Answers
Answer:
Resistance = 0.22 Ohms
Current = 13.63636 A
Explanation:
Total resistance for resistors in parallel is given by:
[tex]\frac{1}{T} =\frac{1}{R1} +\frac{1}{R2} +...+\frac{1}{Rn}[/tex] where n is the number of resistors
[tex]\frac{1}{T} = \frac{1}{1.1} +\frac{1}{1.1} +\frac{1}{1.1} +\frac{1}{1.1} +\frac{1}{1.1}[/tex]
if you solve that you get [tex]\frac{1}{T} = 5/1.1 \\\\T = 1.1/5T = 0.22 Ohms[/tex]
Solve current using V=IR
I=V/R =
I=3/0.22
I = 13.63636 A
Related Questions
Car X is travelling at 30m/s north. Its driver looks at car Y approaching on another road and he estimates it is moving at 15m/s south-west relative to his car. Calculate the velocity of car Y relative to the ground.
Answers
Answer: 22.1 m/s
Explanation:
The velocity of Car traveling 30 m/s towards the north
In vector form it is
[tex]v_x=30\hat{j}[/tex]
The velocity of car Y w.r.t X is
[tex]\Rightarrow v_{yx}=15[-\cos 45^{\circ}\hat{i}-\sin 45^{\circ}\hat{j}][/tex]
Solving this
[tex]\Rightarrow v_{yx}=v_y-v_x\\\Rightarrow v_y=v_{yx}+v_x[/tex]
putting values
[tex]\Rightarrow v_y=15[-\cos 45^{\circ}\hat{i}-\sin 45^{\circ}\hat{j}]+30\hat{j}[/tex]
[tex]\Rightarrow v_y=-10.606\hat{i}+19.39\hat{j}[/tex]
absolute velocity relative to ground is
[tex]\left | v_y\right |=\sqrt{(-10.606)^2+(19.39)^2}\\\left | v_y\right |=22.101\ m/s[/tex]
The Sun is divided into three regions.
True оr False?
Answers
Answer:
false I think
Explanation:
hope that help
so it's not divided in 3 regions
A truck with a mass of 1370 kg and moving with a speed of 12.0 m/s rear-ends a 593 kg car stopped at an intersection. The collision is approximately elastic since the car is in neutral, the brakes are off, the metal bumpers line up well and do not get damaged. Find the speed of both vehicles after the collision in meters per second.
Answers
Answer:
speed of car after collision, v2 =16.1 m/s and of the truck, v1 = 4.6 m/s
Explanation:
Given:
mass of truck M = 1370 kg
speed of truck = 12.0 m/s
mass of car m = 593 kg
collision is elastic therefore,
Applying law of momentum conservation we have
momentum before collision = momentum after collision
1370×12 + 0( initially car is at rest) = 1370×v1+ 593×v2 ....(i)
Also for a collision to be elastic,
velocity of approach = velocity of separation
12 -0 = v2-v1 ....(ii)
using (i) and (ii) we have
So speed of car after collision, v2 =16.1 m/s and of the truck, v1 = 4.6 m/s
A metal pot feels hot to the touch after a short time on the shove. what type of material is the metal pot
Answers
What is your hypothesis (or hypotheses) for this experiment?
(about Thermal Energy Transfer)
Answers
Answer:
I hypothesis that the motion involving the balls in the experiment were moving to create data.
Explanation:
I hope this helps!
In an effort to be the star of the half-time show, the majorette twirls a highly unusual baton made up of four mases fastened to the ends of light rods. Each rod is 1.0 m lone. Find the moment of inertia of the system about an axis perpendicular to the page and passing through the point where the rods cross.
Answers
Answer:
"0.25 kg-m²" is the appropriate answer.
Explanation:
The diagram of the question is missing. Find the attachment of the diagram below.
According to the diagram, the values are:
m₁ = 0.2
m₂ = 0.3
m₃ = 0.3
m₄ = 0.2
d₁ = d₂ = d₃ = d₄ = 0.5 m
As we know,
The moment of inertia is:
⇒ [tex]I=\Sigma M_id_i^2[/tex]
then,
⇒ [tex]I=m_1d_1^2+m_2d_2^2+m_3d_3^2+m_4d_4^2[/tex]
⇒ [tex]=d^2(m_1+m_2+m_3+m_4)[/tex]
On substituting the values, we get
⇒ [tex]=0.5^2\times (0.2+0.3+0.3+0.2)[/tex]
⇒ [tex]=0.25\times 1[/tex]
⇒ [tex]=0.25 \ Kg-m^2[/tex]
Jim and Sally both do identical jobs. Jim works quickly while Sally works slowly. Which of the following is true?
