Answer:
Explanation:
Something that attracts iron...duuhhh
What is the magnitude of the force, directed parallel to the ramp, that he needs to exert on the crate to get it to start moving up the ramp
Complete question is;
Jason works for a moving company. A 75 kg wooden crate is sitting on the wooden ramp of his truck; the ramp is angled at 11°.
What is the magnitude of the force, directed parallel to the ramp, that he needs to exert on the crate to get it to start moving UP the ramp?
Answer:
F = 501.5 N
Explanation:
We are given;
Mass of wooden crate; m = 75 kg
Angle of ramp; θ = 11°
Now, for the wooden crate to slide upwards, it means that the force of friction would be acting in an opposite to the slide along the inclined plane. Thus, the force will be given by;
F = mgsin θ + μmg cos θ
From online values, coefficient of friction between wooden surfaces is μ = 0.5
Thus;
F = (75 × 9.81 × sin 11) + (0.5 × 75 × 9.81 × cos 11)
F = 501.5 N
The higher something is off the ground, the more
potential energy it has.
True
False
Given that water at standard pressure freezes at 0∘C, which corresponds to 32∘F, and that it boils at 100∘C, which corresponds to 212∘F, calculate the temperature difference ΔT in degrees Fahrenheit that corresponds to a temperature difference of 1 K on the Kelvin scale. Give your answer to two significant figures.
Answer:
In two significant figure 360K
Explanation:
The temperature difference (ΔT) can be calculated as the boiling temperature minus the freezing temperature in Fahrenheit.
Hence,
ΔT = 212 - 32
ΔT = 180°F
To convert to °F to kelvin, we use the formula below
= (°F - 32) × 5/9 + 273.15
= (180°F - 32) × 5/9 + 273.15
= 355.37K ⇔ 360K
How can you tell if hair was forcibly removed?
Answer:Explanation:
A microscopic hair examination can also determine if a hair was forcibly removed, artificially treated or diseased. A comparison microscope can be used to compare a questioned hair to a known hair sample in order to determine if the hairs are similar and if they could have come from a common source.
The masses of the Earth and Moon are 5.98×1024kg and 7.35×1022kg respectively, and their centers are separated by 3.84×108m.
At what distance from the center of the Earth is the CM of the Earth-Moon system located?
What can you say about the motion of the Earth-Moon system about the Sun, and of the Earth and Moon separately about the Sun?
A) The Earth-Moon CM follows the orbit around the Sun. Earth and Moon rotate around their CM. The radius of rotation of Earth around the CM is much greater than radius of rotation of Moon around the CM.
B) The Earth and Moon follow separate circular orbits around the Sun.
C) The Earth-Moon CM follows the orbit around the Sun. Earth and Moon rotate around their CM. The radius of rotation of Moon around the CM is much greater than radius of rotation of Earth around the CM.
Answer:
C) The Earth-Moon CM follows the orbit around the Sun. Earth and Moon rotate around their CM. The radius of rotation of Moon around the CM is much greater than radius of rotation of Earth around the CM.
Explanation:
At first we assume that both Earth and Moon can be treated as particles, the center of mass of the Earth-Moon system is obtained by using this formula:
[tex]r_{CM} = \frac{r_{E}\cdot m_{E}+r_{M}\cdot m_{M}}{m_{E}+m_{M}}[/tex] (Eq. 1)
Where:
[tex]r_{E}[/tex] - Location of the center of the Earth, measured in kilometers.
[tex]r_{M}[/tex] - Location of the Moon, measured in kilometers.
[tex]r_{CM}[/tex] - Location of the center of mass, measured in kilometers.
[tex]m_{E}[/tex] - Mass of the Earth, measured in kilograms.
[tex]m_{M}[/tex] - Mass of the Moon, measured in kilograms.
