Answer:
m=57.65 kg
Explanation:
Given Data
Ricardo mass m₁=80 kg
Canoe mass m₂=30 kg
Canoe Length L= 3 m
Canoe moves x=40 cm
When Canoe was at rest the net total torque is zero.
Let the center of mass is at x distance from the canoe center and it will be towards the Ricardo cause. So the toque around the center of mass is given as
[tex]m_{1}(L/2-x)=m_{2}x+m_{2}(L/2-x)[/tex]
We have to find m₂.To find the value of m₂ first we need figure out the value of.As they changed their positions the center of mass moved to other side by distance 2x.
so
2x=40
x=40/2
x=20 cm
Substitute in the above equation we get
[tex]m_{x}=\frac{m_{1}(L/2-x)-m_{2}x }{L/2+x}\\m_{x}=\frac{80(\frac{3}{2}-0.2 )-30*0.2}{3/2+0.2}\\m_{x}=57.65 kg[/tex]
Estimate how long it would take one person to mow a football field using an ordinary home lawn mower. Suppose that the mower moves with a 1 km/h speed, has a 0.5-m width, and a field is 360 ft long and 160 ft wide, 1 m-3.281 ft. (Figure 1) O 110 hours O 11 hours O 110 minutes O 11 minutes
Answer:
option B
Explanation:
given,
Length of field = 360 ft
= 360/3.281 = 109.72 m ∵ 1 m = 3.281 ft
width of field = 160 ft
= 160/ 3.281 = 48.76 m
width of mower = 0.5 m
number of rounds required
= [tex]\dfrac{48.76}{0.5}[/tex]
= 97.5
we know, 1 km/h = 0.278 m/s
time taken for each round is equal to
[tex]t = \dfrac{length}{speed}[/tex]
[tex]t = \dfrac{109.72}{0.278}[/tex]
t = 394.67 s
total time,
T = 394.67 x 94.5 = 37296.91 s
[tex]T = \dfrac{37296.91}{3600}\ hours[/tex]
[tex]T =11\ hours[/tex]
Hence. the correct answer is option B
What condition associated with shocks does passing a large current through the heart fail to cure?
A. Uniform polarization of the heart membranes
B. Return fibers to resting position
C. Start the heart functioning again
D. All of the above is accomplished by passing a large current through the body
Answer:
Uniform polarization of the heart membranesExplanation:
Uniform polarization of the heart membranes is associated with shocks, when passing a large current through the heart fail to cure. Therefore, the correct option is option A.
What is heart?The heart is an organ that acts as a blood pump. In spiders as well as annelid worms, it is a straight tube. In mollusks, it is a little more complex structure with one or even more collecting chambers (atria) and a primary pumping chamber (ventricle).
The heart of fish is indeed a folded tube with three to four enlarged regions that resemble the chambers inside the heart of a mammal. Uniform polarization of the heart membranes is associated with shocks, when passing a large current through the heart fail to cure.
Therefore, the correct option is option A.
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At an oceanside nuclear power plant, seawater is used as part of the cooling system. This raises the temperature of the water that is discharged back into the ocean. The amount that the water temperature is raised has a uniform distribution over the interval from 10° to 25° C. Suppose that a temperature increase of more than 18° C is considered to be potentially dangerous to the environment. What is the probability that at any point in time, the temperature increase is potentially dangerous?
Two rockets are launched at a fireworks display. Rocket A is launched with an initial velocity v0 5 100 m/s and rocket B is launched t1 seconds later with the same initial velocity. The two rockets are timed to explode simultaneously at a height of 300 m as A is falling and B is rising. Assuming a constant acceleration g 5 9.81 m/s2, determine (a) the time t_1, (b) the velocity of B relative to A at the time of the explosion.
Answer
given,
initial speed of the rocket A = 100 m/s
height of explode = 300 m
acceleration due to gravity = 9.8 m/s²
rocket b is launched after t₁ time
now, using equation of motion to calculate time
[tex]s = ut + \dfrac{1}{2}gt^2[/tex]
[tex]300 = 100t + \dfrac{1}{2}(-9.8)t^2[/tex]
4.9 t² - 100 t + 300 = 0
using quadratic equation
[tex]t = \dfrac{-(-100)\pm \sqrt{100^2-4\times 4.9 \times 300}}{2\times 4.9}[/tex]
t₁ = 3.65 s and t₂ = 16.75 s
now, rocket A reaches 300 m on return at 16.75 s
rocket B reaches 300 m after 3.65 s
time difference of launch:
t = 16.75 - 3.65
t = 13.1 s
velocity of rocket A
v_a = u + g t
v_a =100 - 9.8 x 16.75
v_a = -64.15 m/s
velocity of rocket B
v_b = u + g t
v_b =100 - 9.8 x 3.65
v_b =+64.23 m/s
relative velocity of B relative to A at the time of the explosion
[tex]V_{BA} = v_b - V_a[/tex]
[tex]V_{BA} = 64.23 -(-64.15)[/tex]
[tex]V_{BA} = 128.38\ m/s[/tex]
relative velocity is equal to 128.38 m/s
20 J of work is done per cycle on a refrigerator with a coefficient of performance of 4.0.
Part A: How much heat is extracted from the cold reservoir per cycle?
Part B: How much heat is exhausted to the hot reservoir per cycle?
Answer:
A)Qa=80 J
B)Qr= 100 J
Explanation:
Given that
W= 20 J
COP = 4
Heat rejected from cold reservoir = Qa
Heat exhausted to hot reservoir = Qr
The COP of refrigerator is given as
[tex]COP=\dfrac{Qa}{W}[/tex]
[tex]4=\dfrac{Qa}{20}[/tex]
Qa= 4 x 20 J
Qa=80 J
By using first law of refrigerator
Qr= Qa + W
Qr= 80 + 20 J
Qr= 100 J
A)Qa=80 J
B)Qr= 100 J
The heat extracted from the cold reservoir per cycle is 80 J and the heat exhausted to the hot reservoir per cycle is 100 J.
