Answer:
elative magnitude of the two forces is the same and they are applied in a constant direction.
Explanation:
Newton's second law states that the sum of the forces is equal to the mass times the acceleration
∑ F = m a
in this case there are two forces on the x axis
F_applied - fr = 0
since they indicate that the velocity is constant, consequently
F_applied = fr
the relative magnitude of the two forces is the same and they are applied in a constant direction.
A new car manufacturer advertises that their car can go from zero to sixty mph in 8 [s]. This is a description of
Answer:
Acceleration
Explanation:
The fact that new can go from zero to 60mph in 8 secs is a description of its pick-up or in physics, it's called acceleration.
Here initial velocity u= 0
final velocity v = 60 mph = 1m/minute.
or v =1609.344/60 = 26.82m/s
and time taken to do so is 8 sec
Acceleration a = (v-u)/t
a = (26.82-0)/8 = 3.35 m/s^2
Therefore, acceleration of the car a = 3.35 m/s^2.
how many bits are required to sample an incoming signal 4000 times per second using 64 different amplitude level
Answer:
6 bits
Explanation:
The quality of digitized signal can be improved by reducing quantizing error. This is done by increasing the number of amplitude levels, thereby minimizing the difference between the levels and hence producing a smoother signal.
Also, Sampling frequently (also known as oversampling) can help in improving signal quality.
To get the number of bits, we use:
2ⁿ = amplitude level
where n is the number of bits.
Given an amplitude level of 64, hence:
2ⁿ = 64
2ⁿ = 2⁶
n = 6 bits
Which parts of The Action Potential Are Represented On The ECG?
Answer:
The phases of the cardiac action potential correspond to the surface ECG (ECG) (Figure). The P wave reflects atrial depolarization (phase 0), the PR interval reflects the conduction velocity through the AV node, the QRS complex the ventricular depolarization and QT interval the duration potential ventricular action.
a. What do you mean by chromatic aberration in lenses?
A Catapult throws a payload in a circle with an arm that is 65.0 cm long. At a certain instant, the arm is rotating at 8.0 rad/s and the angular speed is increasing at 40.0 rad/s2. For this instant, find the magnitude of the acceleration of the payload.
Answer:
The acceleration of the payload is 26 m/s2.
Explanation:
length, L = 65 cm = 0.65 m
angular acceleration = 40 rad/s^2
The acceleration is given by
a = angular acceleration x length
a = 40 x 0.65
a = 26 m/s^2
Container A and container B hold samples of the same ideal gas. The volume and the pressure of container A is equal to the volume and pressure of container B, respectively. If Container A has half as many molecules of the ideal gas in it as Container B does, then which of the following mathematical statements is correct regarding the absolute temperatures TA and TB in Container A and Container B. respectively?
A. TA = TB/2.
B. TA = 4TB.
C. TA = TB/4.
D. TA = 2TB.
E. TA = TB
Answer:
A. TA = TB/2.
Explanation:
Since container A has half as many molecules of the ideal gas in it as container B. Therefore, container A will have half the volume of gas as in container B:
[tex]V_A = \frac{1}{2}V_B[/tex]
Now, from Charle's Law:
[tex]\frac{V_A}{T_A}=\frac{V_B}{T_B}\\\\\frac{1}{2}\frac{V_B}{T_A}=\frac{V_B}{T_B}\\\\T_A = \frac{T_B}{2}[/tex]
Hence, the correct option is:
A. TA = TB/2.
Topic: Chapter 10: Projectory or trajectile?
Projectile range analysis:
A projectile is launched from the ground at 10 m/s, at
an angle of 15° above the horizontal and lands 5.1 m away.
What other angle could the projectile be launched at, with the same velocity,
and land 5.1 m away?