If we know that [tex]r_{E} = 0\,km[/tex], [tex]r_{M} = 3.84\times 10^{8}\,m[/tex], [tex]m_{E} = 5.98\times 10^{24}\,kg[/tex] and [tex]m_{M} = 7.35\times 10^{22}\,kg[/tex], the location of the center of mass respect to the Earth is:
[tex]r_{CM} = \frac{(0\,km)\cdot (5.98\times 10^{22}\,kg)+(3.84\times 10^{8}\,m)\cdot (7.35\times 10^{22}\,kg)}{5.98\times 10^{24}\,kg+7.35\times 10^{22}\,kg}[/tex]
[tex]r_{CM} = 4.662\times 10^{6}\,m[/tex]
The Earth has a radius of [tex]6.371\times 10^{6}[/tex] meters, we notice that center of mass in located inside the Earth and the radius of rotation of the Earth around the center of mass is much greater than the radius of rotation of the Moon around the center of mass. That center of mass follows an orbit around the sun.
In consequence, correct answer is C.
The distance from the center of the Earth is the Center of Mass of the Earth-Moon system located is 4.662×10⁶ m.
What is center of mass?The center of mass of a object or for a system of object is the point, where the center of distribution of mass in space. The center of mass for the two bodies can be given as,
It can be given as,
[tex]x_{cm}=\dfrac{m_1x_1+m_2x_2}{m_1+m_2}[/tex]
Here, (m) represents the mass and (x) represents the distance.
The masses of the Earth and Moon are 5.98×1024kg and 7.35×1022kg respectively,The centers of the Moon and Earth are separated by 3.84×108m.
Thus the center of mass with respect to Earth using the above formula can be given as,
[tex]x=\dfrac{(5.98\times10^{22})(0)+7.25\times10^{22}(3.84\times10^8)}{5.98\times10^{22}+7.25\times10^{22}}\\x=4.662\times 10^6\rm m[/tex]
The motion of the Earth-Moon system about the Sun is less than the Earth and Moon separately about the Sun is, due to the center of rotation of the earth.
Thus, the Earth-Moon CM follows the orbit around the Sun. Earth and Moon rotate around their CM. The radius of rotation of Moon around the CM is much greater than the radius of rotation of Earth around the CM. Option C is correct.
Thus, the distance from the center of the Earth is the Center of Mass of the Earth-Moon system located is 4.662×10⁶ m.
Learn more about the center of mass here;
https://brainly.com/question/13499822
You are working out on a rowing machine. Each time you pull the rowing bar (which simulates the oars) toward you, it moves a distance of 1.1 m in a time of 1.8 s. The readout on the display indicates that the average power you are producing is 90 W. What is the magnitude of the force that you exert on the handle?
Answer:
147.27N
Explanation:
Power = workdone/time
Power = Force*distance/time
Given
Power = 90Watts
Distance = 1.1m
Time = 1.8secs
Force = ?
Substitute the given parameters into the formula:
[tex]90 = \frac{1.1d}{1.8}\\cross \ multiply\\ 90 \times 1.8 = 1.1F\\162 = 1.1F\\1.1F = 162\\F = \frac{162}{1.1} \\F = 147.27N[/tex]
Hence the magnitude of the force that you exert on the handle is 147.27N
What is the resultant velocity of a plane that is traveling at 245 m/s North and encounters a tailwind of 55 m/s North?
Answer:
b
Explanation:
An astronaut, whose mass on the surface of the Earth is m, orbits the Earth in the space shuttle at an altitude of 450 km. What is her mass while orbiting in the space shuttle?
a. 0.125 m
b. 0.50 m
c. 0.75 m
d. m
Answer:
d. m
Explanation:
Assuming that the shuttle orbits at a low speed compared with the speed of light, the mass must be the same.This is due to the mass is a property of bodies which is the same at any point in the universe, it doesn't depend of the force applied to it or any other thing, as weight does, provided no relativistic effects take place.So, the right answer is d.This question involves the concepts of mass, Law of conservation of mass, and altitude.
