Answer:
a full length of image of a distance tall building can definitely be seen by using telescope
An unmarked police car traveling a constant 85 km/h is passed by a speeder. Precisely 2.00 s after the speeder passes, the police officer steps on the acceleratorIf the police car accelerates uniformly at 2.00 m/s2 and overtakes the speeder after accelerating for 9.00 s , what was the speeder's speed?.
Answer:
observe o gráfico abaixo que mostra o número de internação em um hospital no município no período de 1960
A girl is riding her bicycle. She rides 4 miles in 2.5 hours. What was her average speed?
Explanation:
4/2.5 = 1.6 miles/hour
What is a primary source?
Answer:
provides a first hand account of an event or time period and are considered to be authoritative
Answer:
main source
Explanation:
What is the car's average velocity (in m/s) in the interval between t = 0.5 s
to t = 2 s?
Answer:
[tex]1.0\; \rm m \cdot s^{-1}[/tex].
Explanation:
Consider a time period of duration [tex]\Delta t[/tex]. Let change in the position of an object during that period of time be denoted as [tex]\Delta x[/tex]. The average velocity of that object during that period would be:
[tex]\displaystyle \bar{v} = \frac{\Delta x}{\Delta t}[/tex].
For the toy car in this question, the time interval has a duration of [tex]2.0 \; \rm s - 0.5\; \rm s = 1.5\; \rm s[/tex]. That is: [tex]\Delta t = 1.5\; \rm s[/tex]. (One decimal place, two significant figures.)
On the other hand, what would be the change in the position of this toy car during that [tex]1.5\; \rm s[/tex]?
Note, that from readings on the snapshot in the diagram:
The position of the toy car was [tex]0.1\; \rm m[/tex] at [tex]t = 0.5\; \rm s[/tex] (the beginning of this [tex]1.5[/tex]-second time period.)The position of the toy car was [tex]1.6\; \rm m[/tex] at [tex]t = 2.0\; \rm s[/tex] (the end of this [tex]1.5[/tex]-second time period.)Therefore, the change to the position of this toy car over that time period would be [tex]\Delta x = 1.6\; \rm m - 0.1\; \rm m = 1.5\; \rm m[/tex]. (One decimal place, two significant figures.)
The average velocity of this car over this period of time would thus be:
[tex]\displaystyle \bar{v} = \frac{\Delta x}{\Delta t} = \frac{1.5\; \rm m}{1.5\; \rm s} = 1.0\; \rm m \cdot s^{-1}[/tex]. (Two significant figures.)
The train traveled 500 kilometers north to Odessa in 2 hours. What is the train’s speed? What is it’s velocity?
A woman leaning over the railing of a tall bridge accidentally drops her cell phone. If the bridge is 31.16 m high, how long will it take the phone to hit the river below the bridge?
Answer:
2.52 s
Explanation:
We can use equation of motion to solve this.
[tex]s \: = ut + \frac{1}{2} a {t}^{2} [/tex]
Assuming initial velocity to be 0 and no air resistance,
[tex] s \: = \frac{1}{2} a {t}^{2} \\ 31.16 = \frac{1}{2} (9.81) {t}^{2} \\ t = 2.52s[/tex]
The time of flight is 3.54 s.
What are equations of motion?There are three equation of motions that can be used when motion of the object is under constant acceleration and on a straight path.
They are listed below as:
[tex]v = u + at\\\\s = ut + \dfrac{1}{2} at^2\\\\v^2 = u^2 + 2as[/tex]
where the symbols have following meanings:
u = initial velocity of the considered object
v = final velocity of the object
a = acceleration of the object
s = distance traveled by the object in 't' time.
It is given that A woman leaning over the railing of a tall bridge accidentally drops her cell phone. If the bridge is 31.16 m high, then how long will it take the phone to hit the river below the bridge.
We can use equation of motion to solve this;
[tex]s = ut + \dfrac{1}{2} at^2[/tex]
where:
s = 31.16 m is the vertical displacement of the phone
u = 0 is the initial vertical velocity
Assuming initial velocity to be 0 and no air resistance,
[tex]s = 1/2 (9.8)(t)^2\\31.16 = 1/2 (9.8)(t)^2\\t = 2.52 s[/tex]
Hence, The time of flight is 3.54 s.
Learn more about free fall here:
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Which of the following is a negatively charged particle that is found in "clouds" around the nucleus?
Group of answer choices
proton
quark
electron
neutron
Explain the Law of Conservation of Momentum
What is the speed of an electron traveling 32 cm in 2 ns?