90°
75°
45
50°
30°
Answer:
The other angle is 75⁰
Explanation:
Given;
velocity of the projectile, v = 10 m/s
range of the projectile, R = 5.1 m
angle of projection, 15⁰
The range of a projectile is given as;
[tex]R = \frac{u^2sin(2\theta)}{g}[/tex]
To find another angle of projection to give the same range;
[tex]5.1 = \frac{10^2 sin(2\theta)}{9.81} \\\\100sin(2\theta) = 50\\\\sin(2\theta) = 0.5\\\\2\theta = sin^{-1}(0.5)\\\\2\theta = 30^0\\\\\theta = 15^0\\\\since \ the \ angle \ occurs \ in \ \ the \ first \ quadrant,\ the \ equivalent \ angle \\ is \ calculated \ as;\\\\90- \theta = 15^0\\\\\theta = 90 - 15^0\\\\\theta = 75^0[/tex]
Check:
sin(2θ) = sin(2 x 75) = sin(150) = 0.5
sin(2θ) = sin(2 x 15) = sin(30) = 0.5
Notice that all the initial spring potential energy was transformed into gravitational potential energy. If you compressed the spring to a distance of 0.200 mm , how far up the slope will an identical ice cube travel before reversing directions
Answer:
The correct answer will bs "2.41 m".
Explanation:
According to the question,
M = 50 g
or,
= 0.050 kg
[tex]\Theta = 25^{\circ}[/tex]
k = 25.9 N/m
Δx = 0.200 m
Let the traveled distance be "x".
By using trigonometry, the height will be:
⇒ [tex]h = l Sin \Theta[/tex]
hence,
⇒ [tex]Potential \ energy \ at \ the \ top=Spring \ potential \ energy[/tex]
[tex]Mgh=\frac{1}{2} k(\Delta x)^2[/tex]
By putting the values, we get
[tex]0.050\times 9.8\times lSin 25^{\circ}=\frac{1}{2}\times 25.0\times (0.200)^2[/tex]
[tex]l=2.41 \ m[/tex]
I need help on this physics problem.
Answer:
the speed of the nerve impulse in miles per hour is 201.59 mi/hr
Explanation:
Given;
the speed of the nerve impulse, v = 90.1 m/s
To convert this speed in meters per second to miles per hour, we use the following method;
1,609 meter = 1 mile
3,600 s = 1 hour
[tex]v(mi/h) = 90.1 \ \frac{m}{s} \times \frac{1 \ mile}{1,609 \ m} \times \frac{3,600 \ s}{1 \ hour} = (\frac{90.1 \times 3,600}{1,609} )\frac{mi}{hr} = 201.59 \ mi/hr[/tex]
Therefore, the speed of the nerve impulse in miles per hour is 201.59 mi/hr
An inductor of inductance 0.02H and capacitor of capatance 2uF are connected in series to an a.c. source of frequency 200 Hz- Calculate the Impedance in the circuit . TC
Explanation:
Given:
L = 0.02 H
C = [tex]2\:\mu \text{F}[/tex]
f = 200 Hz
The general form of the impedance Z is given by
[tex]Z = \sqrt{R^2 + (X_L - X_C)^2}[/tex]
Since this is a purely inductive/capacitive circuit, R = 0 so Z reduces to
[tex]Z = \sqrt{(X_L - X_C)^2} = \sqrt{\left(\omega L - \dfrac{1}{\omega C} \right)^2}[/tex]
[tex]\:\:\:\:\:\:\:= \sqrt{\left(2 \pi L - \dfrac{1}{2 \pi f C} \right)^2}[/tex]
[tex]\:\:\:\:\:\:\:= \sqrt{\left[2 \pi (200\:\text{Hz})(0.02\:\text{H}) - \dfrac{1}{2 \pi (200\:\text{Hz})(2×10^{-6}\:\text{F})} \right]^2}[/tex]
[tex]\:\:\:\:\:\:\:= \sqrt{(25.13\:\text{ohms} - 397.89\:\text{ohms})^2}[/tex]
[tex]\:\:\:\:\:\:\:=372.66\:\text{ohms}[/tex]
A tire is filled with air at 22oC to a gauge pressure of 240 kPa. After driving for some time, if the temperature of air inside the tire is 45oC, what fraction of the original volume of air must be removed to maintain the pressure at 240 kPa?
Answer:
7.8% of the original volume.
Explanation:
From the given information:
Temperature [tex]T_1[/tex] = 22° C = 273 + 22 = 295° C
Pressure [tex]P_1[/tex] = 240 kPa
Temperature [tex]T_2[/tex] = 45° C
At initial temperature and pressure:
Using the ideal gas equation:
[tex]P_1V_1 =nRT_1[/tex]
making V_1 (initial volume) the subject:
[tex]V_1 = \dfrac{nRT_1}{P_1}[/tex]
[tex]V_1 = \dfrac{nR*295}{240}[/tex]
Provided the pressure maintained its rate at 240 kPa, when the temperature reached 45° C, then:
the final volume [tex]V_2[/tex] can be computed as:
[tex]V_2 = \dfrac{nR*318}{240}[/tex]
Now, the change in the volume ΔV = V₂ - V₁
[tex]\Delta V = \dfrac{nR*318}{240}- \dfrac{nR*295}{240}[/tex]
[tex]\Delta V = \dfrac{23nR}{240}[/tex]
∴
The required fraction of the volume of air to keep up the pressure at (240) kPa can be computed as:
[tex]= \dfrac{\dfrac{23nR}{240}}{ \dfrac{295nR}{240}}[/tex]
[tex]= {\dfrac{23nR}{240}} \times { \dfrac{240}{295nR}}[/tex]
[tex]= 0.078[/tex]
= 7.8% of the original volume.
An object accelerates from rest, and after traveling 145 m it has a speed of 420 m/s. What was the acceleration of the object?
I am not sure how to calculate acceleration without being given the time directly.
Explanation:
Here,we've been given that,
Initial velocity (u) = 0 m/s (as it starts from rest)Distance (s) = 145 mFinal velocity (v) = 420 m/sWe've to find the acceleration of the object. By using the third equation of motion,
→ v² - u² = 2as
→ (420)² - (0)² = 2 × a × 145
→ 176400 - 0 = 290a
→ 176400 = 290a
→ 176400 ÷ 290 = a
→ 608.275862 m/s² = a
If you know initial speed and final speed, you can find the average speed. Then, knowing distance, you can find the time.
KimYurii posted the first answer to this question.
That answer is well organized, well presented, elegant and correct, and it deserves to be awarded "Brainliest" and several merit badges.
My problem is that I can never remember all the different formulas. I guess I had to work with so many uvum in all the Physics, Geometry, and Calculus classes that I took, I filled up all the memory slots with formulas, and over the years they all eventually merged into a big glob of goo. Now, the only formulas I can remember are the ones I had to use as an Electrical Engineer.
When I see this kind of question, I can only remember one or two simple formulas, and I reason it out like this:
Starting speed . . . zero
Ending speed . . . 420 m/s
Formula: Average speed . . . (1/2)·(0 + 420) = 210 m/s
Distance covered . . . 145 m
Formula: Time taken = (distance) / (average speed) = (145/210) second
(Now you have the time.)
Formula: Distance = (1/2)·(acceleration)·(time²)
145 m = (1/2)·(acceleration)·(145/210 sec)²
Acceleration = 290 m / (145/210 s)²
Acceleration = 608.28 m/s²
A massless, hollow sphere of radius R is entirely filled with a fluid such that its density is p. This same hollow sphere is now compressed so that its radius is R/2, and then it is entirely filled with the same fluid as before. As such, what is the density of the compressed sphere?
a. 8p
b. p/8
c. p/4
d. 4p
Answer:
a. 8p
Explanation:
We are given that
Radius of hollow sphere , R1=R
Density of hollow sphere=[tex]\rho[/tex]
After compress
Radius of hollow sphere, R2=R/2
We have to find density of the compressed sphere.
We know that
[tex]Density=\frac{mass}{volume}[/tex]
[tex]Mass=Density\times volume=Constant[/tex]
Therefore,[tex]\rho_1 V_1=\rho_2V_2[/tex]
Volume of sphere=[tex]\frac{4}{3}\pi r^3[/tex]
Using the formula
[tex]\rho\times \frac{4}{3}\pi R^3=\rho_2\times \frac{4}{3}\pi (R/2)^3[/tex]
[tex]\rho R^3=\rho_2\times \frac{R^3}{8}[/tex]
[tex]\rho_2=8\rho[/tex]
Hence, the density of the compressed sphere=[tex]8\rho[/tex]
Option a is correct.
suppose the tank is open to the atmosphere instead of being closed. how does the pressure vary along
Answer:
Pressure is more in the open container than the closed one.
Explanation:
The pressure due to the fluid at a depth is given by
Pressure = depth x density of fluid x gravity
So, when the container is open, the atmospheric pressure is also add up but when the container is closed only the pressure due to the fluid is there.
So, when the container is open, the pressure is atmospheric pressure + pressure due to the fluid.
hen the container is closed only the pressure due to the fluid is there.
In what direction is the centripetal force directed?
Answer:
towards the center
Explanation:
that is the solution above
A train moving with a uniform speed covers a distance of 120 m in 2 s. Calculate
(i) The speed of the train
(ii) The time it will taketo cover 240 m.
Answer:
(I)
[tex]{ \bf{s = ut + \frac{1}{2}a {t}^{2} }} \\ 120 = (u \times 2) + \frac{1}{2} \times 0 \times {2}^{2} \\ 120 = 2u \\ { \tt{speed = 60 \: {ms}^{ - 1} }}[/tex]
(ii)
[tex]{ \bf{s = ut + \frac{1}{2}a {t}^{2} }} \\ 240 = (60t) \\ { \tt{time = 4 \: seconds}}[/tex]
What is the biggest planet in the solar system
Answer:
Jupiter
Explanation:
Answer:
The answer is Jupiter.
Explanation:
Jupiter is an orange/yellow colored planet.
Find the amount og work done
Answer:
100j
Explanation:
1. Compare and contrast the SI and the English systems of measurement.
Answer:The SI system is based on the number 10 as well as multiples and products of 10. This makes it much easier to use, and so it has been the accepted system in scientific and technical applications. The English system is more complicated as relationships between units of the same quantity aren't uniform.
Explanation:
Answer:
The metric system is an internationally agreed decimal system of measurement while The International System of Units (SI) is the official system of measurement in almost every country in the world
A 1.0 ball moving at 2.0 / perpendicular to a wall rebounds from the wall at 1.5 /. If the ball was in contact with the wall for 0.1 , what force did the wall impart onto the ball?
Answer:
5N
Explanation:
We have a simple problem of momentum here.
ΔMomentum= mΔv= FΔt
Solve for F
mΔv/Δt=F
Plug in givens
1*(2-1.5)/0.1=F
F=5N
The amount of force that the wall imparts on the ball is 5.0N
According to Newton's second law, the formula for calculating the force applied is expressed as:
[tex]F=ma[/tex]
m is the mass of the object
a is the acceleration of the object
Since acceleration is the change in velocity of an object, hence [tex]a=\frac{\triangle v}{t}[/tex]
The applied force formula becomes [tex]F=\frac{m\triangle v}{t}[/tex]
Given the following parameters
m = 1.0kg
[tex]\triangle v=2.0-1.5\\\triangle v=0.5m/s[/tex]
t = 0.1sec
Substitute the given parameter into the formula
[tex]F=\frac{1.0\times 0.5}{0.1}\\F=\frac{0.5}{0.1}\\F=5N[/tex]
Hence the amount of force that the wall imparts on the ball is 5.0N
Learn more here: https://brainly.com/question/17811936
A loop of wire is carrying current of 2 A . The radius of the loop is 0.4 m. What is the magnetic field at a distance 0.09 m along the axis and above the center of the loop
Answer:
[tex]B=2.91\ \mu T[/tex]
Explanation:
Given that,
The current in the loop, I = 2 A
The radius of the loop, r = 0.4 m
We need to find the magnetic field at a distance 0.09 m along the axis and above the center of the loop. The formula for the magnetic field at some distance is given as follows :
[tex]B=\dfrac{\mu_o}{4\pi }\dfrac{2\pi r^2 I}{(r^2+d^2)^{3/2}}[/tex]
Put all the values,
[tex]B=10^{-7}\times \dfrac{2\pi \times 0.4^2 \times 2}{(0.4^2+0.09^2)^{3/2}}\\\\=2.91\times 10^{-6}\ T\\\\or\\\\B=2.91\ \mu T[/tex]
So, the required magnetic field is equal to [tex]2.91\ \mu T[/tex].
A 1.64 kg mass on a spring oscillates horizontal frictionless surface. The motion of the mass is described by the equation: X = 0.33cos(3.17t). In the equation, x is measured in meters and t in seconds. What is the maximum energy stored in the spring during an oscillation?
Answer:
[tex]K.E_{max}=0.8973J[/tex]
Explanation:
From the question we are told that:
Mass [tex]m=1.64kg[/tex]
Equation of Mass
[tex]X=0.33cos(3.17t)[/tex]...1
Generally equation for distance X is
[tex]X=Acos(\omega t)[/tex]...2
Therefore comparing equation
Angular Velocity [tex]\omega=3.17rad/s[/tex]
Amplitude A=0.33
Generally the equation for Max speed is mathematically given by
[tex]V_{max}=A\omega[/tex]
[tex]V_{max}=0.33*3.17[/tex]
[tex]V_{max}=1.0461m/s[/tex]
Therefore
[tex]K.E_{max}=0.5mv^2[/tex]
[tex]K.E_{max}=0.5*1.64*(1.0461)^2[/tex]
[tex]K.E_{max}=0.8973J[/tex]
As you move farther away from a source emitting a pure tone, the ___________ of the sound you hear decreases.
Answer:
frequency
Explanation:
The phenomenon of apparent change in frequency due to the relation motion between the source and the observer is called Doppler's effect.
So, when we move farther, the frequency of sound decreases. The formula of the Doppler's effect is
[tex]f' = \frac{v + v_o}{v+ v_s} f[/tex]
where, v is the velocity of sound, vs is the velocity of source and vo is the velocity of observer, f is the true frequency. f' is the apparent frequency.
A student claimed that thermometers are useless because a
thermometer always registers its own temperature. How would you
respond?
[
Would this pressure difference be greater or smaller if the scuba diver were in seawater (density 1050 kg/m3 ) and went to the same depth you calculated in question D1, took and held his breath, and then returned to the surface
Answer:
Greater.
Explanation:
This pressure difference will be greater if the scuba diver were in seawater and went to the same depth because the seawater have salts which increases the density of water as compared to freshwater. Salt in water increases the density which automatically increases the pressure on the diver so that's why we can say that the pressure will be increases for the scuba diver in seawater as compared to freshwater.
Part AFind the x- and y-components of the vector d⃗ = (4.0 km , 29 ∘ left of +y-axis).Express your answer using two significant figures. Enter the x and y components of the vector separated by a comma.d⃗ = km Part BFind the x- and y-components of the vector v⃗ = (2.0 cm/s , −x-direction).Express your answer using two significant figures. Enter the x and y components of the vector separated by a comma.v⃗ = cm/s Part CFind the x- and y-components of the vector a⃗ = (13 m/s2 , 36 ∘ left of −y-axis).Express your answer using two significant figures. Enter the x and y components of the vector separated by a comma.a⃗ x = m/s2
Solution :
Part A .
Given : The [tex]x[/tex] and [tex]y[/tex] components of the vector, d = [tex]\text{4 km 29}[/tex] degree left of [tex]y[/tex]-axis.
So the [tex]x[/tex] component is = -4 x sin (29°) = -1.939 km
[tex]y[/tex] component is = 4 x cos (29°) = 3.498 km
Part B
Given : The [tex]x[/tex] and [tex]y[/tex] components of the vector, [tex]\text{v = 2 cm/s}[/tex] , [tex]\text{-x direction}[/tex]
So the [tex]x[/tex] component is = -2 cm/s
[tex]y[/tex] component is = 0
Part C
Given : The [tex]x[/tex] and [tex]y[/tex] components of the vector, [tex]\text{a = 13 m/s, 36 degree}[/tex] left of [tex]y[/tex]-axis.
So the [tex]x[/tex] component is = -13 x sin (36°) = -7.6412 [tex]m/S^2[/tex]
[tex]y[/tex] component is = -13 x cos (36°) = -10.517 [tex]m/S^2[/tex]
The x- and y-components of the vectors is mathematically given as as follows for each Part respectively
x= -1.939 km, y= 3.498 km
x= -2 cm/s, 0
y=, x= -7.6412m/s^2, -10.517m/s^2
What are the x- and y-components of the vectors?
Question Parameters:
Generally, we follow a basic principle where
x component= Fsin\theta
y component= Fcos\theta
Therefore
For A
x component is
x= -4 x sin (29°)
x= -1.939 km
y component is
y= 4 x cos (29°)
y= 3.498 km
For B
x component is
x= -2 cm/s
y component is
y= 0
For C
x component is
x= -13 x sin (36°)
x= -7.6412m/s^2
y component is
y= -13 x cos (36°)
y= -10.517m/s^2
Read more about Cartession co ordinate
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PLEASE HELP ME WITH THIS ONE QUESTION
A photon has 2.90 eV of energy. What is the photon’s wavelength? (h = 6.626 x 10^-19, 1 eV = 1.6 x 10^-19 J)
Explanation:
First, we convert the energy from eV to Joules:
[tex]2.90\:\text{eV}×\left(\dfrac{1.6×10{-19}\:J}{1\:\text{eV}} \right)[/tex]
[tex]= 4.64×10^{-19}\:\text{eV}[/tex]
We know from definition that
[tex]E=h\nu = \dfrac{hc}{\lambda}[/tex]
so the wavelength of the photon is
[tex]\lambda = \dfrac{hc}{E} = 4.28×10^8\:\text{m}[/tex]
Two children sit on a seesaw that is in rotational equilibrium. The first child has weight W and sits at distance d from the pivot. If the second child sits at a distance of 7*d from the pivot, what must be the weight of the second child
Answer:
W/7
Explanation:
By principle of moments,
Sum of clockwise moment = sum of anticlockwise moment
Weight × 7d = W × d
Weight = W/7
Since the two children are in rotational equilibrium, the weight of the second child is W/7.
How can the weight of the second child be determined?The weight of the second child can be determined from the principle of moments.
The principle of moments states that for a body in equilibrium, the sum of the clockwise moments and anticlockwise moments about a point is zero.
Let the weight of the second child be X
From the principle of moments:
W × d = 7×d × X
X = W/7
Therefore, the weight of the second child is W/7.
Learn more about principle of moments at: https://brainly.com/question/20298772
the speed of the pulse depends on what?
Answer:
The pulse speed depends on the properties of the medium and not on the amplitude or pulse length of the pulse.
Explanation:
Hope this helps.
A moderate wind accelerates a pebble over a horizontal xy plane with a constant acceleration a with arrow = (4.60 m/s2)i hat + (7.00 m/s2)j. At time t = 0, the velocity is (4.3 m/s)i hat. What are magnitude and angle of its velocity when it has been displaced by 11.0 m parallel to the x axis?
Explanation:
Given
Acceleration of the pebble is
At t=0, velocity is
considering horizontal motion
[tex]\Rightarrow x=ut+0.5at^2 \\\Rightarrow 11=4.3t+0.5(4.6)t^2\\\Rightarrow 2.3t^2+4.3t-11=0\\\Rightarrow (t-1.4435)(t+3.3131)=0\\\Rightarrow t=1.44\ s\quad [\text{Neglecting negative time}]\\[/tex]
Velocity acquired during this time
[tex]\Rightarrow v_x=4.3+4.6\times 1.44\\\Rightarrow v_x=4.3+6.624\\\Rightarrow v_x=10.92\ s[/tex]
Consider vertical motion
[tex]\Rightarrow v_y=0+7(1.44)\\\Rightarrow v_y=10.08\ m/s[/tex]
Net velocity is
[tex]\Rightarrow v=\sqrt{10.92^2+10.08^2}\\\Rightarrow v=\sqrt{220.85}\\\Rightarrow v=14.86\ m/s[/tex]
Angle made is
[tex]\Rightarrow \tan \theta =\dfrac{10.08}{10.92}\\\\\Rightarrow \tan \theta =0.92307\\\\\Rightarrow \theta =42.7^{\circ}[/tex]
