Recall the formulas,
[tex]x_f=x_i+v_it+\dfrac12at^2[/tex]
[tex]v_f=v_i+at[/tex]
(a)
[tex]10.0\,\mathrm m=\left(19.0\dfrac{\rm m}{\rm s}\right)+\dfrac12\left(-3.50\dfrac{\rm m}{\mathrm s^2}\right)t^2[/tex]
[tex]\implies \boxed{t\approx0.555\,\mathrm s}[/tex]
(b)
[tex]v_f=19.0\dfrac{\rm m}{\rm s}+\left(-3.50\dfrac{\rm m}{\mathrm s^2}\right)(0.555\,\mathrm s)[/tex]
[tex]\implies \boxed{v_f\approx17.1\dfrac{\rm m}{\rm s}}[/tex]
Someone plans to float a small, totally absorbing sphere 0.518 m above an isotropic point source of light, so that the upward radiation force from the light matches the downward gravitational force on the sphere. The sphere's density is 22.2 g/cm3, and its radius is 2.12 mm. (a) What power would be required of the light source
Answer:
Explanation:
From the question we are told that
The height is [tex]h = 0.518 \ m[/tex]
The sphere density is [tex]\rho = 22.2g/cm^3 = \frac{22.2 }{1000} * 1*10^{6} = 22200 kg/m^3[/tex]
The radius is [tex]r = 2.12 \ mm =0.00212 \ m[/tex]
Generally the power required is mathematically represented as
[tex]P = \frac{16 * \pi * \rho * r * g * h^2 * c }{3}[/tex]
substituting values
[tex]P = \frac{16 * 3.142 * 22200 * 0.00212 * 9.8 * 0.518^2 * 3.0*10^{8}}{ 3}[/tex]
[tex]P = 6.22*10^{11} \ W[/tex]
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
supposed the Kingfisher Dives with an average speed of 4.6 m / s for 1.4 seconds before hitting the water. What is the height from which the bird dove
Answer:
h = 1.07 m
Explanation:
It is given that,
The average sped of the kingfisher, v = 4.6 m/s
Time, t = 1.4 s
Final speed, v = 0 (as it hits the water)
We need to find the height from which the bird dove. Let the height is h. According to the conservation of energy.
[tex]mgh=\dfrac{1}{2}mv^2\\\\h=\dfrac{v^2}{2g}\\\\h=\dfrac{(4.6)^2}{2\times 9.8}\\\\h=1.07\ m[/tex]
So, the bird dove from a height of 1.07 m.
What is the velocity during the first 4 seconds?
Answer:
6?
Explanation:
Answer:
did you get the answer bc I think its -3
Explanation:
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 small mailbag is released from a helicopter that is descending steadily at 1.20 m/s. (a) After 5.00 s, what is the speed of the mailbag?
Answer:
50.2m/s
Explanation:
Using first equation of motion we have
V= u+gt
V= final speed
U= initial speed
So v = 1.2m/s + 9.8(5s)
50.2m/s
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:
A kayak on a lake is moving at a constant speed of 4.1 m/s in a direction 41.6° north of west. After 10 min, find the following.(a) how far west the kayak has traveled km (b) how far north the kayak has traveled km
Given :
A kayak on a lake is moving at a constant speed of 4.1 m/s in a direction 41.6° north of west.
To Find :
After 10 min :
(a) how far west the kayak has traveled km.
(b) how far north the kayak has traveled km.
Solution :
a) Time taken in seconds :
[tex]t=10\times 60 \ s = 600\ s[/tex] .
Distance travelled in 600 s is :
[tex]D=s\times t\\\\D=4.1\times 600\ m\\\\D=2460\ m[/tex]
Distance in km is :
[tex]D=\dfrac{2460}{1000}=2.46\ km[/tex]
b) Now , distance from the north is given by :
[tex]d=D\times sin\ 41.6^o\\\\d=2.46 \times sin\ 41.6^o\\\\d=1.63\ km[/tex]
Hence , this is the required solution .
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
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!
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.
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]
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 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]
what distance is traveled by a police car that moves at a constant speed of 1.5 km/min for 5.0 min?
Answer:
7.5 meters
Explanation:
Answer:
7.5 kilometers
Explanation:
The distance can be found by multiplying the speed by the time.
[tex]d=s*t[/tex]
The car's speed is 1.5 kilometers per minute. It is traveling for 5.0, or 5 minutes. Therefore,
[tex]s= 1.5 km/min\\t=5 min[/tex]
Substitute the values into the formula.
[tex]d=1.5 km/min * 5 min[/tex]
Multiply. When you multiply, the minutes, or "min" will cancel each other out.
[tex]d= 1.5 km * 5[/tex]
[tex]d=7.5 km[/tex]
The distance the car travels is 7.5 kilometers.
When measuring the diameter of a cylinder the following measurements are obtained.
16.4 millimeters
16.8 millimeters
16.9 millimeters
16.6 millimeters
16.8 millimeters
Find the percent uncertainty in the measurements.
(Keep 2 decimal places in the percentage.)
