The steam from a boiling pot of water is
A: conduction
B: Convection
C: radiation
D: Radiant energy
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!
The speed of a car is decreasing from 35 m/s to 15 m/s in 4s
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
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
Light is described as having a dual wave-particle nature. Which piece of evidence provides support for the model of light as a particle?
. Young’s double slit experiment showed that light waves show interference.
. Light reflects when it hits a surface.
. Light refracts when it moves from one medium to another.
. Light does not need a medium to travel.
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 :]
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
A balloon is launched straight upward from the edge of a 40 m high cliff with a velocity of 30
m/s. The balloon goes up and barely misses the edge of the cliff on the way down and hits the
ground below.
A.) Determine the overall hang time of the balloon.
B.) How high did the balloon travel above the cliff?
C.) Determine the velocity of the balloon at t = 3 s.
D.) Determine the acceleration of the balloon at t = 3 s.
Answer:
A) t = 7.25 sec
B) 45.92 m above the cliff and 85.92 m counting from ground level
C) about 0.6 m/s
D) The only acceleration is always that of gravity : 9.8 m/s^2
Explanation:
The kinematic equation for the position as a function of time is:
y = 40 + 30 * t - (g/2) t^2
The total time the balloon was in the air is calculated for when the final position is y = 0 (when it touches the ground), that is:
0 = 40 + 30 * t - (g/2) t^2
0 = 40 +30 t - 4.9 t^2
which can be solved for "t" using the quadratic formula, and which renders two different times "t", of which we pick the positive answer: t = 7.25 sec.
For the maximum height, we estimate the time it takes for the balloon to reach the maximum height at which the velocity is zero (changes direction of motion):
vf = vi - g * t
0 = 30 - 9.8 * t
t = 30 / 9.8 = 3.06 seconds
Now we use this value in our position equation and get:
ymax = 40 + 30 (3.06) - 4.9 (3.06)^2
ymax = 85.92 m (from ground level
Therefore, from the cliff (that is at 40 m height) the height is 85.92 - 40 = 45.92 m.
The velocity of the balloon at t = 3 seconds can be calculated with the velocity expression we have been using:
v(t) = vi - g * t
v(3) = 30 - 9.8 (3) = 0.6 m/s
The only acceleration acting on the balloon is always the acceleration due to gravity (g = 9.8 m/s^2)
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.
A car with a mass of 1500kg is traveling at a speed of 30m/s. What force must be applied to stop the car in 3 seconds?
Answer:
The answer is 15,000 NExplanation:
To find the force given the mass , velocity and time can be found by using the formula
[tex]f = \frac{m \times v}{t} \\ [/tex]
where
m is the mass
v is the velocity
t is the time
From the question
m = 1500 kg
v = 30 m/s
t = 3 s
We have
[tex]f = \frac{1500 \times 30}{3} = \frac{45000}{3} \\ [/tex]
We have the final answer as
15,000 NHope this helps you
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.
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]
Open Box. Consider a hollow box with the top missing. The sides have negligible thickness and each has length L and mass m. (a) Find the x-coordinate of the center of mass.
Answer:
x_{cm} = L / 2
Explanation:
The center of mass is defined by
[tex]x_{cm}[/tex] = 1 / M ∑ m_{i} x_{i}
where M is the total mass of the system
in this case the system is continuous so, for which we use the density
ρ = dm / dx
dm = ρ dx
substituting
x_{cm} = 1 / M ∫ x ρ dx
x_{cm} = ρ / M ∫ x dx
we integrate and evaluate from x = 0 to x = L
x_{cm} = ρ / M (L² /2 -0)
we introduce the density which is constant
ρ = M / L
x_{cm} = 1 /M (M/L) L² / 2
x_{cm} = L / 2
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
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].
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]
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:
the neuron is considered a (a. Cell. (B.artery. (C. Vein
Answer:
A Cell
Explanation:
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
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).
A cyclist and his bicycle have a combined mass of 88 kg and a combined
weight of 862.4 N. The cyclist accelerates at 1.2 m/s2. After 2 seconds he
reaches a speed of 2.4 m/s. What is his momentum at this point?
A. 36.7 kg m/s
B. 359.3 kg:m/s
C. 105.6 kg-m/s
D. 211.2 kg:m/s
The cyclist accelerates at 1.2 m / s² after 2 seconds he reaches a speed of 2.4 m / s, then the momentum at this point would be 211.2 kg-m/s, therefore the correct answer is option D.
What is momentum?It can be defined as the product of the mass and the speed of the particle, it represents the combined effect of mass and the speed of any particle, and the momentum of any particle is expressed in Kg m/s unit.
As given in the problem a cyclist and his bicycle has a combined mass of 88 kg and a combined weight of 862.4 N. The cyclist accelerates at 1.2 m/s2. After 2 seconds he reaches a speed of 2.4 m/s.
The momentum of the cyclist = 88 × 2.4
= 211.0 kgm/s
Thus, the momentum of the cyclist would be 211.0 kgm/s.
To learn more about momentum from here, refer to the link given below;
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An increase in temperature the kinetic energy and average speed of the gas particles. As a result, the pressure on the walls of the container . Answer Bank What temperature must a gas, initially at 10 ∘C, be brought to for the pressure to triple?
Answer:
a
The pressure will increase
b
[tex]T_2 = 576^oC[/tex]
Explanation:
From the ideal gas law we have that
[tex]PV = nRT[/tex]
We see that the temperature varies directly with the pressure so if there is an increase in temperature that pressure will increase
The initial temperature is [tex]T_i = 10^oC = 10 + 273 = 283 \ K [/tex]
The objective of this solution is to obtain the temperature of the gas where the pressure is tripled
Now from the above equation given that nR and V are constant we have that
[tex]\frac{P}{T} = constant[/tex]
=> [tex]\frac{P_1}{T_1} =\frac{P_2}{T_2}[/tex]
Let assume the initial pressure is [tex]P_1 = 1 Pa[/tex]
So tripling it will result to the pressure being [tex]P_2 = 3 Pa[/tex]
So
[tex]\frac{1}{283} =\frac{3}{T_2}[/tex]
=> [tex]T_2 = 3 * 283[/tex]
=> [tex]T_2 = 3 * 283[/tex]
=> [tex]T_2 = 849 \ K [/tex]
Converting back to [tex]^oC[/tex]
[tex]T_2 = 849 - 273[/tex]
=> [tex]T_2 = 576^oC[/tex]
give three factors which are responsible for the vanishing forest
Answer:
1. Huge wildfires
2. Deforestation
3. Reduced amount of aforestation, etc
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
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]
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.
Learn more about Density here:
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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]
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.
Learn more about the kinematic equation of motion here:
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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
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