ANSWER:
907.2 kilograms
STEP-BY-STEP EXPLANATION:
To convert pounds to kilograms we must resort to the conversion factor between both units.
Where 1 pound is equal to 0.4536 kilograms.
Therefore:
[tex]2000\text{ pounds}\cdot\frac{0.4536\text{ kg}}{1\text{ pound}}=907.2\text{ kg}[/tex]The mass of the elephant is 907.2 kilograms
A woman tries to throw a rock over a wall, releasing the rock at a height of 1.55m abovethe ground. If she throws the rock at 6 m/s, will it reach the top of the wall 3.75m aboveher?
Answer:
[tex]H_{\max }-H_{\text{wall}}=-0.3635m\Rightarrow Negative[/tex]Explanation: We need to find if the rock can reach the height of the wall, provided the initial velocity of 6m/s and it is thrown from the height of 1.55m above the ground., the equations used to solve this problem are as follows:
[tex]\begin{gathered} v(t)=v_o-gt\Rightarrow(1) \\ y(t)=y_o+v_ot-\frac{1}{2}gt^2\Rightarrow(2) \end{gathered}[/tex]Plugging in the known values in the equation (1) and (2) we get the following results:
[tex]\begin{gathered} v(t)=6ms^{-1}-(9.8ms^{-2})t\Rightarrow(3) \\ y(t)=1.55m+(6ms^{-1})t-\frac{1}{2}(9.8ms^{-2})t^2\Rightarrow(4) \\ \end{gathered}[/tex]Setting equation (3) equal to zero gives the time to reach the maximum height as follows:
[tex]\begin{gathered} v(t)=6ms^{-1}-(9.8ms^{-2})t=0 \\ t=\frac{(6ms^{-1})}{(9.8ms^{-2})}=0.61s \\ t=0.61s \end{gathered}[/tex]Substituting the time t calculated above in equation (4) gives the maximum height, the steps for the calculation are shown as follows:
[tex]\begin{gathered} y(0.62)=1.55m+(6ms^{-1})(0.62s)-\frac{1}{2}(9.8ms^{-2})(0.62s)^2 \\ y(0.62)=1.55m+3.72m-1.88356m=3.3865m \\ y(0.62)=3.3865m \\ H_{\max }=3.3865m \\ H_{\max }-H_{\text{wall}}=3.3865m-3.75m=-0.3635m\Rightarrow Negative \\ \\ \end{gathered}[/tex]Therefore, the ball can not reach the height of the wall!
If the average kinetic energy of the molecules in an ideal gas initially at 26 degrees celsius doubles, which of the following is the final temperature of the gas?325 degrees Celsius, 554 degrees Celsius, 52
We will have the following:
The final termperature should then be 52 °C.
There is a distance of 3.60 * 10 ^ 7 m between the Earth and a satellite. The magnitude of the gravitational force of attraction between the and the Earth is 400 N. What would be the magnitude of the gravitational force of attraction if the distance between them changed to 1.80 * 10 ^ 7 meters?
Given:
The initial distance between the earth and the satellite, R₁=3.60×10⁷ m
The initial force between the earth and the satellite, F₁=400 N
The changed distance between the earth and the satellite R₂=1.80×10⁷ m
To find:
The changed force between the earth and the satellite.
Explanation:
From Newton's gravitational law, the gravitational force between two objects is directly proportional to the product of the mass of the objects and inversely proportional to the square of the distance between them.
Let us assume that the mass of the earth is M and the mass of the satellite is m.
Thus,
[tex]\begin{gathered} F_1=\frac{GMm}{R_1^2} \\ \Rightarrow F_1R_1^2=GMm\text{ }\rightarrow\text{ \lparen i\rparen} \end{gathered}[/tex]And, the changed force is given by,
[tex]\begin{gathered} F_2=\frac{GMm}{R_2^2} \\ \Rightarrow F_2R_2^2=GMm\rightarrow\text{ \lparen ii\rparen} \end{gathered}[/tex]From equation (i) and equation (ii),
[tex]\begin{gathered} F_1R_1^2=F_2R_2^2 \\ \Rightarrow F_2=\frac{F_1R_1^2}{R_2^2} \end{gathered}[/tex]On substituting the known values,
[tex]\begin{gathered} F_2=\frac{400\times\left(3.60\times10^7\right?^2}{\lparen1.80\times10^7)^2} \\ =1600\text{ N} \end{gathered}[/tex]Final answer:
The magnitude of the gravitational force of attraction between the earth and the satellite after the distance between them was changed is 1600 N.
At which section does the object move with a positive velocity
The velocity of any objetc is defined in terms of initial and final position of the object is,
[tex]v=\frac{d_f-d_i}{t}[/tex]As in the graph during the motion of object, the final position of the object is decreasing with time. Thus, the final position is always less than the initial position.
[tex]d_f-d_i<0[/tex]Thus, the value of velocity at every point is,
[tex]v<0[/tex]Hence, first option (This object never has a positive velocity) is the correct answer.
I need help finding the area for these acceleration time graphs problem C.
We are given that an object starts with an initial velocity of 5 m/s and that we are required to determine its velocity after 2 seconds. We are given the area of acceleration vs time. In a graph of acceleration vs time, the area under the curve represents the velocity of the object under the given acceleration.
We can determine the velocity reached by the object between t = 0 and t = 2 finding the area under the curve. Since the acceleration is constant the area under the curve is a square, therefore:
[tex]v=(2\frac{m}{s^2})(2s)=4\frac{m}{s}[/tex]This is the velocity after the object starts accelerating the final velocity is determined by adding the initial velocity therefore, the final velocity is:
[tex]v_f=4\frac{m}{s}+5\frac{m}{s}=9\frac{m}{s}[/tex]
For the following calculation, give the answer to the correct number of significant figures.
Given,
[tex]\frac{(71.359\text{ m}-71.357\text{ m})}{(3.2\text{ s}\times3.67\text{ s})}[/tex]When calculating using the numbers with the different number of significant digits, the final answer should contain the same number of significant digits as the number with the least number of significant digits.
On simplifying the above equation,
[tex]\frac{0.002}{12}[/tex]In the above equation, the numerator (0.002) has only one significant digit.
Therefore the answer will have one significant digit
On further simplifying,
[tex]=2\times10^{-4}[/tex]Thus the correct answer is 2×10⁻⁴
the diagram shows four identical magnets that have been dipped into piles of shavings of four different metals. what can be conducted from the diagram above
We can conclude from the experiment that not all the metals can be attracted by a magnet. Since not all the metals could be picked up by the magnet
What is magnetism?A magnet is a material that is able to pick up metallic materials. This implies that the materials can be picked up by the magnets. Metals have magnetic domains in them. These domains get aligned when the magnets have been exposed to a metal.
However, the magnetic domains of all metals do not get aligned when they are exposed to a magnet. As such, not all the metals that we can see are the metals that can be attracted by a magnet.
If we look at the image, we can see that the magnet does attract the cobal and the nickel particles very well but not the aluminium and the copper shavings.
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Missing parts;
.
The diagram shows four identical magnets that have been dipped into piles of shavings of four different metals.
What conclusion can be made about a magnet’s ability to attract metals based on what is shown in this diagram?
answer choices
Magnets do not attract all metals
Magnets only attract aluminum and copper.
Magnets attract all metals.
Steel is the only metal that magnets attract.
If John takes 45 minutes to bicycle a total of four kilometers northwest to his grandmothers house , what is the velocity in km/hr?
time = 45 minutes
distance = 4 kilometers
direction = Northwest
Since the velocity is in km/hr, convert minutes to hours:
1 hour = 60 minutes
45 / 60 = 0.75
45 min = 0.75 hours
Velocity = distance / time = 4 km/ 0.75 h = 5.33 km/ h , northwest.
What does the lower scale read? Answer in units of N
We will have the following:
First, we are given:
*Mass of the breaker: 1.1kg
*Mass of water: 3.3 kg
*Mass of metallic alloy: 4.2kg
*Density of the alloy: 5300kg/m^3
*Density of water: 1000kg/m^3
Now, we find the volume of water displaced by the alloy:
[tex]V_{\text{w}}=4.2\operatorname{kg}\cdot\frac{m^3}{5300\operatorname{kg}}\Rightarrow V_w=\frac{21}{26500}m^3\Rightarrow V_w\approx7.92\cdot10^{-4}m^3[/tex]Then, from the reading in the hanging scale we will have the force experienced by the alloy due to the upthrust when placed in water, that is:
[tex]R=mg-\rho Vg[/tex]So:
[tex]R=(4.2\operatorname{kg})(9.8m/s^2)-(1000\operatorname{kg}/m^3)(\frac{21}{26500}m^3)(9.8m/s^2)[/tex][tex]\Rightarrow R=33.39396226\ldots N\Rightarrow R\approx33.4N[/tex]The reading on the lower scale is due to the weight of the water in the breaker and upthrust on the scale:
[tex]R=g(m_1+m_2)+\rho Vg[/tex]Finally:
[tex]R=(9.8m/s^2)(1.1\operatorname{kg}+3.3\operatorname{kg})+(1000\operatorname{kg}/m^3)(\frac{21}{26500}m^3)(9.8m/s^2)[/tex][tex]\Rightarrow R=50.886003774\ldots N\Rightarrow R\approx50.9N[/tex]So, the readin on the lower scale is approximately 50.9N.
A mass of 10 kg, initially at rest on a horizontal frictionless surface, is acted upon by a horizontal force of 25 N. The speed of the mass after it has moved 5.0 m is:
We will have the following:
[tex]\begin{gathered} W\ast d=\frac{1}{2}mv_f^2-\frac{1}{2}mv_i^2 \\ \\ \end{gathered}[/tex]Then, we will have that:
[tex]\begin{gathered} (25N)(5m)=\frac{1}{2}(10kg)v_f^2-\frac{1}{2}(10kg)(0m/s){}2 \\ \\ \Rightarrow125N\ast m=(5kg)v_f^2\Rightarrow v_f^2=\frac{25N\ast m}{kg} \\ \\ \Rightarrow v_f=5m/s \end{gathered}[/tex]So, the final velocity of the mass after it has moved 5 meters is 5m/s.
3.00 m^3 of water is at 20.0°C.If you raise its temperature to60.0°C, by how much will itsvolume expand?WaterB = 207•10-6 0-1(Unit = m^3)
Given,
Initial volume of the water, V₁=3.00 m³
The initial temperature of the water, T₁=20.0 °C=293.15 K
The final temperature of the water, T₂=60 °C=333.15 K
From Charle's law, we have,
[tex]\frac{V_1}{T_1_{}}=\frac{V_2}{T_2}[/tex]On rearranging the above equation,
[tex]V_2=\frac{V_1T_2}{T_1}[/tex]On substituting the known values in the above equation,
[tex]V_2=\frac{3.00\times333.15}{293.15}=3.41m^3[/tex]Therefore the change in the volume is,
[tex]\Delta V=V_2-V_1[/tex]i.e.,
[tex]\Delta V=3.41-3.00=0.41m^3[/tex]Therefore, the volume of the water will expand by 0.41 m³
If it takes 299 calories to melt a mass of ice into water at a temperature of zero degrees Celsius then what was the mass of the ice in grams? A)0.0471 B)3.75 C)161000
Given:
The heat absorbed by ice to melt into water is Q = 299 calories.
To find the mass of the ice in grams.
Explanation:
When the ice starts melting into water, the temperature of the system does not change, all the heat provided to the system is used up to convert the ice to water. This is known as latent heat of fusion.
The latent heat of fusion for water is
[tex]L_f=79.8\text{ cal/g}[/tex]The formula to calculate mass is
[tex]Q=mL_f[/tex]Substituting the values, the mass will be
[tex]\begin{gathered} m=\frac{Q}{L_f} \\ =\frac{299}{79.8} \\ =3.75\text{ g} \end{gathered}[/tex]Final Answer: The mass of ice is 3.75 g
A ball is thrown upward from an initial height of 2 meters. The ball reaches a height of 5 meters then falls to the ground. What is the total distance traveled by the ball?
We will have the following:
First, we find the distance it traveled upward:
[tex]d_u=5m[/tex]Now, we find the distance it traveled to the ground:
[tex]d_d=5m+2m\Rightarrow d_d=7m[/tex]Now, the total distance is:
[tex]d_T=d_u+d_{\text{d}}\Rightarrow d_T=5m+7m[/tex][tex]\Rightarrow d_T=12m[/tex]So, the ball traveled a total of 12 meters.
Each set of protons and electrons represents a different atom. Place theatoms in order of their overall charge. Order them from most positive to mostnegative.1 30 protons, 26 electrons11 17 protons, 10 electrons10 protons, 10 electrons12 protons, 11 electrons
Firstly, we need to remember how to find the liquid charge for a single atom. We can start by saying that
[tex]Q=n_pq_p+n_eq_e[/tex]Where Q is the total charge, np is the number of protons, qp is the charge of a proton, ne is the number of electrons and qe is the charge of an electron
As we know, the charge of an electron is the opposite of the proton's. If we consider it as -1 and 1, respectively, we're left with
[tex]Q=n_p-n_e[/tex]Then, let us find the charge for each atom.
So, for our atoms, we have
[tex]Q_1=30-26=4[/tex][tex]Q_2=17-10=7[/tex][tex]Q_3=10-10=0[/tex][tex]Q_4=12-11=1[/tex]Then, our answer is
[tex]Q_2>Q_1>Q_4>Q_3[/tex]The virtual image of a lens is always on the same side of the lens as the object. Is this true or false?
The lens follows the law of refraction to form an image of the object.
Thus, for the formation of real image, the image should be formed on the opposite side of the lens.
The virtual image is formed when the image is on the same side of the lens as where the object is present.
Thus, the given statement is true.
Is it possible to get More work out of a machine than you put in?
If we were able to get more work out of a machine than we put in, energy would be created in the process. According to the Law of Conservation of Energy, this is not possible.
Therefore, the answer is:
[tex]\begin{gathered} \text{ No, it is not possible to get more work} \\ \text{ out of a machine than we put in.} \end{gathered}[/tex]URGENT!! ILL GIVE
BRAINLIEST!!!! AND 100 POINTS!!!!!!
Thermal energy can be converted to heat energy.
Option C is correct.
Different Energy Conversion MethodsElectrical energy is a form of chemical energy. Heat energy is a form of thermal energy. Electrical, potential, and other forms of energy can be created from mechanical energy. Heat and light can both be produced with nuclear energy.
What type of energy conversion occurs more during running?Running requires the body to transform potential energy into kinetic energy. A system's internal energy is known as its potential energy. When a system moves horizontally using kinetic energy, potential energy is consumed. Potential energy is held in the form of chemical energy within the human body.
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10 Isabella researched how waves travel through the ground during an earthquake.
She drew a diagram of one, called an S wave, moving through Earth's crust.
Wave motion
Based on her diagram, what kind of wave is an S wave?
A light
B sound
C longitudinal
D transverse
PLEASE REPLY ASAP
How much heat is needed to increase the internal energy of a gas in a piston by 4,258 J if the gas does 801 J of work on the environment by expanding if the process
According to the First Law of Thermodynamics, the change in internal energy (ΔU) of a system is equal to the sum of the heat (Q) added to the system and the work (W) exerted over the system:
[tex]\Delta U=Q+W[/tex]The sign of the heat and the work is positive when energy is transferred to the system and negative when energy is transferred from the system to its surroundings.
In this case, the system is formed by a gas on a piston, which gains 4,258 Joules in internal energy and releases 801 Joules as work. Then:
[tex]+4258J=Q-801J[/tex]Isolate Q from the equation to find the heat needed for this process:
[tex]\Rightarrow Q=4258J+801J=5059J[/tex]Therefore, the amount of heat needed to increase the internal energy of a system by 4,258J if it releases 801 of work to the environment, is 5059J.
If I am finding the x component of acceleration and it’s m/s^3, would the answer be m/s^3
No, the aceleration is alwas m/s^2
If you have something in m/s^3, probablly you are missing a multiplication times the time or maybe you have to integrated the function
[tex]y^{\prime}\left(t\right)=\frac{0.018m}{s^2}[/tex]
(a) How much power (in W) is dissipated in a short circuit of 225 V AC through a resistance of 0.280 Ω? W(b) What current (in A) flows? A
Let's name some variables:
V: voltage; V = 225V
R: resistance; R = 0.28 ohm
I: current
P: power
Let's use this equation first to solve for P:
P = V^2/R
Plugging in variables:
P = 225^2/0.28
P = 180803.57 W = 180.80357 kW
We can use the following to solve for I:
V = IR
225 = I*0.28
I = 803.571 A = 803571 mA
We can also use this method:
Now that we have P, we can use this to solve for I:
P = IV
Plugging in the known variables:
180803.57 = 225*I
I = 803.571 A = 803571 mA
A particularly scary roller coaster contains a loop-the-loop inwhich the car and rider are completely upside down. If the radiusof the loop is 22.7 m with what minimum speed must the cartraverse the loop so that the rider does not fall out while upsidedown at the top? Assume the rider is not strapped to the car.
The forces on the car at the top of the loop are the normal force and the weight, they both points downward. Also the accereation at that point points downward then Newton's second law is:
[tex]-F_N-F_g=-ma[/tex]Now, since this is a circular motion the acceleration is a centripetal one and that is given as:
[tex]a=\frac{v^2}{R}[/tex]where v is the velocity and R is the radius of the circle. Plugging this in Newton's second law we have:
[tex]-F_N-F_g=-m\frac{v^2}{R}[/tex]If the car has the minimum speed to reamin in contact then it would be on the verge of losing contact , this means that Fn=0 at the toop of the loop. Then we have that the equation above:
[tex]\begin{gathered} -F_N-F_g=-m\frac{v^2}{R} \\ 0-mg=-m\frac{v^2}{R} \\ v=\sqrt[]{gR} \end{gathered}[/tex]Plugging the values of the acceleration of gravity and the radius of the loop we have:
[tex]\begin{gathered} v=\sqrt[]{(9.8)(22.7)} \\ v=14.92 \end{gathered}[/tex]Therefore the minimum velocity is 14.92 meters per second.
your calculation.11. What is the kinetic energy of a 2kg rabbit hopping at a speed of 1.25 m/s?
Kinetic energy (KE) = 1/2 m v^2
Where:
m= mass = 2 kg
v= velocity = 1.25 m/s
Replacing:
KE = 1/2 x 2 kg x (1.25 m/s)^2 = 1.5625 J
When you walk across a carpet during the winter you often get a shock when you touch a metal door knob. The reason is that you pick up electrons via friction between your feet and the carpet. If you pick up -97.37 microC of charge, how many electrons did you acquire ?
This is the answer tab
The total charge can be written as
[tex]Q=-97.37*10^{-6}C[/tex]And the charge of an individual electron is
[tex]q=-1.6*10^{-19}C[/tex]Thus, the number of electrons is the total charge acquired, divided by the amount of charge on each individual electron. This gives us:
[tex]n=\frac{Q}{q}=\frac{-97.37*10^{-6}}{-1.6*10^{-19}}=6.085625*10^{14}[/tex]So, we have acquired 6.085625*10^14 electrons
A 12V battery with 0.3 ohm internal resistance is connected to a 24V voltage source charging it.What is the power delivered by the voltage source?
-2880 Watts
Explanation
Step 1
Let
[tex]\begin{gathered} V=24V \\ Internal\text{ resistance of battery=}0.3\Omega \\ \end{gathered}[/tex]
Now, we need aplly the formula
[tex]I(current\text{ in circuit)=}\frac{e.m.f}{\text{total resistance}}=\frac{24\text{v}}{0.3\Omega}=80\text{ Ampers}[/tex]Step 2
now, we need to find the powe delivered, to do that, let's use this formula:
The power delivered by the voltage source is: PV = – I * V =
[tex]\begin{gathered} P=-I\cdot V \\ P=-80\text{ A}\cdot(36)V \\ P=-2880\text{W} \end{gathered}[/tex]hence the answer is -2880 Watt
I hope this helps you
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If you pick a mass of 11.4 kg from the ground up to 0.84 m above the ground and it takes 2.3 s to do so, how much power is expended?
Power is the rate at which energy is transferred with respect to time.
When you pick a mass m from the ground up to a height h, that mass gains potential energy (U), which is given by:
[tex]U=mgh[/tex]Where g is the gravitational acceleration:
[tex]g=9.81\frac{m}{s^2}[/tex]Find the potential energy that the mass of 11.4kg gains when it is raised 0.84m:
[tex]U=(11.4\operatorname{kg})(9.81\frac{m}{s^2})(0.84m)=93.94056J[/tex]Then, divide the total potential energy by 2.3s to find the power expended in doing so:
[tex]P=\frac{U}{t}=\frac{93.94056J}{2.3s}=40.84372174\ldots W\approx41W[/tex]Therefore, the power expended to raise a mass of 11.4kg up to 0.84m above the ground in 2.3 seconds, is 41 Watts.
How do I solve this problem Hint: 1. Draw the forces -rear car: weight downwards, normal force upwards, tension right, drag leftmiddle car: weight downwards, normal force upwards, tension right and left, drag leftlocomotive: weight downwards, normal force upwards, tension left, drag left, thrust/applied right2. The system is moving with a constant speed, so the forces are balanced!3. Start by just looking at the rear car. The tension to the right must balance the drag. This will also be equal to the tension on the middle car. 4. Next, just look at the middle car. The tension to the right must balance the drag and tension to the left. This will also be equal to the tension on the locomotive. 5. Now, just look at the locomotive. The thrust/applied force must balance the drag plus the tension.
The free body diagram of locomotive can be shown as,
Therefore, according to free body diagram, four types of forces are acting on locomotive.
The free body diagram of middle car can be shown as,
Therefore, there are two forces acting on the middle car.
The free body diagram of the last car is shown as,
Therefore, there are three number of forces acting on the last car.
According to free body diagram of locomotive, the tension in cable 2 is equal to net force acting on locomotive. Therefore, the tension in cable 2 is 1000 N.
According to free body diagram of middle car, the tension in cable 1 is equal to tension in cable 2 therefore, the tension in cable 1 is also 1000 N.
The net force generated by locomotive can be given as,
[tex]F=F_{drag}-F_{middle\text{ car}}-F_{\text{rear car}}[/tex]Substitute the known values,
[tex]\begin{gathered} F=1000\text{ N-200 N-200 N} \\ =600\text{ N} \end{gathered}[/tex]Thus, the net force generated by locomotive is 600 N.
Your shopping cart has a mass of 63.9 kilograms. In order to accelerate the shopping cart down an aisle at 0.22 m/s^2, what force (in newtons, N) would you need to apply to the cart?Answer: ________ N
Given:
The mass of the cart is,
[tex]m=63.9\text{ kg}[/tex]The acceleration of the cart is,
[tex]a=0.22\text{ m/s}^2[/tex]To find:
The force applied to the cart
Explanation:
The force on the cart is,
[tex]\begin{gathered} F=ma \\ =63.9\times0.22 \\ =14.06\text{ N} \end{gathered}[/tex]Hence, the force is 14.06 N.
An elevator starts from rest with a constant
upward acceleration and moves 1 m in the first
1.9 s. A passenger in the elevator is holding a
7.4 kg bundle at the end of a vertical cord.
What is the tension in the cord as the elevator accelerates? The acceleration of gravity
is 9.8 m/s
Answer in units of N
The tension in the cord as the elevator accelerates is 76.59 N.
As the elevator starts from rest with a constant upward acceleration and moves 1 m in the first 1.9 s, if the acceleration of the elevator is a, then:
1m = 1/2 a(1.9)²
⇒ a = 2/(1.9²) = 0.55 m/s².
Mass of the bundle; m = 7.4 kg.
Acceleration due to gravity; g = 9.8 m/s².
So, the tension in the cord as the elevator accelerates is = mg + ma
= m(g+a)
= 7.4 ( 9.8 + 0.55) N.
= 76.59 N.
Hence, the tension in the cord is 76.59 N.
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A hypothesis is:an educated guess about a problem or question.the unit of measurement for volume.the official result of an experiment.None of the choices are correct.
Answer:
an educated guess about a problem or question.
Explanation:
An hypothesis is usually concerned with predicting the outcome of an experiment. It is an educated or well structured guess as to what the outcome would be and it is put to test. Thus, the correct option is
an educated guess about a problem or question.