Answer:
Speed, [tex]v=1.6\times 10^8\ m/s[/tex]
Explanation:
Given that,
Distance covered by the electron, d = 32 cm = 0.32 m
Time, t = 2 ns
We need to find the speed of an electron. Speed is equal to distance covered divided by time. So,
[tex]v=\dfrac{0.32}{2\times 10^{-9}}\\\\v=1.6\times 10^8\ m/s[/tex]
So, the speed of the electron is [tex]1.6\times 10^8\ m/s[/tex].
if the earth stops rotating the Apparent value of g on its surface will
Answer:
If the earth stops rotating the apparent weight will increase at most of the places because the upward component of the centrifugal force due to rotation will disappear.
The portion of a uniform violin string that vibrates is from the "nut" to the "bridge" at the end of the finger board, and has length LA and mass m. The string tension F can only be increased or decreased by tightening or loosening the tuning pegs above the nut. Say that a string is tuned to produce a note with fundamental frequency ft.
a) Then, to play a different note with fundamental frequency fp, the violinist uses her finger to push the string against the fingerboard, reducing the length of the vibrating part of the string to L2 (The string now vibrates from her finger to the bridge.)
i) When she makes this change, determine if each of the following quantities increase, decrease, or remain unchanged: (1) string tension, (2) mass density, (3) wave speed, and (4) wavelength. Explain each.
ii) Is fy greater or less than fx ? Explain.
iii) Find an equation for Lg in terms of all or some of the given parameters. Simplify.
b) Find an equation for F in terms of the given parameters. c) Calculate numerical values for L, and F if m= 2.00 g, LA = 60.0 cm, fa = 440 Hz (A note), fB = 494 Hz (B note). -Bridge
Answer:
A)i) 1. constant, 2. constant, 3. constant, 4. decrease
ii) frecuency increase
iii) L = n /2f √T/μ
B) L_b = 0.534 m
Explanation:
We can approximate the violin string as a system of a fixed string at its two ends, therefore we have a node at each end and a maximum in the central part for a fundamental vibration,
λ = 2L / n
where n is an integer
The wavelength and frequency are related
v = λ f
and the speed of the wave is given by
v = √T /μ
with these expressions we can analyze the questions
A)
i) In this case the woman decreases the length of the rope L = L₂
therefore the wavelength changes
λ₂ = 2 (L₂) / n
as L₂ <L₀ the wavelength is
λ₂ < λ₀
The tension of the string is given by the force of the plug as it has not moved, the tension must not change and the density of the string is a constant that does not depend on the length of the string, therefore the speed of the string wave in the string should not change.
ii) how we analyze if the speed of the wave does not change
v = λ f
as the wavelength decreases, the frequency must increase so that the speed remains constant
fy> fx
iii) It is asked to find the length of the chord
let's use the initial equations
λ = 2L / n
v = λ f
v = 2L / n f
v = √ T /μ
we substitute
2 L / n f = √ T /μ
L = n /2f √T/μ
this is the length the string should be for each resonance
b) in this part they ask to calculate the frequency
f = n / 2L √ T /μ
the linear density is
μ = m / L
μ = 2.00 10⁻³ / 60.0 10⁻²
μ = 3.33 10⁻³ kg / m
we assume that the length is adequate to produce a fundamental frequency in each case
f_{a} = 440Hz
λ = 2La / n
λ = 2 0.60 / 1
λ = 1.20 m
v = λ f
v = 1.20 440
v = 528 m / s
v² = T /μ
T = v² μ
T = 528² 3.33 10⁻³
T = 9.28 10² N
Let's find the length of the chord for fb
f_{b} = 494 hz
L_b = 1 /(2 494) √(9.28 10² / 3.33 10⁻³)
L_b = 0.534 m
What is the magnite of the net displacement of the mouse?
Complete Question
A field mouse trying to escape a hawk runs east for 5.0m, darts southeast for 3.0m, then drops 1.0m down a hole into its burrow. What is the magnitude of the net displacement of the mouse?
Answer:
The values is [tex]s = 7.49 \ m[/tex]
Explanation:
From the question we are told that
The distance it travels eastward is [tex]x = 5.0 \ m[/tex]
The distance it travel towards the southeast is [tex]l = 3.0\ m[/tex]
The distance it travel towards the south is [tex]z = 1 \ m[/tex]
Let x-axis be east
y-axis south
z-axis into the ground
The angle made between east and south is [tex]\theta = 45^o[/tex]
The displacement toward x-axis is
[tex]x = 5 + 3cos(45)[/tex]
[tex]x = 7.12[/tex]
The displacement toward the y-axis is
[tex]y = 3 * sin (45)[/tex]
[tex]y = 2.123[/tex]
Now the overall displacement of the rat is mathematically evaluated as
[tex]s = \sqrt{7.12^2 + 2.12^2 + 1^2}[/tex]
[tex]s = 7.49 \ m[/tex]
A snowboarder on a slope starts from rest and reaches a speed of 8.9 m/s after 9.5 s.
