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.
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?
Answer:
false I think
Explanation:
hope that help
so it's not divided in 3 regions
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)
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]
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
Answer:
B as distance increase force decrease, but it is not a linear relationship.
A scientist analyzes the light from a distant galaxy and finds that it is shifted to the longer wavelength of the electromagnetic spectrum. What does this data help to study?
1) the color of the galaxy
2) the distance of the galaxy from Earth
3) the existence of life on any planet in the galaxy
4) the study of the amount of light scattered by dust in space
Answer:
Option 2
Explanation:
As per the relation between the distance of the galaxy and shifting of the light of the galaxy towards any specific wavelength of the electromagnetic spectrum, a galaxy at great distance shifts more towards the red spectra that has the highest wavelength.
Thus, this observation give details about the distance of the galaxy from earth.
Answer:
b
Explanation:
Match the following:
machinery part :nickel or chromium
ornamentation and decoration pieces :silver and gold
processed food :tin coated iron can
bridges and automobiles :zinc metal
distilled water:bad conductor
Answer:
iron metal :chromium
machinery part :nickel or chromium
ornamentation and decoration pieces :silver and gold
processed food :tin coated iron can
bridges and automobiles :zinc metal
distilled water:bad conductor
Explanation:
2.Test the age of your eyes. a.Hold a pencil or ballpoint pen vertically at arm's length. b.Close your left eye and focus on the tip. c.Quickly bring the pencil closer to your eye until it is out of focus. d.Have your partner measure the distance between your eye and the pencil. e.Repeat for both eyes. f.Try it with and without glasses (if you wear glasses). Age of your Eyes Cm91013185083
Answer:
See Explanation
Explanation:
Given
Steps: a - f
Table
[tex]\begin{array}{ccccccc}{cm} & {9} & {10} & {13} & {18} & {50} & {83} \ \\ {Age} & {10} & {20} & {30} & {40} & {50} & {60} \ \end{array}[/tex]
Note that: The question is a practical question and the result may differ base on individuals and environment.
So, I will pick up the question from how to determine the age of the eye after the distance between the eyes and the pencil has been established
In my case, the measurement is:
[tex]Length= 10.4[/tex]
Approximate
[tex]Length= 10[/tex]
From the above table, the corresponding age to 10cm is:
[tex]Age = 20cm[/tex]
If in your measurement, the length is approximately (for example):
[tex]Length = 9cm[/tex]
The age will be:
[tex]Age = 10[/tex]
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?
A ball is dropped off the side of a bridge,
After 1.55 S, how far has it fallen?
(Unit=m)
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.)
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
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
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
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
What is your hypothesis (or hypotheses) for this experiment?
(about Thermal Energy Transfer)
Answer:
I hypothesis that the motion involving the balls in the experiment were moving to create data.
Explanation:
I hope this helps!
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
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
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.
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
plz help me with my career!!!
part one...
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.
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
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.