The height at which the 20 kilogram crate is raised to, in the warehouse is approximately 5.1 meters.
What is the height of the crate?Work done is simply defined as the energy transfer that takes place when an object is either pushed or pulled over a certain distance by an external force.
It is expressed as;
Work done = force × distance
Given the data in the question;
Work done = 1000J = 1000kgm²/s²Mass of the crate m = 20kgHeight ( distance ) h = ?Note that;
Force = mass × acceleration
But since the crate is raised and under gravity.
Force = mass × acceleration due to gravity
F = mg
Hence;
Work done = mg × h
1000kgm²/s² = 20kg × 9.8m/s² × h
1000kgm²/s² = 196kgm/s² × h
h = 1000kgm²/s² ÷ 196kgm/s²
h = 5.1m
Therefore, the height of the crate is 5.1 meters.
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Ricardo, of mass 72 kg, and Carmelita, who is lighter, are enjoying Lake Merced at dusk in a 39 kg canoe. When the canoe is at rest in
the placid water, they exchange seats, which are 3.2 m apart and symmetrically located with respect to the canoe's center. Ricardo
notices that the canoe moves 46 cm horizontally relative to a pier post during the exchange and calculates Carmelita's mass. What is it?
Ricardo notices that the canoe moves 46 cm horizontally relative to a pier post during the exchange and the mass is 49 kg.
What is mass?Mass is defined as a measurement of the fundamental attribute of all stuff, inertia.
Mass is also defined as a physical body's total amount of matter.
As per question,
Let's refer to Ricardo's mass as MR and Carmelita's mass as MC.
Let the two-person system's center of mass, which is believed to be closer to Ricardo, be x meters from the center of the canoe's length L and mass m.
MC(L/2 +x) = mx + MR(L/2-x).
It is derived from the second law of parallel forces, which states that the sum of the moments of mass moving clockwise and counterclockwise is equal.
The canoe's center has now moved twice as far from its initial location.
So, x = 46/2 = 23 cm = 0.23 m
MC = MR (L / 2x) - mx / L / 2 + x
MC = 72 ( 3.2 / 2 - 0.23 ) - 39 x 0.23 / 3.2 / 2 + 0.23
MC = 72 ( 1.6 - 0.23 ) - 39 x 0.23 / 1.6 + 0.23
MC = 72 x 1.37 - 8.97 / 1.83
MC = 98.64 - 8.97 / 1.83
MC = 49 kg
Thus, Ricardo notices that the canoe moves 46 cm horizontally relative to a pier post during the exchange and the mass is 49 kg
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A roller coaster car is traveling through a loop at 13 m/s. If the loop has a 23m radius, what centripetal force will the 53kg rider feel?
Given,
The velocity of the car, v=13 m/s
The radius of the loop, r=23 m
The mass of the rider, m=53 kg
The centripetal force is the force that keeps an object in its circular path.
The centripetal force is given by,
[tex]F=\frac{mv^2}{r}[/tex]On substituting the known values,
[tex]\begin{gathered} F=\frac{53\times13^2}{23} \\ =389.43\text{ N} \end{gathered}[/tex]Therefore, the centripetal force on the rider is 389.43 N
a 30.0 kg child starting from rest slides down a water slide with the vertical height of 10.0 m what is the child speed (a )halfway down the slides vertical distance and (b) 3/4 of the way down
We know that
• The mass is m = 30.0 kg.
,• The vertical height is h = 10.0 m.
(a)We have to use the conservation of energy theorem, which states that mechanical energy is constant all the time. Also, halfway down means a height of 5.0 m. It's important to know that at the top the total energy is potential, while halfway is distributed as kinetic and potential, the expression below shows this
[tex]E_{p1}=E_{k1}+E_{p2}[/tex]Then, using the definition of each energy, we have
[tex]mgh_1=\frac{1}{2}mv^2+mgh_2[/tex]Now, we use the given values to find the speed.