The mass of the astronaut orbiting in the space shuttle is "d. m".
The mass of an object is defined as the quantity of matter contained inside a body. According to the law of conservation of mass, the mass of a body remains constant everywhere in the universe.
The quantity that changes with the change in location, is weight. The change in weight is caused by the change in acceleration due to gravity, which varies from planet to planet and also varies with altitude.
Thus, the mass of the astronaut will remain constant.
Learn more about the law of conservation of mass here:
https://brainly.com/question/13383562?referrer=searchResults
Find the mass of the two-dimensional object. A disk of radius 7 in with density at distance x from origin given by rho ( x ) = √ 10 x
Answer:
1030.3504 units
Explanation:
Disk radius = 7 inches
Density at distance x = p(x) = [tex]\sqrt{10x}[/tex]
Calculate the mass of the two-dimensional object
M = 2[tex]\pi[/tex][tex]\sqrt{10} * (\frac{2}{5} ) *(7)^{\frac{5}{2} }[/tex] = 1030.3504 units
Activity Problem 8.3 This problem asks you to compare the finite current element approximation to the infinite straight wire approximation for the isolated segment of wire carrying current of 2A to the right. The wire is 8cm long. Note, the finite current approximation will be a fairly poor approximation to the actual field which would be found by integration; the infinite wire field is a better approximation in this case. (a)Compute the magnetic field at point C a distance 2cm for the center of the wire using the infinite straight wire approximation. (b)Compute the magnetic field at point C using the finite current element approximation. (c)Compute the magnetic field at point E a distance 2cm from the end using the finite current element approximation.
Answer: (a) B = 2 x 10⁻⁵T
(b) B = 1.94 x 10⁻⁵T
(c) B = 1.8 x 10⁻⁴T
Explanation: A magnetic field due to a current passing through a straight wire is calculated using the Biot-Savart Law:
[tex]dB=\frac{\mu_{0}}{4.\pi} \frac{IdLXR}{r^{3}}[/tex]
where
dL is current length element
[tex]\mu_{0}[/tex] is permeability of free space ([tex]4.\pi.10^{-7}[/tex]T.m/A)
(a) For a infinite straight wire:
[tex]B=\frac{\mu_{0}I}{2.\pi.R}[/tex]
[tex]B=\frac{4.\pi.10^{-7}.2}{2.\pi.2.10^{-2}}[/tex]
B = 2x10⁻⁵T
For an infinite, long and straight wire, magnetic field is 2x10⁻⁵T.
(b) For a finite wire:
[tex]B=\frac{\mu_{0}I}{2.\pi.R}\frac{L}{\sqrt{L^{2}+R^{2}} }[/tex]
[tex]B=\frac{4.\pi.10^{-7}.2}{2.\pi.2.10^{-2}} \frac{8.10^{-2}}{\sqrt{(8.10^{-2})^{2}+(2.10^{-2})^{2}} }[/tex]
B = 1.94x10⁻⁵T
The magnetic field for a finite wire in the same conditionsas infinite wire is 1.94x10⁻⁵T.
(c) For a finite wire at a point distant from the end of the wire:
[tex]B=\frac{\mu_{0}I}{4.\pi.L\sqrt{2} }[/tex]
[tex]B=\frac{4.\pi.10^{-7}.2}{4.\pi.8.10^{-2}\sqrt{2} }[/tex]
B = 0.18x10⁻⁵T
At a point at the end, magnetic field is 1.8x10⁻⁴T.
Brownian motion is observed when looking at smoke particles in a microscope.What causes the smoke particles to move at random?
Answer:
Explanation:
Brownian motion is evidence that air molecules move in random directions, with random speeds (and hence kinetic energy). When these air molecules knock into the smoke particles, they cause the smoke particles to move in random motions.
Why can't we see the air molecules themselves? That's because the air molecules are invisible, even with microscopes
PLS ANSWER WILL MARK BRANLIEST!!!!!!!!!!!!
