Answer:
366 N
Explanation:
τ = Iα
FR = ½mR²α
F = ½mR(Δω/t)
F = ½(207)(1.50)(0.75)(2π) /2.00
F = 365.79919...
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
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 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
If you dive underwater, you notice an uncomfortable pressure on your eardrums due to the increased pressure. The human eardrum has an area of about 70 mm217 * 10-5 m22, and it can sustain a force of about 7 N without rupturing. If your body had no means of balancing the extra pressure (which, in reality, it does), what would be the maximum depth you could dive without rupturing your eardrum
Answer:
[tex]h=10m[/tex]
Explanation:
From the question we are told that:
Area [tex]a=70 x 10^{-6}[/tex]
Force [tex]F=7N[/tex]
Generally the equation for Pressure is mathematically given by
Pressure = Force/Area
[tex]P=\frac{F}{A}[/tex]
[tex]P=\frac{ 7}{(70 * 10^{-6})}[/tex]
[tex]P= 1*10^{5} Pa[/tex]
Generally the equation for Pressure is also mathematically given by
[tex]P=hpg[/tex]
Therefore
[tex]h=\frac{P}{hg}[/tex]
[tex]h=\frac{10000}{1000*9.8}[/tex]
[tex]h=10m[/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
In Young's double slit experiment, 402 nm light gives a fourth-order bright fringe at a certain location on a flat screen. What is the longest wavelength of visible light that would produce a dark fringe at the same location? Assume that the range of visible wavelengths extends from 380 to 750 nm.
Answer:
λ₂ = 357.3 nm
Explanation:
The expression for double-slit interference is
d sin θ = m λ constructive interference
d sin θ = (m + ½) λ destructive interference.
The initial data corresponds to a constructive interference, they indicate that we are in the fourth order (m = 4), let's look for the separation of the slits
d sin θ = m λ₁
now ask for destructive interference for m = 4
d sin θ = (m + ½) λ₂
we match these two expressions
m λ₁ = (m + ½) λ₂
λ₂ = ( m / m + ½) λλ₁
let's calculate
λ₂ =[tex]\frac{4}{(4.000 +0.5) \ 401}[/tex]
λ₂ = 357.3 nm
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]
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
:))))
I need help with this please!!!!
Answer:
1.84 hours
I hope it's helps you
How much amount of water can be decomposed
through electrolysis by passing 2 F charge?
Answer:
So, with 2 Faraday of electricity, we can decompose (2/4 × 2) = 1 mole of water. So 18 grams of water is decomposed.
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
If a bus travels 50 km in 10 hours, how fast was the
bus travelling?
Answer:
5 kilometers per hour
Explanation:
Speed = distance / time
Distance: 50km
Time: 10 hours
Speed = 50/10 = 5kph
Answer:
5kmph
Explanation:
if the bus traveled 50 km in 10 hours, we have to divide 50 by 10 to see how fast it traveled per hour.
50/10 = 5
therefore, the bus was traveling 5 km per hour
hope this helps :)
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).
Convert 385k to temperature of
Answer:
233.33°F
Explanation:
(385K - 273.15) * 9/5 + 32 = 233.33°F
A car accelerates from 0 m/s to 25 m/s in 5 seconds. What is the average acceleration of the car.
Answer:
5 m/s I hope it will help you
Explanation:
mark me as a brainlist answer
Which one of the following physical quantities has its S.I. unit m/s?
(i) Acceleration
(ii) Velocity
(iii) Force
(iv) Density
Answer:
velocity is the answer of this question.
Answer:
Velocity is the right answer ok
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
state the laws of reflection
Answer:
Explanation:
The law of reflection says that the reflected angle (measured from a vertical line to the surface called the normal) is equal to the reflected angle measured from the same normal line.
All other properties of reflection flow from this one statement.
