of the metal;
E = W + K.E
where;
W is the work function of the metal
W = E - K.E
W = 4.417 x 10⁻¹⁹ J - 2.5 x 10⁻²⁰ J
W = 44.17 x 10⁻²⁰ J - 2.5 x 10⁻²⁰ J
W = 41.67 x 10⁻²⁰ J
W = 4.167 x 10⁻¹⁹ J
Therefore, the work function of the metal is 4.167 x 10⁻¹⁹ J .
The work function of the photon is 4.167*10^19J
The energy of this photon can be calculated as
E = hc/λ
Data given;
λ = 450nm = 450*10^-9mh = 6.626*10^-34JSK.E = 2.5*10^-20Jc = 3.0*10^8m/sEnergy of the Photonsubstituting the values into the equation;
[tex]E = hc / y\\E = \frac{6.626*10^-^3^4*3.0*10^8}{450*10^-^9} \\E = 4.42*10^-^1^9J[/tex]
Work FunctionThe work function of the photon can be calculated as;
E = K.E + Ф
4.42*10^-19 = 2.5*10^-20 + Ф
Ф = [tex]4.42*10^-^1^9 - 2.5*10^-^2^0=4.167*10^-^1^9J[/tex]
The work function of the photon is 4.167*10^-19 J
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. Which Of The Following Will Produce Diffuse Reflection Of Light?
[A] Plane mirror and piece of paper
[B] Piece of paper and still water in lake
[C] Plane mirror and leather bag
[D] Piece of paper and leather bag
Answer:
A. Plane mirror and piece of paper
Explanation:
Diffuse reflection is the reflection of light or other waves or particles from a surface such that a ray incident on the surface is scattered at many angles rather than at just one angle as in the case of specular reflection. An ideal diffuse reflecting surface is said to exhibit Lambertian reflection, meaning that there is equal luminance when viewed from all directions lying in the half-space adjacent to the surface.
If a car traveled 100 km in 2 hr, what was its average speed in km/hr?
average speed =
d =
t =
Answer:
Average speed will be 50km/hr
Explanation:
Distance/Time
100/2
50kmph
plz mark as brainliest
Answer:
50km/hr
Explanation:
avgspeed=Δdistance/Δtime
Δd=100km
Δt=2hr
=100km/2hr
=50km/hr
Can anyone help me with these questions? TIA!
(Don’t actually answer please! :) )
[tex]\huge\text{Graphs:}[/tex]
The graph of
• The I-V for Ohmic Metal wire conductor at constant temperature always shows a straight line between the Current(I) plotted at Y axis and Voltage(V) plotted at X axis. Picture 1
• The I-V graph for Diode shows that first the current is zero but as we increase the potential difference(voltage), it results in the increase in the current. Picture 2
_____________________________________ Question 8:[tex]\Large\textbf{Diode:}[/tex]
A diode is a device that allows current to flow in only one direction.
[tex]\Large\textbf{Forward and Reverse Biasing:}[/tex]
Forward Bias, When a diode is forward bias (a voltage in the "forward" direction) then the P-side of the diode is attached to the positive terminal and N-side is fixed to the negative side of the battery which is connected, current flows freely through the device. The forward bias decreases the thickness of potential barrier(The potential barrier barrier in which the charge requires additional force for crossing the region)
Reverse Bias, When a diode is Reverse bias(a voltage in the "backward direction) then the P-side of the diode is connected to the negative terminal and N-side is connected to the positive terminal of the battery which is connected. The reverse bias increases the thickness of the potential barrier resulting in the flow of no current.
[tex]\Large\textbf{Answer to the Question "Resistance"}[/tex]
The Forward bias decreases the resistance of the diode whereas the reversed bias increases the resistance of the diode. As in forward biasing the current is easily flowing through the circuit whereas reverse bias does not allow the current to flow through it.
_____________________________________ Best Regards, 'Borz'Derive Isothermal process through ideal gas.( derive it numerically)
Explanation:
Hope it helps..........
Helium ballon containing 5m3 of Gad at a pressure of 30pka is released. If the temperature is constant calculate the volume the balloon would have when the pressure inside the ballon has fallen to 5kpa answer in metres cubed
Answer:
30m³
Explanation:
Calculation for the volume the balloon would have
First step is to calculate the Constant
Constant=30pka*5m³
Constant=150m³
Now let calculate the Volume using this formula
Balloon Volume=Constant/ Pressure
Let plug in the formula
Balloon Volume=150m³/5kpa
Balloon Volume=30m³
Therefore the volume that the balloon would have when the pressure inside the ballon has fallen to 5kpa is 30m³
A vertical spring with a force constant of 26 N/m has a relaxed length of 35 cm. A 200 g mass is then
attached to the end of the spring and is allowed to come to rest. (a) Find the length of the stretched spring.
(b) How much elastic potential energy is stored in the spring?
I just need the work please the answers is
(a) 45.2 J
(b) 0.0739
Answer:
(a) The length of the stretched string is approximately 42.53846 cm
(b) The elastic potential energy stored in the string is approximately 0.073877 Joules
Explanation:
(a) The given parameter are;
The force constant of the vertical spring, k = 26 N/m
The relaxed length of the spring, L = 35 cm = 0.35 m
The mass attached to the end of the spring, m = 200 g = 0.2 kg
By Hooke's law, we have;
F = k·x
Where;
F = The applied force on the sting = The weight of the attached mass, W
∴ F = W
The weight of the attached mass, W = m × g
Where;
g = The acceleration due to gravity ≈ 9.8 m/s²
Therefore, F = W = 0.2 kg × 9.8 m/s² = 1.96 N
From Hooke's law, we have;
x = F/k = 1.96 N/(26 N/m) ≈ 0.0753846 m
The extension of the stretched spring, x ≈ 0.0753846 m
The length of the stretched string, L[tex]_{stretched}[/tex] = L + x
∴ L[tex]_{stretched}[/tex] ≈ 0.35 m + 0.0753846 m ≈ 0.4253846 m
0.4253846 m = 42.53846 cm
The length of the stretched string ≈ 42.53846 cm
(b) The elastic potential energy stored in the string, U = 1/2·k·x²
By substituting the known values, we get;
U = 1/2 × 26 N/m × (0.0753849 m)² ≈ 0.073877 Joules