How far does the snowboarder travel in this time?
Explanation:
Given that,
Initial speed, u = 0
Final speed, v = 8.9 m/s
Time, t = 9.5 s
Let a is the acceleration of the snowboarder. It can be calculated as follows :
[tex]a=\dfrac{v-u}{t}\\\\a=\dfrac{8.9-0}{9.5}\\\\a=0.936\ m/s^2[/tex]
The distance traveled by the snowboarder in this time is calculated using third equation of motion as follows :
[tex]v^2=u^2=2as\\\\s=\dfrac{v^2}{2a}\\\\s=\dfrac{(8.9)^2}{2\times 0.936}\\\\s=42.31\ m[/tex]
So, 42.31 m of distance is traveled by snowboarder in this time.
What is the direction of the magnetic field around a wire carrying a current perpendicularly into this page
Answer:
Clockwise direction
Explanation:
In a case of a wire carrying a current, the right hand rule is used.
The thumb in the direction of current while the finger curl around in the direction of the magnetic field.
The right hand rule applies to a current in a straight line wire.
If the direction a wire carrying a current perpendicularly into this page, the direction of the magnetic field will be in a clockwise direction .
Tectonic plates are large segments of the Earth's crust that move slowly. Suppose one such plate has an average speed of 8.8 cm per year. (a) What distance does it move in 35 seconds at this speed (note that your answer should be in meters while your given speed is in cm/year)? (b) How many miles will the plate move in one million years at this speed?
Answer:
Jun 29, 2016 · Tectonic plates are large segments of the Earth’s crust that move slowly. Suppose that one such plate has an average speed of 4.0 cm/year. (a) What distance does it move in 1.0 s at this speed? (b) What is its speed in kilometers per million years?
Explanation:
how to find average acceleration?
Answer:
Explanation:
A way to see this is that the definite integral of the acceleration is the change in velocity (i.e. the final velocity minus the initial velocity), and the change in velocity divided by the length of the time interval is the average acceleration on the interval.
Jerry is conducting an experiment to test friction. His setup is shown below. If he wants to test the friction between different types of materials, what should he change in the setup above? A. Replace the mass with a different size mass B. Replace the string with a different length string C. Replace the spring scale with a different kind of scale D. Replace the wooden block with a different kind of block
Answer:
The correct option is D
Explanation:
This question is incomplete because of the absence of the setup which as been attached below. The setup shows/determines/tests the friction of wood (which is a block material), since Jerry wants to test the friction between different types of materials, he will have to replace the wooden block with another type of block material of choice so as to determine the friction of that also.
In order to have a comprehensive experiment, Jerry can use 4-5 different types of block material in the course of the experiment.
Answer:
D
Explanation: i did it and my teacher said it was right
If a truck travels 100 miles in 2 hours, what is its speed?
Answer:50 miles per hour 50/1hr
Explanation:100 divided by 2 is 50, divide 2 by 2 thats 1
Answer:
50 miles/hr
Explanation:
[tex]Distance = 100\: miles\\Time = 2\:hours\\\\Speed = \frac{Distance}{Time} \\\\Speed = \frac{100\:miles}{2\:hours}\\\\ Speed = 50\:miles/hour[/tex]
A body is projected down at an angle of 30° with the horizontal
from the top, of a building 170m high. Its initial speed is 40m/s .
How long will it take before striking the ground.
Answer:
It will take 15 seconds
The quantity represented by vi is a function of time (i.e., is not constant).A. TrueB. False
Answer:
The answer is false
What is the velocity during the first 4 seconds?
Answer:
6?
Explanation:
Answer:
did you get the answer bc I think its -3
Explanation:
Which graph shows a negative acceleration
Answer:
It's d on edg
Explanation:
what will happen to the fieldof view for each resultant magnification as you change objectives from 4 to 10 to 43
Answer:
The field of view is reduced.
Explanation:
Given that,
The field of view for every resultant magnification like you change objectives from 4 to 10 to 43.
We know that,
Field of view :
When the view is observed at a point in a defined field then these field called field of view.
The normal angle of field of view is 90°.
The formula of field of view is define as,
[tex]field\ of\ view = \dfrac{field\ number}{magnification}[/tex]
We can say that,
The field of view is inversely proportional to the magnification.
When magnification is low then field of view will be large.
When magnification is higher then field of view will be small .
According to question,
When the magnification adjust from 4 to 10 to 43, the field of view is reduced.
