With what force (per hand) does a man need to lift to keep the wheelbarrow in balance?The total weight of the cart and the load is 100 kg and at the point shown in the center of gravityAnd the right answer about this is 12.5kg per hand
ANSWER
122.5N per hand
EXPLANATION
First, we have to draw a free body diagram,
A wheelbarrow is an example of a lever but in this case, both forces are on the same side of the pivot - which is the wheel. To keep the wheelbarrow in balance, the output force, Fo, must be equal to the weight of the wheelbarrow,
[tex]F_o=m\cdot g=100\operatorname{kg}\cdot9.8m/s^2=980N[/tex]Using the mechanical advantage relation between the input and output forces and the lever arms,
[tex]\frac{F_o}{F_i}=\frac{l_i}{l_o}_{}[/tex]We have to find Fi. To solve for Fi, first raise both sides to the exponent of -1 - in other words, flip both fractions,
[tex]\frac{F_i}{F_o}=\frac{l_o}{l_i}[/tex]And multiply both sides by Fo,
[tex]F_i=F_o\cdot\frac{l_o}{l_i}[/tex]The output force is the one found before, 980N. Then, the length of the lever arm of the output force is the distance from the center of gravity to the wheel, 0.5m, and the length of the lever arm of the input force is the distance from the man's hands to the wheel, 2m,
[tex]F_i=980N\cdot\frac{0.5m}{2m}=\frac{980N}{4}=245N[/tex]The total force the man has to make is 245N. Assuming each hand can support the same weight, then each hand should be doing 122.5N of force to keep the wheelbarrow in balance.
A neutral cloth and a neutral rod are rubbed together. Electrons are transferred from the cloth to the rod. What is the final charge on the overall cloth-rod system?A. PositiveB. NeutralC. NegativeD. Not enough info
Given that a neutral cloth and a neutral rod are rubbed together and electrons are transferred from the cloth to the rod.
Let's find the final charge on the overall cloth-rod system.
Here, since a neutral cloth and a neural rod are rubbed together, there is a transfer of charges between both materials.
The metal rod is a conductor while the cloth is an insulator.
As both materials are rubbed together, there is a transfer of electrons. This is due to friction on the materials.
Now, we know that electrons are negatively charged.
As electrons transfer, the rod becomes positively charged while the cloth becomes negatively charged.
Since both materials have opposite charges, the final charge on the overall cloth-rod system will be a NEUTRAL charge.
When a system is neutral, it means that the system has an equal amount of positive and negative cha
• ANSWER:
B. Neutral
How many kilocalories must be removed from 920 grams of water at a temperature of zero degrees Celsius to transform it into ice at zero degrees Celsius? A)2760 B)497000 C)73.3
Given:
The mass of the water, m=920 g
The temperature of the water, T=0 °C
To find:
The heat required to be removed to transform water to ice.
Explanation:
The latent heat of the fusion of ice is L=79.7 Cal/g.
The latent heat of fusion is the heat that is needed to be removed from the water to convert it into ice.
The amount of the heat that is needed to be removed is given by,
[tex]Q=mL[/tex]On substituting the known values,
[tex]\begin{gathered} Q=920\times79.7 \\ =73324\text{ Cal} \\ \approx73.3\text{ kCal} \end{gathered}[/tex]Final answer:
The amount of heat that is to be removed to convert the given amount of water into ice is 73.3 kCal.
Therefore the correct answer is option C.
Question 3 of 25Which of the following statements is true of the phases of a given substance?A. The particles of a gas are in a fixed location.B. The particles of liquids have more kinetic energy than the particles of gases.C. Gases have strong intermolecular forces between the atoms or molecules.D. The particles of solids have less kinetic energy than the particles of liquids or gases.
ANSWER
D. The particles of solids have less kinetic energy than the particles of liquids or gases.
EXPLANATION
We want to identify the statement that is true about the phases of a given substance.
A gas is a state of matter that expands freely so as to fill the container which contains it. This implies that gases have no fixed shape or location.