[tex]\begin{gathered} \text{mgh}_1=m(\frac{1}{2}v^2+gh_2) \\ gh_1=\frac{1}{2}v^2+gh_2 \\ 9.81m/s^2\cdot10m=\frac{1}{2}v^2+9.81m/s^2\cdot5m \\ 98.1m^2/s^2=\frac{1}{2}v^2+49.05m^2/s^2 \\ 98.1m^2/s^2-49.05m^2/s^2=\frac{1}{2}v^2 \\ 2\cdot49.05m^2/s^2=v^2 \\ v=\sqrt[]{98.1m^2/s^2} \\ v\approx9.9m/s \end{gathered}[/tex]Therefore, the speed of the child halfway down is 9.9 meters per second.(b)In this case, we just have to use as the second height of the equation the magnitude 2.5 meters because that's 3/4 of the way down. So, let's use the same process and expression
[tex]\begin{gathered} gh_1=\frac{1}{2}v^2+gh_2 \\ 9.81m/s^2\cdot10m=\frac{1}{2}v^2+9.81m/s^2\cdot2.5m \\ v=\sqrt[]{2(98.1m^2/s^2-24.53m^2/s^2)} \\ v\approx12.1m/s \end{gathered}[/tex]Therefore, the speed of the child 3/4 of the way down is 12.1 meters per second.A ball rolls off a table and falls 0.75m to the floor, landing with a speed of 4 m/s. (A) What is the acceleration of the ball just before it strikes the ground? (B) What was the initial speed of the ball? (C) What initial speed must the ball have if it is to land with a speed of 5 m/s?
The acceleration of the ball is 21.33m/s²,
We are given that,
The ball fall's from height = d = 0.75m
The final speed of the ball = Vf = 4m/s
So that to know the acceleration of the ball we can calculate by the equation of motion in term of velocity and acceleration i.e. given as,
V = u + at
Where, V is the final velocity , u is the initial velocity , t is the time and a is the acceleration of the object.
t = d/v
t = (0.75m)/(4m/s)
t = 0.1875s
Thus, the value of t, V, and initial velocity is zero putting in equation of motion to get acceleration,
a = (4m/s)/(0.1875s)
a = 21.33m/s²
The acceleration of the ball would be 21.33m/s²
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How many kilometers does the space shuttle have to travel to complete one orbit? In terms of a circle, what is this distance called? Explain.
40,840.7 kilometers the space shuttle have to travel to complete one orbit. In terms of a circle, this distance is termed as the circumference of the circle.
The circumference of a circle is the length measured around its edge. The diameter of a circle is the distance from the center to the outside.
Here we need to find the distance of the space shuttle that completed one circle, ie, the circumference of the orbit. The Circumference or distance covered by the space shuttle can be denoted by [tex]C_{SS}[/tex] and can be calculated by application of the below formula,
[tex]C_{SS}[/tex] = 2πr
where r is the radius of earth ie, 6500Km
Therefore, the equation becomes:
[tex]C_{SS}[/tex] = 2×π×r
= 2×π×6500
=40,840.7
So, the kilometer required to travel is 40,847.7Km
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Which of the following best explains why Venus has a higher temperature than Mercury? And Why?
A. The atmosphere on Venus is thicker than the atmosphere on Mercury.
B. The atmosphere on Venus is thinner than the atmosphere on Mercury.
C. The length of time it takes Venus to revolve around the Sun is shorter than Mercury.
D. The length of time it takes Venus to revolve around the Sun is longer than that of Mercury.
Answer: The reason why Venus is hotter than Mercury is because it has a thick atmosphere primarily made up of carbon dioxide, which is a greenhouse gas that helps retain the heat from the Sun. In comparison to this, Mercury has almost no atmosphere, so any heat that beats down on the planet isn’t retained.
Explanation: Searched
The potential of a 5.0 cm radius conducting sphere is -100 V. What is the charge density on its surface?Group of answer choices-1.8x10-8 C/m23.5x10-8 C/m22.2x10-8 C/m2-2.2x10-8 C/m2-3.5x10-8 C/m2
First lets calculate the surface area
[tex]S=4\cdot\pi\cdot r^2=4\pi\cdot0.05^2m=0.0314m^2[/tex]Now to know the surface charge density we need the next formula:
[tex]CD=\frac{q}{A}[/tex]But we are missing the amount of charge, we only have the potential
So in this case, we going to apply a different formula
[tex]V=\frac{q}{4\cdot\pi\cdot\xi\cdot r}[/tex]q=5.56*10^-10
[tex]CD=-5.56\cdot\frac{10^{-10}}{0.0314}=-1.77\cdot\frac{10^{-8}C}{m^2}[/tex]The anwer might be -1.8x10-8
In Young's double-slit experiment, two slits are separated by 5.0 mm and illuminated by light with a wavelength of 480 nm. The screen is 3.0 m from the plane of the slits. Calculate the separation between the eighth bright fringe and the third dark fringe observed with respect to the central bright fringe.
We are given the following information.
Seperation between slits: d = 5.0 mm
Wavelength of light: λ = 480 nm
Distance from the plane of slits: D = 3.0 m
We are asked to calculate the separation between the 8th bright fringe and the 3rd dark fringe observed with respect to the central bright fringe.
The position of the 8th bright fringe is given by
[tex]\begin{gathered} x_n=\frac{n\lambda D}{d} \\ x_8=\frac{8\cdot480\times10^{-9}\cdot3}{5\times10^{-3}} \\ x_8=2.304\times10^{-3}\;m \end{gathered}[/tex]The position of the 3rd dark fringe is given by
[tex]\begin{gathered} x_n=(\frac{2n-1}{2})\frac{\lambda D}{d} \\ x_3=(\frac{2\cdot3-1}{2})\frac{480\times10^{-9}\cdot3}{5\times10^{-3}} \\ x_3=7.2\times10^{-4}\;m \end{gathered}[/tex]Finally, the separation between the 8th bright fringe and the 3rd dark fringe is
[tex]\begin{gathered} x_8-x_3=2.304\times10^{-3}-7.2\times10^{-4} \\ x_8-x_3=1.584\times10^{-3}\;m \end{gathered}[/tex]Therefore, the separation between the eighth bright fringe and the third dark fringe observed with respect to the central bright fringe is 1.584×10⁻³ m.
The x-component of a force on a 46-g golf ball by a 7-iron versus time is plotted in the following figure: a. Find the x-component of the impulse during the intervals i. [0, 50 ms], and ii. [50 ms, 100 ms] b. Find the change in the x-component of the momentum during the intervals iii. [0, 50 ms], and iv. [50 ms, 100 ms]
The x-component of the impulse during the intervals [0, 50 ms] is 750 Nms.
The x-component of the impulse during the intervals [50, 100 ms] is 1,500 Nms.
The change in the x-component of the momentum during the intervals [0, 50 ms] is 0.75 kgm/s.
The change in the x-component of the momentum during the intervals [50, 100 ms] is 1.5 kgm/s.
What is the impulse experienced by the ball?
The impulse experienced by the ball is calculated from the product of force and time of motion of the ball.
J = Ft
where;
F is the applied forcet is the time of motionThe x-component of the impulse during the intervals i. [0, 50 ms] is calculated as follows;
From the diagram, the impulse between (0, 50 ms) is the area of the triangle.
Jₓ = ¹/₂(b)(h)
where;
b is the base of the triangle = 50 ms h is the height of the triangle = 30 NJₓ = ¹/₂(50 ms)(30 N) = 750 N.ms
The impulse during [50 ms, 100 ms] is the area of the rectangle,
Jₓ = Lb
where;
L is the length = 100 ms - 50 ms = 50 msb is the breadth = 30 NJₓ = 50 ms x 30 N
Jₓ = 1,500 Nms
Impulse is the change in momentum of an object.
