Answer:
The drill's angular displacement during that time interval is 24.17 rad.
Explanation:
Given;
initial angular velocity of the electric drill, [tex]\omega _i[/tex] = 5.21 rad/s
angular acceleration of the electric drill, α = 0.311 rad/s²
time of motion of the electric drill, t = 4.13 s
The angular displacement of the electric drill at the given time interval is calculated as;
[tex]\theta = \omega _i t \ + \ \frac{1}{2}\alpha t^2\\\\\theta = (5.21 \ \times \ 4.13) \ + \ \frac{1}{2}(0.311)(4.13)^2\\\\\theta = (21.5173 ) \ + \ (2.6524)\\\\\theta =24.17 \ rad[/tex]
Therefore, the drill's angular displacement during that time interval is 24.17 rad.
3
4
Lucy runs 4 meters to the east, then 5 meters south. What is the magnitude of her displacement?
Show Your Work
Answer:
The displacement is 6.4m
Explanation:
Step one:
given
we are told that Lucy runs 4 meters to the east,
then 5 meters south.
let the distance east be the displacement in the x-direction, and south be the y-direction
Step two:
The resultant of the x and y displacement is the magnitude of the total displacement z
applying Pythagoras theorem we have
z=√x^2+y^2
z=√4^2+5^2
z=√16+25
z=√41
z=6.4m
PLEASE ans The question's in the pictures, please don't answer what already has answers. Only answer if you can finish both pages completely PLEASE I NEED HELP :(( if ur ans is relevant I will mark brainliest
[2.21] Please help me find a) and b)
Answer:
A. 28.42 m/s
B. 41.21 m.
Explanation:
A. Determination of the initial velocity of the ball:
Time (t) to reach the maximum height = 2.9 s
Final velocity (v) = 0 (at maximum height)
Acceleration due to gravity (g) = –9.8 m/s² (since the ball is going against gravity)
Initial velocity (u) =?
Thus, we can obtain the initial velocity of the ball as follow:
v = u + gt
0 = u + (–9.8 × 2.9)
0 = u – 28.42
Collect like terms
u = 0 + 28.42
u = 28.42 m/s
Therefore, the initial velocity of the ball is 28.42 m/s.
B. Determination of the maximum height reached.
Final velocity (v) = 0 (at maximum height)
Acceleration due to gravity (g) = –9.8 m/s² (since the ball is going against gravity)
Initial velocity (u) = 28.42 m/s.
Maximum height (h) =?
Thus, we can obtain the maximum height reached by the ball as follow:
v² = u² + 2gh
0² = 28.42² + (2 × –9.8 × h)
0 = 807.6964 + (–19.6h)
0 = 807.6964 – 19.6h
Collect like terms
0 – 807.6964 = – 19.6h
– 807.6964 = – 19.6h
Divide both side by – 19.6
h = –807.6964 / –19.6
h = 41.21 m
Therefore, the maximum height reached by the ball is 41.21 m
How high does a rocket have to go above the earth's surface to be subject to a gravitational field from the earth that is 50.0 percent of its value at the earth's surface?
A) 2.650 km
B) 3,190 km
C) 9.020 km
D) 12.700 km
Answer:
A) 2.650 km
Explanation:
The relationship between acceleration of gravity and gravitational constant is:
[tex]g = \frac{Gm}{R^2}[/tex] ---- (1)
Where
[tex]R = 6,400 km[/tex] -- Radius of the earth.
From the question, we understand that the gravitational field of the rocket is 50% of its original value.
This means that:
[tex]g_{rocket} = 50\% * g[/tex]
[tex]g_{rocket} = 0.50 * g[/tex]
[tex]g_{rocket} = 0.5g[/tex]
For the rocket, we have:
[tex]g_{rocket} = \frac{Gm}{r^2}[/tex]
Where r represent the distance between the rocket and the center of the earth.
Substitute 0.5g for g rocket
[tex]0.5g = \frac{Gm}{r^2}[/tex] --- (2)
Divide (1) by (2)
[tex]\frac{g}{0.5g} = \frac{Gm}{R^2}/\frac{Gm}{r^2}[/tex]
[tex]\frac{g}{0.5g} = \frac{Gm}{R^2}*\frac{r^2}{Gm}[/tex]
[tex]\frac{1}{0.5} = \frac{1}{R^2}*\frac{r^2}{1}[/tex]
[tex]2 = \frac{r^2}{R^2}[/tex]
Take square root of both sides
[tex]\sqrt 2 = \frac{r}{R}[/tex]
Make r the subject
[tex]r = R * \sqrt 2[/tex]
Substitute [tex]R = 6,400 km[/tex]
[tex]r = 6400km * \sqrt 2[/tex]
[tex]r = 6400km * 1.414[/tex]
[tex]r = 9 049.6\ km[/tex]
The distance (d) from the earth surface is calculated as thus;
[tex]d = r - R[/tex]
[tex]d = 9049.6\ km - 6400\ km[/tex]
[tex]d = 2649.6\ km[/tex]
[tex]d = 2650\ km[/tex] --- approximated
Which of the following is the recommended amount of fats per meal for male clients
Answer:
44 grams- 55 grams through the whole day. Probably about 14.6 grams per meal.
Explanation:
Answer:
2 thumbs (ISSA Guide)
Explanation:
A merry-go-round on a playground consists of a horizontal solid disk with a weight of 805 N and a radius of 1.58 m. A child applies a force 49.5 N tangentially to the edge of the disk to start it from rest. What is the kinetic energy of the merry-go-round disk (in J) after 2.95 s?
Answer:
The value is [tex]KE = 259.6 \ J[/tex]
Explanation:
From the question we are told that
The weight of the horizontal solid disk is [tex]W = 805 \ N[/tex]
The radius of the horizontal solid disk is [tex]r = 1.58 \ m[/tex]
The force applied by the child is [tex]F = 49.5 \ N[/tex]
The time considered is [tex]t = 2.95 \ s[/tex]
Generally the mass of the horizontal solid disk is mathematically represented as
[tex]m_h = \frac{W}{ g}[/tex]
=> [tex]m_h = \frac{805}{ 9.8 }[/tex]
=> [tex]m_h = 82.14 \ N[/tex]
Generally the moment of inertia of the horizontal solid disk is mathematically represented as
[tex]I = \frac{1}{2} * m * r^ 2[/tex]
=> [tex]I = \frac{1}{2} * 82.14 * 1.58^ 2[/tex]
=> [tex]I = 102.5 \ kg \cdot m^2[/tex]
Generally the net torque experienced by the horizontal solid disk is mathematically represented as
[tex]T = I * \alpha = F * r[/tex]
=> [tex]\alpha = \frac{ F * r }{ I }[/tex]
=> [tex]\alpha = \frac{ 49.5 * 1.58 }{ 102.53 }[/tex]
=> [tex]\alpha = 0.7628[/tex]
Gnerally from kinematic equation we have that
[tex]w = w_o + \alpha t[/tex]
Here [tex]w_o[/tex] is the initial angular velocity velocity of the horizontal solid disk which is [tex]w_o = 0\ rad/s[/tex]
So
[tex]w = 0 + 0.7628 * 2.95[/tex]
=> [tex]w = 2.2503 \ rad/s[/tex]
Generally the kinetic energy is mathematically represented as
[tex]KE = \frac{1}{2} * I * w^2[/tex]
=> [tex]KE = \frac{1}{2} * 102.53 * 2.2503 ^2[/tex]
=> [tex]KE = 259.6 \ J[/tex]
A vertical tube one meter long is open at the top. It is filled with 50 cm of water. If the velocity of sound is 344 m/s, what will the fundamental resonant frequency be (in Hz)?
Answer:
The fundamental resonance frequency is 172 Hz.
Explanation:
Given;
velocity of sound, v = 344 m/s
total length of tube, Lt = 1 m = 100 cm
height of water, hw = 50 cm
length of air column, L = Lt - hw = 100 cm - 50 cm = 50 cm
For a tube open at the top (closed pipe), the fundamental wavelength is given as;
Node to anti-node (N ---- A) : L = λ / 4
λ = 4L
λ = 4 (50 cm)
λ = 200 cm = 2 m
The fundamental resonance frequency is given by;
[tex]f_0 = \frac{v}{\lambda}\\\\f_0 = \frac{344}{2}\\\\f_0 = 172 \ Hz\\\\[/tex]
Therefore, the fundamental resonance frequency is 172 Hz.
When weather predictions are incorrect what is the most likely cause
A: measurements of the initial conditions may have been very in accurate
B: small differences in models can lead to large differences in complex systems
C: The person predicting the weather may have had a bias
D: The elevation of different landforms I have been significantly in accurate
Answer:small differences in models can lead to large differences in complex systems
Explanation: this is the most accurate phrase
A rotating heavy wheel is used to store energy as kinetic energy. If it is designed to store 1.00 x 106 J of kinetic energy when rotating at 64 revolutions per second, find the moment of inertia (rotational inertia) of the wheel. (Hint: Start with the expression for rotational kinetic energy.)
We know, [tex]1\ rpm = \dfrac{2\pi}{60} \ rad/s[/tex] .
[tex]64\ rpm\ is = \dfrac{2\pi}{60}\times 64\ rad/s\\\\= \dfrac{32\pi}{15}\ rad/s[/tex]
We know, kinetic energy is given by :
[tex]K.E = \dfrac{I\omega^2}{2}\\\\I = \dfrac{2(K.E)}{\omega^2}\\\\I = \dfrac{2\times 10^6}{\dfrac{32}{15}\times \pi}\\\\I = 298415.52 \ kg \ m^2[/tex]
Hence, this is the required solution.
name the force that help us to walk
Compute the specific heat capacity at constant volume of nitrogen (N2) gas. The molar mass of N2 is 28.0 g/mol.
Answer:
724.3J/Kg.K
Explanation:
CHECK THE COMPLETE QUESTION BELOW
Compute the specific heat capacity at constant volume of nitrogen (N2) gas.and compare with specific heat of liquid water. The molar mass of N2 is 28.0 g/mol.
The specific heat capacity can be computed by using expression below
c= CV/M
Where c= specific heat capacity
M= molar mass
CV= molar hear capacity
Nitrogen is a diatomic element, the Cv can be related to gas constant with 5/2R
Where R= 8.314J/mol.k
Molar mass= 28 ×10^-3Kg/mol
If we substitute to the expression, we have
c= (5R/2)/(M)
=5R/2 × 1/M
=(5×8.314) /(2×28 ×10^-3)
=724.3J/Kg.K
Hence, the specific heat capacity at constant volume of nitrogen (N2) gas is
724.3J/Kg.K
The specific heat of liquid water is about 4182 J/(K kg) which is among substance with high specific heat, therefore specific heat of Nitrogen gas is 724.3J/Kg.K which is low compare to that of liquid water.
Psychologists often talk of the nature-nurture controversy. Which of these
concepts supports the "nurture" over the "nature" part?
O A. Freudian concepts
O B. James-Lange
OC. Tabula rasa
OD. Biological origins
Answer:
O A. Freudian concepts
Freud (1905) stated that events in our childhood have a great influence on our adult lives, shaping our personality. He thought that parenting is of primary importance to a child's development, and the family as the most important feature of nurture was a common theme throughout twentieth-century psychology (which was dominated by environmentalists theories).
Explanation:
Hope this helped!
O A. Freudian concepts. Freud (1905) stated that events in our childhood have a great influence on our adult lives, shaping our personality.
Who is O.A. Freudian?He thought that parenting is of primary importance to a child's development, and the family as the most important feature of nurture was a common theme throughout twentieth-century psychology (which was dominated by environmentalists theories).
One of the case studies most closely linked with the Austrian psychotherapist Sigmund Freud is the hysterics and treatment of Anna O.
Despite the fact that Freud is intimately identified with Anna O, it is thought that he never actually treated her; instead, Breuer saw the patient.
Therefore, O A. Freudian concepts. Freud (1905) stated that events in our childhood have a great influence on our adult lives, shaping our personality.
To learn more Psychology, refer to the link:
https://brainly.com/question/29436774
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How to calculate net radiation
Answer:
(1) R n = ( 1 − α ) R si − L ↑ + L ↓ where Rn is the net radiation (W m−2), Rsi is the solar radiation (W m−2), α is the soil surface albedo (α = 0–1
Explanation:
Hey guys this is Ap physics please help I need this to pass i will mark brainliest for a good attempt
Split up the forces into components acting parallel to and perpendicular to the slope. See the attached picture for the reference axes.
The box stays on the surface of the plane, so that the net force acting perpendicular to it is 0, and the only acceleration is applied in the parallel direction.
Let m be the mass of the box, θ the angle the plane makes with the ground, and a the acceleration of the box. By Newton's second law, we have
• net parallel force
∑ Force (//) = W (//) - F = m a
(that is, the net force in the parallel direction is the sum of the parallel component of the weight W and the friction F which acts in the negative direction)
• net perpendicular force
∑ Force (⟂) = W (⟂) + N = 0
Notice that
W (//) = W sin(θ) … … … which is positive since it points down the plane
W (⟂) = -W cos(θ) … … … which is negative since it points opposite the normal force N
So the equations become
W sin(θ) - F = m a
-W cos(θ) + N = 0
Solving for a gives
a = (W sin(θ) - F ) / m
which is good enough if you know the magnitude of the friction force.
If you don't, you can write F in terms of the coefficient of kinetic friction between the box and plane, µ, as
F = µ N
so that
a = (W sin(θ) - µ N ) / m
and the normal force itself has a magnitude of
N = W cos(θ)
so that
a = (W sin(θ) - µ W cos(θ) ) / m
The weight W has magnitude m g, where g is the magnitude of the acceleration due to gravity, so
a = (m g sin(θ) - µ m g cos(θ) ) / m
a = g (sin(θ) - µ cos(θ))
one newton equals 0.225
in the case shown below, the 1 kg rock rides on a horizontal disk that rotates at constant speed 5m/s
Answer:25 N
Explanation:
in the case shown below, the 1 kg rock rides on a horizontal disk that rotates at constant speed 5m/s is 25N
what is speed ?Speed is the ratio of distance with respect to the time in which the distance was covered. Speed is a scalar quantity as it does not have magnitude only have direction
The formula of speed can be represented as s=d/t, Where, s is the speed in m.s-1, d is the distance traveled in m, t is the time taken in s
Uniform speed is defined when the object covers equal distance at equal time intervals, variable speed is defined as when the object covers a different distance at equal intervals of times.
Average speed is defined as the total distance travelled by an object to the total time taken by the object.
Instantaneous speed is defined as when the object is move with variable speed, then the speed at any instant of time is known as instantaneous speed.
For more details regarding speed, visit
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A force Ě = F, î + Fy h acts on a particle
that undergoes a displacement of g = s, it
Sy Ĵ where F, = 10 N, F, = -1 N, 8z = 4 m,
and sy=1 m.
Find the work done by the force on the
particle.
Answer in units of J.
Answer:
Pt 1: [tex]W=39J[/tex]
Pt 2: θ = [tex]19.7[/tex]
Explanation:
Part 1:
[tex]W=FS[/tex]
[tex]W=F_{_xs_x}+F{_ys_y}[/tex]
imagine the "i"and the "j" with the hat
[tex]W=(10i-1j)(4i+3j)[/tex]
[tex]W=(10*4)-(1*1)[/tex]
[tex]W=40-1[/tex]
[tex]W=39J[/tex]
Part 2:
[tex]|F|=\sqrt{10^2+(-1)^2}[/tex]
[tex]|F|=\sqrt{101}[/tex]
[tex]|S|=\sqrt{4^2+1^2}[/tex]
[tex]|S|=\sqrt{17}[/tex]
[tex]W=|F||S|cos[/tex]θ
[tex]39=(\sqrt{101})(\sqrt{17} )cos[/tex]θ
θ = [tex]19.7[/tex]
Hope it helps
Light travels from a
laser across an 8 m
room where it reflects
off a mirror. Uniform or non uniform velocity
Answer:
Non-uniform velocity as the laser light beam has got reflected by the mirror
And as the light got reflected there is a change in velocity making it non-uniform velocity
The mass and coordinates of three objects are given below: m1 = 6.0 kg at (0.0, 0.0) m, m2 = 1.5 kg at (0.0, 4.1) m, and m3 = 4.0 kg at (1.9, 0.0) m. Determine where we should place a fourth object with a mass m4 = 7.9 kg so that the center of gravity of the four-object arrangement will be at (0.0, 0.0) m
Answer:
The location of the center of gravity of the fourth mass is [tex]\vec r_{4} = (-0.961\,m,-0.779\,m)[/tex].
Explanation:
Vectorially speaking, the center of gravity with respect to origin ([tex]\vec r_{cg}[/tex]), measured in meters, is defined by the following formula:
[tex]\vec r_{cg} = \frac{m_{1}\cdot \vec r_{1}+m_{2}\cdot \vec r_{2}+m_{3}\cdot \vec r_{3}+m_{4}\cdot \vec r_{4}}{m_{1}+m_{2}+m_{3}+m_{4}}[/tex] (1)
Where:
[tex]m_{1}[/tex], [tex]m_{2}[/tex], [tex]m_{3}[/tex], [tex]m_{4}[/tex] - Masses of the objects, measured in kilograms.
[tex]\vec r_{1}[/tex], [tex]\vec r_{2}[/tex], [tex]\vec r_{3}[/tex], [tex]\vec r_{4}[/tex] - Location of the center of mass of each object with respect to origin, measured in meters.
If we know that [tex]\vec r_{cg} = (0,0)\,[m][/tex], [tex]\vec r_{1} = (0,0)\,[m][/tex], [tex]\vec r_{2} = (0, 4.1)\,[m][/tex], [tex]\vec r_{3} = (1.9,0.0)\,[m][/tex], [tex]m_{1} = 6\,kg[/tex], [tex]m_{2} = 1.5\,kg[/tex], [tex]m_{3} = 4\,kg[/tex] and [tex]m_{4} = 7.9\,kg[/tex], then the equation is reduced into this:
[tex](0,0) = \frac{(6\,kg)\cdot (0,0)\,[m]+(1.5\,kg)\cdot (0,4.1)\,[m]+(4.0\,kg)\cdot (1.9,0)\,[m]+(7.9\,kg)\cdot \vec r_{4}}{6\,kg+1.5\,kg+4\,kg+7.9\,kg}[/tex]
[tex](6\,kg)\cdot (0,0)\,[m]+(1.5\,kg)\cdot (0,4.1)\,[m]+(4\,kg)\cdot (1.9,0)\,[m]+(7.9\,kg)\cdot \vec r_{4} = (0,0)\,[kg\cdot m][/tex]
[tex](7.9\,kg)\cdot \vec r_{4} = -(6\,kg)\cdot (0,0)\,[m]-(1.5\,kg)\cdot (0,4.1)\,[m]-(4\,kg)\cdot (1.9,0)\,[m][/tex]
[tex]\vec r_{4} = -0.759\cdot (0,0)\,[m]-0.190\cdot (0,4.1)\,[m]-0.506\cdot (1.9,0)\,[m][/tex]
[tex]\vec r_{4} = (0, 0)\,[m] -(0, 0.779)\,[m]-(0.961,0)\,[m][/tex]
[tex]\vec r_{4} = (-0.961\,m,-0.779\,m)[/tex]
The location of the center of gravity of the fourth mass is [tex]\vec r_{4} = (-0.961\,m,-0.779\,m)[/tex].
What is the name for family labeled #4 (Yellow)?
#3
#5
#2
#
341 sud-
lasa 1
17:55
Alkaline Earth Metals
Metalloids
Transition Metals
Alkali Metals
Answer:
transition metals im sorry if this was too late
What is the initial vertical velocity of the ball?
A.
0 m/s
B.
9.81 m/s
C.
20.0 m/s
D.
60.0 m/s
A horse has a momentum of 1200 kg·m/s. If the horse has a mass of 313 kg, what is the speed of the horse?
Answer:
3.83 m/sExplanation:
The speed of the horse can be found by using the formula
[tex]v = \frac{p}{m} \\ [/tex]
p is the momentum
m is the mass
From the question we have
[tex]v = \frac{1200}{313} \\ = 3.83386..[/tex]
We have the final answer as
3.83 m/sHope this helps you
Does anyone knows this for physics?
find the vector parallel to the resultant of the vector A=i +4j-2k and B=3i-5j+k
Answer:
2008
Explanation:
2000+3+5======2008
Answer:
[tex]8\hat i-2\hat j-2\hat k[/tex]
Explanation:
Vectors in 3D
Given a vector
[tex]\vec P = P_x\hat i+P_y\hat j+P_z\hat k[/tex]
A vector [tex]\vec Q[/tex] parallel to [tex]\vec P[/tex] is:
[tex]\vec Q = k.\vec P[/tex]
Where k is any constant different from zero.