The particles of a liquid are more closely packed than gases, implying that motion is restricted for liquids. Since kinetic energy is dependent on motion, it implies that less motion equals less kinetic energy and so liquids have lesser kinetic energy than gases.
In gases, the intermolecular forces are very weak and can be said to be negligible since the particles move freely.
As stated above, kinetic energy is dependent on motion. This means that the more movement that occurs, the more the kinetic energy. The particles of solids are more closely packed than in liquids and gases. This implies that these particles possess less kinetic energy.
Therefore, the correct option is option D.
A typical adult ear has a surface area of 2.61× 10-3 m². The sound intensity during a normal conversation is about 2.65 x 10-6 W/m² at the listener's ear. Assume that the sound strikes the surface of the ear perpendicularly. How much power is intercepted by the ear?
ANSWER:
6.9 x 10^-9 W
STEP-BY-STEP EXPLANATION:
Given:
Area (A) = 2.61× 10^-3 m²
Intensity (I) = 2.65 x 10^-6 W/m²
We can determine the power by the following formula:
[tex]\begin{gathered} P=A\cdot I \\ \\ \text{ We replacing:} \\ \\ P=2.61\cdot10^{-3}\cdot2.65\cdot10^{-6} \\ \\ P=6.9\cdot10^{-9}\text{ W} \end{gathered}[/tex]The power is 6.9 x 10^-9 W
A toy that Is 0.5 ft long is used to model the actions of an actual car that is 15 ft long. Which ratio shows the relationship between the sizes of the model and the actual car????
The length of the toy is,
[tex]l=0.5\text{ ft}[/tex]The actual length is given as,
[tex]l^{\prime}=15\text{ ft}[/tex]The ratio of these two values is,
[tex]\begin{gathered} \frac{l}{l^{\prime}}=\frac{0.5}{15} \\ \frac{l}{l^{\prime}}=\frac{1}{30} \end{gathered}[/tex]Thus, the model is 1/30 times size of an actual car.
Determine the mass of an object when the period of oscillation is 11 s and spring constant is 10 N/m.
Answer:
30.65 Kg.
Explanation:
The period of oscillation T, the spring constant k, and the mass m are related by the following equation.
[tex]T=2\pi\sqrt[]{\frac{m}{k}}[/tex]So, solving for m, we get:
[tex]\begin{gathered} \frac{T}{2\pi}=\sqrt[]{\frac{m}{k}} \\ \frac{T^2}{4\pi^2}=\frac{m}{k} \\ \frac{T^2k}{4\pi^2}=m \\ m=\frac{T^2k}{4\pi^2} \end{gathered}[/tex]Therefore, replacing T = 11 s and k = 10 N/m, we get:
[tex]m=\frac{(11s)^2(10\text{ N/m)}}{4\pi^2}=30.65\text{ kg}[/tex]Then, the mass of the object is 30.65 Kg.
URGENT!! ILL GIVE
BRAINLIEST!!!! AND 100 POINTS!!!!!!
Answer:
a - high thermal conductivity
Answer:
A. high thermal conductivity
Explanation:
high thermal conductivity means a material has the ability to easily take up heat from their environment, in this case the metal pan is taking in heat from it's environment and therefore the mitt is needed.
hope this helps
A water heater uses a resistor of 16 Ω to boil water. The heater is connected to a 240 V power source. If the heater is used for 2 hours a day, calculate the cost to run the heater for a year (365 days). The electricity cost by TNB is given as 21.8 sen/kWh.
First, calculate the power used each day. Use the following formula:
[tex]P=\frac{V^2}{R}[/tex]where,
V: voltage = 240V
R: resistor = 16 Ω
Replace the previous values into the formula for P:
[tex]P=\frac{(240V)^2}{16\Omega}=3600W=3.6kW[/tex]Then, consider that the energy used in kWh, each day, is:
[tex]E=P\cdot t=3.6kW\cdot2h=7.2\text{kWh}[/tex]And in a year, the energy used is:
[tex]E^{\prime}=7.2\text{kWh}\cdot365=2628kWh[/tex]And the cost will be:
[tex]\text{cost}=(2628kWh)(\frac{21.8sen}{\text{kWh}})=57290.4sen[/tex]Hence, the cost to run the heater in one year is 57290.4sen
What is the index of refraction of a substance if the critical angle for a ray going into Water (n=1.33) is 69.30?1.422.2652.11.24
First, remembering Snell's law, we know that:
[tex]sin(\theta)=\frac{n_{entering}}{n_{leaving}}[/tex]So, we know that n_entering is 1.33, as the ray is entering water, and we have the angle, thus:
[tex]sin(69.30°)=\frac{1.33}{n_{leaving}}[/tex]And therefore:
[tex]n_{leaving}=\frac{1.33}{sin(69.30°)}=1.42[/tex]A car, initially traveling at 2.0 miles per hour, comes to rest in 8.6 s. What was its acceleration in this time?
Given:
the speed of the car is
[tex]u=2.0\text{ mi/h}[/tex]Time during which it comes to rest
[tex]t=8.6\text{ s}[/tex]Required: acceleration is to be calculated
Explanation:
first, convert the speed from miles per hour to meters per second
in doing so we use a conversion factor
that is given as
[tex]1\text{ mi/h= 0.447 m/s}[/tex]Then the speed is ,
[tex]\begin{gathered} u=2.0\text{ }\times0.447 \\ u=0.894\text{ m/s} \end{gathered}[/tex]now calculate the acceleration,
we know that
[tex]v=u+at[/tex]where v is the final velocity that is zero because finally, the car comes to rest.
plugging all the values in the above relation, we get
[tex]\begin{gathered} 0=0.894\text{ m/s }+a\times8.6\text{ s} \\ a=-\frac{0.894}{8.6} \\ a=-0.10\text{ m/s}^2 \end{gathered}[/tex]negative sign shows that acceleration in the opposite direction
Thus, the acceleration is
[tex]0.10\text{ m/s}^2[/tex]Suppose two cars are headed toward each other. One is headed north with a mass of1,415 kg at a velocity of 22.9 m/s. The other is headed south with a mass of 819 kg at avelocity 11 m/s. They collide and stick together.What is their total velocity after the collision in the northern direction?
The initial momentum of the system is,
[tex]p_i=m_1u_1+m_2u_2[/tex]The final momentum of the system is,
[tex]p_f=(m_1+m_2)v[/tex]According to conservation of momentum,
[tex]p_i=p_f[/tex]Plug in the known expressions,
[tex]\begin{gathered} m_1u_1+m_2u_2=(m_1+m_{2_{}})v \\ v=\frac{m_1u_1+m_2u_2}{m_1+m_2} \end{gathered}[/tex]Substitute the known values,
[tex]\begin{gathered} v=\frac{(1415\text{ kg)(22.9 m/s)+(819 kg)(11 m/s)}}{(1415\text{ kg+819 kg)}} \\ =\frac{32403.5\text{ kgm/s+}9009\text{ kgm/s}}{2234\text{ kg}} \\ =\frac{41412.5\text{ kgm/s}}{2234\text{ kg}} \\ \approx18.5\text{ m/s} \end{gathered}[/tex]Therefore, the total velocity after the collision is 18.5 m/s
Can you explain the second law of Kepler? Simple terms please
According to Kepler's second law.
The planets move in an elliptical orbit. An elliptical orbit is non-circular which means its radius is not uniform as Sun is present in one of the foci.
According to Kepler's law
Areal speed is constant. This means that when the planet is near the sun, its speed increases and when it is farther from the sun its speed decreases.
When speed is more, the time taken is less and when the speed is less, the time taken will be more.
Thus, areal speed always remains constant.
A 45.8 kg block is placed on an inclined plane that is 44.2 degrees from the horizontal. What is the acceleration of the block? Ignore friction.I have to do the following:1. Draw a free body diagram2.Identify Givens and Unknowns3.Identify the Equations4.Set up the equation using the givens and unknowns5.Solve
The acceleration of the block = 6.83 m/s²
Explanation:The free body diagram representing the given description is:
The mass of the block, m = 45.8 kg
The angle of inclination, θ = 44.2°
Acceleration due to gravity, g = 9.8 m/s²
Applying Newton's second law of motion
∑F = ma
mgsinθ = ma
Divide both sides by m
[tex]\begin{gathered} \frac{mg\sin \theta}{m}=\text{ }\frac{ma}{m} \\ a\text{ = gsin}\theta \end{gathered}[/tex]a = 9.8 sin (44.2)
a = 9.8(0.697)
a = 6.83 m/s²
A girl ties a rope securely to the windowsill of a tower and climbs down to the base of it. She is with a group of friends and wants them to come down too, but easier. She ties the other end of the rope to a tree 30m down and 60m away horizontally, creating a zipline. The first person down the zip line weighs 80 kg. How quickly does he accelerate, and how long does it take for him to reach the end of the zipline?
Answer:
Acceleration = 4.38 m/s²
Time = 5.54 s
Explanation:
We can represent the situation as follows:
So, first, we need to find the angle θ. Using trigonometric functions, we get:
[tex]\begin{gathered} \tan \theta=\frac{opposite}{adjacent} \\ \text{tan}\theta=\frac{60}{20} \\ \tan \theta=2 \\ \theta=tan^{-1}(2)=63.43 \end{gathered}[/tex]Then, the net force in the direction of the rope is equal to:
[tex]\begin{gathered} F_{\text{net}}=mg\cos \theta \\ F_{\text{net}}=(80\operatorname{kg})(9.8m/s^2)\cos (63.43) \\ F_{\text{net}}=350.62N \end{gathered}[/tex]By the second law of Newton, this force is equal to mass times acceleration, so we can solve for acceleration as follows:
[tex]\begin{gathered} F_{\text{net}}=ma \\ a=\frac{F_{net}}{m}=\frac{350.62N}{80\operatorname{kg}}=4.38m/s^2 \end{gathered}[/tex]So, the first person accelerates at 4.38 m/s².
Now, we need to find the length of the rope. Using the Pythagorean theorem, we get:
[tex]\begin{gathered} L=\sqrt[]{60^2+30^2} \\ L=\sqrt[]{3600+900} \\ L=\sqrt[]{4500}=67.08\text{ m} \end{gathered}[/tex]Then, using a kinetic equation, we get:
[tex]\begin{gathered} x=v_it+\frac{1}{2}at^2 \\ x=\frac{1}{2}at^2 \\ 2x=at^2 \\ \frac{2x}{a}=t^2 \\ t=\sqrt[]{\frac{2x}{a}} \end{gathered}[/tex]Where x is the distance traveled, vi is the initial velocity, which is 0 m/s, a is the acceleration and t is the time.
Now, we can replace x by the length of the rope 67.08m and a by 4.38 to get:
[tex]t=\sqrt[]{\frac{2(67.08)}{4.38}}=5.54\text{ s}[/tex]So, the first person takes 5.54 s to reach the end of the zipline.
Therefore, the answers are
Acceleration = 4.38 m/s²
Time = 5.54 s
hi good afternoon pls I need solution to a problem the question goes like this : The horizontal floor of a water reservoir is said to be 1.0m deep when viewed vertically from above .if the refractive index of water is 1.35 calculate the real depth of the reservoir a)2.35mb)1.35mc)1.00md)0.35m
First, you have to use Snell's law to calculate the refractive angle. Remember the light enters at a 90 angle and the refractive index of air is 1.
[tex]\begin{gathered} h\cdot sin\theta2=1m \\ h=\frac{1}{sin(47.79)}=1.35m \end{gathered}[/tex]A 0.09 kg beetle climbs vertically a distance of 0.35 m up a wall at a speed of 0.02 m/s. What is the beetle’s power output?
The power output of the beetle of mass 0.09 kg is 0.0176 W.
What is power?Power is the rate at which work is done per seconds
To calculate the power output of the beetle, we use the formula below.
Formula:
P = mgv............. Equation 1Where:
P = Power output of the beetlem = Mass of the beetlev = Velocity of the beetleg = Acceleration due to gravityFrom the question,
Given:
m = 0.09 kgg = 9.8 m/s²v = 0.02 m/sSubstitute these values into equation 1
P = 0.09×9.8×0.02P = 0.0176 W.Hence, the beetle's power output is 0.0176 W.
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A 2-newton force applied to a 2-kilogram object caused it to move in a straight line 2 meters during an interval of 2 seconds. The object gains kinetic energy K during this interval. In which of the following cases will the object gain the same kinetic energy K? A. The same force is applied to a 4-kilogram object for the same time. B. The same force is applied to a 4-kilogram object for the same distance.C. A 4-Newton force is applied to a 4-kilogram object for the same time.D. A 4-Newton force is applied to a 4-kilogram object for the same distance.E. The same force is applied to a 4-kilogram object for 4 seconds.
By the kinetic energy theorem we know that:
[tex]W=\Delta K[/tex]where W is the work done by the force and delta K is the change in kinetic energy.
In the original scenario the work done by the force is:
[tex]W=Fd=(2)(2)=4[/tex]Which means that the object gained 4 J in kinetic energy.
Now we need to determine in which scenario given the work done is the same, from the options we notice that in option B we apply the same force for the same distance, which means that the work will be the same and hence the object will gain the same kinetic energy (note that this does not mean that the final velocity will be the same, just that the object will gain the same amount of kinetic energy). In all the other cases we can't conclude the same.
Therefore, the answer is option B.
find the drag force on a car traveling at 15 m/s. assume that the cars' cross-sectional area is 3.5 m^2, and that air has a density of 1.2 kg/m^3
The drag force on a car traveling at 15 m/s. assume that the cars' cross-sectional area is 3.5 m^2, and that air has a density of 1.2 kg/m^3 will be 472.5 C(d) Newton
Drag force = 1/2 * density of fluid * [tex](speed)^{2}[/tex] * drag coefficient * cross sectional area
= 1/2 * 1.2 * [tex]15^{2}[/tex] * C (d) * 3.5
= 472.5 C(d) Newton
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An atom of carbon-14 isotope has more protons than an atom of carbon-12 isotope. Is this true or false?
Carbon-12 and carbon-14 are isotopes of each other. Isotopes are the elements that contain the same number of protons but a different number of neutrons.
CArbon-12 contains six protons and six neutrons. While carbon-14 contains six protons and eight neutrons.
Thus the given statement is false.
Determine the order of magnitude of the following small number by entering the appropriate exponent n below: 4,870×1021kg≈10nkg .
We can see that the order of magnitude of the problem is:
[tex]n=21[/tex]A rectangular painting is 120 cm wide and 80 cm high. At what speed must the painting move parallel to its width if it is to appear to be square?
Given
A rectangular painting is 120 cm wide and 80 cm high.
To find
At what speed must the painting move parallel to its width if it is to appear to be square?
Explanation
According to length contraction, the proper length of an object contracts when it moves with a velocity.
Thus,
[tex]\begin{gathered} l=l_o\sqrt{1-\frac{v^2}{c^2}} \\ \Rightarrow80=120\sqrt{1-\frac{v^2}{c^2}} \\ \Rightarrow\frac{2}{3}=\sqrt{1-\frac{v^2}{c^2}} \\ \Rightarrow\frac{4}{9}=1-\frac{v^2}{c^2} \\ \Rightarrow\frac{v^2}{c^2}=1-\frac{4}{9} \\ \Rightarrow\frac{v^2}{c^2}=\frac{5}{9} \\ \Rightarrow v=\frac{\sqrt{5}}{3}c \end{gathered}[/tex]Conclusion:
The required velocity is
[tex]\frac{\sqrt{5}c}{3}\frac{m}{s}[/tex]Which of the following best describes refraction of a wave?A. As a wave moves around a corner or through an opening, itsdirection of motion changes.B. As a wave reaches the interface between two different media, itsdirection of motion changes and it remains in the first medium.C. As a wave moves from one medium to another, the energy of thewave is converted to other forms of energy.O D. As a wave moves from one medium into another in which itsspeed is different, the wave changes direction.
From the given answer choices, let's select the best statement that best describes refraction of a wave.
Refraction of a wave can be said to be a change in direction of waves when they move from one medium to another.
As wave move from
5. What is the net force on an object with a friction of 50 N toward the left and an applied force of 80 N towards the right?130 N right130 N left30 N left30 N right
Given,
The frictional force is 50 N towards left.
The applied force is 80 N towards right.
Thus the net force is:
F=80N-50N=30 N toards right.
The answer is 30 N towards right.
I think this is true I’m not really sure can you explain it?
Assuming there was gravitational force, an object flying upwards would decelerate due its weight. This would cause the velocity to decrease. In this scenario, the area is free of gravitational forces. The velocity is likely to remain the same. Therefore, the engine must not run at some maximum level in order to maintain the ships velocity. Thus, the answer is False
A truck is driving 500 meters/minute. He slows down to a complete stop over 0.25minutes. Calculate his acceleration.
Given
The velocity is v=500 meters/minute
The time taken is t=0.25 minute
To find
The acceleration
Explanation
We know,
The acceleration is the ratio of velocity to time.
Thus,
[tex]\begin{gathered} a=\frac{v}{t} \\ \Rightarrow a=\frac{500}{0.25} \\ \Rightarrow a=2000\text{ m/min}^2 \end{gathered}[/tex]Conclusion
The acceleration is
[tex]2000\text{ m/min}^2[/tex]This week, you completed a simulation in which you tested conductivity of various materials to see if you could complete the circuit and light the bulb. The screenshot above shows you a simple circuit, along with an opening for different objects to be tested. Among the objects provided in the simulation are the following:A dollar billA coinA paper clipAn eraserQuestion to answer:Which two objects listed would act as conductors if clipped into the circuit, and which two objects listed would act as insulators? Please include a sentence to summarize how you'd know if each object is considered a conductor or insulator based on the way the circuit is arranged.hi, can someone help me with this? you don’t have to give me an answer or anything cause i know it’s a lot but maybe tell me the steps on how to find the answer?
We will have the following:
From the options provided we can see that the two objects that could act as conductors in order to complete the circuit are the two metallic objects: the coin and the paper clip; reason being that metals are most of the time much better conductors than other materials.
The two objects then that could potentially act as insulators are the dollar bill and the eraser; reason being in part process of elimination and the fact that paper and the material of the eraser are most of the times bad conductors and present a high resistance value due to the nature of the material.
Light from red, green and blue spotlights shine on asheet of paper. The paper contains cyan pigment. Whichof the three primary light colors will be absorbed by thepigment?
Given data:
The given colors are red, green, and blue.
The cyan pigment is capable of absorbing red color. The
The correct answer is red light.
Which picture correctly shows the path of the reflected light rays given an object outside the focal point?Select one:a. Ab. Bc. Cd. D
We will have that the graph that describes the scenario is given by graph B.
The once-popular LP (long-play) records were 12 in. in diameter and turned at a constant 33 1/3 rpm.
The angular speed of the long-play (LP) in rad/s is 3.49rads/s and its period in seconds is 1.80 seconds.
How is angular speed and period calculated?To convert the angular speed from revolution per minutes (rpm) to seconds the following formula is used:
ω = (k revolution/60 seconds) x (2π/1 revolution)
ω = 2πk/60seconds (k x 2π/T in seconds)
where, k = constant in rpm, 33 1/3rpm = 33.3333rpm
ω = 2π33.3333/60 = 209.4374/60
ω = 3.49rad/s
For period in seconds, ω = 2π/T
where, T = period and ω = angular speed, 3.49rad/s
To find T, T = 2π/ω
T = 2π/3.49rad/s
T = 1.80s
Using the formula, ω = 2πk/60seconds, the angular speed in rad/s of the LP is 3.49rad/s and the time period for the records using T = 2π/ω is 1.80s.
The full question is:
The once-popular LP (long-play) records were 12 inches in diameter and turned at a constant 33(1/3) rpm. Find the angular speed of the LP in rad/s and its period in seconds.
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