Explanation:The equation for the coefficient of performance (COP) of a refrigerator is COP = Q_c / W, where Q_c is the heat extracted from the cold reservoir and W is the work done. Given the COP and the work done, we can rearrange this equation to find Q_c: Q_c = COP * W. Substituting the given values: Q_c = 4.0 * 20 J = 80 J.
The heat exhausted to the hot reservoir (Q_h) for a refrigerator can be found using the equation: Q_h = W + Q_c. Substituting the values we found: Q_h = 20 J + 80 J = 100 J.
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A uniform solid disk with a mass of 24.3 kg and a radius of 0.314 m is free to rotate about a frictionless axle. Forces of 90 and 125 N are applied to the disk in the same horizontal direction but one is applied to the top and the other is applied to the bottom. What is the angular acceleration of the disk (in rad/s2)?
Answer:
α = 9.18 rad/s²
Explanation:
given,
mass of the solid disk = 24.3 Kg
radius of the disk = 0.314 m
Force, F₁ = 90 N
F₂ = 125 N
net force acting on the disk
F = 125 - 90
F = 35 N
Torque
τ = F . r
τ = 35 x 0.314
τ = 11 N.m
we know that
τ = I α
moment of inertia of the solid disk
[tex]I = \dfrac{1}{2}MR^2[/tex]
[tex]I = \dfrac{1}{2}\times 24.3\times 0.314^2[/tex]
I = 1.198 kg.m²
now,
11 = 1.198 x α
α = 9.18 rad/s²
the angular acceleration of the disk is equal to 9.18 rad/s²
To find the angular acceleration of the disk, use the formula: angular acceleration = (net torque) / (moment of inertia). Calculating the torques exerted by the forces and the moment of inertia will allow us to find the answer.
Explanation:The angular acceleration of the disk can be found using the formula:
angular acceleration = (net torque) / (moment of inertia)
where the moment of inertia of a disk is given by:
moment of inertia = (1/2) * mass * radius^2
In this case, the net torque on the disk is the difference between the torques exerted by the two forces:
net torque = torque(top force) - torque(bottom force)
Each torque can be calculated using the formula:
torque = force * radius
Substituting the given values into these formulas, we can find the angular acceleration of the disk.
So, the angular acceleration of the disk is approximately 18.26 rad/s^{2}
.
On a straight road with the +x axis chosen to point in the direction of motion, you drive for 5 hours at a constant 20 miles per hour, then in a few seconds, you speed up to 60 miles per hour and drive at this speed for 1 hour.
What was the x component of average velocity for the 6-hour period, using the fundamental definition of average velocity, which is the displacement divided by the time interval?
Answer:
v = 26.7 mph
Explanation:
During the first 5 hours, at a constant speed of 20 mph, we find the total displacement to be as follows:
Δx₁ = v₁*t₁ = 20 mph*5 h = 100 mi
Assuming we can neglect the displacement during the speeding up from 20 to 60 mph, we can find the the total displacement at 60 mph as follows:
Δx₂ = v₂*t₂ = 60 mph*1 h = 60 mi
So, the total displacement during all the trip wil be:
Δx = Δx₁ + Δx₂ = 100 mi + 60 mi = 160 mi
So we can find the the average velocity during the 6-hour period, applying the definition of average velocity, as follows:
v = Δx / Δt = 160 mi / 6 h = 26.7 mph
When gravitational, magnetic and any forces other than static electric forces are not present, electric field lines in the space surrounding a charge distribution show
a. the directions of the forces that exist in space at all times.
b. Only the directions in which static charges would accelerate when a points on those lines.
c. only the directions in which moving charges would accelerate when at points on those lines.
d. tangents to the directions in which either static or moving charges would accelerate when passing through points on those lines.
e. the paths static or moving charges would take.
A very useful way to graphically schematize a field is to draw lines that go in the same direction as that field at several points. This is done through the electric field lines, which are imaginary lines that describe, if any, changes in the direction of the electric field when passing from one point to another, so that these lines are tangent, in each point of the space where the electric field is defined, to the direction of the electric field at that point.
According to Newton's first law, the force acting on a particle produces a change in its velocity; therefore, the movement of a charged particle in a region will depend on the forces acting on it at each point in that region.
Therefore the electric field lines in space surrounding a charge distribution will show tangents to the directions in which either static or moving charges would accelerate when passing through points on those lines.
The correct answer is D.
Foraging bees often move in straight lines away from and toward their hives. Suppose a bee starts at its hive and flies 500 m due east, then flies 430 m west, then 670 m east. How far is the bee from the hive
Answer:
The distance of the bee from the hive is 740 m.
Explanation:
Given that,
Bee starts fly 500 m due east, 430 m west and 670 m east.
The direction of the bee
500 m in positive direction
430 m in negative direction
670 m in positive direction
We need to calculate the net distance
Using formula of distance
[tex]D=500-430+670[/tex]
[tex]D=740\ m[/tex]
Hence, The distance of the bee from the hive is 740 m.
Consider two charges placed a fixed distance apart.If the charge on each of two small spheres is halved, the force of attraction between the spheres will be ___________.
A. doubled.
B. quadrupled.
C. halved.
D. quartered.
E. the same as before.
Final answer:
When the charge on each of two small spheres is halved, the force of attraction between them, according to Coulomb's Law, will be quartered (Option D). This is because the force is proportional to the product of the charges, which would be reduced to a quarter.
Explanation:
The question is related to Coulomb's Law, which describes the electrostatic force between two charges. According to Coulomb's Law, the force F between two charges is proportional to the product of the charges (q1 and q2) and inversely proportional to the square of the distance (r^2) between them: F = k * (q1*q2) / r^2, where k is Coulomb's constant.