A) Sally uses more energy.
B) Jim uses more energy.
C) Jim uses more power.
D) Sally uses more power.
Answers
A typical laboratory centrifuge rotates at 4000 rpm. Testtubes have to be placed into a centrifuge very carefully because ofthe very large accelerations.
Part A) What is the acceleration at the end of a test tubethat is 10 cm from the axis of rotation?
Part B) For comparison, what is the magnitude of theacceleration a test tube would experience if dropped from a heightof 1.0 m and stopped in a 1.0-ms-long encounter with a hardfloor?
Answers
Answer:
A) a_c = 1.75 10⁴ m / s², B) a = 4.43 10³ m / s²
Explanation:
Part A) The relation of the test tube is centripetal
a_c = v² / r
the angular and linear variables are related
v = w r
we substitute
a_c = w² r
let's reduce the magnitudes to the SI system
w = 4000 rpm (2pi rad / 1 rev) (1 min / 60s) = 418.88 rad / s
r = 1 cm (1 m / 100 cm) = 0.10 m
let's calculate
a_c = 418.88² 0.1
a_c = 1.75 10⁴ m / s²
part B) for this part let's use kinematics relations, let's start looking for the velocity just when we hit the floor
as part of rest the initial velocity is zero and on the floor the height is zero
v² = v₀² - 2g (y- y₀)
v² = 0 - 2 9.8 (0 + 1)
v =√19.6
v = -4.427 m / s
now let's look for the applied steel to stop the test tube
v_f = v + a t
0 = v + at
a = -v / t
a = 4.427 / 0.001
a = 4.43 10³ m / s²
Most stars are _____ compared to the sun.
slightly smaller
much bigger
much smaller
slightly bigger
Answers
plz help me with my career!!!
part one...
Answers
Answer:
#1 Yes
Explanation: #1: The rest of them are used mainley by farmers, and crops are used by common citizens in the world.
Question 1: Crops.
Question 2: Diagnostic Services.
Question 3: A cable company needs to lay new fiber optic cable to reach its customers across a large lake.
Question 4: A bachelor's degree in energy research.
Question 5: Environmental Resources.
If any of these answers are incorrect, please tell me, so I can fix my mistake. Thank you.
Part D Here is one last question as a final check on your understanding of your work for this problem, looking at this problem as an example of the Conservation of Energy. The action in this problem begins at location A , with the block resting against the uncompressed spring. The action ends at location B, with the block moving up the ramp at a measured speed of 7.35 m/s . From A to B, what has been the work done by non-conservative forces, and what has been the change in the mechanical energy of the block-Earth system (the ramp is a part of the Earth)
Answers
Answer:
The answer is "39.95 J".
Explanation:
Please find the complete question in the attached file.
[tex]\to W_{AC}=(\mu \ m \ g \ \cos \theta ) d[/tex]
[tex]=(0.45 \times 1.60 \times 9.8 \times \cos 26^{\circ}) 6.30 \\\\=(7.056 \times \cos 26^{\circ}) 6.30 \\\\=6.34189079\times 6.30\\\\=39.95 \ J\\\\[/tex]
[tex]\therefore \\\\\bold{\Delta E =39.95 \ J}[/tex]
A spring in a toy gun has a spring constant of 10 N/m and can be compressed 4 cm.
It is then used to shoot a 1 g ball out of the gun. Find the velocity of the ball as it
leaves the gun
Answers
We know from this that a=10,000m/s2
V=at
X=vt
You end up with v^2=ax
Plug in 10,000 and 0.04 and solve for v =20m/s
Sound waves rely on matter to transmit their energy. They cannot ravel in a vacuum. True or false
Answers
Answer:
false
Explanation:
Hello,
QUESTION)True,Sound travels in a material medium, in space, there is no matter, so sound cannot propagate.
why do the stars rotate
Answers
Answer:
Angular momentum
Explanation:
Stars are formed as a result of a collapse of a low-temperature cloud of gas and dust. During the colapse conservation of angular momentum causes any small net rotation of the cloud to increase thus forcing the material into rotating
A ball is dropped off the side of a bridge,
After 1.55 S, how far has it fallen?
(Unit=m)
Answers
Answer:
Distance S = 11.77 m (Approx.)
Explanation:
Given:
Time t = 1.55 Second
Gravity acceleration = 9.8 m/s²
Find:
Distance S
Computation:
S = ut + (1/2)(g)(t)²
S = (0)(1.55) + (1/2)(9.8)(1.55)²
S = (0)(1.55) + (1/2)(9.8)(1.55)²
Distance S = 11.77 m (Approx.)
A continental polar air mass forms in
a. the Pacific Ocean.
b. northern Canada.
c.
the Gulf of Mexico.
the desert Southwest.
d.
Please select the best answer from the choices provided
А
B
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С
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my uq xbbbnxnjjxjxusjhhhwhhhnn he c x. Yes suhsjjdhhehy yes eirui
M I was going ask m
Iqijjm.ndjbh
Answers
Answer:
B. Northern Canada
Explanation:
A continental polar air mass can form over the land during the winter months. In the Northern Hemisphere, it originates in northern Canada or Alaska. As it moves southward, it brings dry weather conditions to the United States. Temperature and humidity levels are both low. Hope this helps :)
A 40.0 kg block of lead is heated from -25°C to 200°C.
How much heat is absorbed by the lead block?
A. 2,354,000 J
B. 1,170,000 J
C. 56,891 J
D. 10,650 J
Answers
Answer:
B. 1,170,000 J
Explanation:
Given;
mass of lead block, m = 40 kg
initial temperature, t₁ = -25 ⁰C
final temperature, t₂ = 200 ⁰C
The heat absorbed the lead block is calculated as;
H = mcΔt
where;
c is the specific heat capacity of lead = 130 J/kg⁰C
H = 40 x 130 x (200 - (-25))
H = 40 x 130 x (200 + 25)
H = 40 x 130 x 225
H = 1,170,000 J
Therefore, the heat absorbed the lead block is 1,170,000 J
Select the correct answer.
In general, how does an increase in distance from the Sun affect a planet?
ОА. .
The amount of sunlight the planet receives decreases.
OB.
The gravitational force exerted by the Sun increases.
ОС.
The thickness of a planet's atmosphere increases.
OD. The planet's atmosphere absorbs more heat.
Answers
Answer:
a.
gravity decreases by distance
thickness doesn't change
Less heat.
Fifty grams of ice at 0◦ C is placed in a thermos bottle containing one hundred grams of water
at 6◦ C. How many grams of ice will melt? The heat of fusion of water is 333 kJ/kg and the
specific heat is 4190 J/kg · K.Immersive Reader
Answers
Answer:
7.55 g
Explanation:
Using the relation :
Δt = temperature change = (6° - 0°) = 6°
Q = quantity of heat
C = specific heat capacity = 4190 j/kg/k
1000 J = 1kJ
333 KJ = 333000 j
The quantity of ice that will melt ;
= 0.419 * 6 * 100 / 333000
= 2514000 / 333000
= 7.549 g
The mass of ice that will melt :
2.514 / 0.333
= 7.549 g
Build a second circuit with a battery and a light bulb but this time add a switch. Your circuit might look something like the one at right. When a switch is open in a circuit, it means the two ends are disconnected and current cannot flow between them. When a switch is closed in a circuit, it means the two ends are connected and current can flow between them. Play with the switch to check how it affects the flow of current. With the switch closed, compare the brightness of the bulb and the flow of current in this circuit, with that of your first circuit. Did increasing the length of wire in the circuit change the brightness of the bulb or the curr
Answers
Answer:
the resistance of the wire has no effect on the brightness of the bulb.
Explanation:
Let's apply ohm's law for your light bulb circuit plus wires plus switch
V = I R_{bulb} + I R_ {wire}
the current in a series circuit is constant
V = I (R_{bulb} + R_{wire})
To know the effect of the wires on the brightness of the bulb, we must look for the value of the typical resistance of these elements.