Describe the life cycle of a star before it collapses into a black hole.
Describe the life cycle of a star before it becomes a black dwarf.
What is the likely outcome of our sun? *
The sun will supernova and become a black hole.
The sun will swell, encompassing the inner planets and collapses into a dwarf star.
The sun will become a pulsar.
How Do You Know?
P.S. the how do you know is only for the last question
answer:
uhm this makes no sense like at all so chile anyways
explanation:
Let k be the Boltzmann constant. If the thermodynamic state of gas at temperature T changes isothermally and reversibly to a state with three times the number of microstates as initially, the energy input to gas as heat is:______
a. Q = 0
b. Q = 3kT
c. Q = −3kT
d. kT ln 3
e. −kT ln 3
Answer:
d. kT ln 3
Explanation:
Given;
k as Boltzmann constant
Let initial initial microstate, = Фi
Let the final microstate, = Фf = 3Фi
at constant temperature, T
The energy input to gas as heat is given by;
Q = TΔS = Tk(lnФf - lnФi)
[tex]Q= kT*\frac{ln \phi _f}{ln \phi_i} \\\\Q= kT*\frac{3ln \phi _i}{ln \phi_i}\\\\Q= kT ln3[/tex]
Therefore, the energy input to gas as heat is kT ln 3
plzzz help its for my civics class what answer is it????????????
Answer:
the correct answer is The English Bill of Rights
Explanation:
This is the answer because i have taken the test.
In the figure below, a block of 1.67 slides on a track with different levels, which has friction only at the highest point where the kinetic coefficient of friction is uk = 0.35. If the block has an initial speed V0 = 7.5m/s and the highest point of the track is at ℎ = 2.1 above the initial position of the block, calculate the distance where the friction force for the block is.
Answer:
2.0 m
Explanation:
Energy is conserved.
Initial KE = Final PE + Work done by friction
½ mv² = mgh + Fd
½ mv² = mgh + mgμd
½ v² = gh + gμd
½ v² − gh = gμd
d = (½ v² − gh) / (gμ)
d = (½ (7.5 m/s)² − (10 m/s²) (2.1 m)) / ((10 m/s²) (0.35))
d = 2.0 m
1. What are the two (2) classes of matter?
A. Atom and molecule
B. Element and compound
C. Plant and animal cells
D. Substance and mixture
2. Which is true regarding water?
A. Compound of element
B. Element of gases
C. Mixture of
gases
D. Substances of gases
3. What is the difference between a substance and a mixture?
A. Substance can be separated while mixture cannot be separated.
B. Substance has many components while mixture has one only.
C. Substance is heterogeneous while mixture is homogeneous.
D. Substance is pure while mixture is impure.
Answer:
substance and mixture
All biomes don’t have the same level of biodiversity. What seems to be the optimal conditions for high biodiversity?
Answer:
See the answer below
Explanation:
The optimal conditions for high biodiversity seem to be a warm temperature and wet climates.
The tropical areas of the world have the highest biodiversity and are characterized by an average annual temperature of above 18 [tex]^oC[/tex] and annual precipitation of 262 cm. The areas are referred to as the world's biodiversity hotspots.
Consequently, it follows logically that the optimal conditions for high biodiversity would be a warm temperature of above 18 [tex]^oC[/tex] and wet environment with annual precipitation of not less than 262 cm.
The variation in temperature and precipitation across biomes can thus be said to be responsible for the variation in the level of biodiversity in them.
A man with a mass of 86.5 kg stands up in a 61-kg canoe of length 4.00 m floating on water. He walks from a point 0.75 m from the back of the canoe to a point 0.75 m from the front of the canoe. Assume negligible friction between the canoe and the water. How far does the canoe move?