If the charge on each of the two small spheres is halved, we have new charges q1/2 and q2/2. Substituting these into the formula, we get the new force F' as F' = k * ((q1/2)*(q2/2)) / r^2. Simplifying this, F' = (1/4) * k * (q1*q2) / r^2, which is one quarter of the original force F. Therefore, the force of attraction between the spheres will be quartered when the charge on each sphere is halved.
Calculate the linear momentum per photon,energy per photon, and the energy per mole of photons for radiation of wavelength; (a) 600 nm (red), (b) 550 nm (yellow), (c) 400 nm (violet), (d) 200 nm (ultraviolet), (e) 150 pm (X-ray), and (f ) 1.0 cm (microwave).
Answer:
The detailed explanations is attached below
Explanation:
What is applied is the De brogile equation and the equation showing a relationship between Energy, speed of light and wavelength.
The explanation is as attached below.
Why does the large number of hydrogen atoms in the universe suggest that other elements?
Answer:
Explanation:
The abundance of the chemical elements is a measure of the occurrence of the chemical elements relative to all other elements in a given environment. Abundance is measured in one of three ways: by the mass-fraction (the same as weight fraction); by the mole-fraction (fraction of atoms by numerical count, or sometimes fraction of molecules in gases); or by the volume-fraction. Volume-fraction is a common abundance measure in mixed gases such as planetary atmospheres, and is similar in value to molecular mole-fraction for gas mixtures at relatively low densities and pressures, and ideal gas mixtures. Most abundance values in this article are given as mass-fractions.
For example, the abundance of oxygen in pure water can be measured in two ways: the mass fraction is about 89%, because that is the fraction of water's mass which is oxygen. However, the mole-fraction is about 33% because only 1 atom of 3 in water, H2O, is oxygen. As another example, looking at the mass-fraction abundance of hydrogen and helium in both the Universe as a whole and in the atmospheres of gas-giant planets such as Jupiter, it is 74% for hydrogen and 23–25% for helium; while the (atomic) mole-fraction for hydrogen is 92%, and for helium is 8%, in these environments. Changing the given environment to Jupiter's outer atmosphere, where hydrogen is diatomic while helium is not, changes the molecular mole-fraction (fraction of total gas molecules), as well as the fraction of atmosphere by volume, of hydrogen to about 86%, and of helium to 13%.[Note 1]
The abundance of chemical elements in the universe is dominated by the large amounts of hydrogen and helium which were produced in the Big Bang. Remaining elements, making up only about 2% of the universe, were largely produced by supernovae and certain red giant stars. Lithium, beryllium and boron are rare because although they are produced by nuclear fusion, they are then destroyed by other reactions in the stars.[1][2] The elements from carbon to iron are relatively more abundant in the universe because of the ease of making them in supernova nucleosynthesis. Elements of higher atomic number than iron (element 26) become progressively rarer in the universe, because they increasingly absorb stellar energy in their production. Also, elements with even atomic numbers are generally more common than their neighbors in the periodic table, due to favorable energetics of formation.
The abundance of elements in the Sun and outer planets is similar to that in the universe. Due to solar heating, the elements of Earth and the inner rocky planets of the Solar System have undergone an additional depletion of volatile hydrogen, helium, neon, nitrogen, and carbon (which volatilizes as methane). The crust, mantle, and core of the Earth show evidence of chemical segregation plus some sequestration by density. Lighter silicates of aluminum are found in the crust, with more magnesium silicate in the mantle, while metallic iron and nickel compose the core. The abundance of elements in specialized environments, such as atmospheres, or oceans, or the human body, are primarily a product of chemical interactions with the medium in which they reside.
If the pressure of gas is doubled and its absolute temperature is quadrupled, the volume is what factor times the original?
a. 2
b. 1/2
c. 1/4
d. 4
Answer:
Volume will increase by factor 2
So option (A) will be correct answer
Explanation:
Let initially the volume is V pressure is P and temperature is T
According to ideal gas equation [tex]PV=nRT[/tex], here n is number of moles and R is gas constant
So [tex]V=\frac{nRT}{P}[/tex]....................eqn 1
Now pressure is doubled and temperature is quadrupled
So new volume [tex]V_{new}=\frac{nR4T}{2P}=\frac{2nRT}{P}[/tex] ........eqn 2
Now comparing eqn 1 nad eqn 2
[tex]V_{new}=2V[/tex]
So volume will increase by factor 2
So option (A) will be correct answer
A bucket of water with a mass of 2.0 kg is attached to a rope that is wound around a cylinder. The cylinder has a mass of 4.0 kg and is mounted horizontally on frictionless bearings. The bucket is released from rest. (a) Find its speed after it has fallen through a distance of 1.50 m
Answer:
Explanation:
Given mass of bucket is [tex]m=2\ kg[/tex]
mass of cylinder [tex]M=4\ kg[/tex]
Suppose T is the tension in the rope
For bucket [tex]mg-T=ma[/tex]
where a=acceleration
For cylinder with Radius R
[tex]I\times \alpha =T\cdot R[/tex]
[tex]\frac{MR^2}{2}\times \frac{a}{R}=T\times R[/tex]
[tex]T=\frac{Ma}{2}[/tex]
[tex]a=\frac{mg}{m+0.5M}[/tex]
[tex]a=4.9\ m/s^2[/tex]
Using [tex]v^2-u^2=2a s[/tex] for bucket
v=final velocity
u=initial velocity
s=displacement
[tex]v^2-0=2\times 4.9\times 1.5[/tex]
[tex]v=\sqrt{14.7}[/tex]
[tex]v=3.83\ m/s[/tex]
Final answer:
The speed of the bucket after falling a distance of 1.50 m is 3.83 m/s.
Explanation:
To find the speed of the bucket after falling a distance of 1.50 m, we can use the principle of conservation of energy. The initial potential energy of the bucket is given by the equation PE = mgh, where m is the mass of the bucket, g is the acceleration due to gravity, and h is the height from which the bucket is released. In this case, the initial potential energy is PE = (2.0 kg)(9.8 m/s^2)(1.50 m).
