Answer:
light reflects when it hits the surface
Explanation:
Youngs double slit is a evidence for wave nature,
The properties refraction are attributed as properties of waves. The phenomena of interference and diffraction also fall in this category.
So,
the answer must be B
Answer: Light does not need a medium to travel.
Explanation: I took the test and got it right :]
the neuron is considered a (a. Cell. (B.artery. (C. Vein
Answer:
A Cell
Explanation:
Take the regular compass and hold it so the case is vertical. Now use it to investigate the direction of the coil’s magnetic field at locations other than the central axis. What happens as you move away from the center axis toward the coil? What happens above the coil? Outside the coil? Below the coil?
Answer:
Please find the answer in the explanation
Explanation:
Take the regular compass and hold it so the case is vertical. Now use it to investigate the direction of the coil’s magnetic field at locations other than the central axis.
What happens as you move away from the center axis toward the coil? The direction of the magnetic compass needle will move in an opposite direction since the direction of the induced voltage is reversed.
What happens above the coil?
the needle on the magnetic compass will be deflected. Since compasses work by pointing along magnetic field lines
Outside the coil? The magnetic compass needle will experience no deflection. Since there is no induced voltage or current.
Below the coil?
The needle will move in an opposite direction.
You measure the radius of a sphere as (6.45 ± 0.30) cm, and you measure its mass as (1.79 ± 0.08) kg. What is the density and uncertainty in the density of the sphere, in kilograms per cubic meter?
Answer:
[tex](1630.13\pm 300.10)\ kg/m^3[/tex]
Explanation:
Given that,
The radius of a sphere is (6.45 ± 0.30) cm
Mass of the sphere is (1.79 ± 0.08) kg
Density = mass/volume
For sphere,
[tex]d=\dfrac{m}{V}\\\\d=\dfrac{m}{\dfrac{4}{3}\pi r^3}\\\\d=\dfrac{1.79\ kg}{\dfrac{4}{3}\pi (6.4\times 10^{-2}\ m)^3}\\\\d=1630.13\ kg/m^3[/tex]
We can find the uncertainty in volume as follows :
[tex]\dfrac{\delta V}{V}=3\dfrac{\delta r}{r}\\\\=3\times \dfrac{0.3\times 10^{-2}}{6.45\times 10^{-2}}\\\\=0.1395[/tex]
Uncertainty in mass,
[tex]\dfrac{\delta m}{m}=\dfrac{0.08}{1.79}\\\\=0.0446[/tex]
Now, the uncertainty in density of sphere is given by :
[tex]\dfrac{\delta d}{d}=\dfrac{\delta m}{m}+\dfrac{\delta V}V}\\\\=0.0446+0.1395\\\\\dfrac{\delta d}{d}=0.1841\\\\\delta d=0.1841\times d\\\\\delta d=0.1841\times 1630.13\\\\\delta d = 300.10\ kg/m^3[/tex]
Hence, the density pf the sphere is [tex](1630.13\pm 300.10)\ kg/m^3[/tex]
What is the solution?
Answer:
1) x = 30 - 8 t
2) x = -10
3) x = -10 + 5 t
4) x = -10 - 4 t
Explanation:
Motion 1: constant negative velocity calculated via the points (0, 30) and (5, -10) rendering the equation of motion x = 30 - 8 t
Motion 2: constant position over time, so the object is not moving, and the equation of motion is x = -10
Motion 3: constant positive velocity estimated via the points (0, -10) and (2, 0), and the equation of motion is: x = -10 + 5 t
Motion 4: constant negative velocity estimated via the points (0, -10) and (5, -30), and the equation of motion is: x = -10 - 4 t
The starting position of motion 1 is 30 meters
the starting position for the other 3 motions is - 10 meters. And none of them is accelerated (acceleration = zero for all).
color code of electrical resistors
Answer:
Tolerance: [tex]\pm 10\%[/tex]
Explanation:
Resistor Color Codes
Resistor Color Coding uses colored bands to quickly identify the resistive value or resistors and its percentage of tolerance.
Since the question does not provide a specific color table, we'll use the table attached below.
The colors of the resistor shown in the question are:
First band: orange
Second band: blue
Third band: brown
Fourth band: silver
The colors relate to the following numbers respectively:
3, 6, 10Ω, [tex]\pm 1\%[/tex]
The first two colors form the number 36
The third color is the multiplier: 36*10Ω = 360Ω
And the fourth color is the tolerance or the possible variation of the resistance [tex]\pm 1\%[/tex]
Resistance: 360Ω
Tolerance: [tex]\pm 10\%[/tex]
The Jamaican Bobsled Team is sliding down a hill in a toboggan at a rate of 5 m/s when he reaches an even steeper slope. If he accelerates at 2 m/s2 for the 5 m slope, how fast is he traveling when he reaches the bottom of the 5 m slope?
Answer:
6.7 m/s
Explanation:
Given:
Δx = 5 m
v₀ = 5 m/s
a = 2 m/s²
Find: v
v² = v₀² + 2aΔx
v² = (5 m/s)² + 2 (2 m/s²) (5 m)
v = 6.7 m/s
In a spy movie, the hero, James, stands on a scale that is positioned horizontally on the floor. It registers his weight as 810 N . Unknown to our hero, the floor is actually a trap door, and when the door suddenly disappears, James and the scale fall at the acceleration of gravity, down towards an unknown fate. As James falls, he looks at the scale to see his weight. What does he see
Answer:
His weight would be zero on the scale i.e he is weightless at that instance.
Explanation:
weight = mg
where m is the mass of the object, and g is the acceleration of gravity.
⇒ 810 = mg
During free fall, the weight of an object can be determined by:
W = mg - ma (provided that acceleration of gravity is greater than acceleration of the object)
where a is the acceleration of the object.
But since James fall at the acceleration of gravity, then:
g = a
mg = ma = 810 N
So that;
W = 810 - 810
= 0 N
Therefore though the weight of James is 810 N, but the scale reads 0 N. this condition is referred to as weightlessness.
The particles of a more dense substance are closer together
than the particles of a less dense substance.
TRUE
FALSE
The particles of a more dense substance are closer together than the particles of a less dense substance. Thus, the given statement is true.
What is density of particles?Density of the particles is the substance's mass per unit of volume. The symbol which is most often used for the density is ρ (rho), although the Latin letter D can also be used to denote density.
Density is the mass of a unit volume of a material substance or particle. The formula for density is d = M/V (mass per unit volume), where d is density, M is the mass of particle, and V is the volume. Density is commonly expressed in the units of grams per cubic centimeters.
The S.I. unit of density is made up of the mass of the particle which is kg and that of volume is meter cube. Hence, the S.I. unit of density is kg/m³.
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A car is traveling south at 8.77 m/s. It then begins a uniform acceleration until it reaches a velocity of 47.8 m/s over a period of 3.84s. What is the car's acceleration?
Please help !
Answer:
The acceleration of the car is 10.16m/s²
Explanation:
Given parameters:
Initial velocity = 8.77m/s
Final velocity = 47.8m/s
Time duration = 3.84s
Unknown:
Acceleration of the car = ?
Solution:
To find the acceleration, we must bear in mind that this physical quantity is the change in velocity with time;
Acceleration = [tex]\frac{V - U}{T}[/tex]
V is the final velocity
U is the initial velocity
T is the time taken
Input the parameters and solve for acceleration;
Acceleration = [tex]\frac{47.8 - 8.77}{3.84}[/tex] = 10.16m/s²
The acceleration of the car is 10.16m/s²
21. A toy car starts from rest and begins to
accelerate at 11.0 m/s2. What is the toy car's
final velocity after 6.0 seconds?