A solenoid has a length , a radius , and turns. The solenoid has a net resistance . A circular loop with radius is placed around the solenoid, such that it lies in a plane whose normal is aligned with the solenoid axis, and the center of the outer loop lies on the solenoid axis. The outer loop has a resistance . At a time , the solenoid is connected to a battery that supplies a potential . At a time , what current flows through the outer loop
This question is incomplete, the complete question is;
A solenoid has a length 11.34 cm , a radius 1.85 cm , and 1627 turns. The solenoid has a net resistance of 144.9 Ω . A circular loop with radius of 3.77 cm is placed around the solenoid, such that it lies in a plane whose normal is aligned with the solenoid axis, and the center of the outer loop lies on the solenoid axis. The outer loop has a resistance of 1651.6 Ω. At a time of 0 s , the solenoid is connected to a battery that supplies a potential 34.95 V. At a time 2.58 μs , what current flows through the outer loop?
Answer:
the current flows through the outer loop is 1.3 × 10⁻⁵ A
Explanation:
Given the data in the question;
Length [tex]l[/tex] = 11.34 cm = 0.1134 m
radius a = 1.85 cm = 0.0185 m
turns N = 1627
Net resistance [tex]R_{sol[/tex] = 144.9 Ω
radius b = 3.77 cm = 0.0377 m
[tex]R_o[/tex] = 1651.6 Ω
ε = 34.95 V
t = 2.58 μs = 2.58 × 10⁻⁶ s
Now, Inductance; L = μ₀N²πa² / [tex]l[/tex]
so
L = [ ( 4π × 10⁻⁷ ) × ( 1627 )² × π( 0.0185 )² ] / 0.1134
L = 0.003576665 / 0.1134
L = 0.03154
Now,
ε = d∅/dt = [tex]\frac{d}{dt}[/tex]( BA ) = [tex]\frac{d}{dt}[/tex][ (μ₀In)πa² ]
so
ε = μ₀n [tex]\frac{dI}{dt}[/tex]( πa² )
ε = [ μ₀Nπa² / [tex]l[/tex] ] [tex]\frac{dI}{dt}[/tex]
ε = [ μ₀Nπa² / [tex]l[/tex] ] [ (ε/L)e^( -t/[tex]R_{sol[/tex]) ]
I = ε/[tex]R_o[/tex] = [ μ₀Nπa² / [tex]R_o[/tex][tex]l[/tex] ] [ (ε/L)e^( -t/[tex]R_{sol[/tex]) ]
so we substitute in our values;
I = [ (( 4π × 10⁻⁷ ) × 1627 × π(0.0185)²) / (1651.6 ×0.1134) ] [ ( 34.95 / 0.03154)e^( -2.58 × 10⁻⁶ / 144.9 ) ]
I = [ 2.198319 × 10⁻⁶ / 187.29144 ] [ 1108.116677 × e^( -1.7805 × 10⁻⁸ )
I = [ 1.17374 × 10⁻⁸ ] × [ 1108.116677 × 0.99999998 ]
I = [ 1.17374 × 10⁻⁸ ] × [ 1108.11665 ]
I = 1.3 × 10⁻⁵ A
Therefore, the current flows through the outer loop is 1.3 × 10⁻⁵ A
Answer:
1.28 *10^-5 A
Explanation:
Same work as above answer. Needs to be more precise
An ideal double slit interference experiment is performed with light of wavelength 640 nm. A bright spot is observed at the center of the resulting pattern as expected. For the 2n dark spot away from the center, it is known that light passing through the more distant slit travels the closer slit.
a) 480 nm
b) 600 nm
c) 720 nm
d) 840 nm
e) 960 nm
Answer:
960 nm
Explanation:
Given that:
wavelength = 640 nm
For the second (2nd) dark spot; the order of interference m = 1
Thus, the path length difference is expressed by the formula:
[tex]d sin \theta = (m + \dfrac{1}{2}) \lambda[/tex]
[tex]d sin \theta = (1 + \dfrac{1}{2}) 640[/tex]
[tex]d sin \theta = ( \dfrac{3}{2}) 640[/tex]
dsinθ = 960 nm
1 Poin Question 4 A 85-kg man stands in an elevator that has a downward acceleration of 2 m/s2. The force exerted by him on the floor is about: (Assume g = 9.8 m/s2) А ON B 663 N C) 833 N D) 1003 N
Answer:
D) 1003 N
Explanation:
Given the following data;
Mass of man = 85 kg
Acceleration of elevator = 2 m/s²
Acceleration due to gravity, g = 9.8 m/s²
To find the force exerted by the man on the floor;
Force = mg + ma
What Are the type's of Tidal turbines?
Answer:
Types of tidal turbines
Axial turbines.
Crossflow turbines.
Flow augmented turbines.
Oscillating devices.
Venturi effect.
Tidal kite turbines.
Turbine power.
Resource assessment.
Answer:
Axial turbines
Crossflow turbines
flow augmented turbines
name a device that converts mechanical energy into electrical energy.
Answer:
Electric generator is the device that converts mechanical energy into electrical energy
When the lightbulbs were used as the resistors, you observed only a flash of light, as opposed to a continuous glow. Explain why that behavior is expected. After all, the light bulb is directly connected to the power supply.
Solution :
Whenever the lightbulbs are used as resistors, we throw the switch to the left. This allows the current to flow through the circuit which causes the bulb to glow and also the capacitor gets charged. When the capacitor gets fully charged, the electric field becomes constant between its two plates. Now there is no displacement current induced in the plates of the capacitor. The capacitor works as an open switch and the bulb gets switched off.
And thus the bulb flashes for the moment as opposed to continuous glow.
If a boy lifts a mass of 6kg to a height of 10m and travels horizontally with a constant velocity of 4.2m/s, calculate the work done? Explain your answer.
Answer:
W = 641.52 J
Explanation:
The work done here will be the sum of potential energy and the kinetic energy of the boy. Here potential energy accounts for vertical motion part while the kinetic energy accounts for the horizontal motion part:
[tex]Work\ Done = Kinetic\ Energy + Potential\ Energy\\\\W = K.E +P.E\\\\W = \frac{1}{2}mv^2+mgh\\\\[/tex]
where,
W = Work Done = ?
m = mass = 6 kg
v = speed = 4.2 m/s
g = acceleration dueto gravity = 9.81 m/s²
h = height = 10 m
Therefore,
[tex]W = \frac{1}{2}(6\ kg)(4.2\ m/s)^2+(6\ kg)(9.81\ m/s^2)(10\ m)[/tex]
W = 52.92 J + 588.6 J
W = 641.52 J
How do you know that a liquid exerts pressure?
Answer:
The pressure of water progressively increases as the depth of the water increases. The pressure increases as the depth of a point in a liquid increases. The walls of the vessel in which liquids are held are likewise subjected to pressure. The sideways pressure exerted by liquids increases as the liquid depth increases.
Two identical loudspeakers 2.30 m apart are emitting sound waves into a room where the speed of sound is 340 m/s. Abby is standing 4.50 m in front of one of the speakers, perpendicular to the line joining the speakers, and hears a maximum in the intensity of the sound.
Required:
What is the lowest possible frequency of sound for which this is possible?
Answer:
Abby is standing (4.5^2 + 2.3^2)^1/2 from the far speaker
D2 = 5.05 m from the far speaker
The difference in distances from the speakers is
5.05 - 4.5 = .55 m (Let y be wavelength, lambda)
n y = 4.5
(n + 1) y = 5.05 for the speakers to be in phase at smallest wavelength
y = .55 m subtracting equations
f = v / y = 340 / .55 = 618 / sec should be the smallest frequency
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
2- A student ran 135 meters in 15 seconds. What was the student's velocity?
*
7.5 m/s
9 m/s
12 m/s
15 m/s
Answer:
9 Brainly hahaha ............huh
Parallel Wires: Two long, parallel wires carry currents of different magnitudes. If the current in one of the wires is doubled and the current in the other wire is halved, what happens to the magnitude of the magnetic force that each wire exerts on the other?
Answer:
Explanation:
Given force between 2 currents carrying
wires = F₀
Magnetic force between the2 wires =F₀= (μ₀/4π) x ( 2 (μ₀/4π) x ( 2I₁I₂ / μ) x L
where I₁=Current in wire 1
I₂= Current in wire 2
L= Length of the wire
when one current is doubled and the other is halved
I₁= 2 I₁
I₂= I₂/2
F₀ = (μ₀/4π) x ( 2× (2I₁) (I₂/2) / μ) x L