The final kinetic energy of the bucket is given by the equation KE = 0.5mv^2, where m is the mass of the bucket and v is the speed of the bucket. Since the bucket starts from rest, the initial kinetic energy is zero. Therefore, the final kinetic energy is equal to the initial potential energy. We can set up the equation (2.0 kg)(9.8 m/s^2)(1.50 m) = 0.5(2.0 kg)v^2 and solve for v.
Simplifying the equation, we get ([tex]29.4 kg*m^2/s^2[/tex]. Dividing both sides by (2.0 kg), we find v^2 = 14.7 m^2/s^2. Taking the square root of both sides, we find v = 3.83 m/s.
An object falls a distance h from rest. If it travels 0.560h in the last 1.00 s, find (a) the time and (b) the height of its fall.
Answer:
(a) t = 2.97s
(b) h = 43.3 m
Explanation:
Let t be the time it takes to fall a distance h, then t - 1 (s) is the time it takes to fall a distance of h - 0.56h = 0.44 h
For the ball to fall from rest a distance of h after time t
[tex]h = gt^2/2[/tex]
Also for the ball to fall from rest a distance of 0.44h after time (t-1)
[tex]0.44h = g(t-1)^2/2[/tex]
We can substitute the 1st equation into the 2nd one
[tex]0.44gt^2/2 = g(t-1)^2/2[/tex]
and divide both sides by g/2
[tex]0.44t^2 = (t-1)^2[/tex]
[tex]0.44t^2 = t^2 - 2t + 1[/tex]
[tex]0.56t^2 - 2t + 1 = 0[/tex]
[tex]t= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}[/tex]
[tex]t= \frac{2\pm \sqrt{(-2)^2 - 4*(0.56)*(1)}}{2*(0.56)}[/tex]
[tex]t= \frac{2\pm1.33}{1.12}[/tex]
t = 2.97 or t = 0.6
Since t can only be > 1 s we will pick t = 2.97 s
(b) [tex]h = gt^2/2 = 43.3 m[/tex]
In Houston, Earth's B→ field has a magnitude of 5.2 × 10−5 Tand points in a direction 57∘ below a horizontal line pointing north.Part ADetermine the magnitude of the magnetic force exerted by the magnetic field on a 11-m-long vertical wire carrying a 11-A current straight upward. Express your answer to two significant figures and include the appropriate units.Part BDetermine the direction of the magnetic force.a. to the northb. to the eastc. to the southd. to the west
Answer:
F = 0.0034 N
Explanation:
Given:
[tex]B = 5.2*10^(-5) T\\Q = 57 degrees\\I_{wire} = 12 A\\L_{wire} = 10 m[/tex]
The angle between B and wire = 90 - 57 = 33 degrees
Using formula:
[tex]F = B*I*L*sin (90-Q)\\F = (5.2*10^(-5)*(12)*(10)*sin (33)\\F = 0.0034 N[/tex]
(A) The magnetic force exerted on the wire is 3.4×10⁻³N
(B) The direction of the force is to the west.
Magnetic force:Given that the magnetic field B = [tex]5.2\times10^{-5}T[/tex] which points in the direction 57° below a horizontal line pointing north.
Length of the wire L = 11m
current in the wire I = 11A
The angle between the wire and the magnetic field is θ = (90-57) = 33°
(A) The magnetic force on a finite wire of length L carrying a current I is given by:
[tex]F=BILsin\theta\\\\F=5.2\times10^{-5}\times11\times11\sin33\\\\F=3.4\times10^{-3}N[/tex]
(B) The direction of the force is given by dl×B, now B is at 57° with the north direction and the wire is verticle, so the direction of the field will be to the west.
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A triangle with equal sides of length 14 cm has -2.5-nC charged objects at each corner. Determine the direction of the electrical force exerted on the object at the top corner due to the two objects at the horizontal base of the triangle.
(A) vertically down
(B) Vertically upward
(C) horizontally rightward
(D) horizontally leftward
Answer:
(B) Vertically upward
Explanation:
r = Side of triangle = 14 cm
q = Charge = -2.5 nC
Electrical force is given by
[tex]F=\dfrac{kq_1q_2}{r^2}\\\Rightarrow F=\dfrac{8.99\times 10^9\times (2.5\times 10^{-9})^2}{0.14^2}\\\Rightarrow F=2.86671\times 10^{-6}\ N[/tex]
For the top charge
Net force on both charges is given by
[tex]F_n=2Fcos\theta\\\Rightarrow F_n=2\times 2.86671\times 10^{-6}\times cos30\\\Rightarrow F_n=4.96529\times 10^{-6}\ N[/tex]
The net force acting on the top charge is [tex]4.96529\times 10^{-6}\ N[/tex]
Here the forces are symmetrical hence, the net force is along +y axis i.e., upward
The direction of the electrical force exerted on the object at the top corner due to the two objects at the horizontal base of the triangle is vertically upward.
Explanation:The direction of the electrical force exerted on the object at the top corner due to the two objects at the horizontal base of the triangle can be determined using Coulomb's Law. Since the objects at the base of the triangle have a negative charge and the object at the top corner has a negative charge as well, the force between them will be repulsive. As a result, the direction of the electrical force on the top object will be vertically upward.
You are pushing a crate horizontally with 100 N across a 10 m factory floor. If the force of friction on the crate is a steady 70 N, how much kinetic energy is gained by the crate
Answer:
K E = 300 J
Explanation:
given,
Force of pushing crate, F = 100 N
distance of push = 10 m
frictional force = 70 N
kinetic energy is gained by the crate = ?
using work energy theorem
K E = work done
K E = F . s
K E = (100 - 70) x 10
K E = 30 x 10
K E = 300 J
kinetic energy is gained by the crate is equal to 300 J.
