ANSWER:
0.045 joules
STEP-BY-STEP EXPLANATION:
Given:
mass (m) = 10 g = 0.01 kg
velocity (v) = 3 m/s
The kinetic energy is given by the following formula:
[tex]K_E=\frac{1}{2}mv^2[/tex]We replacing:
[tex]\begin{gathered} K_E=\frac{1}{2}\cdot0.01\cdot3^2 \\ K_E=0.045\text{ J} \end{gathered}[/tex]The kinetic energy is 0.045 joules.
A mass falls from rest to the growl in 15 secs. Find y, the vertical distance it travelled
We will determine the distance fallen as follows:
[tex]d=v_it+\frac{1}{2}at^2[/tex]Thus:
[tex]d=(m/s)(15s)+\frac{1}{2}(9.8m/s^2)(15s)^2\Rightarrow d=1102.5m[/tex]So, it traveled 1102.5 meters.
- (True/False) If there is no friction, the efficiency of a machine can be greater than 100%.
ANSWER:
False
STEP-BY-STEP EXPLANATION:
Even without friction, the efficiency cannot be greater than 100%, because the output cannot be greater than the input, therefore, the statement is false.
The binding energy of a nucleus is always negative.Question 4 options:TrueFalse
ANSWER
False.
EXPLANATION
Nuclear binding energy is the energy that is required to split the nucleus (of an atom) into nucleons: protons and neutrons.
The binding energy of a nucleus is always positive because nuclei require energy to separate them and it is impossible for nuclei to gain energy by being separated.
Therefore, the answer is false.
If raindrops are falling vertically at 7.90 m/s, what angle from the vertical do they make for a person jogging at 2.47 m/s? (Enter your answer in degrees.)
17.36 degrees
Explanation
Step 1
Diagram:
so, we have a rigth triangle then let
[tex]\begin{gathered} angle\text{ =}\theta \\ opposite\text{ side= 2.47} \\ adjacent\text{ side =7.9} \end{gathered}[/tex]Step 2
now, we need a function that relates those values to find the missin angle , it is
[tex]tan\theta=\frac{opposite\text{ side}}{adjacent\text{ side}}[/tex]replace ans solve for the angle
[tex]\begin{gathered} tan\theta=\frac{2.47}{7.9} \\ tan\theta=0.31 \\ inverse\text{ tan function in both sides} \\ \tan^{-1}(tan\theta)=\tan^{-1}(0.31) \\ \theta=17.36\text{ \degree} \end{gathered}[/tex]therefore, the answer is
17.36 degrees
I hope this helps you
McKenna is performing an experiment. She is testing the relationship between the temperature of water, and how long it takes for rust to form on iron plates. If she finds that higher temperatures lead to lower times for rusting, what is the qualitative relationship?positive correlationno correlationnegative correlationinverse square correlation
As with the increase in the temperature, the time taken to form the rust on the iron plate decreases.
This correlation is known as a negative correlation.
Thus, the relationship between the temperature of the water and the time taken to form rust on the iron plate is a negative correlation.
Hence, the third option is the correct answer.
Hello, i just need to check this problem, thanks in advance!
Given:
The mass of the bucket is m = 0.25 kg
The length of the string from the center is L = 0.4 m
The speed of the bucket is v = 5 m/s
To find The tension in the string at the bottom of the circle.
Explanation:
Forces acting on the bottom of the circle are tension due to the string and force due to gravity.
The tension in the string at the bottom of the circle can be calculated by the formula
[tex]T=\frac{mv^2}{L}+mg[/tex]Here, g = 9.8 m/s^2 is the acceleration due to gravity.
Substituting the values, the tension in the string at the bottom of the circle is
[tex]\begin{gathered} T\text{ = }\frac{0.25\times(5)^2}{0.4}+0.25\times9.8 \\ =18.075\text{ N} \end{gathered}[/tex]
Final Answer: Thus the tension in the string at the bottom of the circle is 18.075 N
In this example, the system gains the ability to cause change the higher the brick is raised above the earth. Is this true or false?
ANSWER
True
EXPLANATION
The higher the brick is, the more potential energy it has. Therefore, it is true that it gains the ability to cause change the higher it is raised above the Earth.
A block of mass of 1kg is hit by a force of magnitude 5 N which causes the block to have an acceleration of magnitude a. If the same block is hit by a force of magnitude 40 N, then which of these is the new magnitude of acceleration of the block?A) 2.8284aB) one eighth of aC) one/ 64 of aD) 8aE) 64a
The acceleration is related to the force and mass by Newton's second law which states:
[tex]a=\frac{F}{m}[/tex]for the first mass we have:
[tex]a_1=\frac{5}{1}=5[/tex]If the same block is hit by a force of 40 N then the new acceleration is:
[tex]a=\frac{40}{1}=40[/tex]Which is eight times the original acceleration; therefore, the answer is D.
A uniform disk of radius 0.473 m and unknown mass is constrained to rotate about a perpendicular axis through its center. A ring with the same mass as the disk is attached around the disk's rim. A tangential force of 0.203 N applied at the rim causes an angular acceleration of 0.125 rad/s2. Find the mass of the disk.
From the calculations, the mass of the object is 3.43 Kg.
What is the mass of the rod?We know that the term circular motion has to do with motion along a circular path. The force that acts on the body in this case is tangential to the circle. This tangential force can be obtained from;
F = mrα
F = tangential force
m = mass of the object
r = radius of the object
α = angular acceleration
When we make m the subject of the formula;
m = F/rα
m = 0.203 N/0.473 m * 0.125 rad/s2
m = 3.43 Kg
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y=a on x Represents_____________
Answer:
a
Explanation:
to be linear equation, x needs to be multiplied by a constant but xy = a doesn't satisfy that
A speedboat increases its speed from 13.5 m/s to 27.4 m/s in a distance of 241 m. Determine the time over which this acceleration occurs?
V² = U² + 2aS
27.4² = 13.5² + 2*241a
750.76 = 182.25 + 482a
750.76 - 182.25 = 482a
568.51/482
a = 1.18m/s^2
A pitcher throws a 0.14-kg baseball with a speed of 28.8 m/s. The batter strikes it with an average force of 5000 N, which results in the ball traveling with an initial speed of 37.0 m/s toward the pitcher. For how long we’re the bat and ball in contact?
Given data:
* The velocity of the ball before hitting is,
[tex]u=28.8\text{ m/s}[/tex]* The velocity of the ball after hitting the bat is,
[tex]v=-37\text{ m/s}[/tex]Here, the negative sign is indicating the direction of motion of ball is in opposite direction to the direction of motion before hitting by the bat.
* The mass of the ball is,
[tex]m=0.14\text{ kg}[/tex]* The average force on the ball by the bat is,
[tex]F=5000\text{ N}[/tex]Solution:
The change in the momentum of the ball after hitting is,
[tex]\Delta p=p_2-p_1[/tex]where p2 is the momentum of the ball after hitting from the bat and p1 is the momentum of the ball before hitting from the bat,
The change in momentum in terms of mass is,
[tex]\begin{gathered} \Delta p=mv-m\text{u} \\ =m\mleft(v-u\mright) \end{gathered}[/tex]Substituting the known values,
[tex]\begin{gathered} \Delta p=0.14(-37-28.8) \\ \Delta p=-0.14\times65.8 \\ \Delta p=-9.212\text{ kgm/s} \end{gathered}[/tex]Thus, the time at which the bat and ball were in contact is,
[tex]\begin{gathered} F=\frac{|\Delta p|}{\Delta t} \\ \Delta t=\frac{|\Delta p|}{F} \end{gathered}[/tex]Substituting the known values,
[tex]\begin{gathered} \Delta t=\frac{9.212}{5000} \\ \Delta t=0.0018\text{ s} \\ \Delta t=1.8\times10^{-3}\text{ s} \end{gathered}[/tex]Thus, the ball and bat were in contact for 0.0018 seconds.
What are some simple machines that decrease mechanical advantage used to do?
Omultiply distance to increase speed
Odivide distance to decrease speed
O decrease work
O increase work
Some simple machines decrease mechanical advantage and are used to A) multiply distance to increase speed.
Define Mechanical Advantage.
When analyzing the forces in straightforward machines like levers and pulleys, mechanical advantage—a measure of the ratio of output force to input force in a system—is utilized. The conservation of energy still holds true and the output energy is equal to the input energy even when the forces being applied change.
The mechanical advantage is typically described in terms of perfect systems, also known as 100% efficient systems, where there are no energy losses between the input and output times.Through energy efficiency:Fi×di=Fo×do
The input energy (or work put into a machine) is on the left, and the output energy (or work that comes out of a machine, for no-loss energy) is on the right.
By extending the distance across which the effort force is applied, the equations demonstrate how a basic machine can produce the same amount of work while using less effort force.
Hence, Option A) is correct.
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If a bullet leaves the muzzle of a rifle with a speed of 600 m/s, and the barrel of the rifle is 0.800 m long, at what rate is the bullet accelerated while in the barrel?That is the question I am having trouble with
Given that the final velocity of the bullet is
[tex]V_f=600\text{ m/s}[/tex]The initial velocity of the bullet is
[tex]V_i=0\text{ m/s}[/tex]The distance traveled by the bullet is x = 0.8 m
We have to find the acceleration.
Let the acceleration be denoted by a.
The equation which can be used to calculate acceleration is
[tex](V_f)^2-(V_i)^2=2ax[/tex]Substituting the values, acceleration will be
[tex]\begin{gathered} (600)^2-0^2=2\times a\times0.8 \\ a=\frac{360000}{1.6} \\ =225000m/s^2 \end{gathered}[/tex]Thus the acceleration will be 225000 m/s^2.
Identify the Action and Reaction Forces
A pair of forces that will always move in conflict with one another is referred to as an action-reaction force.
What is newton's third law?The third law of Newton asserts because when bodies interact, they exert forces that are equal in size and directed in the reverse way. Another name for the newton's third is the law of action and reaction. This equation is crucial for understanding issues with static balance, in which all forces are all in balance, but it also holds true for bodies moving at a steady or accelerated speed.
Hammer hitting a nail : In this, the hitting by the hammer is the action and the nail is going into the surface is the reaction force.
Book placed on the table : In this book is applying gravitational force that is action force and the reaction force is applied by the table.
Hitting a tennis ball : In this, hitting the ball is the action and the ball is going in the direction of hitting is the reaction force.
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The correct unit for the frequency of oscillation is asecond/wavewave/secondHertzsecond
To find the unit of frequency.
Explanation:
The SI unit of frequency is Hertz (Hz). It is named after Heinrich Hertz.
It is equivalent to
[tex]\begin{gathered} frequency\text{ =}\frac{1}{time\text{ period}} \\ 1\text{ Hz}=\frac{1}{s} \\ =s^{-1} \end{gathered}[/tex]
3. The mass of an electron is 9.1x10^-31 kg. The mass of a proton is 1.7x10^-27 kg.The gravitational force between them in the hydrogen atom is 1.0x10^-47 N.What is the separation distance between them?
Given:
The mass of the electron is,
[tex]m_e=9.1\times10^{-31}\text{ kg}[/tex]The mass of the proton is,
[tex]m_p=1.7\times10^{-27}\text{ kg}[/tex]The gravitational force between the electron and the proton in the hydrogen atom is,
[tex]1.0\times10^{-47}\text{ N}[/tex]To find:
the separation between them
Explanation:
The gravitational force between two masses is,
[tex]F=\frac{Gm_1m_2}{r^2}[/tex]Here, the universal gravitational constant is,
[tex]G=6.67\times10^{-11}\text{ Nm}^2kg^{-2}[/tex]Substituting the values we get,
[tex]\begin{gathered} 1.0\times10^{-47}=\frac{6.67\times10^{-11}\times9.1\times10^{-31}\times1.7\times10^{-27}}{r^2} \\ r^2=\frac{6.67\times10^{-11}\times9.1\times10^{-31}\times1.7\times10^{-27}}{1.0\times10^{-47}} \\ r^2=1.03\times10^{-20} \\ r=\sqrt{1.01\times10^{-20}} \\ r=1.01\times10^{-5}\text{ m} \end{gathered}[/tex]Define electricity in detail.
Electricity is the phenomenon of the flow of electric charges due to the potential difference.
The flow of electric charges across the conductor is known as electric current.
Unit of electric current is Ampere.
Potential difference is the difference of potential between two charges.
There are two charges namely positive and negative.
Uneven distribution of charges causes the potential difference.
Suppose A and B are two points on a conductor where A has a 5 C charge and B has a 3 C charge.
Thus, the current will flow from higher potential to lower potential which is from A to B.
Resistance is the opposing property of the conductor which opposes the flow of current.
The unit of resistance is ohm.
b. How many square meters does an 80. ft2 rug occupy? (1 ft = 12 in.)
b.
First squared the following conversion factor: 1 ft = 12 in, as follow:
[tex]\begin{gathered} 1ft=12in \\ (1ft)^2=(12in)^2 \\ 1ft=144in^2 \end{gathered}[/tex]Then, use the previous result as a conversion factor to convert 80.0ft^2, as follow:
[tex]80.0ft^2\cdot\frac{144in^2}{1ft^2}=11520in^2[/tex]Now, consider that 1 m = 39.3701 in, then, if you squared you obtain:
[tex]\begin{gathered} 1m=39.3701in \\ (1m)^2=(39.37in)^2 \\ 1m^2=1550.004774in^2 \end{gathered}[/tex]use the previous result as a conversion factor:
[tex]11520in^2\cdot\frac{1m^2}{1550.004774in^2}\approx7.43m^2[/tex]Hence, 80.0ft^2 is approximately 7.43 m^2
The frequency of a sound wave changes if the source of the sound is moving relative to the listener. What is this called?
Given
The frequency of a sound wave changes if the source of the sound is moving relative to the listener.
To find
What is this called?
Explanation
For an observer, when any source of sound is moving towards it or away from it then the sound becomes higher and lower respectively.
Thus the moment of the source changes the frequency of the sound. This is known as doppler effect.
Conclusion
The given effect is known as Doppler effect.
Two spherical objects have masses of 1.5 x 105 kg and 8.5 x 102 kg. Their centers are separated by a distance of 2500 m. Find the gravitational attraction between them.
Given data
*The mass of the first spherical object is m = 1.5 × 10^5 kg
*The mass of the second spherical object is M = 8.5 × 10^2 kg
*The separated distance between the two spherical objects is r = 2500 m
The expression for the gravitational attraction between them is given as
[tex]F=\frac{GmM}{r^2}[/tex]*Here G = 6.67 × 10^-11 N.m^2/kg^2 is the value of the gravitational constant
Substitute the known values in the above expression as
[tex]\begin{gathered} F=\frac{(6.67\times10^{-11})(1.5\times10^5)(8.5\times10^2)}{(2500)^2} \\ =1.36\times10^{-9}\text{ N} \end{gathered}[/tex]Hence, the gravitational attraction between them is F = 1.36 × 10^-9 N
A jet is goes from 180km/ to 139km/s in. 22 seconds what is its acceleration Vi-Vf-T-A-
Given
Initial velocity,
[tex]v_i=180\text{ km/s}[/tex]Final velocity,
[tex]v_f=139\text{ km/s}[/tex][tex]Time\text{ taken , t=22 s}[/tex]To find
The acceleration
Explanation
Acceleration is the ratio of change in velocity to that of the time taken
Thus,
[tex]\begin{gathered} a=\frac{v_f-v_i}{t} \\ \Rightarrow a=\frac{180-139}{22} \\ \Rightarrow a=1.86\text{ km/s}^2 \end{gathered}[/tex]Conclusion
The acceleration is
[tex]1.86\text{ km./s}^2[/tex]A hair dryer draws 6.0 A when plugged into a 120-V line. (a) What is its resistance? (b) How much charge passes through it in 15 min?
We will have the following:
a)
[tex]R=\frac{120V}{6A}\Rightarrow R=20\Omega[/tex]b)
[tex]Q=(6A)(15min\ast\frac{60s}{1min})\Rightarrow Q=5400W[/tex]So, the charge is 5400 W in 15 minutes.
A full tea cup has a mass of 0.40 kg. If the full cup applies a pressure of 1000.the radius of the circular ring imnrinted on the table?
Given:
The mass of the cup, m=0.40 kg
The pressure applied by the cup, P=1000 N/m²= 1000 Pa
To find:
The radius of the ring imprinted on the table.
Explanation:
The pressure is defined as the force per unit area.
Thus the pressure applied by the cup is given by,
[tex]\begin{gathered} P=\frac{F}{A} \\ =\frac{mg}{\pi r^2} \end{gathered}[/tex]Where A is the area of the ring, g is the acceleration due to gravity, and r is the radius of the ring.
On rearranging the above equation,
[tex]r=\sqrt{\frac{mg}{P\pi}}[/tex]On substituting the known values,
[tex]\begin{gathered} r=\sqrt{\frac{0.40\times10}{1000\pi}} \\ =0.036\text{ m} \\ =3.6\text{ cm} \end{gathered}[/tex]Final answer:
Thus the radius of the ring imprinted on the table is 3.6 cm
Therefore the correct answer is option 3.
Charge q1 = 2.00 x 10-9 C is located +0.020 m from the origin along the x-axis. Charge q2 =-3.00 x 10-9 C is located +0.040 m from the origin along the x-axis. What is the electric force exerted by these two charges on a third charge, q3 = 5.00 x 10-9 C, located at the origin?
Given,
Charge q1 = 2.00 x 10-9 C is located +0.020 m from the origin along the x-axis. Charge q2 =-3.00 x 10-9 C is located +0.040 m from the origin along the x-axis.
There is a third charge q3 = 5.00 x 10-9 C, located at the origin.
Now the force acting on the third charge due to other two charge is:
[tex]\begin{gathered} F=k\frac{2\times10^{-9}\times5\times10^{-9}}{0.02^2}+k\frac{3\times10^{-9}\times5\times10^{-9}}{0.04^2} \\ \Rightarrow F=0.0002225+0.0000834=0.000305C \end{gathered}[/tex]
Can you please answer the questions and explain how you got it?
Given data:
The mass of the block on table is M=500 g.
The hanging mass is m=300 g.
The kinetic friction is μ=0.25.
Part (a)
The free body diagram of the block on table is shown below,
The free body diagram of hanging mass is shown below,
Part (b)
The equation according Newton's second law for 500 g mass will be,
[tex]\begin{gathered} F_{net}=Ma \\ T-f=Ma \\ T-\mu N=Ma \\ T-\mu Mg=Ma \\ T=M(a+\mu g)\ldots\ldots(1) \end{gathered}[/tex]Here, f is the friction force, N is the normal force, and T is the tension in the string.
Part (c)
The equation according Newton's second law for 300 g mass will be,
[tex]\begin{gathered} F_{net}=ma \\ mg-T=ma \\ T=mg-ma \\ T=m(g-a)\ldots\ldots(2) \end{gathered}[/tex]Part (d)
Equating equation (1) and (2) to solve for acceleration,
[tex]\begin{gathered} M(a+\mu g)=m(g-a) \\ 500(a+0.25(9.81))=300(9.81-a) \\ a+2.45=0.6(9.81-a) \\ a+2.45=5.886-0.6a \\ a+0.6a=5.886-2.45 \\ 1.6a=3.436 \\ a=2.14m/s^2 \end{gathered}[/tex]Thus, the acceleration of the blocks are 2.14 m/s².
Part (e)
Substitute the value of acceleration in equation (1) to calculate the tension,
[tex]\begin{gathered} T=M(a+\mu g) \\ T=500(2.14+(0.25)(9.81)) \\ T=2296gm/s^2\times\frac{1\text{ N}}{1000gm/s^2} \\ T=2.296\text{ N} \end{gathered}[/tex]Thus, the tension in the sting is 2.296 N.
Q4. A 5.0-kg bowling ball rolls down africtionless 2.5 m tall ramp and strikes astationary mass at the bottom of the ramp in aperfectly elastic collision. To whatheight back up the ramp does the first bowlingtravel after the collision ita) the stationary mass is also 5.0 kgb) the stationary mass is 10.0 kgC) the stationary mass is 500.00 kg
a)
Using conservation of energy for ball 1:
[tex]\begin{gathered} E1=E2 \\ K1+U1=K2+U2 \\ 0+m1gh=\frac{1}{2}m1v1^2 \end{gathered}[/tex]Solve for v1:
[tex]\begin{gathered} v1=\sqrt[]{2gh} \\ v1=\sqrt[]{2(9.8)(2.5)} \\ v1=7\frac{m}{s} \end{gathered}[/tex][tex]v1=v2^{\prime}-v1^{\prime}[/tex]Using conservation of momentum:
[tex]\begin{gathered} m1v1=m1(v2^{\prime}-v1^{\prime})+m2v2^{\prime} \\ v2^{\prime}=\frac{2m1v1}{(m1+m2)} \\ \end{gathered}[/tex]a)
m2 = 5 kg
[tex]v2^{\prime}=7\frac{m}{s}[/tex]so:
[tex]\begin{gathered} v1^{\prime}=7-7 \\ v1^{\prime}=0 \end{gathered}[/tex]so:
[tex]\begin{gathered} m1gh^{\prime}=\frac{1}{2}m1(v1^{\prime})^2 \\ h^{\prime}=0 \end{gathered}[/tex]If the stationary mass is 5.0 kg the height back up the ramp is 0 meters
b)
m2 = 10
[tex]v2^{\prime}=\frac{14}{3}\frac{m}{s}[/tex]so:
[tex]v1^{\prime}=-\frac{7}{3}\frac{m}{s}[/tex][tex]\begin{gathered} m1gh^{\prime}=\frac{1}{2}m1(v1^{\prime})^2 \\ h^{\prime}\approx0.2778m \end{gathered}[/tex]If the stationary mass is 10.0 kg the height back up the ramp is 0.2778 meters
c)
m2 = 500.00kg
[tex]v2^{\prime}=\frac{14}{101}\frac{m}{s}[/tex]so:
[tex]v1^{\prime}\approx-\frac{693}{101}\frac{m}{s}[/tex][tex]\begin{gathered} m1gh^{\prime}=\frac{1}{2}m1(v1^{\prime})^2 \\ h^{\prime}\approx2.4m \end{gathered}[/tex]If the stationary mass is 500.0 kg the height back up the ramp is 2.4 meters
Layla was hiking in the Grand Canyon with her best friend Saige. The temperature when they started gut was only 4.0° C. She shouted at the canyon wall and 2.8 s later she heard her own voice echoing back. How far away was the canyon wall? a) 4.6 x 10² m b) 46.6 m c) 47 m d) 466.7 m
Given:
The temperature of the surrounding is 4 degrees Celsius.
The time taken for the echo to reach the ears is t = 2.8 s
Required:
The distance from the canyon wall.
Explanation:
The distance traveled by the sound is 2d as it travels up to the wall and bounces back.
The distance can be calculated by the formula
[tex]d=\frac{v\times t}{2}[/tex]Here, the speed of sound at 4 degrees Celsius is
[tex]v=331.3\text{ m/s}[/tex]On substituting the values, the distance traveled will be
[tex]\begin{gathered} d=\frac{331.3\times2.8}{2} \\ =4.6\times10^2\text{ m} \end{gathered}[/tex]Final Answer: The distance traveled is 4.6 x 10^2 m
Assume that the ammeter in the figure below is removed and the current that flows through the 4.0Ω path, I3, is unknown. Determine all the currents in the circuit.
The circuit is in parallel connection
Equivalent resistance = 1/Req = 1/R1 + 1/R2 + 1/R3
From the information given,
R1 = 5
R2 = 2
R3 = 4
1/Req = 1/5 + 1/2 + 1/4 = (4 + 10 + 5)/20 = 19/20
Req = 20/19 = 1.053 ohms
I = V/R
Given that V = 12,
Current flow through circuit = 12/1.053 = 11.4 A
I1 + I2 + I3 = 11.4
I1 = 12/5 = 2.4 A
I2 = 12/2 = 6 A
I3 = 12/4 = 3A
wut is phisics\and what do the do in physica
Physics is the branch of science that deals with the energy and matter and how they are relate to each other. Physics is simply as a natural science that depends on the approximation, measurements and mathematical analysis.