Answer:
300
Explanation:
450Newton × 2Meter ÷ 3sec
The feeling of weightlessness occurs because _____________________.
there is no supporting force under your mass.
there is no gravity present.
there is only a small amount of gravity present.
Answer:
there is only a small amount of gravity present.
Explanation:
this is because the only force acting upon your body during free fall is the force of gravity which is a non contact force.
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]
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]
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!
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.
Super Mario and Bowser Jr. are racing around a track when Baby Bowser launches a green shell at Mario, bringing him to rest. Bowser Jr. then passes Mario at his top speed of 30 blocks/h, moving down the track in a straight line. Mario quickly accelerates and reaches his top speed of 40 blocks/h in order to catch back up and pass Bowser Jr., but by this point he has opened up a 75 block head start down the track. Mario world measure distance using the units of blocks, with 1 block = 0.47 m.
a) What are Mario and Bowser Jr.'s speeds in m/s?
Assuming both Mario and Bowser Jr. race to the finish in a straight line at their top speeds,
b) How long does it take for Mario to catch Bowser Jr.?
c) How far down the track is Mario from the point at which he reaches his top speed?
Answer:
(a). Mario's speed in m/s = 5.2 × 10^-3 m/s.
Bowser Jr.'s speeds in m/s = 3.92 × 10^-3 m/s.
(b). 27001.2 seconds(s)..
(c). 141 metre(m).
Explanation:
So, the following data or parameters or information was given in the question above. These informations are going to help us in solving this question or problem;
=>" Bowser Jr. then passes Mario at his top speed = 30 blocks/h.
=> " Mario quickly accelerates and reaches his top speed of 40 blocks/h in order to catch back up and pass Bowser Jr., but by this point he has opened up a 75 block head start down the track."
=> "Mario world measure distance using the units of blocks, with 1 block = 0.47 m"
Therefore, the solution is given below;
(1). For the first part, we are to determine or calculate Mario and Bowser Jr.'s speeds in m/s.
Therefore, Mario's speed in m/s = 40 × 0.47) ÷ 3600 = 5.2 × 10^-3 m/s.
Also, Bowser Jr.'s speeds in m/s = ( 30 × 0.47) ÷ 3600 = 3.92 × 10^3 m/s.
(2). So, the next thing to do now.is to determine or calculate how long it took for Mario to catch Bowser Jr.
Thus, the time it took for Mario to catch Bowser Jr. Can be related as below;
[ ( 5.2 × 10^-3 m/s) - (3.92 × 10^-3 m/s) × (time,t taken for Mario to catch Bowser Jr.) = 75 × 0.47.
Therefore, the time it took for Mario to catch Bowser Jr. = 27001.2 seconds.
(3). Now, we calculate How far down the track Mario from the point at which he reaches his top speed.
The distance = 5.2 × 10^-3 m/s × 27001.2m = 141m
A 849-kg car starts from rest on a horizontal roadway and accelerates eastward for 5.00 s when it reaches a speed of 35.0 m/s. What is the average force exerted on the car during this time
Answer:
The average force exerted on the car during this time is 5,943 N
Explanation:
Given;
mass of the car, m = 849 kg
initial velocity of the car, u = 0
time of motion of the car, t = 5.00 s
final velocity of the car, v = 35 m/s
The average force exerted on the car during this time is given by;
[tex]F = ma \\\\F = \frac{m(v-u)}{t}\\\\F = \frac{849(35-0)}{5}\\\\F = \frac{849(35)}{5}\\\\ F = 849*7\\\\F = 5,943 \ N[/tex]
Therefore, the average force exerted on the car during this time is 5,943 N
Answer:
5943N
Let's say (+x) = eastward
Average horizontal acceleration
ax = vx -v0x/5.00s
= 35.0m/s-0/5.00s
= +7.09m/s
From here we apply the second law of newton
During this period average horizontal force acting on car
Summation x = max = (849kg)(+7.09m/s²)
= 5943N
+5.943x10³N
= 5.94kN east ward.
The precision value of measuring tape is
1)0.1cm
2)0.1mm
3)1cm
4)0.01cm
C.1cm
Explanation:
precision is how close two or more measurements are to each other
4?
Explanation:
sorry im not sure but
You can always take a metre ruler as a starting point. Your metre ruler has the same precision as your 15.0cm or 30.0cm ruler, so bring a ruler during exams as they'll come in handy ;)
The order goes like this:
rulers: 0.1cm or 1mm
measuring tape: 0.01cm or 0.1mm
vernier calipers: 0.001cm or 0.01mm
micrometer screw gauge: 0.0001cm or 0.001mm
((if i'm not wrong))
A single-turn circular loop of radius 9.4 cm is to produce a field at its center that will just cancel the earth's field of magnitude 0.7 G directed at 70degrees below the horizontal north direction. Find the current in the loop.
Answer:
The current in the loop is 10.5 A.
Explanation:
Given that,
Radius = 9.4 cm
Magnetic field = 0.7 G
Angle = 70°
We know that,
The magnetic field due to the current in a loop is
[tex]B_{I}=\dfrac{\mu_{0}NI}{2r}[/tex]
The magnetic field due to the current is equal to the magnetic field of earth.
[tex]B_{E}=B_{I}=\dfrac{\mu_{0}NI}{2r}[/tex]
We need to calculate the current in the loop
Using formula of magnetic field
[tex]B=\dfrac{\mu_{0}NI}{2r}[/tex]
[tex]I=\dfrac{2rB}{\mu_{0}N}[/tex]
Put the value into the formula
[tex]I=\dfrac{2\times9.4\times10^{-2}\times0.7\times10^{-4}}{4\pi\times10^{-7}\times1}[/tex]
[tex]I=10.5\ A[/tex]
Hence, The current in the loop is 10.5 A.
A cannonball is fired horizontally from a 10 m high cliff at 20 m/s. How long will the cannonball be in the air? How far away will the cannonball strike the ground?
Answe ¡Buenos días! –
#3 ¡Buenas tardes! –
#4 ¡Bienvenid
What resistance must be connected in parallel with a 633-Ω resistor to produce an equivalent resistance of 205 Ω?
Answer:
303 Ω
Explanation:
Given
Represent the resistors with R1, R2 and RT
R1 = 633
RT = 205
Required
Determine R2
Since it's a parallel connection, it can be solved using.
1/Rt = 1/R1 + 1/R2
Substitute values for R1 and RT
1/205 = 1/633 + 1/R2
Collect Like Terms
1/R2 = 1/205 - 1/633
Take LCM
1/R2 = (633 - 205)/(205 * 633)
1/R2 = 428/129765
Take reciprocal of both sides
R2 = 129765/428
R2 = 303 --- approximated
The steam from a boiling pot of water is
A: conduction
B: Convection
C: radiation
D: Radiant energy
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]
A water balloon launcher with a mass of 2.2 kg is suspended on a wire. It fires a 0.85 kg balloon to the north at a velocity of 13.0 m/s. What is the resulting velocity of the launcher if the net force on the launcher is equal to the reaction force?
Answer:
5.0 m/s south
Hope this Helps!
Answer:
5.0 m/s south
Explanation:
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
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.
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 crane uses a single cable to lower a steel girder into place. The girder moves with constant speed. The cable tension does work WT and gravity does work WG. Which statement is true
Explanation:
Work done by a force is given by :
[tex]W=Fd\cos\theta[/tex]
Where
F is force, d is displacement and [tex]\theta[/tex] is the angle between F and d.
In this problem, a crane is moving in downward direction, the force gravity is in downward direction and the tension is in upward direction.
We know that if force and displacement is in same direction, work is positive while if force and displacement is in oposite direction, work is negative.
I would mean that, [tex]W_g[/tex] is positive, because gravity is parallel to the displacement and [tex]W_t[/tex] is negative, because the tension is opposite to the displacement.
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].
the neuron is considered a (a. Cell. (B.artery. (C. Vein
Answer:
A Cell
Explanation:
A 1870 kg car traveling at 13.5 m/s collides with a 2970 kg car that is initally at rest at a stoplight. The cars stick together and move 1.93 m before friction causes them to stop. Determine the coefficient of kinetic friction between the cars and the road, assuming that the negative acceleration is constant and all wheels on both cars lock at the time of impact.