Final answer:
The kinetic energy gained by the crate is 300 J, calculated as the net work done on the crate, which is the work done by the applied force (1000 J) minus the work done by friction (700 J).
Explanation:
To determine how much kinetic energy is gained by the crate, we can use the work-energy theorem which states that the work done on an object is equal to the change in kinetic energy of the object. In this scenario, the net work done on the crate is the work done by the applied force minus the work done by friction. The work done by the applied force is force times distance, which is 100 N x 10 m = 1000 J. The work done by friction is 70 N x 10 m = 700 J. Therefore, the net work done on the crate, which is also the kinetic energy gained by the crate, is 1000 J - 700 J = 300 J.
In the parts that follow select whether the number presented in statement A is greater than, less than, or equal to the number presented in statement B. Be sure to follow all of the rules concerning significant figures.Statement A: 2.567 km, to two significant figures.Statement B: 2.567 km, to three significant figuresso, statement A is ........... statement BStatement A: (2.567 km+ 3.146 km), to two significant figures.Statement B: (2.567 km, to two significant figures) + (3.146 km, to two significant figures).so, statement A is .............statement B
Answer:
a) Therefore 2.6km is greater than 2.57km.
Statement A is greater than statement B.
b) Therefore 5.7km is equal to 5.7km
Statement A is equal to statement B
Explanation:
a) Statement A : 2.567km to two significant figures.
2.567km 2. S.F = 2.6km
Statement B : 2.567km to three significant figures.
2.567km 3 S.F = 2.57km
Therefore 2.6km is greater than 2.57km.
Statement A is greater than statement B.
b) statement A: (2.567 km + 3.146km) to 2 S.F
(2.567km + 3.146km) = 5.713km to 2 S.F = 5.7km
Statement B : (2.567 km, to two significant figures) + (3.146 km, to two significant figures).
2.567km to 2 S.F = 2.6km
3.146km to 2 S.F = 3.1km
2.6km + 3.1km = 5.7km
Therefore 5.7km is equal to 5.7km
Statement A is equal to statement B
Considering significant figures, statement A: 2.567 km is less than Statement B: 2.567 km, since it rounds to 2.6 km. For the addition, Statement A (5.713 km rounded to 5.7 km) is equal to Statement B (adding 2.6 km and 3.1 km).
Explanation:When we consider significant figures, statement A: 2.567 km round off to two significant figures would be 2.6 km and Statement B: 2.567 km is the same to three significant figures. Therefore, statement A is less than statement B because 2.6 km is less than 2.567 km. Similarly, for the next case, if we add 2.567 km and 3.146 km, it gives us 5.713 km, rounded off to two significant figures it becomes 5.7 km for statement A. Whereas statement B, 2.567 km rounded off becomes 2.6 km and 3.146 km becomes 3.1 km. Adding both gives 5.7 km. Hence, statement A is equal to statement B.
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In a simple model of the hydrogen atom, the electron moves in a circular orbit of radius 0.053nm around a stationary proton.How many revolutions per second does the electron make? Hint: What must be true for a force that causes circular motion?Ans: ___ Hz
Answer:
Frequency, [tex]f=6.57\times 10^{15}\ Hz[/tex]
Explanation:
It is given that, the electron moves in a circular orbit of radius 0.053 nm around a stationary proton. The electric force acting on the electron is balanced by the centripetal force as :
[tex]\dfrac{kq^2}{r^2}=\dfrac{mv^2}{r}[/tex]
v is the speed of electron
[tex]v=\sqrt{\dfrac{ke^2}{mr}}[/tex]
[tex]v=\sqrt{\dfrac{9\times 10^9\times (1.6\times 10^{-19})^2}{9.1\times 10^{-31}\times 0.053\times 10^{-9}}}[/tex]
[tex]v=2.18\times 10^6\ m/s[/tex]
The speed of electron is given by :
[tex]v=\dfrac{2\pi r}{t}[/tex]
[tex]t=\dfrac{2\pi r}{v}[/tex]
[tex]t=\dfrac{2\pi \times 0.053\times 10^{-9}}{2.18\times 10^6}[/tex]
[tex]t=1.52\times 10^{-16}\ s[/tex]
We know that the number of revolutions per second is called frequency of electron. It is given by :
[tex]f=\dfrac{1}{t}[/tex]
[tex]f=\dfrac{1}{1.52\times 10^{-16}}[/tex]
[tex]f=6.57\times 10^{15}\ Hz[/tex]
So, the total number of revolutions per second make by the electron is [tex]f=6.57\times 10^{15}\ Hz[/tex]. Hence, this is required solution.
Two blocks of ice, one four times as heavy as theother, are at rest on a frozen lake. A person pushes each block thesame distance d.Ignore friction and assume that an equal force F_vec is exerted on each block.
this is what i have sofar:
PartA
Which of the following statements istrue about the kinetic energy of the heavier block after thepush?
It is equalto the kinetic energy of the lighter block.
PartB
Compared to the speed of the heavierblock, how fast does the light block travel?
twice asfast
This is what i cant figure out:
PartC
Now assume that both blocks have thesame speed after being pushed with the same force F_vec. What can be said about the distances the two blocks arepushed?
a The heavyblock must be pushed 16 times farther than the lightblock.
b The heavyblock must be pushed 4 times farther than the lightblock.
c The heavyblock must be pushed 2 times farther than the lightblock.
d The heavytblock must be pushed the same distance as the lightblock.
e The heavyblock must be pushed half as far as the light block.
Answer:
A. [tex]KE_h=\frac{1}{16}\times KE_l[/tex]
B. the lighter block travels at a speed 4 times faster than the heavier block.
C. b The heavy block must be pushed 4 times farther than the light block.
Explanation:
mass of lighter block, [tex]m[/tex]mass of the heavier block, [tex]4m[/tex]Given that the blocks are acted upon by equal force.
