Answer:
0.3 N
Explanation:
Electromagnetic force is F= Kq1q2/r^2, where r is the distance between charges. If r is doubled then the force will be 1/4F which is 0.3 N.
The magnitude of the force exerted on Q by q when the distance between them is doubled is 0.3 N
Coulomb's law equationF = Kq₁q₂ / r²
Where
F is the force of attraction K is the electrical constant q₁ and q₂ are two point charges r is the distance apart Data obtained from the question Initial distance apart (r₁) = rInitial force (F₁) = 1.2 NFinal distance apart (r₂) = 2rFinal force (F₂) =? How to determine the final forceFrom Coulomb's law,
F = Kq₁q₂ / r²
Cross multiply
Fr² = Kq₁q₂
Kq₁q₂ = constant
F₁r₁² = F₂r₂²
With the above formula, we can obtain the final force as follow:
F₁r₁² = F₂r₂²
1.2 × r² = F₂ × (2r)²
1.2r² = F₂ × 4r²
Divide both side by 4r²
F₂ = 1.2r² / 4r²
F₂ = 0.3 N
Learn more about Coulomb's law:
https://brainly.com/question/506926
11. From this lab we can conclude that a) the heat transferred when objects are rubbed together creates an energy that can cause objects to move towards or away from each other. b) objects such as balloons and sweaters have a natural affinity towards each other. They will attract each other whether they are rubbed together or not. c) charges exert forces on other charges. do) charges do not exert forces on other charges.
Answer:
c) charges exert forces on other charges.
Explanation:
When two different materials are rubbed together, there is a transfer of electrons from one material to the other material so this causes one object to become positively charged and the other object is negatively charged so they will attract each other not repel each other. Charges exert forces on other charges i.e. opposite charges attract each other whereas similar charges repel each other so in both cases force are exerted on one another.
explain what would happen if the cell was disconnected from the circuit ( please help me)
Most of the time, "cell" means a battery, and if it gets disconnected, all voltage and current in the circuit goes away.
I know this is a lame answer, but it kinda depends on what the cell was doing in the circuit, and what else is in the circuit besides the cell, and you haven't told us anything about these details, so that's really all we can guess.
Although your question is incomplete a general answer within the concept of your question is provided: when the cell is disconnected from the Circuit the flow of voltage across the circuit will be halted.
A cell/battery is often used as a voltage source in electrical circuits, since the cell is a power source, when the cell is connected to a circuit the cell discharges some of its voltage to the components of the circuit ( such as capacitors ) to keep the circuit functional. so when the cell is disconnected from the circuit the flow of voltage is halted
Hence we can conclude that without a voltage source ( cell ) in a circuit there will be no flow of voltage across the components of the circuit.
learn more : https://brainly.com/question/16598952
Convert 385k to temperature of
Answer:
233.33°F
Explanation:
(385K - 273.15) * 9/5 + 32 = 233.33°F
if 6000j of energy is supplid to a machine to lift a load of 300N through a vvertical height of 1M calculatework out put
Answer:
300J
Explanation:
Work done = Force x the distance travelled in the direction of the force
=300 x 1
=300J
A regulation soccer field for international play is a rectangle with a length between 100 m and a width between 64 m and 75 m. What are the smallest and largest areas that the field could be?
Answer:
The smallest and largest areas could be 6400 m and 7500 m, respectively.
Explanation:
The area of a rectangle is given by:
[tex] A = l*w [/tex]
Where:
l: is the length = 100 m
w: is the width
We can calculate the smallest area with the lower value of the width.
[tex] A_{s} = 100 m*64 m = 6400 m^{2} [/tex]
And the largest area is:
[tex] A_{l} = 100 m*75 m = 7500 m^{2} [/tex]
Therefore, the smallest and largest areas could be 6400 m and 7500 m, respectively.
I hope it helps you!
Answer:
the largest areas that the field could be is [tex]A_l[/tex]=7587.75 m
the smallest areas that the field could be is [tex]A_s[/tex]=6318.25 m
Explanation:
to the find the largest and the smallest area of the field measurement error is to be considered.
we have to find the greatest possible error, since the measurement was made nearest whole mile, the greatest possible error is half of 1 mile and that is 0.5m.