Answer:
a. 1.80 %
b. 0.60 %
c. 1.20 %
d. 0.60 %
e. 0.60 %
Explanation:
In order to calculate percent uncertainties in each case, we forst need to calculate the average value:
Average Diameter = (16.4 mm + 16.8 mm + 16.9 mm + 16.6 mm +16.8 mm)/5
Average Diameter = 16.7 mm
Now, the formula for percent uncertainty is:
% Uncertainty = (Uncertainty/Average) * 100 %
where,
Uncertainty = |Value - Average|
For Each Case:
a. 16.4 mm:
Uncertainty = |16.4 mm - 16.7 mm| = 0.3 mm
Therefore,
% Uncertainty = (0.3 mm/16.7 mm) * 100%
% Uncertainty = 1.80 %
b. 16.8 mm:
Uncertainty = |16.8 mm - 16.7 mm| = 0.1 mm
Therefore,
% Uncertainty = (0.1 mm/16.7 mm) * 100%
% Uncertainty = 0.60 %
c. 16.4 mm:
Uncertainty = |16.9 mm - 16.7 mm| = 0.2 mm
Therefore,
% Uncertainty = (0.2 mm/16.7 mm) * 100%
% Uncertainty = 1.20 %
d. 16.6 mm:
Uncertainty = |16.6 mm - 16.7 mm| = 0.1 mm
Therefore,
% Uncertainty = (0.1 mm/16.7 mm) * 100%
% Uncertainty = 0.60 %
e. 16.8 mm:
Uncertainty = |16.8 mm - 16.7 mm| = 0.3 mm
Therefore,
% Uncertainty = (0.1 mm/16.7 mm) * 100%
% Uncertainty = 0.60 %
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.
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 .
Hypothesis If an object rolls over _______________ (type of material), then it will _____________ (describe the prediction of distance it will travel)
Answer:
smooth
go far
Which dartboard represents high accuracy and low precision?
1. A car travels at 55 km/h for 6.0 hours. How far does it travel?
Answer:
330 km
55*6= 330 km. An easy formula for this is multiplying time with speed.
Explanation:
330 kilometros
55 * 6 = 330 km. Una fórmula fácil para esto es multiplicar el tiempo por la velocidad.
Answer:330 km
Explanation:55*6= 330 km. An easy formula for this is multiplying time with speed.
What unit is time usually measured in?
If the kinetic energy of the ball decreases then the potential energy will____
Hey There!! ~
The answer to this is: Complete these sentences.
If a bouncing ball has a total energy of 20 J and a kinetic energy of 5 J, the ball’s potential energy is 15J.
If the kinetic energy of the ball decreases, then the potential energy will Increase.
Hope It Helped!~
[tex]ItsNobody[/tex]~
If the velocity of a proton is straight up (thumb pointing up) then RHR2 shows that the force points to the left. What would the direction of the force be if the velocity were
Answer:
a) to the right
b) up
c) down
d) there will be no force on the proton
e) there will be no force on the proton
Explanation:
The complete question is
If the velocity of a proton is straight up(thumb pointing up) then RHR2 (right hand rule) shows that the force points to the left. What would the direction of the force be if the velocity were a)down, b)to the right, c)to the left, d)into the page, and e)out of the page?
The right hand rule for a positive charge states that...
Hold the thumb of the right hand at right angle to the rest of the fingers, and the rest of the fingers parallel to one another, and pointing away from the body. If the thumb shows the velocity of a positive charge in a magnetic field, and the fingers all point in the direction of the magnetic field, then the palm will push in the direction of the force on the positive charge (proton).
From this, we can deduce from the original statement about this proton that the direction of the field is into the screen of this computer. with that field direction held constant, we can work out that
a) if the thumb point down, the force will be to the right
b) if the thumb points to the right, the force will be upwards
c) if the thumb points to the left, the force is downwards
d) if the thumb points into the page, then there will be no force on the proton since the proton must travel perpendicularly to the magnetic field for a force to be induced on it
e) explanation is the same as foe option d.
What is the electric field produced by this charge at point P, which is on the x-axis, 9.83 mm from the origin
Given :
Distance of point P from the origin is , d = 9.83 mm .
To Find :
The electric field produced by this charge at point P , if charge is placed at origin .
Solution :
Let , charge on origin is q .
Electric field at point P is given by :
[tex]E=\dfrac{kq}{r^2}[/tex] ..... ( 1 )
Here , k is constant , [tex]k=9\times 10^{9}\ N \ m^2/C^2[/tex] .
Putting value of k and r in above equation , we get :
[tex]E=\dfrac{q\times 9\times 10^9}{(9.83\times 10^{-3})^2}\ N/C\\\\E=(9.31\times 10^{13})q\ N/C[/tex]
Hence , this is the required solution .
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:
brainly.com/question/1748290
#SPJ2
according to the law , naturalized amrican citizens
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
Lightning occurs when there is a flow of electric charge (principally electrons) between the ground and a thundercloud. The maximum rate of charge flow in a lightning bolt is about 20,000 C/s this lasts for 100 μs or less.1. How much charge flows between the ground and the cloud in this time? (Q= ? C)2. How many electrons flow during this time? (n_e = ?)
Answer:
a
The charge is [tex]Q = 2.0*10^8 \ C[/tex]
b
The number of electrons [tex]N = 1.25 *10^{27} \ electrons[/tex]
Explanation:
From the question we are told that
The maximum rate of charge flow is [tex]\frac{dq}{dt} = I = 20000 \ C/s[/tex]
Here I means current
The time interval is [tex]t=100 \mu s = 100*10^{-6}[/tex]
Generally the amount of charge is mathematically represented as
[tex]Q = \frac{I}{t}[/tex]
=> [tex]Q = \frac{20000}{ 100*10^{-6}}[/tex]
=> [tex]Q = 2.0*10^8 \ C[/tex]
The number of electrons that flow during that time is evaluated as
[tex]N = \frac{Q}{e}[/tex]
Here e is the charge on a single electron with value [tex]e= 1.60*10^{-19} \ C[/tex]
So
[tex]N = \frac{2.0*10^8}{ 1.60*10^{-19}}[/tex]
[tex]N = 1.25 *10^{27} \ electrons[/tex]
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: