Answer:
θ = 12.5 rotations
Explanation:
The number of rotations can be found by using the second equation of motion:
[tex]\theta = \omega_i t + \frac{1}{2}\alpha t^2\\\\[/tex]
where,
[tex]\theta[/tex] = angular displacement = ?
ωi = initial angular speed = 0 rad/s
t = time = 5 s
α = angular acceleration = 2π rad/s²
Therefore,
[tex]\theta = (0\ rad/s)(5\ s)+\frac{1}{2}(2\pi\ rad/s^2)(5\ s)^2\\\\\theta = 78.54\ rad[/tex]
converting it to no. or rotations:
[tex]\theta = (78.54\ rad)(\frac{1\ rotation}{2\pi\ rad})[/tex]
θ = 12.5 rotations
Use a scientific calculator to perform the operation below.
5.92 x 107 + 2.11 x 106
A. 6.13 x 107
B. 2.81 x 101
C. 1.25 x 1014
D. 5.71 x 107
SUBMIT
Answer:
A. 6.13 x 107
Explanation:
Given the expression
5.92 x 10^7 + 2.11 x 10^6
First, we need to convert to the whole number
5.92 x 10^7 = 59200000
2.11 x 10^6 = 2110000
Add both values
59200000 + 2110000
= 61,310,000
Express in standard form
= 6.13 * 10^7
This gives the required result
A certain electric stove has a 16 Ω heating element. The current going through the element is 15 A. Calculate the voltage across the element.
Answer:
V = 240V
Explanation:
V = I*R
V = 15A*16ohms
V = 240V
A total positive charge of 12.00 mC is evenly distributed on a straight thin rod of length 6.00 cm.
A positive point charge, Q = 4.00 nC, is located a distance of 5.00 cm above the midpoint of the
rod. What will be the electrical force on the point charge?
A box with mass 25.14 kg is sliding at rest from the top of the slope with height 13.30 m
and slope angle 30 degree, suppose the coefficient of friction of the slope surface is
0.25, find (neglect air resistance,take g=10 m/s^2)
The friction force experienced by the box.
00) The acceleration of the box along the slope.
(1) The time T required for the object to reach the bottom of the slope from the slope top.
Answer:
Explanation:
The first thing we are asked to find is the Force experienced by the box. That is found in the formula:
F - f = ma where F is the force exerted by the box, f is the friction opposing the box, m is the mass, and a is the acceleration (NOT the same as the pull of gravity). But F can be rewritten in terms of the angle of inclination also:
[tex]wsin\theta-f=ma[/tex] where w is the weight of the box. We will use this version of the formula because it will help us answer the second question, which is to solve for a. Filling in:
First we need the weight of the box. Having the mass, we find the weight:
w = mg so
w = 25.14(10) so
w = 251.4 N (I am not paying any attention at all to the sig fig's here, since I noticed no one on this site does!) Now we have the weight. Filling that in:
251.4sin(30) - f = ma Before we go on to fill in for f, let's answer the first question. F = 251.4sin(30) so
F = 125.7 And in order to answer what a is equal to, we find f:
f = μ[tex]F_n[/tex] where Fn is the weight of the object.
f = .25(251.4) so
f = 62.85. Filling everything in now altogether to solve for a, the only missing value:
125.7 - 62.85 = 25.14a and
62.85 = 25.14a so
a = 2.5 m/s/s
Now we have to move on to another set of equations to answer the last part. The last part involves the y-dimension. In this dimension, what we know is that
a = -10 m/s/s
v₀ = 0 (it starts from rest)
Δx = -13.30 m (negative because the box falls this fr below the point fro which it started). Putting all that together in the equation for displacement:
Δx = v₀t + [tex]\frac{1}{2}at^2[/tex] and we are solving for time:
[tex]-13.30=0t+\frac{1}{2}(-10)t^2[/tex] and
[tex]t=\sqrt{\frac{2(-13.30)}{-10} }[/tex] so
t = 1.6 seconds to reach the bottom of the slope from 13.30 m high.
If ATM is 102 kPa, what force does the atmosphere exert on the palm of your hand which has an area of 0.016 meters?