We are given the vectors:
[tex]\vec A = \hat i+4\hat j-2\hat k[/tex]
[tex]\vec B = 3\hat i-5\hat j+\hat k[/tex]
It's not specified what the 'resultant' is about, we'll assume it's the result of the sum of both vectors, thus:
[tex]\vec A +\vec B = \hat i+4\hat j-2\hat k + 3\hat i-5\hat j+\hat k[/tex]
Adding each component separately:
[tex]\vec A +\vec B = 4\hat i-\hat j-\hat k[/tex]
To find a vector parallel to the sum, we select k=2:
[tex]2(\vec A +\vec B )= 8\hat i-2\hat j-2\hat k[/tex]
Thus one vector parallel to the resultant of both vectors is:
[tex]\mathbf{8\hat i-2\hat j-2\hat k}[/tex]
a car traveling at 28.4 m/s undergoes a constant deceleration of 1.92 m/s2 when the breaks are applied. How many revolutions does each tire make before the car comes to a stop
Complete Question
a car traveling at 28.4 m/s undergoes a constant deceleration of 1.92 m/s2 when the breaks are applied. How many revolutions does each tire make before the car comes to a stop? Assume that the car does not skid and that each tire has a radius of 0.307 m. Answer in units of rev.
Answer:
The value is [tex]N = 109 \ rev[/tex]
Explanation:
From the question we are told that
The speed of the car is [tex]u = 28.4 \ m/s[/tex]
The constant deceleration experienced is [tex]a = 1.92 \ m/s^2[/tex]
The radius of the tire is [tex]r = 0.307 \ m[/tex]
Generally from kinematic equation we have that
[tex]v^2 = u^2 + 2as[/tex]
Here v is the final velocity which is 0 m/s
So
[tex]0^2 = 28.4^2 + 2 * 1.92 * s[/tex]
=> [tex]s = 210.04 \ m[/tex]
Generally the circumference of the tire is mathematically represented as
[tex]C = 2 \pi r[/tex]
=> [tex]C = 2 * 3.142 * 0.307[/tex]
=> [tex]C = 1.929 \ m[/tex]
Generally the number of revolution is mathematically represented as
[tex]N = \frac{ s}{C}[/tex]
=> [tex]N = \frac{210.04}{1.929}[/tex]
=> [tex]N = 109 \ rev[/tex]
_____ are group of tissue working together to perform a certain job.
Answer:
organ
Explanation:
Answer:
an organ
Explanation:
cell -> tissue -> organ -> organ system -> organism
In order to prevent injury in a car crash, it is recommended that you _______.
A) Increase the time of the collision.
B) Increase the change in momentum of the collision.
C) Increase the force in the collision.
D) Increase the initial velocity of the collision.
Increase of momentum of the collision will have the car in a unsettling position, creating an unsettling spot for it. Moreover increasing this would most likely take worse effect, it is better than increasing the time, which will only create the car faster. Let me show you what I mean...
A) Creates the car go faster, creating an even worse tragedy. B) save this for later...C) Increasing force will only result in worse damage.And finally, D) Increasing the velocity is basically increasing speed, once again, making things worse.So overall in this piece, the answer may very well end up being B. I sincerely hope this helped you by whatever means possible. It's logic that helps in real life situations, so take this as a little lesson- I guess :3How long does it take a plane, traveling at a constant speed of 123 m/s, to fly once around a circle whose radius is 4330 m?
Answer:
3.7 minExplanation:
Step one:
given data
speed = 123m/s
radius of circle= 4330m
Step two:
We need to find the circumference of the circle, it represents the distance traveled
C=2πr
C= 2*3.142*4330
C= 27209.72m
Step three:
We know that velocity= distance/time
time= distance/velocity
time= 27209.72/123
time=221.2 seconds
in minute = 221.2/60
time= 3.7 min
HELPP physics final will give brainliest
The students look through the side of the aquarium.
They notice that the image of the tongs appears to break as the tongs enter the water.
Which property of light are the students observing in this situation?
Answer:
light refraction
Explanation: