Answer:
This measurement would be a measure of the speed of an object(the car ) in motion
Explanation:
A driver looks down at their speedometer and sees they are moving at 45 mph. This measurement would be about the speed of the car or about how fast the car is moving.
This means, the car is travelling the distance of 45 miles in every hour.
This measurement would also be about the instantenous speed of the car which is 45mph.
Two cylindrical resistors are made from the same material. The shorter one has a length LL and diameter DD . The longer one has a length 16L16L and diameter 4D4D . How do their resistances compare? The resistance of the longer resistor is four times the resistance of the shorter resistor. The resistance of the longer resistor is twice the resistance of the shorter resistor. The resistance of the longer resistor is the same as the resistance of the shorter resistor. The resistance of the longer resistor is half the resistance of the shorter resistor. The resistance of the longer resistor is a quarter of the resistance of the shorter resistor.
Answer:
The resistance of the longer resistor is a quarter of the resistance of the shorter resistor.Explanation:
If Two cylindrical resistors are made from the same material, then their resistivity will be the same. Formula for calculating resistivity of a material is expressed as;
[tex]\rho = \frac{RA}{L} \ where \ A = \frac{\pi d^2}{4}[/tex] where;
R is the resistance
A is the cross sectional area of the material
L is the length of the material
For the shorter cylinder:
Length = L
diameter = D
[tex]\rho = \dfrac{R_s(\frac{\pi D^2}{4})}{L} \\\\\rho = \dfrac{R_s{\pi D^2}}{4L}[/tex]
For the longer cylinder:
Length = 16L
diameter = 4D
[tex]\rho = \dfrac{R_l(\frac{\pi (4D)^2}{4})}{16L} \\\\\\\rho = \dfrac{R_l(\frac{\pi (16D^2)}{4})}{16L} \\\\\rho = \dfrac{R_l{16\pi D^2}}{16L}\\\\\rho = \dfrac{R_l{\pi D^2}}{L}[/tex]
Since their resistivity are the same then;
[tex]\dfrac{R_s{\pi D^2}}{4L} = \dfrac{R_l{\pi D^2}}{L} \\\\ \dfrac{R_s}{4} = {R_l} \\\\R_s = 4R_l\\\\R_l = \frac{R_s}{4}[/tex]
Hence the resistance of the longer resistor is a quarter of the shorter resistor.
The x vector component of a displacement vector has a magnitude of 94.8 m and points along the negative x axis. The y vector component has a magnitude of 149 m and points along the negative y axis. Find (a) the magnitude and (b) direction of . Specify the direction as a positive angle with respect to the negative x axis.
Answer:
We use pythagorean theorem
So that
r = √(x² + y²)
r = √(94.8)² + (-149)² )
r =√ 9682.6 + 22201
= 178.6
direction we use
စ = tan^-1(98.4/149)
= tan^-1 (0.6604)°
If you walk 5 km north and then 12 km east. Your resultant displacement is____.a. magnitude: 1.1km; direction: 53.1 degrees east of north.b. magnitude: 2.00; direction: 53.1 degrees east of north.c. magnitude: 2.00; direction: 36.9 degrees east of north.d. magnitude: 1.1km; direction: 36.9 degrees east of north.
Answer:
Resultant = 13km
Direction = 67.38° East of North
Explanation:
Given the following :
5km North ; 12km East
Resultant Displacement (r) :
r² = 5² + 12²
r² = 25 + 144
r² = 169
r = √169
r = 13
Direction:
Tangent = opposite / Adjacent
Tanθ = opposite / Adjacent
Opposite = 12 ; adjacent = 5
Tanθ = (12/5)
Tanθ = 2.4
θ = tan^-1(2.4)
θ = 67.38° east of north
Which wavelength of light is NOT absorbed by riboflavin?
Answer:
525 nm and 550nm
Explanation:
The absorption spectrum of any compound is a plot of absorption against wavelength. Sometimes, the absorption is plotted against frequency or wave number. This plot appears as humps. The peak of the highest hump is the wavelength of maximum absorption. The concentration of each solution may be indicated above the hump when the instrument is calibrated using different concentrations of the analyte.
The absorption spectrum of riboflavin indicates that its wavelength of maximum absorption is 450nm and it does not absorb at light 525 nm and 550nm wavelengths respectively.
A long, thin, insulated wire carries a current I1=1A out of the plane of the paper. The wire is surrounded by a long cylinder of radius a=12cm. The cylinder carries a current I2=9A that is unformly distributed over its cross section and flows into the plane of the paper.
Find the radius r at which the amgnetic field is zero.
Answer:
The magnetic field is zero at a radius r = 4 cm.
Explanation:
The magnetic field due to the wire is given by B₁ = μ₀I₁/2πr where I₁ = current in wire = 1 A and r = distance of point where magnetic field is zero from wire.
Now, since the current is uniformly distributed in the cross section of the cylinder, its current density is constant.
So, with current I₂ = 9 A flowing in the cylinder and radius, a = 12 cm. Let I' = current at radius r where the magnetic field is zero. So,
I'/πr² = I₂/πa²
I' = I₂r²/a²
Using Ampere's Law, the magnetic field B₂ at the distance r is given by
∫B.ds = μ₀I'
∫Bdscos0 = μ₀I' (since the magnetic field is parallel to the path)
B∫ds = μ₀I' ∫ds = 2πr
B2πr = μ₀I'
2Bπr = μ₀I₂r²/a²
B₂ = μ₀I₂r/2πa²
So, when the net magnetic field is zero, B₁ = B₂
So, μ₀I₁/2πr = μ₀I₂r/2πa²
I₁/r = I₂r/a²
I₁/I₂ = r²/a²
r² = I₁/I₂a²
taking square root of both sides,
r = a√(I₁/I₂)
substituting the values of the variables, we have
r = 12√(1/9)
r = 12/3
r = 4 cm
The magnetic field is zero at a radius r = 4 cm.
Suppose a radio signal (light) travels from Earth and through space at a speed of 3 × 108/ (this is the speed of light in vacuum). How far (in meters) into space did the signal travel during the first 10 minutes?
Answer:
18*10^10 meters
Explanation:
V= d/t 10 mins = 600 seconds
3*10^8 = d/600s
(3*10^8)*(6*10^2) = d
d = 18*10^10 m
(b) If you decrease the length of the pendulum by 25%, how does the new period TN compare to the old period T?
Answer:
The new period will be reduced by 50%
Explanation:
The period of pendulum is given by;
[tex]T= 2\pi\sqrt{\frac{L}{g} }\\\\\frac{T}{2\pi} = \sqrt{\frac{L}{g} }\\\\(\frac{T}{2\pi} )^2 = {\frac{L}{g}}\\\\\frac{T^2}{4\pi ^2} = {\frac{L}{g}}\\\\T^2(\frac{g}{4\pi ^2}) = L\\\\ \frac{g}{4\pi ^2}= \frac{L}{T^2}\\\\\frac{L_1}{T_1^2} = \frac{L_2}{T_2^2}[/tex]
When the length is decreased by 25%, the new length L₂ is given by;
L₂ = 25/100(L₁)
L₂ = 0.25L₁
[tex]\frac{L_1}{T_1^2} = \frac{L_2}{T_2^2}\\\\T_2^2 = \frac{T_1^2L_2}{L_1} \\\\T_N^2 = \frac{T^2(0.25L_1)}{L_1}\\\\ T_N^2 =0.25T^2\\\\T_N = \sqrt{0.25T^2}}\\\\T_N = 0.5 T[/tex]
Thus, the new period will be reduced by 50%
A car drives 10km with a speed of 72 km/hr and then runs out of gas. Then you walk 2km for the next 30 min until you find a gas station. (A) What is the displacement of the total trip? (B) How long does the entire trip take? (C) What is the average velocity of the entire trip?
Answer:
A- 12 km
B- 42 minutes
C- 17.14 km per hour
Explanation:
A- The displacement of the total trip is 12 kilometers, which emerges by adding the 10 kilometers traveled by car to the 2 kilometers traveled on foot.
B- The trip, in total, took 42 minutes, which arises from adding the 30 minutes of the journey on foot, plus 12 minutes of travel by car at 72 km / h (72/60 x 10 = 12).
C- While 12 km were covered in 42 minutes, the average speed of the trip was 17.14 km / h. This arises from the following calculation:
42 = 12
60 = X
(60 x 12) / 42 = X
17.14 = X
Why is it important not to present a biased argument as a public speaker?
A.
Because it is unconvincing
B.
Because it is immoral
C.
Because it is unfair
D.
Because it is pointless
Answer:
Explanation:
a) because it's immoral as you're trying to convince people of your views when you should be giving both sides of the story so people are able to come up with their own opinions on what you're talking about
65°F to degrees Celsius
Answer:
18.3 C
Explanation:
Answer: 55/3 °C
Explanation:
Formula:
C=5/9(F-32)
Given:
F=65
Solve:
C=5/9(F-32)
C=5/9(65-32)
C=5/9(33)
C=55/3
If an automobile with a velocity of 4.0 m/s accelerates at a rate of 4.0 m/s2
for 2.5 s, what is the final velocity?
Answer:
V=u+at
=4+4*2.5
=14m/s
Explanation:
Initial velocity =4m/s
acceleration =4m/(second) squared
time =2.5 seconds
v=u+at
v=4+4*2.5
v=14m/s
What the answers due in 2 min
Hello ┬┴┬┴┤◕3◕)づ├┬┴┬┴
The answer is 3 m
the gap in between 4s and 6s is 5s
you follow your finger up to the purple like, and follow the purple line in a straight line to the left. It ends at 3m and that is your answer.
The speed of x-rays is 300 000. m/s. In scientific notation, and with the correct number of significant figures, this speed is… *
Answer:
3.0 x 10 and the exponent on ten is 5.
Explanation:
You are moving the decimal point over to the left 5 times, making the exponent a positive 5 and then you would put a decimal after 3. Making it 3.0 x 10, and the exponent 5.
The speed of x-rays in scientific notation is 3.0 x 10⁶ m/s.
The given parameters;
speed of the x-ray, v = 300,000 m/s
The standard form of the given number represents the scientific notation of the number.
To represent a number in standard form, multiply the number in terms of power or exponent of 10. The value of the exponent represents the number of available zeros present.In scientific notation the speed of the x-ray can be expressed as follows;
300,000 m/s = 3.0 x 10⁶ m/s
Thus, the speed of x-rays in scientific notation is 3.0 x 10⁶ m/s.
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A stone is thrown vertically downward with an initial speed of 12.0 m/s from the top of a
building. The stone takes 1.54 s to reach the ground.
The question is incomplete; However, the height from which the stone is thrown is most likely to be what's required of the question
Answer:
See Explanation
Explanation:
Given
[tex]Initial\ Speed, u = 12.0m/s[/tex]
[tex]Time, t = 1.54\ s[/tex]
Required
Determine the height
This question will be answered using the following equation of motion;
[tex]S = ut + \frac{1}{2}gt^2[/tex]
Take g as 9.8
This gives:
[tex]S = 12 * 1.54 + \frac{1}{2} * 9.8 * 1.54^2[/tex]
[tex]S = 12 * 1.54 + \frac{1}{2} * 9.8 * 2.3716[/tex]
[tex]S = 18.48 + 11.62084[/tex]
[tex]S = 30.10084[/tex]
[tex]S = 30.1\ m[/tex] (Approximated)
Hence; Height = 30.1m
A corvette starts from rest and travels 69.0 meters in 50 s. What is its acceleration?
Answer:
0.0552 m/s²
Explanation:
Given:
Δx = 69.0 m
v₀ = 0 m/s
t = 50 s
Find: a
Δx = v₀ t + ½ at²
69.0 m = (0 m/s) (50 s) + ½ a (50 s)²
a = 0.0552 m/s²
A watermelon is dropped from the edge of the roof of a build- ing and falls to the ground. You are standing on the sidewalk and see the watermelon falling when it is 30.0 m above the ground. Then 1.50 s after you first spot it, the watermelon lands at your feet. What is the height of the building
Answer:
The hight of the building is 38.16 m
Explanation:
These two pieces of information given, first, the watermelon is 30 m above the ground and after 1.50 s the watermelon has been spotted. Now we are required to find the height of the building.
Use the below formula to find the height of buildings.
S = ut + ½ gt^2
30 =1.5u + (1/2) × 9.8 (1.5)^2
u = 12.65 m/sec
v^2 – u^2 = 2gs
(12.65)^2 = 2×9.8 s’
S’ = 8.16 m
h = s + s’
h = 30 + 8.16 = 38.16 m
The hight of the building is 38.16 m.
The height of the building is 38.16 m.
Given data:
The height above the ground is, h = 30.0 m.
The time interval after observation of first spot is, t = 1.50 s.
We need to find the height of building. And since two pieces of information given, first, the watermelon is 30 m above the ground and after 1.50 s the watermelon has been spotted. So, using the second kinematic equation of motion as,
[tex]h = ut + \dfrac{1}{2}gt^{2}[/tex]
Here, u is the initial speed. Solving as,
[tex]30 = (u \times 1.50) + \dfrac{1}{2} \times 9.8 \times (1.50)^{2}\\\\u =12.65 \;\rm m/s[/tex]
Now landing distance (s') is calculated using the third kinematic equation of motion as,
[tex]v^{2} =u^{2}+2(-g)s\\\\0^{2} =12.65^{2}+2(-9.8)s\\\\s = 8.16 \;\rm m[/tex]
Then the height of building is given as,
H = h + s
H = 30 m + 8.16 m
H = 38.16 m
Thus, we can conclude that the height of the building is 38.16 m.
Learn more about the kinematic equations of motion here:
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A ball with mass kg is thrown upward with initial velocity m/s from the roof of a building m high. Neglect air resistance. Use m/s2. Round your answers to one decimal place. (a) Find the maximum height above the ground that the ball reaches. meters (b) Assuming that the ball misses the building on the way down, find the time that it hits the ground. s
Answer:
Explanation:
The question is incomplete. Here is the complete question.
A ball with mass m kg is thrown upward with initial velocity 22 m/s from the roof of a building 17 m high. Neglect air resistance. Use g=9.8 m/s2. Round your answers to one decimal place. (a) Find the maximum height above the ground that the ball reaches. xmax= meters (b) Assuming that the ball misses the building on the way down, find the time that it hits the ground.
a) Using the equation of motion formula;
v² = u²+2gH where;
u is the initial velocity
v is the final velocity
theta is the angle of launch
g is the acceleration due to gravity.
H is the maximum height reached by the ball
Since the ball is thrown upwards, the acceleration due to gravity will be negative. The equation then becomes;
v² = u²-2gH
Given
v = 0m/s
u = 22m/s
g = 9.8m.s²
0² = 22²-2(9.8)H
-22² = -19.6H
H = -22²/-19.6
H = 24.69m
If the biuliding is 17m high, the maximum height above the ground that the ball reaches will be;
Hmax = 24.69+17
Hmax = 41.69m
b) The time it takes to hit the ground can be expressed using the formula
v = u-gt
0 = 22-9.8t
-22 = -9.8t
t = -22/9.8
t = 2.45secs
27.
In a graph showing how temperature of a material changes over time, temperature
change is the:
A. dependent variable
C. variable with the smallest range
B. independent variable
D. variable with the largest range
Answer:
A. Dependent variable
Explanation:
In a graph showing how temperature of a material changes over time, temperature is taken on y-axis and time is taken on x-axis. It shows how temerature altered as the time changes.
In temperature-time graph, temperature is directly dependent on time. With the increase in time, temperature rises, falls or remains constant. It implies that temperature is dependent variable that depend on time.
Hence, the correct option is (A).
Q1. My brakes stop me at -5.5m/s^2 when I lock them up. A puppy runs in front of my car, and I stop just in time 3.5 seconds later. How far did I skid? Q2. I floor the pedal and accelerate from 7.0m/s to 35m/s over a fouth of a mile (0.40km). What is my accelleration?
please explain step by step, much thanks :)
Explanation:
Q1. Given:
v = 0 m/s
a = -5.5 m/s²
t = 3.5 s
Find: Δx
Δx = vt − ½ at²
Δx = (0 m/s) (3.5 s) − ½ (-5.5 m/s²) (3.5 s)²
Δx ≈ 33.7 m
Q2. Given:
Δx = 400 m
v₀ = 7.0 m/s
v = 35 m/s
Find: a
v² = v₀² + 2aΔx
(35 m/s)² = (7.0 m/s)² + 2a (400 m)
a = 1.47 m/s²
c) A sample of substance of volume 10 cm3 was brought back to Earth from the Moon.
The weight of the sample on the Moon was 0.13 N. The gravitational field strength
on the Moon is 1.6 N kg-1.
(1)
Calculate the density of the sample.
Answer:
8125
Explanation:
[tex]P=0,13 N\\a=1,6 N/kg\\m=P/a=0,13/1,6=0,08125\\V=10(cm)^{3} =10^{-5} meters^{3} \\p=m/V=0,08125/10^{-5} =0,08125*100000=8125kg/meters^{3}[/tex]
Enter the expression 2cos2(θ)−1 , where θ is the lowercase Greek letter theta.
Answer:
[tex]2cos^2(\theta) - 1 = cos(2\theta)[/tex]
Explanation:
Given
[tex]2cos^2(\theta) - 1[/tex]
Required
Simplify
In trigonometry:
[tex]sin^2(\theta) + cos^2(\theta) = 1[/tex]
So; the given expression becomes
[tex]2cos^2(\theta) - (sin^2(\theta) + cos^2(\theta))[/tex]
Open Bracket
[tex]2cos^2(\theta) - sin^2(\theta) - cos^2(\theta)[/tex]
Collect Like Terms
[tex]2cos^2(\theta) - cos^2(\theta)- sin^2(\theta)[/tex]
[tex]cos^2(\theta)- sin^2(\theta)[/tex]
In trigonometry:
[tex]cos(\theta + \theta) = cos^2(\theta)- sin^2(\theta)[/tex]
This implies that:
[tex]cos^2(\theta)- sin^2(\theta) = cos(\theta + \theta)[/tex]
=
[tex]cos(\theta + \theta)[/tex]
[tex]cos(2\theta)[/tex]
Hence:
[tex]2cos^2(\theta) - 1 = cos(2\theta)[/tex]
Three different groups each measured the diameter of a CD three times (the actual diameter of a CD is 12.00cm). The class data is shown in the table. Which group(s) was the most accurate in their measurements? Group 1 only Group 1 only Group 2 only Group 2 only Group 3 only Group 3 only Groups 1 & 3 were equally accurate
What exerts the thrust force acting on the cart?