therefore to find the largest possible area we add the error in the mix of the formular for finding the perimeter with the largest width as shown below:
[tex]A_l[/tex]= (L+0.5)(W+0.5)
(100+0.5)(75+0.5) = (100.5)(75.5) = 7587.75 m
To find the smallest length we will have to subtract instead of adding the error factor value of 0.5 as shown below:
[tex]A_s[/tex]= (L-0.5)(W-0.5)
(100-0.5)(64-0.5) = (99.5)(63.5) = 6318.25 m
2. The vector sums of and the Ark witar must se rue our the directions and maintedes at an Bit CB? What meast le tue about the directions and magnitudes and it cor
Check attached photo
Check attached photo
PleasePlease help me solve these articles with me
A car is driving towards an intersection when the light turns red. The brakes apply a constant force of 1,398 newtons to bring the car to a complete stop in 25 meters. If the weight of the car is 4,729 newtons, how fast was the car going initially
Answer:
the initial velocity of the car is 12.04 m/s
Explanation:
Given;
force applied by the break, f = 1,398 N
distance moved by the car before stopping, d = 25 m
weight of the car, W = 4,729 N
The mass of the car is calculated as;
W = mg
m = W/g
m = (4,729) / (9.81)
m = 482.06 kg
The deceleration of the car when the force was applied;
-F = ma
a = -F/m
a = -1,398 / 482.06
a = -2.9 m/s²
The initial velocity of the car is calculated as;
v² = u² + 2ad
where;
v is the final velocity of the car at the point it stops = 0
u is the initial velocity of the car before the break was applied
0 = u² + 2(-a)d
0 = u² - 2ad
u² = 2ad
u = √2ad
u = √(2 x 2.9 x 25)
u =√(145)
u = 12.04 m/s
Therefore, the initial velocity of the car is 12.04 m/s
A projectile, fired with unknown initial velocity, lands 20sec later on side of hill, 3000m away horizontally and 450m vertically above its starting point. a) what is the vertical component of its initial velocity? b) what is the horizontal component of velocity?
Explanation:
Given:
t = 20 seconds
x = 3000 m
y = 450 m
a) To find the vertical component of the initial velocity [tex]v_{0y}[/tex], we can use the equation
[tex]y = v_{0y}t - \frac{1}{2}gt^2[/tex]
Solving for [tex]v_{0y}[/tex],
[tex]v_{0y} = \dfrac{y + \frac{1}{2}gt^2}{t}[/tex]
[tex]\:\:\:\:\:\:\:=\dfrac{(450\:\text{m}) + \frac{1}{2}(9.8\:\text{m/s}^2)(20\:\text{s})^2}{(20\:\text{s})}[/tex]
[tex]\:\:\:\:\:\:\:=120.5\:\text{m/s}[/tex]
b) We can solve for the horizontal component of the velocity [tex]v_{0x}[/tex] as
[tex]x = v_{0x}t \Rightarrow v_{0x} = \dfrac{x}{t} = \dfrac{3000\:\text{m}}{20\:\text{s}}[/tex]
or
[tex]v_{0x} = 150\:\text{m/s}[/tex]
the specific heat capacity of a substance is 500J/kg/oC. Find the heat required to rise the temperature of 10 quintial of the substance by 3 degree celcius
[tex]\boxed{\sf Q=mc\Delta T}[/tex]
[tex]\\ \sf\longmapsto Q=1000(5000)(3)[/tex]
[tex]\\ \sf\longmapsto Q=15000000J[/tex]
[tex]\\ \sf\longmapsto Q=1.5\times 10^7J[/tex]
Calculate the equivalent of 30 degrees Celsius and 50 degrees Celsius on a Kelvin
[tex]\boxed{\sf 1°C=273K}[/tex]
Sol:-1
[tex]\\ \sf\longmapsto 30°C[/tex]
[tex]\\ \sf\longmapsto 273+30[/tex]
[tex]\\ \sf\longmapsto 303K[/tex]
Sol:-2
[tex]\\ \sf\longmapsto 50°C[/tex]
[tex]\\ \sf\longmapsto 50+273[/tex]
[tex]\\ \sf\longmapsto 323K[/tex]
how does laser works ?
Explanation:
Lasers produce a narrow beam of light in which all of the light waves have very similar wavelengths. The laser's light waves travel together with their peaks all lined up, or in phase. This is why laser beams are very narrow, very bright, and can be focused into a very tiny spot.
Identify the factors that affect the intensity of radiation detected from a radioactive source. Select one or more: The color of the source Type of emission from the source Distance of the detector from the source Type of materials between the source and the detector
The intensity of radiation is the defined as amount of energy per surface angle which can be used to determine the amount of energy emitting from a source that will hit another surface.
The factors that affect the intensity of radiation are
Type of emission from the source :This can be alpha, gamma, beta or electromagnetic rays etc
Distance of the detector from the source: The shorter the distance between the source and the detector, the more the effect and vice versa for the longer the distance.
Type of materials between the source and the detector: The type of material between the source and the detector will tell how absorbing and penetrating the radiation is.
Read more on Radiation Intensity here: https://brainly.com/question/10148635
Trình bày những hiểu biết của em về đại lượng vận tốc dài, vận tốc góc(định nghĩa, công thức, ý nghĩa, đơn vị, loại đại lượng).