The change in the x-component of the momentum during the intervals [0, 50 ms] is calculated as follows;
ΔP = Jₓ = ¹/₂(50 ms)(30 N) = 750 N.ms = 0.75 Ns = 0.75 kgm/s
For interval of [50 ms, 100 ms];
ΔP = 1,500 Nms = 1.5 Ns = 1.5 kgm/s
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Which of the following is not a unit of a force?
Answer: B
Explanation:
One of the common units of force is Newtons, N
Recall,
Force = ma
where
m is the mass of the body
a is the acceleration of the body
If acceleration, a = ms^-2 and mass = kg, then
Force = kgms^-2
Also, joule is the unit of work
recall,
work = force x distance = J
If we divide work by distance(meters), it becomes joule/meter
Thus,
jm^-1 is a unit of force
Thus, the option that is not a unit of force is
B. Js^-1
Two forces F1 = -6.00i + 7.90j and F2 = 6.80i + 5.30j are acting on an object with a mass of m = 4.10 kg. The forces are measured in newtons, i and j are the unit vectors. What is the magnitude of the object's acceleration?
The magnitude of object's acceleration is 3.26m/s².
The mass of the body is 4.10 lg.
The two forces that are acting on the object are F₁ = -6i + 7.9j newton and F₂ = 6.8i + 5.3j Newton.
We know that the force acting on an object is,
F = Ma
Where,
F is the force acting,
M is the mass of the object and,
a is the acceleration of the object.
As we can see, two forces are acting on the body,
We can simplify the forces in x direction and y direction,
The forces are F₁ = -6i + 7.9j N and F₂ = 6.8i + 5.3j N.
So, the total force in x-direction,
Fₓ = (-6+6.8)i
Fₓ = 0.8i
Fᵧ = (7.9+5.3)j
Fᵧ = 13.2j
So, the net force Fₙ on the object is Fₙ = (0.8i + 13.2j) N
Now, putting value of force and mass in the formula,
F = Ma
0.8i + 13.2j = 4.1a
a = 0.19i + 3.21j m/s².
The magnitude of acceleration is,
|a| = √[(0.19)²+(3.21)²]
|a| = 0.361 +10.3
|a| = 3.26m/s².
So, the magnitude of acceleration is 3.26m/s².
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Increasing the mass wouldIncreasing the mass would_______can increase or decrease the value of the gravitational force depending on the initial valuehave no effect on the gravitational forceincrease the gravitational forcedecrease the gravitational force
We will have that:
Increasing the mass would increase the gravitational force.
Exercise 1 :On a circuit, a pilot covers 600 m in 7.2 s.1. Calculate its speed in m / s.2. Convert this speed to km / h in two different ways.
Given data
*The distance covers by the pilot is d = 600 m
*The given time is t = 7.2 s
(1)
The formula for the speed is given as
[tex]s=\frac{d}{t}[/tex]Substitute the known values in the above expression as
[tex]\begin{gathered} s=\frac{600}{7.2} \\ =83.3\text{ m/s} \end{gathered}[/tex]Hence, the speed is s = 83.3 m/s
(2)
The speed converted into km/h as
[tex]\begin{gathered} s=83.3\times(\frac{5}{18}) \\ =299.88\text{ km/h} \end{gathered}[/tex]The second way to convert the speed into kilometer per hour as,
[tex]undefined[/tex]When light goes from a slower medium to a faster medium; what way does the light bend relative to the normal?
Answer: A.
It moves away from the normal.
Explanation: ed mentum or plato
A spring with spring constant 175 N/m has 20 J of EPE stored in it. How much is it compressed?
Given:
• Spring constant = 175 N/m
,• Energy = 20 J
Let's find by how much it is compressed.
Apply the formula:
[tex]E=\frac{1}{2}kx^2[/tex]Where:
E is the energy = 20 J
k is the sring constant = 175 N/m
x is the compression in meters
Rewrite the formula for x:
[tex]x=\sqrt{\frac{2E}{k}}[/tex]Input values and solve for x:
[tex]\begin{gathered} x=\sqrt[]{\frac{2\times20}{175}} \\ \\ x=\sqrt[]{\frac{40}{175}} \\ \\ x=\sqrt[]{0.2285} \\ \\ x=0.48\text{ m} \end{gathered}[/tex]ANSWER:
0.48 m
Compare(how they are the same) and contrast(how they are different) psychoanalysis and behaviorism as two of the early schools of psychology. Offer the major names associated with each and explain how each explained behavior.
Abstract of American Intercontinental University
This paper will compare and contrast three of the 10 main early psychology views. This assignment's three approaches are behavioral, humanistic, and cognitive. Three Early Psychology Perspectives are compared and contrasted. Psychology, like anything else, offers a plethora of theories and methods. One idea may be beneficial to one patient while being ineffective to another. The trick is to discover the optimal method for each patient. The concept that behaviors arise as a result of conditioning is known as behaviorism. This theory does not acknowledge the presence of interior mental factors such as thoughts, feelings, and moods, nor does it take free will into account.
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-%deviation from expected value (show calculations) in relation between frequency and radius.
When an object moves in a circular trajectory with a uniform speed, a centripetal force is needed to keep it in that trajectory. In the experiment, the hanging mass is used to keep the tension of the string constant. The tension is equal to the weight of the hanging mass, and it is responsible for keeping the rubber stopper in a circular trajectory.
The centripetal acceleration of a particle in a circular trajectory with radius r and frequency f is:
[tex]a_c=(2\pi f)^2r[/tex]On the other hand, if the mass of the hanging object is M and the mass of the rubber stopper is m, according to Newton's Second Law of Motion, the relationship between the tension of the string and the centripetal acceleration is:
[tex]\begin{gathered} \Sigma F=ma_c \\ \\ \Rightarrow T=m(2\pi f)^2r \\ \\ \Rightarrow Mg=m(2\pi f)^2r \end{gathered}[/tex]If we isolate f from that expression, we get a theoretical relationship between the radius and the frequency:
[tex]\begin{gathered} \Rightarrow(2\pi f)^2=\frac{Mg}{mr} \\ \Rightarrow2\pi f=\sqrt{\frac{Mg}{mr}} \\ \\ \therefore f=\frac{1}{2\pi}\sqrt{\frac{Mg}{mr}} \end{gathered}[/tex]To find the expected value for the frequency for different values of r, replace M=200g, m=15.3g, g=9.81m/s^2 and the different values of r:
[tex]\begin{gathered} f_{10cm}=\frac{1}{2\pi}\sqrt{\frac{(200g)(9.81\frac{m}{s^2})}{(15.3g)(0.10m)}}\approx5.70Hz \\ \\ f_{15cm}=\frac{1}{2\pi}\sqrt{\frac{(200g)(9.81\frac{m}{s^2})}{(15.3g)(0.15m)}}\approx4.65Hz \\ \\ f_{20cm}=\frac{1}{2\pi}\sqrt{\frac{(200g)(9.81\frac{m}{s^2})}{(15.3g)(0.20m)}}\approx4.03Hz \\ \\ f_{25cm}=\frac{1}{2\pi}\sqrt{\frac{(200g)(9.81\frac{m}{s^2})}{(15.3g)(0.25m)}}\approx3.60Hz \\ \\ f_{30cm}=\frac{1}{2\pi}\sqrt{\frac{(200g)(9.81\frac{m}{s^2})}{(15.3g)(0.30m)}}\approx3.29Hz \end{gathered}[/tex]On the other hand, we can find the experimental measurements for the frequency depending on the radius using the information provided in the table.
To do so, remember that the frequency is equal to the average number of spins from the three trials over the time.
The experimental values for the frequencies are:
[tex]\begin{gathered} f_{10cm-exp}=\frac{35+34+32}{3\times10s}\approx3.37Hz \\ \\ f_{15cm-exp}=\frac{31+29+27}{3\times10s}\approx2.90Hz \\ \\ f_{20cm-exp}=\frac{27+27+25}{3\times10s}\approx2.63Hz \\ \\ f_{25cm-exp}=\frac{25+23+24}{3\times10s}\approx2.40Hz \\ \\ f_{30cm-exp}=\frac{26+23+20}{3\times10s}\approx2.30Hz \end{gathered}[/tex]To find the percent deviation from the expected value, use the following formula:
[tex]\%dev=\frac{f_{exp}-f}{f}\times100\%[/tex]Then, the percent deviations from the expected values are:
[tex]\begin{gathered} \%dev_{10cm}=\frac{3.37Hz-5.70Hz}{5.70Hz}\times100\%\approx-40.9\% \\ \\ \%dev_{15cm}=\frac{2.90Hz-4.65Hz}{4.65Hz}\times100\%\approx-37.6\% \\ \\ \%dev_{20cm}=\frac{2.63Hz-4.03Hz}{4.03Hz}\times100\%\approx-34.7\% \\ \\ \%dev_{25cm}=\frac{2.40Hz-3.60Hz}{3.60Hz}\times100\%\approx-33.3\% \\ \\ \%dev_{30cm}=\frac{2.30Hz-3.29Hz}{3.29Hz}\times100\%\approx-30.1\% \end{gathered}[/tex]When doing a chin-up, a student lifts her body with a force of 400N and a distance of 0.25 meters in 2 seconds. What is the power delivered by the students biceps?
Given:
The student lifts her body with a force F = 400 N
The distance is d = 0.25 N
The time is t = 2 s
To find: The power delivered by student's biceps.
Explanation:
The formula to calculate power is
[tex]P=\frac{F\times d}{t}[/tex]Substituting the values, the power will be
[tex]\begin{gathered} P=\frac{400\times0.25}{2} \\ =\text{ 50 W} \end{gathered}[/tex]Final Answer: The power delivered by student's biceps is 50 W.
if an astronaut weighs 981 N on Earth and only 160 N on the moon, then what is the mass on the moon
If an astronaut weighs 981 N on Earth and only 160 N on the Moon, then his mass on the Moon will be 98.1 kg.
Let's calculate the mass of Earth as per the Earth's acceleration due to gravity.
Now, considering the acceleration due to gravity as 10m/s².
Mass = Weight/Acceleration due to gravity
Mass = 981/10
Mass = 98.1 kg
Now, as per the established fact, the mass is independent of acceleration due to the gravity of the planet i.e. The mass of the person on earth and on the moon is same.
Hence, his mass on the Moon will be 98.1 kg.
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1) Prove that for the adiabatic process of gas pVr =const ; r=Cp/Cv *
2) Write down the expression for the number of molecules of N2 gas having speeds in the interval (100m/s ; 102m/s) for one mole of gas at T=300K
3) Demonstrate how to obtain the Boltzmann Distribution of Energies from the Maxwell-Boltzmann distribution of speeds.*
For an adiabatic process of gas PV(^r)= Constant, where r = Cp/Cv , Cp, and Cv are constants for a particular gas.
What is an ideal gas?It is an imaginary gas for which the volume occupies by it is negligible, this gas does not exist in a practical situation and the concept of an ideal gas is only the theoretical one.
Universal Gas Equation for the ideal gas,
PV=nRT
By differentiating on both sides: P.dV + V.dP = n.R.dT…. (1)
For the adiabatic process, the expansion or compression of gas occurs at a very fast rate which makes no heat transfer. Therefore, dQ= 0
A/c to First Law of TD: dQ=dU+ dW, implies dU= -dW=-P.dV ….. (2)
But dU= n.Cv. dT …. (3) Combining 2 and 3, we get: n.dT= -P.dV/Cv …..(3)
Using R= Cp-Cv, the gas law gives: from 1 and 3.
n.dT= (P.dV + V.dP)/(Cp-Cv) = -PdV/Cv
On rearranging terms we get, (dP/P)+((Cp/Cv)*(dV/V))=0
Cp and Cv are constants for a particular gas, hence integrating,
Ln(P)+ r.Ln(V)=constant (replacing Cp/Cv = r or Gamma)
Hence, P.V(^r)= constant is termed an adiabatic equation.
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Fill in the blanks on machines efficiency
Efficiency = ____________ energy _________ / ____________ energy ______________
Machine Efficiency = output energy/ input energy
The percent of input work that becomes work done by the machine is called efficiency. The output work is always less as compared to the input work because some of the input work gets used in overcoming friction, therefore the efficiency is always less than 100 percent.
Efficiency do not have units. It is written as a decimal or as a percentage. Energy efficiency is the use of less amount of energy to perform the same task. Energy-efficient homes and buildings use lesser energy to heat, cool, and run any appliances and energy-efficient manufacturing facilities use lesser energy to produce goods.
Efficiency measures work or energy that can conserved in a process. We can also say that efficiency is about comparing the output of the energy to the input of the energy.
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What is the electric field strength at a distance of 0.9 m from a charge of 5.71 x 10^-6 C?
Given:
[tex]\begin{gathered} Q=5.71\times10^{-6}\text{ C} \\ r=0.9\text{ m} \end{gathered}[/tex]The electric field strength is given as,
[tex]E=\frac{KQ}{r}[/tex]Here, K is the electrostatic constant.
Putting the values,
[tex]\begin{gathered} E=\frac{9\times10^9\times5.71\times10^{-6}}{(0.9)^2} \\ =63444.44\text{ N/C} \end{gathered}[/tex]Therefore, the electric field strength is 63444.44 N/C.
Which of these does sound travel the fastest Through?A. steel SolidB. water LiquidC. Air gasD. Sound travels the same through all mediums
Speed of sound is maximum on solid.
Correct option.
A) steel Solid
QUESTION 2 (NOVEMBER 2014) Two blocks of masses 20 kg and 5 kg respectively are connected by a light inextensible string, P. A second light inextensible string, Q, attached to the 5 kg block, runs over a light frictionless pulley. A constant horizontal force of 250 N pulls the second string as shown in the diagram below. The magnitudes of the tensions in P and Q are T, and T, respectively. Ignore the effects of air friction.
2.3 Calculate the magnitude of the tension T, in string P. (6)
The magnitude of tension T in string P is 250 N when a constant horizontal force of 250 N pulls the second string.
What is tension and how the tension is calculated out to be 250 N ?Tension is equivalent to pull force , used in most of kinematic questions.Tension can be best explained when you pull a rope or you pull an object during the time period under consideration.Here in this question given, mass of first object is 20 kg , and mass of the second object is 5 kg .Using the equation m1a = T2 - m2a , (m1+m2)a = T2 , (20 +5)250/25 = T2.From this comes the second tension on the string P is 250 N .To know more about tension visit:
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Emily and Gemma did a Reaction time lab. Emily dropped the ruler while Gemma tried to catch it. She caught the ruler 5 times and her average catching distance is 0.12 m. What is Gemma's reaction time?
The average reaction time of Gemma is 0.1564 seconds.
As we know, Gemma is catching the scale and Emily is dropping the scale.
The whole experiment is taking place under gravity, so the acceleration is constant.
As we know, the scale is dropped, it means that the initial velocity of the scale is zero.
We can use the equation of motion,
The equation is,
S = Ut + 1/2at²
Where,
S is the displacement, which is 0.12 m in our case,
U is initial velocity which is 0m/s because the stone is dropped,
t is the time taken, this is equal tot he reaction time here,
a is the acceleration due to gravity whose value is 9.8m/s.
Now, putting all the values,
0.12 = 1/2(9.8)(t)²
t² = 0.24/9.8
t = 0.1564 seconds.
Gemma reacts in 0.1564 seconds to catch the scale.
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In a parallel circuit with a 12 V battery and three 6 Ohm resistors, what is the total current in the entire circuit? Select one:a.36 Ampsb.6 Ampsc.4 Ampsd.2 Amps
First, let's find the equivalent resistance. Since they are in parallel, we can find it as follows:
[tex]\begin{gathered} \frac{1}{R_{eq}}=\frac{1}{R1}+\frac{1}{R2}+\frac{1}{R3} \\ so: \\ \frac{1}{R_{eq}}=\frac{1}{6}+\frac{1}{6}+\frac{1}{6}=\frac{1}{2} \\ Hence: \\ R_{eq}=2\Omega \end{gathered}[/tex]Now, we can use ohm's law to calculate the current:
[tex]\begin{gathered} V=IR \\ so: \\ I=\frac{V}{R}=\frac{12}{2}=6 \end{gathered}[/tex]Answer:
b.
6 Amps
Remember the experiment done by Arthur Holly Compton that demonstrated the particle nature of light (X-rays) definitively. The reaction was:γ+e→γ+e (1) where the outgoing gamma was an X-ray of aa. higher?b. lower?frequency than the initial gamma. Circle your choice.
The correct option is (b)
The outgoing gamma rays are of lower frequency than that of the initial gamma-ray. While investigating the scattering of X-rays, Compton observed that the outgoing rays lose some of their energy in the scattering process and emerge with slightly decreased frequency.
Gravitation always does negative work. Is this true or false?
The given statement 'Gravitation always does negative work' is false.
The work done by the gravitation depends upon the reference level of the system. Therefore, according to reference level it is taken as positive or negative.
A flat coil of wire has an area of 0.020 m2 and contains 50 turns. Initially the coil is oriented so that the normal to its surface is parallel to and in the same direction as a constant magnetic field of 0.18 T. The coil is then rotated through an angle of 30o in a time of 0.10 s. What is the average induced emf? -0.44 V +0.44 V +0.24 V
ANSWER:
3rd option: +0.24 V
STEP-BY-STEP EXPLANATION:
Given:
N = 50
Area = 0.020 m^2
B = 0.18 T
θf = 30°
time = 0.10 s
We can calculate the average induced emf by the following formula
[tex]\epsilon=N\cdot B\cdot A\cdot\left(\frac{\cos\theta_i-\cos\theta_f}{t}\right)[/tex]We replacing:
[tex]\begin{gathered} \epsilon=\left(50\right)\left(0.18\right)\left(0.02\right)\left(\frac{\cos\:0\degree\:-\cos\:30\degree}{\:0.1}\right) \\ \epsilon=0.241\cong0.24\text{ V} \end{gathered}[/tex]The correct answer is 0.24V
- If we start an experiment with 191.3 g of a substance, how much should we end with?
More than 191.3 g
191.3 g
Less than 191.3g PLEASE NOW
If we start an experiment with 191.3 g of a substance, we should be ending with 191.3 g
We know that matter cannot be created out of nothing. So that is why the number of atoms of products must be equal to the number of atoms of reactants. That is why a chemical reaction is written as a balanced reaction equation. This is the law of conservation of mass.
During a combustion reaction or during decomposition reaction, the products mass might seem to be different than that of reactants either higher or lower. This is due to the involvement of environment. So while considering mass environment should also be included.
Therefore, if we start an experiment with 191.3 g of a substance, we should be ending with 191.3 g
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