Answer:
v = 66 m/s
Explanation:
Given that,
The initial velocity of a car, u = 0
Acceleration of the car, a = 11 m/s²
We need to find the final velocity of the toy after 6 seconds.
Let v is the final velocity. It can be calculated using first equation of motion. It is given by :
v = u +at
v = 0 + 11 m/s² × 6 s
v = 66 m/s
So, the final velocity of the car is 66 m/s.
A 10-ohm resistor has a constant current. If 1200 C of charge flow through it in 4 minutes what
is the value of the current?
A. 3.0 A
B 5.0 A
C. 11 A
D. 15 A
E. 20A
Answer:
B 5.0 A .
Explanation:
Hello.
In this case, since we know the charge (1200 C), time (4 min =240 s) and resistance (10Ω) which is actually not needed here, we compute the current as follows:
[tex]I=\frac{Q}{t}[/tex]
Then, for the given data, we obtain:
[tex]I=\frac{1200C}{4min}*\frac{1min}{60s}\\\\I=5A[/tex]
Therefore, answer is B 5.0 A .
Best regards!
A child and sled with a combined mass of 53.9 kg slide down a frictionless slope. If the sled starts from rest and has a speed of 5.71 m/s at the bottom, what is the height of the hill
Answer:
1.66m
Explanation:
Using the conservation law
PE = KE
mgh = 1/2mv²
gh = V²/2
g is the acceleration due to gravity = 9.81m/s²
h is the height of the hill
V is the velocity = 5.71m/s
Substitute
9.81h = 5.71²/2
Cross multiply
2×9.81h = 5.71²
19.62h = 32.6041
h = 32.6041/19.62
h = 1.66m
Hence the height of the hill is 1.66m
A particle is moved along the x-axis by a force that measures 10/(1+x)^2 pounds at a point x feet from the origin. Find the work (in ft-lb) done in moving the particle from the origin to a distance of 9 feet.
Answer:
9 ft*lb
Explanation:
super simple but you just have to understand that the integral is going with the curve
work = integral a to b of f(x)dx = integral 0 to 9 of 10/(1+x)^2dx = 9ft*lb
If the power supply is to be made safe by increasing its internal resistance, what should the internal resistance be for the maximum current in the above situation to be Imax
The complete question is;
A person with body resistance between his hands of 10 kΩ accidentally grasps the terminals of a 16-kV power supply. What is the power dissipated in his body?
A) If the internal resistance of the power supply is 1600 Ω , what is the current through the person's body?
B) What is the power dissipated in his body?
C) If the power supply is to be made safe by increasing its internal resistance, what should the internal resistance be for the maximum current in the above situation to be I_max = 1.00mA or less?
Answer:
A) I = 1.379 A
B) P = 19016.41 W
C) r = 15990000 Ω
Explanation:
A) We are given;
Internal resistance of the power supply; r = 1600 Ω
Body resistance between hands; R = 10kΩ = 10000 Ω
Power supply voltage; E =16 kV = 16000 V
Formula for the current through the person's body with internal resistance is given by;
I = E/(R + r)
Thus;
I = 16000/(10000 + 1600)
I = 1.379 A
B) Formula for power dissipated is;
P = I²R
P = 1.379² × 10000
P = 19016.41 W
C) Now, we are told that the maximum current should be I_max = 1.00mA or less. So, I_max = 0.001 A
Thus, from I = E/(R + r) and making r the subject, we have;
r = (E/I) - R
r = (16000/0.001) - 10000
r = 15990000 Ω
If a rock is skipped into a lake at 24 m/s2, with that what force was the rock thrown if it was 1.75kg?
Answer: f= M×A
1.75kg×24= 42N
Explanation:
Because to find force you do Mass times acceleration so I did 1.75 kg times 24 would equal 42 Newtons!
Write a haiku
poem
explaining
why graphing
is useful.
If you are
able, share
your poem
with others.
Answer:
Explanation:
graphing is helpful
helps visualize the line
of your equation
A mountain climber, in the process of crossing between two cliffs by a rope, pauses to rest. She weighs 555 N. As the drawing shows, she is closer to the left cliff than to the right cliff, with the result that the tensions in the left and right sides of the rope are not the same. Find the tension in the rope to the left of the mountain climber.
Complete Question
The diagram for this question is shown on the first uploaded image
Answer:
The tension in the rope on the left of the mountain climber is [tex] T_a = 1106 \ N [/tex]
Explanation:
From the question we are told that
The weight of the mountain climber is m = 555 N
Generally from the diagram , the total amount of force acting on the rope along the vertical axis at equilibrium is mathematically represented as
[tex]T_a* cos 65 -555 + T_b * cos(85) = 0[/tex]
Here [tex]T_a, T_b[/tex] are the tension of the rope on the left and on the right hand side
So
[tex]0.423T_a + 0.0871T_b = 555[/tex]
=> [tex] 0.0871T_b = 555 - 0.423T_a[/tex]
=> [tex] T_b = \frac{555 - 0.423T_a}{0.0871}[/tex]
Generally from the diagram , the total amount of force acting on the rope along the horizontal axis at equilibrium is mathematically represented as
[tex]T_a* sin 65 - T_b * sin(85) = 0[/tex]
=> [tex] 0.9063T_a - 0.9962T_b = 0[/tex]
=> [tex] 0.9063T_a = 0.9962T_b [/tex]
=> [tex] 0.9063T_a = 0.9962[\frac{555 - 0.423T_a}{0.0871}] [/tex]
=> [tex] 0.9063T_a = [\frac{552.891 - 0.421T_a}{0.0871}] [/tex]
=> [tex] 0.0789T_a = [552.891 - 0.421T_a[/tex]
=> [tex] 0.4999T_a = 552.891 [/tex]
=> [tex] T_a = 1106 \ N [/tex]
I WILL MARK YOU AS BRAINLIEST IF RIGHT
What is the magnitude of the net force acting on this object? And what direction?
Answer:
The magnitude of the net force acting on an object is equal to the mass. and the direction is in 20N
Explanation:
An object moving 20 m/s
experiences an acceleration of 4 m/s' for 8
seconds. How far did it move in that time?
Variables:
Equation and Solve:
Answer:
We are given:
initial velocity (u) = 20m/s
acceleration (a) = 4 m/s²
time (t) = 8 seconds
displacement (s) = s m
Solving for Displacement:
From the seconds equation of motion:
s = ut + 1/2 * at²
replacing the variables
s = 20(8) + 1/2 * (4)*(8)*(8)
s = 160 + 128
s = 288 m
PLEASE HELP EASY MULTIPLE CHOICE!!!!!!!!!!!
Answer:
options C is correct
Explanation:
asking questions is super in this education life
Answer:
option c should be the answer
A car traveling at 27 m/s slams on its brakes to come to a stop. It decelerates at a rate of 8 m/s2 . What is the stopping distance of the car?
v² - u² = 2 a ∆x
where u = initial velocity (27 m/s), v = final velocity (0), a = acceleration (-8 m/s², taken to be negative because we take direction of movement to be positive), and ∆x = stopping distance.
So
0² - (27 m/s)² = 2 (-8 m/s²) ∆x
∆x = (27 m/s)² / (16 m/s²)
∆x ≈ 45.6 m
The stopping distance of car achieved during the braking is of 45.56 m.
Given data:
The initial speed of car is, u = 27 m/s.
The final speed of car is, v = 0 m/s. (Because car comes to stop finally)
The magnitude of deacceleration is, [tex]a = 8\;\rm m/s^{2}[/tex].
In order to find the stopping distance of the car, we need to use the third kinematic equation of motion. Third kinematic equation of motion is the relation between the initial speed, final speed, acceleration and distance covered.
Therefore,
[tex]v^{2}=u^{2}+2(-a)s[/tex]
Here, s is the stopping distance.
Solving as,
[tex]0^{2}=27^{2}+2(-8)s\\\\s = 45.56 \;\rm m[/tex]
Thus, we can conclude that the stopping distance of car achieved during the braking is of 45.56 m.
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The steam from a boiling pot of water is
A: conduction
B: Convection
C: radiation
D: Radiant energy
The speed of a car is decreasing from 35 m/s to 15 m/s in 4s
The interaction between electrical energy and magnetism has been an important
topic in 20th century science, Which term describes this interaction?
Answer:
Maybe
Explanation:
I say maybe because it will help them still but not quite
A car starts from rest and accelerates uniformly over a time of 18 seconds for a distance of 390 m. Determine the acceleration of the car.
Answer:
[tex]a=2.4\ m/s^2[/tex]
Explanation:
Given that,
The initial speed of a car, u = 0
Time, t = 18 s
Distance, d = 390 m
We need to find the acceleration of the car. Let it is a. Using the second equation of motion to find it.
[tex]d=ut+\dfrac{1}{2}at^2[/tex]
or
[tex]d=\dfrac{1}{2}at^2\\\\a=\dfrac{2d}{t^2}\\\\a=\dfrac{2\times 390}{(18)^2}\\\\a=2.4\ m/s^2[/tex]
So, the acceleration of the car is [tex]2.4\ m/s^2[/tex].
Calculate the work WC done by the gas during the isothermal expansion. Express WC in terms of p0, V0, and Rv.
Complete Question
The complete question is shown on the first and second uploaded image
Answer:
The expression is [tex]W_c = P_o V_o ln (R_v)[/tex]
Explanation:
Generally smallest workdone done by a gas is mathematically represented as
[tex]dW = PdV[/tex]
Generally for an isothermal process
[tex]PV = nRT = constant [/tex]
=> [tex]P = \frac{nRT}{V}[/tex]
Generally the total workdone is mathematically represented as
[tex]W_c = \int\limits^{v_f}_{V_o} {\frac{nRT}{V} } \, dV[/tex]
=> [tex]W_c = nRT \int\limits^{V_f}_{V_o} {\frac{1}{V} } \, dV[/tex]
=> [tex]nRT [lnV] | \left \ {V_f}} \atop {V_o}} \right.[/tex]
=> [tex]W_c = nRT [ln(V_f) - ln(V_o)][/tex]
=> [tex]W_c = nRT ln \frac{V_f}{V_o}[/tex]
From the question [tex]\frac{V_f}{V_o } = R_v[/tex]
=> [tex]W_c = P Vln (R_v)[/tex]
at initial state
[tex]W_c = P_o V_o ln (R_v)[/tex]
give three factors which are responsible for the vanishing forest
Answer:
1. Huge wildfires
2. Deforestation
3. Reduced amount of aforestation, etc
A block of mass m1 = 18.5 kg slides along a horizontal surface (with friction, μk = 0.22) a distance d = 2.3 m before striking a second block of mass m2 = 7.25 kg. The first block has an initial velocity of v = 8.25 m/s.
Assuming that block one stops after it collides with block two, what is block two's velocity after impact in m/s?
How far does block two travel, d2 in meters, before coming to rest after the collision?
Answer:
19.5 m/s
87.8 m
Explanation:
The acceleration of block one is:
∑F = ma
-m₁gμ = m₁a
a = -gμ
a = -(9.8 m/s²) (0.22)
a = -2.16 m/s²
The velocity of block one just before the collision is:
v² = v₀² + 2aΔx
v² = (8.25 m/s)² + 2 (-2.16 m/s²) (2.3 m)
v = 7.63 m/s
Momentum is conserved, so the velocity of block two just after the collision is:
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
m₁u₁ = m₂v₂
(18.5 kg) (7.63 m/s) = (7.25 kg) v
v = 19.5 m/s
The acceleration of block two is also -2.16 m/s², so the distance is:
v² = v₀² + 2aΔx
(0 m/s)² = (19.5 m/s)² + 2 (-2.16 m/s²) Δx
Δx = 87.8 m
The velocity of block 2 and the distance traveled by it prior to being at rest post-collision are 19.5 m/s and 87.8 m. Check the calculations below:
FrictionGiven that,
[tex]m_{1}[/tex] = 18.5 kg
d = 2.3m
To find,
Acceleration of block 1:
∑[tex]F = ma[/tex]
⇒ -m₁gμ = m₁a
⇒ a = -gμ
⇒ a [tex]= -(9.8 m/s^2) (0.22)[/tex]
∵ a [tex]= -2.16 m/s^2[/tex]
Now,
To determine the velocity of block one prior to the collision:
We know,
The initial velocity of block 1 = 8.25 m/s
⇒ [tex]v^2 = v_{o}^2 + 2[/tex]aΔx
⇒ [tex]v^2 = (8.25 m/s)^2 + 2 (-2.16 m/s^2) (2.3 m)[/tex]
∵ [tex]v = 7.63 m/s[/tex]
We also know,
[tex]m_{2}[/tex] = 7.25 kg
Now,
The velocity of block 2 post collision:
⇒ [tex]m_{1} u_{1} + m_{1} u_{1} = m_{1} v_{1} + m_{2} v_{2}[/tex]post-collision
Through this,
⇒ [tex](18.5 kg) (7.63 m/s) = (7.25 kg) v[/tex]
∵[tex]v = 19.5 m/s[/tex]
The distance can be found through:
⇒ [tex]v^2 = v_{o} ^{2} + 2[/tex][tex]a[/tex]Δ[tex]x[/tex]
⇒ [tex](0 m/s)^2 = (19.5 m/s)^2 + 2 (-2.16 m/s^2)[/tex]Δ[tex]x[/tex]
∵ Δ[tex]x = 87.8 m[/tex]
Thus, 19.5 m/s and 87.8 m are the correct answers.
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g A child bounces a 50 g super ball on the sidewalk. The velocity change of the super bowl is from 27 m/s downward to 17 m/s upward. If the contact time with the sidewalk is 1 800 s, what is the magnitude of the average force exerted on the superball by the sidewalk
Answer:
The average force exerted on the superball by the sidewalk is 0.00122 N.
Explanation:
Given;
mass of the super ball, m = 50 g = 0.05 kg
initial velocity of the super bowl, u = -27 m/s (assuming downward motion to be negative)
final velocity of the super bowl, u = 17 m/s (assuming upward motion to be positive)
time of motion, t = 1800 s
The average force exerted on the superball by the sidewalk is given by;
[tex]F = ma\\\\F = \frac{m(v-u)}{t} \\\\F = \frac{0.05(17-(-27))}{1800}\\\\ F = \frac{0.05(44)}{1800}\\\\F = 0.00122 \ N[/tex]
Therefore, the average force exerted on the superball by the sidewalk is 0.00122 N.
Converting compound units
You would like to know whether silicon will float in mercury and you know that can determine this based on their densities. Unfortunately, you have the density of mercury in units of kilogram/meter3 and the density of silicon in other units: 2.33 gram/centimeter3. You decide to convert the density of silicon into units of kilogram/meter3 to perform the comparison. By which combination of conversion factors will you multiply 2.33 gram/centimeter3 to perform the unit conversion?
Answer:
Explanation:
Given the density of silicon as 2.33g/cm³
We are to convert this to kg/cm³
We will be using the following conversion factors
1000g = 1kg
2.33g = x
Cross multiply
1000x = 2.33
x = 2.33/1000
x = 0.00233kg
Also we need to convert 1cm³ to 1m³
1cm = 0.01m
1cm³ = 0.01×0.01×0.01
1cm³ = 0.000001m³
Substituting into the density value of silicon
2.33g/cm³ = 0.00233kg/0.000001m³
= 2330kg/m³