Incandescent bulb
Power 60 W
let's use the power ratio
P = V I = V2 / R
R = V2 / P
the voltage value for this power is V = 120 V
R = 120 2/60
R_bulb = 240 Ω
Resistance of a 14 gauge copper wire (most used), we look for it on the internet
R = 8.45 Ω/ km
in a laboratory circuit approximately 2 m is used, so the resistance of our cable is
R = 8.45 10⁻³ 2
R_wire = 0.0169 Ω
let's buy the two resistors
R_{bulb} = 240
R_{wire} = 0.0169
[tex]\frac{R_{bulb} }{R_{wire} } = \frac{240 }{ 0.0169}[/tex]
[tex]\frac{ R_{bulb} }{ R_{wire} } = 1.4 \ 10^4[/tex]
therefore resistance of the bulb is much greater than that of the wire, therefore almost all the power is dissipated in the bulb.
In summary, the resistance of the wire has no effect on the brightness of the bulb.
Help me with both questions please?
Answers
Answer:
1. They all accelerate at the same rate.
2.The object travels at a constant velocity throughout the fall.
Explanation:
Earths gravitational pull is at a constant 9.08 m/s^2. so when objects are free falling, the objects in question can only fall so fast before it would break gravity so to speak.
Which DOES NOT happen during the phase change from liquid to solid?
Kinetic energy decreases
Particles move slower
Freezing occurs
Temperature increases
Answers
Answer:
Temperature increases
During the phase transition from liquid to solid, kinetic energy diminishes, particles move slower, and freezing occurs.
Phase change:Extreme heat is more likely to occur as a result of rising temperatures, and it will last longer. Heatwaves can indeed be deadly, resulting in manifestations including heat kinks and heat exhaustion, as well as death.
Warmer temperatures have the potential to trigger a cascade of other changes around the world.The greenhouse effect is worsened by these gas emissions, which cause the earth's crust temperature to rise. Burning fossil fuels has the greatest impact on climate change of any human activity.Since in changement from liquid to solid the temperature will decrease.Therefore, the final answer is "Temperature increases".
Find out more about the phase change here:
brainly.com/question/11490613
Read the following statements. Which statement provides the correct definition of weather? Select your answer from the options below. * 1 point
Atmospheric conditions at a specific time in a specific area
Atmospheric conditions over a period of time in a specific area
Average atmospheric conditions over a period of time in a region
Average atmospheric conditions across the world
Answers
Answer:
In one day, a store sells 14 pairs of jeans. The 14 jeans represent 20% of the total number of items sold that day. How many items did the store sell in one day? Explain or show how you got your answer.
plz help with the qsn
Explanation:
Let's assume raspberries are 10 wt% protein solids and the remainder water. When making jam, raspberries are crushed and mixed with sugar, in a 45:55 berry to sugar ratio, by mass. Afterward, the mixture is heated, boiling off water until the remaining mixture is 0.4 weight fraction water, resulting in the final product, jam. How much water, in kilograms, is boiled off per kilogram of raspberries processed
Answers
Answer:
The mass of water boiled off is [tex]0.0 \overline{185}[/tex] kg
Explanation:
The given percentage by weight of protein solids in raspberries = 10 weight%
The ratio of sugar to raspberries in ja-m = 45:55
The mass of the mixture after boiling = 0.4 weight fraction water
Let 's' represent the mass of sugar in the mixture, and let 'r' represent the mass of raspberry
The mass of raspberry, r = 1 kg
The percentage by weight of water in raspberry = 90 weight %
The mass of water in 1 kg of raspberry = 90/100 × 1 kg = 0.9 kg
The ratio of the mass of sugar to the mass of raspberry in jam = r/s = 45/55
∴ s = 1 kg × 55/45 = 11/9 kg
The mass of the mixture before boiling = 1 kg + 11/9 kg = 20/9 kg
The weight fraction of water in the remaining mixture after boiling = 0.4 weight fraction
Let 'w' represent the mass of water boiled off, we have;
(0.9 - w)/(20/9 - w) = 0.4
(0.9 - w) = 0.4 × (20/9 - w)
0.9 - w = 8/9 - 0.4·w
9/10 - 8/9 = w - 0.4·w = 0.6·w = (6/10)·w
(81 - 80)/(90) = (6/10)·w
1/90 = (6/10)·w
w = ((10/6) × 1/90) = 1/54
w = 1/54
The mass of water boiled off, w = (1/54) kg = [tex]0.0 \overline{185}[/tex] kg
explain why radiation is dangerous for humans
Answers
Answer:
because it affects the attom in living things
A cat dozes on a stationary merry-go-round, at a radius of 7.0 m from the center of the ride. The operator turns on the ride and brings it up to its proper turning rate of one complete rotation every 6.9 s. What is the least coefficient of static friction between the cat and the merry-go-round that will allow the cat to stay in place, without sliding
Answers
Answer:
0.6
Explanation:
Given that :
Radius, R = 7m
Period, T = 6.9s
The Coefficient of static friction, μs can be obtained using the relation :
μs = v² / 2gR
Recall, v = 2πR/T
μs becomes ;
μs = (2πR/T)² / 2gR
μs = (4π²R² / T²) ÷ 2gR
μs = (4π²R² / T²) * 1/ 2gR
μs = 4π²R / T²g
μs = 4π²*7 / 6.9^2 * 9.8
μs = 28π² / 466.578
μs = 276.34892 / 466.578
μs = 0.5922887
μs = 0.6
Attempt 2 You have been called to testify as an expert witness in a trial involving a head-on collision. Car A weighs 15151515 lb and was traveling eastward. Car B weighs 11251125 lb and was traveling westward at 42.042.0 mph. The cars locked bumpers and slid eastward with their wheels locked for 17.517.5 ft before stopping. You have measured the coefficient of kinetic friction between the tires and the pavement to be 0.7500.750 . How fast (in miles per hour) was car A traveling just before the collision
Answers
Answer:
v = 28.98 ft / s
Explanation:
For this problem we must solve it in parts, let's start by looking for the speed of the two cars after the collision
In the exercise they indicate the weight of each car
Wₐ = 1500 lb
W_b = 1125 lb
Car B's velocity from v_b = 42.0 mph westward, car A travels east
let's find the mass of the vehicles
W = mg
m = W / g
mₐ = Wₐ / g
m_b = W_b / g
mₐ = 1500/32 = 46.875 slug
m_b = 125/32 = 35,156 slug
Let's reduce to the english system
v_b = 42.0 mph (5280 foot / 1 mile) (1h / 3600s) = 61.6 ft / s
We define a system formed by the two vehicles, so that the forces during the crash have been internal and the moment is preserved
we assume the direction to the east (right) positive
initial instant. Before the crash
p₀ = mₐ v₀ₐ - m_b v_{ob}
final instant. Right after the crash
p_f = (mₐ + m_b) v
the moment is preserved
p₀ = p_f
mₐ v₀ₐ - m_b v_{ob} = (mₐ + m_b) v
v = [tex]\frac{ m_a \ v_{oa} - m_b \ v_{ob} }{ m_a +m_b}[/tex]
we substitute the values
v = [tex]\frac{ 46.875}{82.03} \ v_{oa} - \frac{35.156}{82.03} \ 61.6[/tex]
v = 0.559 v₀ₐ - 26.40 (1)
Now as the two vehicles united we can use the relationship between work and kinetic energy
the total mass is
M = mₐ + m_b
M = 46,875 + 35,156 = 82,031 slug
starting point. Jsto after the crash
K₀ = ½ M v²
final point. When they stop
K_f = 0
The work is
W = - fr x
the negative sign is because the friction forces are always opposite to the displacement
Let's write Newton's second law
Axis y
N-W = 0
N = W
the friction force has the expression
fr = μ N
we substitute
-μ W x = Kf - Ko
-μ W x = 0 - ½ (W / g) v²
v² = 2 μ g x
v = [tex]\sqrt{ 2 \ 0.750 \ 32 \ 17.5}[/tex]Ra (2 0.750 32 17.5
v = 28.98 ft / s
Activities:
1. Name the instrument that is used to measure Air Pressure.
2.Explain what is Cyclone and Anticyclone
Answers
Answer: barometer.
A cyclone is a storm or system of winds that rotates around a center of low atmospheric pressure. An anticyclone is a system of winds that rotates around a center of high atmospheric pressure.
Which graph represents the relationship between the magnitude of the gravitational force exerted by earth on a spacecraft the distance between the center of the spacecraft the center of earth
Answers
Answer:
B as distance increase force decrease, but it is not a linear relationship.
A hot air balloon is rising at a speed of 10 km/hr. One hour later, the balloon
is still rising at 10 km/hr. What is its acceleration?
Answers
Acceleration is the rate of change of speed and there’s no change