Answer:
Force = 1.632 Newton
Explanation:
Given the following data;
Pressure = 102 kPa
Area = 0.016 m²
To find what force the atmosphere exert on the palm of your hand;
Mathematically, pressure is given by the formula;
[tex] Pressure = \frac {Force}{area} [/tex]
Force = 102 * 0.016
Force = 1.632 Newton
Answer the following using equations, number substitution and keep units. 1. What is the speed of an object that travels 5m in 10s. 2. What force is on a 10kg mass that accelerates at 3m/s/s. 3. What is the potential energy of a 7kg object 4m off the ground *
show all your work please
Explanation:
1. Distance, d = 5 m
Time, t = 10 s
Speed = distance/time
[tex]v=\dfrac{5}{10}=0.5\ m/s[/tex]
2. Mass, m = 10 kg
Acceleration, a = 3 m/s³
Force, F = mass (m) × acceleration (a)
F = 10 × 3
= 20 N
3. Mass, m = 7 kg
Height, h = 4 m
Potential energy, E = mgh
E = 7 × 9.8 × 4
E = 274.4 J
Hence, this is the required solution.
A person rolls a 7 kg bowling ball down a lane in a bowling alley. The lane is
18 m long. The ball is traveling at 7 m/s when it leaves the person's hand.
What is the ball's kinetic energy at this point?
Answer:
171.5J
Explanation:
K=1/2 *m *U²
K=1/2 *7 *7²
K=171.5 J
The equation of damped oscillations is given in the form x=0.05e^-0.25sin½πt (m). Find the velocity of an oscillating point at the moments of time: 0, T, 2T, 3T and 4T.
Explanation:
Finding the (maximum) respective prime powers would yield the answer. Also we need not ... Is perfectly divisible by 720^n? ... So we can say that for any positive value of n it not divisible.
an object falls freely from rest the total distance covered by it in 2s will be
Answer:
Distance, S = 19.6 meters
Explanation:
Given the following data;
Time = 2 seconds
We know that acceleration due to gravity is equal to 9.8 m/s².
Also, the initial velocity of the object is equal to zero because it's starting from rest.
To find the total distance covered by the object, we would use the second equation of motion;
[tex] S = ut + \frac {1}{2}at^{2}[/tex]
Where;
S represents the displacement or height measured in meters. u represents the initial velocity measured in meters per seconds. t represents the time measured in seconds. a represents acceleration measured in meters per seconds square.Substituting into the equation, we have;
[tex] S = 0*2 + \frac {1}{2}*(9.8)*2^{2}[/tex]
[tex] S = 0 + 4.9*4[/tex]
[tex] S = 4.9*4 [/tex]
Distance, S = 19.6 meters
Therefore, the total distance covered by the object is 19.6 meters.
A car's bumper is designed to withstand a 5.04 km/h (1.4-m/s) collision with an immovable object without damage to the body of the car. The bumper cushions the shock by absorbing the force over a distance. Calculate the magnitude of the average force on a bumper that collapses 0.255 m while bringing a 830 kg car to rest from an initial speed of 1.4 m/s.
Answer:
the magnitude of the average force on the bumper is 3189.8 N
Explanation:
Given the data in the question;
In terms of force and displacement, work done is;
W =[tex]F^>[/tex] × [tex]x^>[/tex]
W = [tex]Fxcos\theta[/tex] ------- let this be equation 1
where F is force applied, x is displacement and θ is angle between force and displacement.
Now, since the displacement of the bumper and force acting on it is in the same direction,
hence, θ = 0°
we substitute into equation 1
W = [tex]Fxcos([/tex] 0° [tex])[/tex]
W = [tex]Fx[/tex] ------- let this be equation 2
Now, using work energy theorem,
total work done on the system is equal to the change in kinetic energy of the system.
[tex]W_{net[/tex] = ΔKE
= [tex]\frac{1}{2}[/tex]mv² - [tex]\frac{1}{2}[/tex]mu² --------- let this be equation 3
where m is mass of object, v is final velocity, u is initial velocity.
from equation 2 and 3
[tex]Fx[/tex] = [tex]\frac{1}{2}[/tex]mv² - [tex]\frac{1}{2}[/tex]mu²
we make F, the subject of formula
F = [tex]\frac{m}{2x}[/tex]( v² - u² )
given that mass of car m = 830 kg, x = 0.255 m, v = 0 m/s, and u = 1.4 m/s
so we substitute
F = [tex]\frac{830}{(2)(0.255)}[/tex]( (0)² - (1.4)² )
F = 1627.45098 ( 0 - 1.96 )
F = 1627.45098 ( - 1.96 )
F = -3189.8 N
The negative sign indicates that the direction of the force was in opposite compare to the direction of the velocity of the car.
Therefore, the magnitude of the average force on the bumper is 3189.8 N
planet smaller than earth but larger than mercury
Answer:
venus......................
Part C
When only one color of light reflects from a piece of paper, what happens to the other colors of light?
Remember that light is energy, and energy cannot be created or destroyed.
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Answer:
the other colours get absorbed by the paper
Answer:
The other colors are absorbed by the paper and not reflected.
A crucible (container) of molten metal has an open top with an area of 5.000 m^2. The molten metal acts as a blackbody radiator. The intensity spectrum of its radiation peaks at a wavelength of 320 nm. What is the temperature of that blackbody?
Answer:
T = 9056 K
Explanation:
In the exercise they indicate that the body can be approximated by a black body, for which we can use the Wien displacement relation
λ T = 2,898 10⁻³
where lam is the wavelength of the maximum emission
T = 2,898 10⁻³ /λ
let's calculate
T = 2,898 10⁻³ / 320 10⁻⁹
T = 9.056 10³ K
T = 9056 K
A balloon is filled with 80 liters of gas on a day where the temperature was 34 degrees at sea level which is 101.3 kPa and released. As the balloon rises to a certain altitude, the temperature drops to 0 degrees celsius and the balloon doubles in volume. What is the atmospheric pressure at that altitude?
Answer:
0.444atm
Explanation:
Using the combined gas law equation;
P1V1/T1 = P2V2/T2
Where;
P1 = initial pressure (
P2 = final pressure (
V1 = initial volume (L)
V2 = final volume (L)
T1 = initial temperature (K)
T2 = final temperature (K)
According to this question,
P1 = 101.3 kPa = 101.3 × 0.00987 = 0.999atm
P2 = ?
V1 = 80L
V2 = 160L (double of V1)
T1 = 34°C = 34 + 273 = 307K
T2 = 0°C = 0 + 273 = 273K
Using P1V1/T1 = P2V2/T2
0.999 × 80/307 = P2 × 160/273
79.92/307 = 160P2/273
Cross multiply
307 × 160P2 = 79.92 × 273
49120P2 = 21818.16
P2 = 21818.16 ÷ 49120
P2 = 0.444
P2 = 0.444atm
What is this sport ⚽⚾
Answer:
sports are all forms of physical activity that contribute to physical fitness, mental well-being and social interaction.
hope it is helpful to you
A conducting sphere of radius R carries an excess positive charge and is very far from any other charges. Draw the graphs that best illustrates the potential (relative to infinity) produced by this sphere as a function of the distance r from the center of the sphere?
Answer:
See annex
Explanation:
By convention potential at ∞ V(∞ ) = 0
As the distance from the sphere decreases the potential increases up to the point d = R ( R is the radius of the sphere. That potential remains constant while d = R and becomes 0 inside the sphere where there is not free charges and therefore the electric field is 0 and so is the potential.
I am sorry I could not make a better graph
The graph that best illustrates the potential (relative to infinity) produced by this sphere as a function of the distance r from the center of the sphere is attached as an image below
[tex]V = \frac{KQ}{R}[/tex]
for r <= R
[tex]V = \frac{KQ}{r}[/tex]
for r > R
Therefore the graph will be
For more information on potentials as function of distance
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a brick of mass 0.8 kg is accidentally dropped from a high scaffolding. it reaches the ground with a kinetic energy of 240 J. How high is scaffolding ?(Take acceleration due to gravity g be 10 m s-¹)
Answer:
30 m
General Formulas and Concepts:
Energy
Gravitational Potential Energy: [tex]\displaystyle U_g = mgh[/tex]
m is mass (in kg)g is gravityh is height (in m)Kinetic Energy: [tex]\displaystyle KE = \frac{1}{2}mv^2[/tex]
m is mass (in kg)v is velocity (in m/s²)Law of Conservation of Energy
Explanation:
Step 1: Define
Identify variables
[Given] m = 0.8 kg
[Given] g = 10 m/s²
[Given] U = 240 J
[Solve] h
Step 2: Solve for h
[LCE] Substitute in variables [Gravitational Potential Energy]: (0.8 kg)(10 m/s²)h = 240 JMultiply: (8 kg · m/s²)h = 240 JIsolate h [Cancel out units]: h = 30 mWhat innovation did jethro wood add to plows in the 1800s?
Built by 1800, it was the home of inventor Jethro Wood (1774-1834), whose 1819 invention of an iron moldboard plow revolutionized American agriculture.
a car runs of a road and collides with a tree. glass pieces from the windscreen are projected forward and are found an average distance of 12m from the car. the average height of the windscreen is 1.2m.
Establish the speed of the car at the time of impact. assume g=10 m/s²
Answer:
v₀ₓ = 24.24 m / s
Explanation:
This is a projectile launching exercise, where the windshield comes out with a horizontal initial velocity.
Y axis
initial vertical velocity is zero
y = y₀ + v_{oy} t - ½ g t²
when it reaches the ground its height is zero and the initial height is y₀=1.2m
0 = y₀ + 0 - ½ g t²
t = [tex]\sqrt{2 y_o/g}[/tex]
t = [tex]\sqrt{2 \ 1.2 / 9.8}[/tex]
t = 0.495 s
X axis
x = v₀ₓ t
v₀ₓ = x / t
v₀ₓ = 12 / 0.495
v₀ₓ = 24.24 m / s
Question 1 of 25
Which equation is an example of a synthesis reaction?
A. HNO3 + KOH → KCI + H20
B. 2Li+ CaCl2 - 2LiCl + Ca
O C. S+ 02 - S02
7
O D. CH4 + 202 - 2H2O + CO2
Answer:
C. S + 02 → S02
Explanation:
A synthesis or combination reaction is that reaction involving two elements as reactants to form a single compound as product.
In the reaction given below;
S + 02 → S02
Sulphur and oxygen are elemental substances that combine to synthesize sulfur IV oxide (SO2), and hence it is an example of synthesis reaction.
Consider a 92.0 kg ice skater who is spinning on the ice. What is the moment of inertia of the skater, if the skater is approximated to be a solid cylinder that has a 0.140 m radius and is rotating about the center axis of the cylinder.
Answer:
[tex]I=0.902kg*m^2[/tex]
Explanation:
From the question we are told that:
Mass [tex]m=92.0kg[/tex]
Radius [tex]r=0.140m[/tex]
Generally the equation for moment of Inertia is mathematically given by
[tex]I = 0.5*m*r^2[/tex]
[tex]I=0.5*(92)*0.14^2[/tex]
[tex]I=0.902kg*m^2[/tex]
____ are the foundation of psychoanalytic theory.
Answer:
an unconscious needs rooted in childhood are the foundation of psychoanalytic theory.
Explanation:
can someone please help me
F = 153 N
[tex]\theta = 11.3°[/tex]
Explanation:
Let us define first our directional convention. Anything pointing up or to the right is considered positive and anything pointing down or to the left is considered negative. Now let's look at the components [tex]F_{x}[/tex] and [tex]F_{y}[/tex]:
[tex]F_{x}[/tex] = 350 N - 200 N = 150 N
[tex]F_{y}[/tex] = 180 N - 150 N = 30 N
The magnitude of the resultant force F is given by
[tex]F = \sqrt{F_{x}^{2}+F_{y}^{2}}[/tex]
[tex]\:\:\:\:\:\:= \sqrt{(150\:N)^{2}+(30\:N)^{2}}[/tex]
[tex]\:\:\:\:\:\:=153\:N[/tex]
To find the direction [tex]\theta[/tex], we use
[tex]\tan \theta = \dfrac{F_{y}}{F_{x}}=\dfrac{30\:N}{150\:N}=0.2[/tex]
or
[tex]\theta = \tan^{-1}(0.2) = 11.3°[/tex]
A skateboarder is inside of a half pipe, shown here. Explain her energy transformations as she jumps off at point A, slides to point B, and finally reaches point C.
You take your pulse and observe 80 heartbeats in a minute. What is the period of your heartbeat? What is the frequency of your heartbeat?
Answer:
120 beats per minute.
Explanation:
If I take your pulse and observe 80 heartbeats in a minute. Then the period of your heartbeat is 0.8 s and frequency is 1.3Hz.
What is Heartbeat ?A pulse is the term used in medicine to describe the tactile arterial palpation of the cardiac cycle (heartbeat) by skilled fingertips. Any location where an artery can be compressed close to the surface of the body, such as the carotid artery in the neck, the radial artery in the wrist, the femoral artery in the groyne, the popliteal artery behind the knee, the posterior tibial artery near the ankle joint, and on the foot, can be used to palpate the pulse (dorsalis pedis artery). Heart rate may be determined by monitoring pulse, or the number of arterial pulses per minute. Auscultation, which is the process of counting the heartbeats while listening to the heart using a stethoscope, is another way to determine the heart rate. Typically, three fingers are used to gauge the radial pulse.
Given,
heart beat = 80 beats/min = 1.3 beats/s
Frequency is nothing but how much beats is heart having in one second and that is 1.3 beats/s. Hence frequency of heart is 1.3Hz.
The Period is reciprocal of frequency,
T = 1/f = 0.8 s
To know more about frequency :
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#SPJ2.
A boxer punches a sheet of paper in midair from rest to a speed of 20 m/s in 0.05 s. If the mass of the paper is 0.01 kg, the force of the punch on the paper is
A) 0.08 N.
B) 4.0 N.
C) 8.0 N.
D) 40 N.
What must be true if energy is to be transferred as heat between two bodies in physical contact?
1-The two bodies must have different volumes.
2-The two bodies must be at different temperatures.
3-The two bodies must have different masses.
4-The two bodies must be in thermal equilibrium.
Answer:
answer is d
Explanation:
i hope this helps you
. A 79 g sample of water at 21oC is heated until it becomes steam with a temperature of 143oC. Find the change in heat content of the system.
Answer:
40479.6 J
Explanation:
Applying,
q = cm(t₂-t₁).................... Equation 1
Where q = change in heat content of the system, c = specific heat capacity of the system, m = mass of the system, t₁ = initial temperature, t₂ = final temperature.
From the question,
Given: m = 79 g = 0.079 kg, t₁ = 21°C, t₂ = 143°C
Constant: c = 4200 J/kg.°C
Substitute these values into equation 1
q = 4200(0.079)(143-21)
q = 331.8(122)
q = 40479.6 J
A 620 N physics student stands on a bathroom scale in an elevator that is supported by a cable. The combined mass of student plus elevator is 870 kg. As the elevator starts moving, the scale reads 450 N.
Required:
a. Find the acceleration of the elevator (magnitude and direction).
b. What is the acceleration if the scale reads 670 N?
c. If the scale reads zero, should the student worry? Explain.
d. What is the tension in the cable in parts (a) and (c)?
Answer:
(a) 9.28 m/s2
(b) 9.03 m/s2
(c) 9.8 m/s2
(d) 450 N, 670 N
Explanation:
mass of elevator + student, m = 870 kg
Reading of scale, R = 450 N
(a) When the elevator goes down, the weight decreases.
Let the acceleration is a.
By the Newton's second law
m g - R = m a
870 x 9.8 - 450 = 870 a
a = 9.28 m/s2
(b) R = 670 N
Let the acceleration is a.
870 x 9.8 - 670 = 870 a
a = 9.03 m/s2
(c) If the scale reads zero, it mean the elevator is falling freely. The acceleration is downwards and its value is 9.8 m/s2.
(d) Tension in cable is 450 N and 670 N.
what affects our utility
Answer:
Energy Bill fluctuations are inevitable and depend on a variety of different factors. Two of the most important are the current weather your home is experiencing and the current price per Kilowatt Hour (which fluctuates more than you might think).