A.
Then the kinetic energy of the blocks depends on their individual velocity.
And velocity is related to momentum through Newton's second law of motion:
[tex]\frac{d}{dt} .p=F[/tex]
[tex]\frac{d}{dt} (m.v_l)=F[/tex]
considering that the time for which the force acts on each mass is equal.
[tex]dv_l=\frac{F.dt}{m}[/tex]
For the heavier block:
[tex]dv_{_h} =\frac{F.dt}{4m}[/tex]
Therefore:
Kinetic energy of lighter block:
[tex]KE_l=\frac{1}{2}\times m.(\frac{F.dt}{m} )^2[/tex]
[tex]KE_l=\frac{1}{2m} \times (F.dt)^2[/tex]
Kinetic energy of heavier block:
[tex]KE_h=\frac{1}{2} \times m.(\frac{F.dt}{4m} )^2[/tex]
[tex]KE_h=\frac{1}{16}\times (\frac{1}{2m} \times (F.dt)^2)[/tex]
[tex]KE_h=\frac{1}{16}\times KE_l[/tex]
B.
From the above calculations and assumptions we observe that the lighter block travels at a speed 4 times faster than the heavier block.
C.
Since the lighter block is having the speed 4 times more than the heavier block so it must be pushed 4 times farther because the speed is directly proportional to the distance.
A) The statement that is true from the attached link about the kinetic energy of the heavier block after the push is;
B: the kinetic energy of the heavier block is equal to the kinetic energy of the lighter block
B) Compared to the speed of the heavierblock, the speed of the light block travel is; twice as fast.
C) If we assume that both blocks have the same speed after being pushed with the same force F_vec, what we can say about the distances the two blocks are pushed is;
The heavyblock must be pushed 4 times farther than the lightblock.
A) The formula for the work done on each block is gotten from the formula;
W = F × d
We are told that the two blocks were pushed the same distance and with the same force F_vec being exerted on them. Thus, the work done on each of the blocks will be the same.
From work-energy theorem, we recall that the work done on an object is equal to its change in kinetic energy. Thus;
W = ΔKE
Where;
ΔKE = Final KE - Initial KE
Thus;
W = Final KE - Initial KE
Since the blocks were initially at rest, then it means that;
Initial KE = 0
Thus;
W = Final KE - 0
W = Final KE
The work done on each block is the same and as a result their final kinetic energies will be the same.
B) We are told that one of the blocks is ¼ times as heavy as the other block.
Thus;
½(m)v_h² = ½(¼mv_l²)
v_h² = ¼v_l²
Where;
v_h is speed of heavy block
v_l is speed of light block
Thus, taking square root of both sides gives;
v_h = ½v_l
Thus, the speed of the light block is twice as fast.
C) We assume the blocks have the same speed after being pushed with the same force. Thus;
It means that heavier block must be pushed four times farther than lighter block.
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A highway patrol car traveling a constant speed of 105 km/h is passed by a speeding car traveling 140 km/h. Exactly 1.00 s after the car passes, the highway patrol officer steps on the accelerator; of the patrolman's car accelerates at 3.50 m/s2, how much time passes after the car passes before the patrol car overtakes the speeder (assume the speeder is moving at constant speed)?
Answer:
The elapsed time from when the speeder passes the patrol car until it is caught is 9.24 s.
Explanation:
Hi there!
The position of the patrol car at a time "t" can be calculated using this equation:
x = x0 + v0 · t + 1/2 · a · t²
Where:
x = position of the patrol car at a time "t"
x0 = initial position.
v0 = initial velocity.
t = time.
a = acceleration.
For the speeding car, the equation is the same only that the acceleration is zero. Then, the equation gets reduced to this:
x = x0 + v · t
Where "v" is the constant velocity.
First, let´s convert the velocity units into m/s:
140 km/h · 1000 m / 1 km · 1 h / 3600 s = 38.9 m/s
105 km/h · 1000 m / 1 km · 1 h / 3600 s = 29.2 m/s
We have to find how much time it takes the patrol car to catch the speeder after the speeder passes the patrol car.
When the patrol car catches the speeder, the position of both cars is the same:
position of the patrol car = position of the speeder
x0 + v0 · t + 1/2 · a · t² = x0 + v · t
if we place the origin of the frame of reference at the point where the patrol car starts accelerating (1 s after the speeder passes the patrol car) then, the initial position of the patrol car will be zero, while the initial position of the speeder will be the traveled distance in 1 s:
x = v · t
x = 38.9 m/s · 1 s = 38.9 m
When the patrol car accelerates, the speeder is 38.9 m ahead of it. Then:
x0 + v0 · t + 1/2 · a · t² = x0 + v · t
0 + 29.2 m/s · t + 1/2 · 3.50 m/s² · t² = 38.9 m + 38.9 m/s · t
Let´s agrupate terms and equalize to zero:
-38.9 m - 38.9 m/s · t + 29.2 m/s · t + 1.75 m/s² · t² = 0
-38.9 m - 9.70 m/s · t + 1.75 m/s² · t² = 0
Solving the quadratic equation for t using the quadratic formula:
t = 8.24 s (the other solution is discarded because it is negative)
The elapsed time from when the speeder passes the patrol car until it is caught is (8.24 s + 1.00) 9.24 s.
(c) The driver of a car traveling at a speed of 17 m/s slams on the brakes and comes to a stop in 3 s. If we assume that the speed changed at a constant rate (constant net force), what was the average speed during this 3 s interval?
Answer:
[tex]v_{avg}=8.5\ m/s[/tex]
Explanation:
given,
initial speed of the car, v₁ = 17 m/s
final speed of the car, v₂ = 0 m/s
car stops in time = 3 s
we need to calculate average speed
[tex]v_{avg}=\dfrac{v_1 + v_2}{2}[/tex]
[tex]v_{avg}=\dfrac{17 + 0}{2}[/tex]
[tex]v_{avg}=\dfrac{17}{2}[/tex]
[tex]v_{avg}=8.5\ m/s[/tex]
average speed of the car during interval of 3 s is 8.5 m/s
What is the magnitude of an electric field in which the electric force it exerts on a proton is equal in magnitude to the proton's weight?