Answer:
This question appears incomplete
Explanation:
The question appears incomplete. However, a thrust can be described as a force or a push. A cart can be described as a vehicle (usually with two wheels) that is used to transport load from one point/place to another.
Newton first law of motion (law of inertia) states that a body will continue to be in a state of rest or if in motion, remains in motion (at constant velocity) unless acted upon by an external force. Hence, a cart will continue to be in a state of rest unless acted on by an external force which can be exerted by animals or humans (depending on the type of cart).
What is the speed of an ocean wave if it’s wavelength is 5.0 m and it’s frequency is 3/s?
Answer:
15 m/s
Explanation:
We know that [tex]v = f * d[/tex] where f = frequency & d = wavelength .
So here.
Wavelength = 5 m
Frequency = 3 s⁻¹
Hence Speed = 5 * 3 = 15 m/s
Which phrase describes a scientific law?
A. A statement that matter cannot be created or destroyed
B. A claim that experiments cannot verify whether matter has been
destroyed
O C. An explanation for why matter cannot be created or destroyed
D. A prediction of how much matter exists in the universe
SUBMIT
In
Answer:
C- An explanation for why matter cannot be created or destroyed
Explanation:
Answer: A. A statement that matter cannot be created or destroyed.
Explanation:
Apex
Juan, a biologist, notices that a particular group of hummingbirds leaves during the colder winter months and returns during the warmer spring months. Juan hypothesizes that the birds fly south to locations with warmer weather and blooming flowers during the winter months. Which procedure should Juan use to test his hypothesis?
Answer:
C. Tag the hummingbirds with transmitters and track their movement.
Explanation:
Hypothesis refers to the conclusion of the analysis found on the grounds of certain observations. The hypothesis is further tested and verified by performing certain experiments. In the given excerpt, Juan hypothesizes that certain group of hummingbirds fly to certain locations depending upon the season. To test the hypothesis, Juan can experiment with finding and tracking the hummingbirds and the direction in which they fly. Tagging the hummingbirds with transmitters and tracking their movement would help Juan concluding about the hypothesis.
Answer:
Tag the hummingbirds with transmitters and track their movement
Explanation:
what are two examples of population distribution?
If you push a book 1.5 m across a table using a constant force of 10.0 N, how much work do you
do on the book?
Answer:
The answer is
15.0 JExplanation:
To calculate the work done by a body we use the formula
work done = force × distanceFrom the question
force = 10 N
distance covered = 1.5 m
So the work done is
work done = 10 × 1.5
We have the final answer as
15.0 JHope this helps you
Which of the following is not one of the emotional expressions that are universally recognized?
OA. Happiness
OB. Disgust
OC. Sadness
OD. Love
HELPP!
Answer:
B
Explanation:
The answer should be B.
An object, initially at rest, moves 250 m in 17 s. What is its acceleration?
Answer:
1.73 m/s²
Explanation:
Given:
Δx = 250 m
v₀ = 0 m/s
t = 17 s
Find: a
Δx = v₀ t + ½ at²
250 m = (0 m/s) (17 s) + ½ a (17 s)²
a = 1.73 m/s²
The acceleration of this object is 1.730 meter per seconds square.
Given the following data:
Initial velocity = 2.5 m/s (since the object is starting from rest).Time = 17 seconds.To find the acceleration of this object, we would use the second equation of motion.
Mathematically, the second equation of motion is given by the formula;
[tex]S = ut + \frac{1}{2} at^2[/tex]
Where:
S is the displacement or distance covered.u is the initial velocity.a is the acceleration.t is the time measured in seconds.Substituting the given values into the formula, we have;
[tex]250 = 0(17) + \frac{1}{2} (a)(17^2)\\\\250 = \frac{1}{2} (289)a\\\\250 = 144.5a\\\\a = \frac{250}{144.5}[/tex]
Acceleration, a = 1.730 [tex]m/s^2[/tex]
Therefore, the acceleration of this object is 1.730 meter per seconds square.
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