Hence, The field of view is reduced.
farah claps his 600times in 60seconds. what is the friquency and periodb
Answer:
[tex]\huge\boxed{\sf Frequency = 10\ Hz}[/tex]
[tex]\huge\boxed{\sf Time\ Period = 0.1\ secs}[/tex]
Explanation:
Given:
No. of times = 600 times
Time = 60 seconds
Required:
Frequency = f = ?
Time Period = T = ?
Solution:
Frequency = No. of Times / TimeFrequency = 600 / 60
Frequency = 10 Hz
Time Period = 1 / FrequencyTime Period = 1 / 10
Time Period = 0.1 secs
Hypothesis If an object rolls over _______________ (type of material), then it will _____________ (describe the prediction of distance it will travel)
Answer:
smooth
go far
A rigid rod of length L rotates about an axis perpendicular to the rod, with one end of the rod fixed to the axis. Which of the following must always be equal at all points on the rod?
1. Angular position.
2. Angular velocity.
3. Angular acceleration.
4. Tangential acceleration.
A. 1 and 2.
B. 1, 2 and 4.
C. 1 and 3.
D. 1, 2, 3 and 4.
Answer:
D
Explanation:
The Angular Position: The angular position is also called the angular displacement, and it is the angle through which a point revolves around the centre line that has been rotated in a particular manner about a given axis.
The Angular Velocity: The angular velocity of an object is the measure of the speed, or say, how fast an object rotates at one point with respect to a point.
The Angular Acceleration: This is very simple as it's a derivative of angular velocity. They both share the same relationship velocity and acceleration share. Angular acceleration is the rate of change of angular velocity.
The only options that are equal at all points on the rod are;
1. Angular position.
2. Angular velocity.
3. Angular acceleration.
Let us look at each of the options;
1) Angular Position: This is also referred to as angular displacement. In centripetal motion, it is defined as the angle through which a a body has been rotated from the initial position or reference position. This must always be equal at all points on the rod.
2) Angular Velocity: This is defined as the speed with which an object is moving in a circular motion. This must always be equal at all points on the rod.
3) Angular Acceleration: This is simply defined as the rate of change of angular velocity with time. Now, since angular velocity must be equal at all points on the rod, then angular acceleration must also be equal at all points on the rod.
4) Tangential acceleration; This is defined as the rate of change of tangential velocity with time. Tangential velocity is the linear component of the speed and it is not equal at all points on the rod and as such the tangential acceleration will also not be equal at all points on the rod.
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What unit is time usually measured in?
A typical atom has a diameter of about 1.32 x 10-10 m. (a) what is this in inches (b) approximately how many atoms arethere along a 1.32 cm line.
Answer:
A ) [tex]5.197 \times 10 ^{-9} inches[/tex]
b) There are approximately 100,000,000 atoms on the 1.32 cm lines
Explanation:
To ensure the accuracy of our answer, we will have to make sure we work in the appropriate units.
A
To convert [tex]1.32 \times 10^{-10} m[/tex] to inches, we can use this factor.
1 m = 39.3701 inches
Therefore [tex]1.32 \times 10^{-10} m[/tex] =
B
To find the number of atoms on a 1.32 cm line, we will first of all need to convert 1.32 cm to meters.
1.32 cm = 0.0132 metres (divided 1.32 by 100 to convert to metres)
Assuming all the atoms are arranged side by side, with their edges touching on the 0.0132m line, the number of atoms present will be
[tex]0.0132/1.32\times 10^{-10}[/tex] = 100,000,000 atoms
A 68 kg deer has a momentum of 952 kg∙m/s. What is its velocity?
All we need to do is divide the momentum and weight to find the velocity:
v = m / w
Solve using the values we are given.
v = 952/68 = 14 m/s
Best of Luck!
You are driving along the New York State Thruway in a line of cars all travelling at a constant speed of 108.5 km/hr. The car in front of you applies its brakes for maximum acceleration. You then apply your brakes to achieve the same maximum acceleration after only a 1 s delay due to reaction time. What distance behind the car in front of you must you be to avoid a collision?
Answer:
Car first should be 30.13 m behind the second car.
Explanation:
Given that,
Constant speed = 108.5 Km/hr
Time = 1 sec
Let the distance covered by second car S and by first car S'
We need to calculate the distance covered by second car
Using equation of motion
[tex]S=ut-\dfrac{1}{2}at^2[/tex].....(I)
The distance covered by first car to avoid collision
[tex]S''=S'+S[/tex]
Put the value into the formula
[tex]S'+S=ut+ut-\dfrac{1}{2}at^2[/tex]...(II)
We need to calculate the distance covered by first car
Using equation (I) and (II)
[tex]S'=ut[/tex]
Put the value into the formula
[tex]S'=108.5\times\dfrac{5}{18}\times1[/tex]
[tex]S'=30.13\ m[/tex]
Hence, Car first should be 30.13 m behind the second car.