Answer:
The value is [tex] \mu = 0.72 [/tex]
Explanation:
From the question we are told that
The mass of the first car is [tex]m_1 = 1870\ kg[/tex]
the initial speed of the car is [tex]u = 13.5 \ m/s[/tex]
The mass of the second car is [tex]m_2 = 2970\ kg[/tex]
The distance move by both cars is s = 1.93 m
Generally from the law of momentum conservation
[tex]m_1 * u_1 + m_2 * u_2 = (m_1 + m_2 ) * v_f[/tex]
Here [tex]u_2 = 0[/tex] because the second car is at rest
and [tex]v_f[/tex] is the final velocity of the the two car
So
[tex]1870* 13.5+ 0= ( 1870 + 2970 ) * v_f[/tex]
=> [tex]v_f = 5.22\ m/s[/tex]
Generally from kinematic equation
[tex]v_f^2 = u_2^2 + 2as[/tex]
here a is the deceleration
So
[tex]5.22^2 = 0 + 2 *a * 1.93[/tex]
=> [tex]a = 7.06 \ m/s^2 [/tex]
Generally the frictional force is equal to the force propelling the car , this can be mathematically represented as
[tex]F_f = F[/tex]
Here F is mathematically represented as
[tex]F = (m_1 + m_2) * a[/tex]
[tex]F = (1870 + 2970) * 7.06 [/tex]
[tex]F =34170.4 \ N[/tex]
and
[tex]F_f = \mu * (m_1 + m_2 ) * g[/tex]
[tex]F_f = 47432 * \mu [/tex]
So
[tex] 47432 * \mu = 34170.4 [/tex]
=> [tex] 47432 * \mu = 34170.4 [/tex]
=> [tex] \mu = 0.72 [/tex]
give three factors which are responsible for the vanishing forest
Answer:
1. Huge wildfires
2. Deforestation
3. Reduced amount of aforestation, etc
A uniform magnetic field of magnitude 0.72 T is directed perpendicular to the plane of a rectangular loop having dimensions 8.2 cm by 14 cm. Find the magnetic flux through the loop.
Answer:
Explanation:
Magnetic flux is expressed as the product of magnetic field and cross sectional area.
Φ = BAsintheta
Given
B = 0.72T
A = 8.2cm×14cm
A = 0.082m × 0.14m
Area = 0.01148m²
Theta = 90°
Substitute into the formula
Φ = BAsintheta
Φ = 0.72(0.01148)sin90°
Φ = 0.72(0.01148)(1)
Φ = 0.0082656
Hence the magnetic flux through the loop is 8.2656 × 10^-3 Weber
The speed of a car is decreasing from 35 m/s to 15 m/s in 4s
"What will the pressure inside the container become if the piston is moved to the 1.60 L mark while the temperature of the gas is kept constant?"
This question is incomplete, the complete question is;
The Figure shows a container that is sealed at the top by a moveable piston, Inside the container is an ideal gas at 1.00 atm. 20.0°C and 1.00 L.
"What will the pressure inside the container become if the piston is moved to the 1.60 L mark while the temperature of the gas is kept constant?"
Answer:
the pressure inside the container become 0.625 atm if the piston is moved to the 1.60 L mark while the temperature of the gas is kept constant
Explanation:
Given that;
P₁ = 1.00 atm
P₂ = ?
V₁ = 1 L
V₂ = 1.60 L
the temperature of the gas is kept constant
we know that;
P₁V₁ = P₂V₂
so we substitute
1 × 1 = P₂ × 1.60
P₂ = 1 / 1.60
P₂ = 0.625 atm
Therefore the pressure inside the container become 0.625 atm if the piston is moved to the 1.60 L mark while the temperature of the gas is kept constant
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
Chris races his Audi north down a road for 1000 meters in 20 seconds, what is his velocity?
Answer:
I think it would be 50 I am not really sure
Explanation:
I think you would have to divid 1000 by 20 Again I'm not sure
Design a tension member and slip-critical splice to carry a factored load of 500 kips. Please use a wide-flange section for the tension member. Please use A572 Gr. 50 steel plates for the splice plates. Please use Group B, A490 bolts for the splice connection. The splice connection should be slip-critical, and have adequate strength after slip occurs as well. Please make any other assumptions you need in order to complete the problem. Provide detailed sketches and drawings for your design.
Answer:
Kindly check the explanation section.
Explanation:
For the design we are asked for in this question/problem there is the need for us to calculate or determine the strength in fracture and that of the yield. Also, we need to calculate for the block shear strength.
From the question, we have that the factored load = 500kips. Also, note that the tension splice must not slip.
Also, the shear force are resisted by friction, that is to say shear resistance = 1.13 × Tb × Ns.
Assuming our db = 3/4 inches, then the slip critical resistance to shear service load = 18ksi(refer to AISC manual for the table).
If db = 7/8 inches, then the shear force resistance for n bolt = 10.2kips, n > 49.6.
The yielding strength = 0.9 × Aj × Fhb= 736 kips > 500
The fracture strength = .75 × Ah × Fhb = 309 kips.
The bearing strength of 7/8 inches bolt at the edge hole and other holes = 46 kips and 102 kips.
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 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]