Final answer:
The magnitude of an electric field in which the electric force equals a proton's weight is calculated to be approximately 102 N/C, considering the proton's charge and mass along with Earth's gravitational acceleration.
Explanation:
To determine the magnitude of an electric field in which the electric force equals a proton's weight, we must equate the electric force to the gravitational force (weight) acting on the proton. The electric force (FE) experienced by a charge in an electric field (E) is given by FE = qE, where q is the charge of the proton. The weight of the proton (W) can be found using W = mg, where m is the mass of the proton and g is the acceleration due to gravity (approximately 9.81 m/s2 on Earth). The charge of a proton is approximately 1.6 x 10-19 C, and its mass is approximately 1.67 x 10-27 kg.
Setting FE equal to W, we have qE = mg. Solving for E, the electric field magnitude, gives us E = mg/q. Thus, substituting the known values, we find:
E = (1.67 x 10-27 kg * 9.81 m/s2) / 1.6 x 10-19 C,
this gives us E approximately equal to 102 N/C. This is the magnitude of the electric field in which the electric force on a proton is equal to the proton's weight.
Final answer:
The magnitude of the electric field can be found by setting the electric force equal to the proton's weight and solving for the electric field. The magnitude of the electric field in this case is approximately 1.03 x 10¹² N/C.
Explanation:
The magnitude of the electric field in which the electric force it exerts on a proton is equal in magnitude to the proton's weight can be found by setting the two forces equal to each other:
FE = mg
Where FE is the electric force, m is the mass of the proton, and g is the acceleration due to gravity. Since the proton's mass is known to be approximately 1.67 x 10⁻²⁷ kg, and the acceleration due to gravity is approximately 9.8 m/s², we can calculate the magnitude of the electric field as follows:
E = mg/q
Where q is the charge of the proton, which is approximately 1.6 x 10⁻¹⁹ C. Plugging in the values:
E = (1.67 x 10⁻²⁷ kg) x (9.8 m/s²) / (1.6 x 10⁻¹⁹ C)
E ≈ 1.03 x 10¹² N/C
What experimental evidence supports the idea that conducting materials have freely moving electrically charged particles inside them
Answer:
Layer of glass rod rubbed with silk.
Explanation:
Some of the atoms in the surface layer of a glass rod positively charged by rubbing it with a silk cloth have lost electrons, leaving a net positive charge because of the unneutralized protons of their nuclei. A negatively charged object has an excess of electrons on its surface.
Final answer:
Conducting materials like metals have freely moving electrons that enable the flow of charge, supported by experimental evidence like studies on conductivity and direct experiments confirming the presence of electrons as charge carriers.
Explanation:
Conducting materials such as metals have freely moving electrically charged particles within them. These free electrons can move through the material allowing the flow of charge. Experimental evidence supporting this idea includes studies on conductivity, Ohm's law for electromagnetics, and direct experiments like the one by Tolman and Stewart confirming that charge carriers in metals are indeed electrons.
A 5.7 g bullet is fired into a 1.5 kg ballistic pendulum. The bullet emerges from the block with a speed of 154 m/s, and the block rises to a maximum height of 11 cm . Find the initial speed of the bullet. The acceleration due to gravity is 9.8 m/s 2 . Answer in units of m/s.
Answer:
540.8m/s
Explanation:
From the information giving, the total energy is conserved and the momentum is conserved.
To determine the speed of the ball after the collision, we use the energy conservation rule, I.e
Kinetic Energy of ball after collision = energy to rise to attain height
1/2mv²=mgh
where m,mass of ballistic pendulum=1.5kg,
v=velocity of ballistic pendulum after collision,
g=gravitational acceleration
h=height attain=11cm=0.11m
if we substitute we arrive at
v=√(2gh)
v=√(2*9.8*0.11)
v=1.47m/s.
since we have determine the velocity of the ballistic pendulum after collision, we now use conservation of momentum to determine the initial speed of the bullet.
since
initial momentum=final momentum
mₓ₁vₓ₁+mₐ₁vₐ₁=mₓ₂vₓ₂+mₐ₂vₐ₂
were mₓ₁vₓ₁,mₓ₂vₓ₂ =mass and velocity of ballistic pendulum before and after collision
mₐ₁vₐ₁,mₐ₂vₐ₂=mass and velocity of bullet before and after collision
if we substitute values,we arrive at
(1.5kg*0m/s)+(0.0057kg*vₐ₁)=(1.5kg*1.47m/s)+(0.0057kg*154m/s)
vₐ₁= 3.0828/0.0057
vₐ₁=540.8m/s
A Geiger–Muller tube is a type of gas‑filled radiation detector. It can detect particles like X‑rays, alpha particles, and beta rays (electrons). This is useful in quantizing the activity of a radioactive source or determining if an area containing radioactive material is safe to enter. Assuming that you have 1 mol of gas, if a Geiger counter is filled with 9846 Pa of argon gas at room temperature ( T = 21.1 °C ) , what is the density, rho , of the gas in this Geiger tube?
Answer:
[tex]0.16098\times 10^{-3}\ g/cm^3[/tex]
Explanation:
P =Pressure = 9846 Pa
V = Volume
n = Amount of substance = 1
T = Temperature = 21.1°C
[tex]\rho[/tex] = Density
R = Gas constant = 8.314 J/mol K
M = Molar mass of argon = 40 g/mol
From ideal gas law we have the relation
[tex]PV=nRT[/tex]
Multiply density on both sides
[tex]PV\rho=nR\rho T\\\Rightarrow PM=nR\rho T\\\Rightarrow \rho=\dfrac{PM}{nRT}\\\Rightarrow \rho=\dfrac{9846\times 40\times 10^{-3}}{8.314\times (21.1+273.15)}\\\Rightarrow \rho=0.16098\ kg/m^3\\\Rightarrow \rho=0.16098\times 10^{-3}\ g/cm^3[/tex]
The density of argon gas is [tex]0.16098\times 10^{-3}\ g/cm^3[/tex]
To find the density of argon gas inside a Geiger tube, we use the Ideal Gas Law and convert the given units, plugging in these values yields 1.65 g/L.
Explanation:The density, ρ, of a gas can be calculated following the Ideal Gas Law formula, which is PV = nRT, where P is pressure, V is volume, n is the number of moles of the gas, R is the universal gas constant, and T is the temperature in Kelvin. To solve for density, we can use the equation ρ = m/V, where m is the mass of the gas and V is the volume. We can also express this in terms of the Ideal Gas Law, leading to ρ = (n×M)/(RT/P), where M is molar mass.
Given the molar mass of Argon is approximately 39.948 g/mol and the universal gas constant R is 8.314462618 J/(mol*K), first convert the temperature from Celsius to Kelvin (T = 21.1 °C + 273.15 = 294.25K), and the pressure from Pascal to atm (1 Pa = 0.00000986923 atm, thus 9846 Pa = 9846× 0.00000986923 = 0.0972 atm).
Plugging the values into the density formula, we get ρ = (1 mol× 39.948 g/mol) / ((8.314462618 J/(mol×K)×294.25K)/ 0.0972 atm) = 1.65 g/L.
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The small piston of a hydraulic lift has a cross-sectional of 3 00 cm2 and its large piston has a cross-sectional area of 200 cm2. What downward force of magnitude must be applied to the small piston for the lift to raise a load whose weight is Fg = 15.0 kN?
Q: The small piston of a hydraulic lift has a cross-sectional of 3.00 cm2 and its large piston has a cross-sectional area of 200 cm2. What downward force of magnitude must be applied to the small piston for the lift to raise a load whose weight is Fg = 15.0 kN?
Answer:
225 N
Explanation:
From Pascal's principle,
F/A = f/a ...................... Equation 1
Where F = Force exerted on the larger piston, f = force applied to the smaller piston, A = cross sectional area of the larger piston, a = cross sectional area of the smaller piston.
Making f the subject of the equation,
f = F(a)/A ..................... Equation 2
Given: F = 15.0 kN = 15000 N, A = 200 cm², a = 3.00 cm².
Substituting into equation 2
f = 15000(3/200)
f = 225 N.
Hence the downward force that must be applied to small piston = 225 N
The downward force of magnitude must be applied to the small piston for the lift to raise a load whose weight is Fg = 15.0 kN is 225N.
Given that,
The small piston of a hydraulic lift has a cross-sectional of 3.00 cm^2.And its large piston has a cross-sectional area of 200 cm^2.Based on the above information, the calculation is as follows:
Here we use the Pascal's principle,
[tex]F\div A = f\div a[/tex] ........... (1)
Here
F denoted Force exerted on the larger piston,
f denoted force that applied to the smaller piston,
A denoted cross-sectional area of the larger piston,
And, a denoted cross-sectional area of the smaller piston.
Now
[tex]f = 15000 \times (3\div 200)[/tex]
= 225 N
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If the gas in a container absorbs 275 Joules of heat, has 125 Joules of work done on it, then does 50 Joules of work, what is the increase in the internal energy of the gas?
Answer:
The increase in the internal energy = 350 J
Explanation:
Given that
Q= 275 J
W= - 125 J
W' = 50 J
W(net)= -125 + 50 = -75 J
Sign -
1.Heat rejected by system - negative
2.Heat gain by system - Positive
3.Work done by system = Positive
4.Work done on the system-Negative
Lets take change in the internal energy =ΔU
We know that
Q= ΔU + W(net)
275 = ΔU -75
ΔU= 275 + 75 J
ΔU=350 J
The increase in the internal energy = 350 J
The internal energy of the gas is the energy contained by the gas. The increase in the internal energy of the gas in the given container is 350 J.
The internal energy of the gas can be calculated by the formula
Q= ΔU + W(net)
Where,
Q - energy absorbed by the system = 275 J
ΔU - internal energy of the gas= ?
W(net)= W - W' = -125 + 50 = -75 J
Put the values in the formula,
275 = ΔU -75
ΔU= 275 + 75 J
ΔU=350 J
Therefore, the increase in the internal energy of the gas in the given container is 350 J.
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One ball is dropped vertically from a window. At the same instant, a second ball is thrown horizontally from the same window.
1. Which ball has the greater speed at ground level?
Answer:
The second ball
Explanation:
Both balls are under the effect of gravity, accelerating with exactly the same value. The first ball is dropped, therefore its initial velocity is zero. Since the second ball has horizontal and vertical velocity components, its initial velocity is given by:
[tex]v=\sqrt{v_x^2+v_y^2}[/tex]
The vertical component is zero, however, it has a horizontal velocity, so its initial speed is not zero, therefore the secong ball has the greater speed at ground level.
The second ball will reach a greater total speed at ground level due to its initial horizontal speed added to its vertical speed. But, both balls will hit the ground at the same time if the horizontal speed of the second ball is not too high.
Explanation:In the described situation, gravity is the only vertical force acting on both balls, hence they will reach the ground at the same speed in the vertical direction. But, the second ball has additional horizontal speed. Therefore, with this horizontal component added to the vertical detachment speed due to gravity, the second ball will have a greater total speed by the Pythagorean theorem: (total speed)^2 = (vertical speed)^2 + (horizontal speed)^2. However, the question of which hits the ground first depends on the initial horizontal speed of the second ball. If the horizontal speed is not too high, both balls should hit the ground at the same time.
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