Given:
• Mass of bag, m =25.0 kg
,• Height, h = 15.0 m
,• c = 0.20 cal/g•°C
Let's find by how much the temperature will increase.
Apply the Law of Conservation of Energy:
[tex]mc\Delta T=mgh[/tex]Where:
• m is the mass
,• c is the specific heat capacity
,• g is acceleration due to gravity
,• h is the height.
,• ΔT is the temperature change.
Thus, we have:
[tex]\begin{gathered} mc\Delta T=mgh \\ \\ Eliminate\text{ m on both sides:} \\ c\Delta T=gh \end{gathered}[/tex]Now, plug in the values and solve for ΔT:
[tex]\begin{gathered} \Delta T=\frac{gh}{c} \\ \\ \Delta T=\frac{9.8*15.0}{0.20\times10^3\times4.2} \\ \\ \Delta T=\frac{147}{840} \\ \\ \Delta T=0.175^o\text{ C} \end{gathered}[/tex]Therefore, the temperature change is 0.175° C.
• ANSWER:
0.175° C
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 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
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 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 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
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
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]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.
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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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
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
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
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.
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.
ulietta and Jackson are playing miniature golf. Julietta's ball rolls into a long, straight upward incline with a speed of 3.24 m/s and accelerates at -0.234 m/s/s for 2.56 seconds until it reaches the top of the incline and then continues along an elevated section. Determine the length of the incline?
The distance that is covered by the ball is 22.4 m.
What is the length of the incline?We know that the length of the incline is the distance that have been travelled by the ball as it rolls up the incline. We have to use the formula for motion which is;
v^2 = u^2 + 2as
v = final velocity
u = initial velocity
a = acceleration
s = distance
Now;
3.24^2 = 0^2 - 2(-0.234) * s
3.24^2 = - 2(-0.234) * s
s = 3.24^2/- 2(-0.234)
s = 22.4 m
Thus the ball would roll up a distance of about 22.4 m.
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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.
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]Current is measured through a resistor during a laboratory experiment. 1.833 C of charge is measured to pass through the resistor in 1,538 s. What is the current of the circuit? (6 decimal places)
Given,
The charge passed through the resistor, q=1.833 C
The time, t=1,538 s
The electric current can be defined as the time rate of flow of charge. Thus the electric current that passes through the resistor is given by,
[tex]I=\frac{q}{t}[/tex]On substituting the known values,
[tex]\begin{gathered} I=\frac{1.833}{1538} \\ =1.191807\times10^{-3}\text{ A} \\ =1.191807\text{ mA} \end{gathered}[/tex]Thus the current of the circuit is 1.191807 mA.
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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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.
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.
One end of a thin rod is attached to a pivot about which it can rotate without friction. Air resistance is absent. The rod has a length of 0.48 m and is uniform. It is hanging vertically straight downward. The end of the rod nearest the floor is given a linear speed V so that the rod begins to rotate upward about the pivot. What must be the value of V such that the rod comes to a momentary halt in a straight up orientation, exactly opposite to its initial orientation?
ANSWER:
5.32 m/s
STEP-BY-STEP EXPLANATION:
We can determine the velocity from the following equation (initial kinetic energy:
[tex]\frac{1}{6}\cdot v^2_0=g\cdot L[/tex]Where L is the length of the rod and g would be gravity, we replace and calculate the value of the initial velocity:
[tex]\begin{gathered} v^2_0=6\cdot9.81\cdot0.48 \\ v^{}_0=\sqrt[]{28.2528} \\ v_0=5.32\text{ m/s} \end{gathered}[/tex]The velocity is 5.32 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.
What is the magnitude of a force that is required to keep a spring with a spring constant 205 N/m compressed 0.4 m from its starting point?
By Hooke's law we have that the magnitude of the force exerted by a pricng is given by:
[tex]F=kx[/tex]where k is the spring constant and x is the length of extension or compression, then we have:
[tex]F=205\cdot0.4=82[/tex]Therefore the force is 82 N
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.
18.
An object with a mass of 3.2 kg has a force
What is the resulting acceleration of the object?
of 7.3 newtons
applied to it.
2.28125 ms^-2 is the resulting acceleration of the object.
Force = ma
a=F/m
a=7.3 newtons/ 3.2 kg
a=2.28125 ms^-2
Acceleration is the rate of change in a moving object's speed and direction over time. When anything moves faster or slower, it is considered to be accelerating. Motion on a circle increases even while the speed is constant because the direction is always changing.
Since acceleration has both a magnitude and a direction, it is a vector quantity. Velocity is a vector quantity as well. The definition of acceleration is the change in velocity vector during a time interval divided by the time interval.
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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]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]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]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.