A student has a large number of coins of different diameters, all made of the same metal. She wishes to find the density of the metal by a method involving placing the coins in water.
a) State the formula needed to calculate the density.
b) Describe how the measurements of the required quantities are carried out.
Answer:
a)density = mass /volume
b)to find volume put water into a container .measure the level of water , put the coins into the beaker containing water , measure the level of water again, subtract the new volume withe the first one . the result is the volume of coins
A crude approximation is that the Earth travels in a circular orbit about the Sun at constant speed, at a distance of 150,000,000 km from the Sun. Which of the following is the closest for the acceleration of the Earth in this orbit?
A. exactly 0 m/s2.
B. 0.006 m/s2.
C. 0.6 m/s2.
D. 6 m/s2.
E. 10 m/s2.
Answer:
The answer is "Option B".
Explanation:
[tex]r=15\times 10^{7}\ km\ = 15\times 10^{10}\ m\\\\w=\frac{2\pi}{1\ year}\\\\=\frac{2\pi}{1\times 365.24 \times 24 \times 60 \times 60\ sec}\\\\a=w^2r\\\\=(\frac{2\pi}{1\times 365.24 \times 24 \times 60 \times 60\ sec})^2 \times 15 \times 10^{10}\ \frac{m}{s^2}\\\\[/tex]
[tex]=5.940 \times 10^{-3} \ \frac{m}{s^2}\\\\=6 \times 10^{-3} \ \frac{m}{s^2}\\\\=0.006\ \frac{m}{s^2}\\\\[/tex]
Express 6revolutions to radians
Answer:
About 37.70 radians.
Explanation:
1 revolution = 2[tex]\pi[/tex] radians
∴ 6 revolutions = (6)(2[tex]\pi[/tex] radians)
6 revolutions = 37.6991 or ≈ 37.70 radians
Use the pressure meter to read the pressure in Fluid A at the bottom of the tank. Do not move the pressure meter. Switch to Fluid B and read the pressure in fluid B. Based on the two readings, compare the density of fluid B to the density of fluid A. Which statement is correct?
Answer:
[tex]P_b = \frac{\rho_b}{\rho_a} \ P_a[/tex]
Explanation:
The pressure at a depth of a fluid is
P = ρ g y
where ρ is the density of the fluid, y the depth of the gauge measured from the surface of the fluid.
In this case the pressure for fluid A is
Pa = ρₐ g y
the pressure for fluid B is
P_b = ρ_b g y
depth y not changes as the gauge is stationary
if we look for the relationship between these pressures
[tex]\frac{P_a}{P_b} = \frac{ \rho_a}{\rho_b}[/tex]
[tex]P_b = \frac{\rho_b}{\rho_a} \ P_a[/tex]
therefore we see that the pressure measured for fluid B is different from the pressure of fluid A
if ρₐ < ρ_b B the pressure P_b is greater than the initial reading
ρₐ> ρ_b the pressure in B decreases with respect to the reading in liquid A
An object is made of glass and has the shape of a cube 0.13 m on a side, according to an observer at rest relative to it. However, an observer moving at high speed parallel to one of the object's edges and knowing that the object's mass is 3.3 kg determines its density to be 8100 kg/m3, which is much greater than the density of glass. What is the moving observer's speed (in units of c) relative to the cube
Answer:
[tex]v=0.9833\ c[/tex]
Explanation:
The density changes means that the length in the direction of the motion is changed.
Therefore,
[tex]$\text{Density} = \frac{m}{lwh}$[/tex]
Given :
Side, b = h = 0.13 m
Mass, m = 3.3 kg
Density = 8100 [tex]kg/m^3[/tex]
So,
[tex]$8100=\frac{3.3}{l \times 0.13 \times 0.13}$[/tex]
[tex]$l=\frac{3.3}{8100 \times 0.13 \times 0.13}$[/tex]
l = 0.024 m
Then for relativistic length contraction,
[tex]$l= l' \sqrt{1-\frac{v^2}{c^2}}$[/tex]
[tex]$0.024= 0.13 \sqrt{1-\frac{v^2}{c^2}}$[/tex]
[tex]$0.184= \sqrt{1-\frac{v^2}{c^2}}$[/tex]
[tex]$0.033= 1-\frac{v^2}{c^2}}$[/tex]
[tex]$\frac{v^2}{c^2}= 0.967$[/tex]
[tex]$\frac{v}{c}=0.9833$[/tex]
[tex]v=0.9833\ c[/tex]
Therefore, the speed of the observer relative to the cube is 0.9833 c (in the units of c).
What will be the acceleration of a body moving with uniform velocity?
Answer:
so the acceleration of the body would be zero because there is no change in velocity
The question is in the photo.
Answer:
heyaa thereeee
see temperature is rising in interval of
0 to 4 minutes
8 to 10 minutes
but 8 to 10 is NOT in options
so answer is option a) 0 to 4 minutes
:))))
a 0.0780 kg lemming runs off a 5.36 m high cliff at 4.84 m/s. what is its kinetic energy when it's 2.00 m above the ground
Answer:
KE_2 = 3.48J
Explanation:
Conservation of Energy
E_1 = E_2
PE_1+KE_1 = PE_2+KE_2
m*g*h+(1/2)m*v² = m*g*h+(1/2)m*v²
(0.0780kg)*(9.81m/s²)*(5.36m)+(.5)*(0.0780kg)*(4.84m/s)² = (0.0780kg)*(9.81m/s²)*(2m)+KE_2
4.10J+0.914J = 1.53J + KE_2
5.01J = 1.53J + KE_2
KE_2 = 3.48J
find the upward force in Newton when each of these is under water(density of 1g/cm3) a lump of iron of volume 2000cm3
Answer:
Upthrust = 19.6 N
Explanation:
When an object is immersed under water, the upward force it experience is called an upthrust. An upthrust is a force which is applied on any object in a fluid which acts in an opposite direction to the direction of the weight of the object.
Upthrust = density of liquid x gravitational force x volume of object
i.e U = ρ x g x vol
Given: ρ = 1g/[tex]cm^{3}[/tex] (1000 kg/[tex]m^{3}[/tex]), volume = 2000 c[tex]m^{3}[/tex] (0.002 [tex]m^{3}[/tex]) and g = 9.8 m/[tex]s^{2}[/tex]
So that;
U = 1000 x 9.8 x 0.002 (kg/[tex]m^{3}[/tex] x [tex]m^{3}[/tex] x m/[tex]s^{2}[/tex])
= 19.6 Kg m/[tex]s^{2}[/tex]
U = 19.6 Newtons
The upthrust on the iron is 19.6 N.
Under normal circumstances: _________
a. Fetal Hb binds to oxygen more tightly than Mb binds.
b. Fetal Hb binds oxygen more tightly in the absence of 2,3-BPG.
c. Fetal Hb does not bind to oxygen.
d. Adult Hb has the lowest affinity for oxygen of the 3.
e. More than one of these statements is correct.
Answer:
Fetal Hb binds oxygen more tightly than adult Hb (not option a)
an aluminum atom has an atomic number of 13 and a mass number of 27,how many
a)protons
b) electrons
pls write the formula too
Element is
[tex]\boxed{\sf {}^{27}Al_{13}}[/tex]
Atomic number=13Mass number=27[tex]\\ \sf\longmapsto No\:of\:Protons=Atomic \:Number=13[/tex]
And[tex]\\ \sf\longmapsto No\:of\:Neutrons=Mass\:number-Atomic\:Number[/tex]
[tex]\\ \sf\longmapsto No\:of\:Neutrons=27-13[/tex]
[tex]\\ \sf\longmapsto No\:of\:Neutrons=14[/tex]
And
[tex]\\ \sf\longmapsto No\:of\:electrons=No\:of\:Protons=13[/tex]
find the equivalent resistance of this circuit
Answer:
Req = 564 Ω
Explanation:
The equivalent resistance between R1 and R2:
1/R =1/R1 + 1/R2
1/R =1/960 + 1/640
1/R = 1/384
R = 384
Now, the equivalent resistance between R and R3:
Req = 384 + 180
Req = 564 Ω
name a device that converts mechanical energy into electrical energy.
Answer:
Electric generator is the device that converts mechanical energy into electrical energy
Data related to Meena’s and Malini’s journey is given below, plot a graph of their
respective journey on a graph paper. You have already plotted Meena’s Journey during the
summer vacation. On the same graph paper, now plot Malini’s Journey.
PDF task 2
Please do this for me urgent I can give you extra points if someone answers this in less than 1 hour.
Answer:
download the pdf
Explanation:
which characteristic of nuclear fission makes it hazardous?
Answer:The radioactive waste
Explanation:Fission is the splitting of a heavy unstable nucleus into two Lighter nuclei
Assume the speed of sound is 343 m/s. You are sitting 150 m away from home plate at a baseball game. How much time in seconds elapses between the batter hitting a home run and the moment you actually hear the batter hitting the ball
Answer:
t = 0.437 s
Explanation:
Sound is a wave so its speed is constant
v = x / t
t = x / v
indicates that the distance is x = 150 m
t = 150/343
t = 0.437 s
this is the time it takes to hear the hit
To see the blow it is almost instantaneous since the speed of light is much greater c = 3 10⁸ m / s