Answer:
The displacement of the canoe is 1.46 m
Explanation:
Given that,
Mass of canoe = 61 kg
Mass of man = 86.5 kg
Length = 4 m
Let the the displacement of the canoe is x'
We need to calculate the displacement of the man
Using formula of displacement
[tex]x=x_{2}-x_{1}[/tex]
Put the value into the formula
[tex]x=4-(0.75+0.75)[/tex]
[tex]x=2.5\ m[/tex]
We need to calculate the displacement of the canoe
Using conservation of momentum
[tex]M_{m}v_{m}=(M_{c}+M_{m})v_{c}[/tex]
[tex]M_{m}\dfrac{x}{t}=(M_{c}+M_{m})\dfrac{x'}{t}[/tex]
[tex]86.5\times2.5=(61+86.5)\times x'[/tex]
[tex]x'=\dfrac{86.5\times2.5}{61+86.5}[/tex]
[tex]x'=1.46\ m[/tex]
Hence, The displacement of the canoe is 1.46 m
as a result, the net electric force experienced by each negatively charged particle is reduced to F/2. The value of q is
Answer:
The value of q is [tex]\dfrac{Q}{8}[/tex]
Explanation:
Given that,
Each charge = -Q
Distance between charges = L
Reduced force = [tex]\dfrac{F}{2}[/tex]
Suppose, Two particles each with a charge -Q are fixed a distance L apart as shown above. Each particle experiences a net electric force F. A particle with a charge +q is now fixed midway between the original two particles.
We know that,
The force on each end is
[tex]F=\dfrac{kQ^2}{L^2}[/tex]...(I)
If the charge q is placed at mid point then
The force on each end charge is
[tex]\dfrac{F}{2}=F+F'[/tex]....(II)
We need to calculate the value of q
Using equation (II)
[tex]\dfrac{F}{2}=F+F'[/tex]
Put the value of F into the formula
[tex]\dfrac{\dfrac{kQ^2}{L^2}}{2}=k\dfrac{Q^2}{L^2}+k\dfrac{q\times(-Q)}{(\dfrac{L}{2})^2}[/tex]
[tex]\dfrac{kq(-Q)}{(\dfrac{L}{2})^2}=-\dfrac{kQ^2}{2L^2}[/tex]
[tex]\dfrac{q}{\dfrac{1}{4}}=\dfrac{Q}{2}[/tex]
[tex]q=\dfrac{Q}{8}[/tex]
Hence, The value of q is [tex]\dfrac{Q}{8}[/tex]
If two identical objects are dropped one after 1 s delay with respect to another then in the absence of the air resistance *
A. distance between two falling objects will keep increasing
B. distance between two falling objects will keep decreasing
C. will stay the same
D. All of the above
Answer:
A.No. Assuming no other factors (such as air resistance) the first object will have a velocity of 32 feet/second when the other is dropped. Since they will both have the same acceleration, the first distance between them will increase by 32 feet per second.
Explanation:
What normal force does a horizontal table exert on a 3.5 kg book that sits at rest?
Answer:
34.3N
Explanation:
The normal force, denoted by F(N), is the force exerted on an object by a surface. The normal force in this case is acting in an opposite manner but equal magnitude to the force of gravity.
F(N) = Fg
Where;
F(N) = Normal force
Fg = gravitational force
Fg = mg
Where;
m = mass (kg)
g = acceleration due to gravity
F(N) = mg
F(N) = 3.5kg × 9.8m/s²
F(N) = 34.3N
Hence, the normal force acting on the object is 34.3N.
3- For given three vectors a, b and c, c = a x b, then the vector c is:
Answer:
VB
Explanation:
How long would it take to travel 50 km traveling at a speed of 10 km/hr
Answer:
it would take 5 hours
Explanation:
plz vote this up i promise its correct
6.7.5. While you sleep, your body is using energy at a rate of 77 W. How many food calories are used during an eight hour period
Answer:
6.38×10¯⁴ Calorie.
Explanation:
The following data were obtained from the question:
Power (P) = 77 W
Time (t) = 8 h
Energy (E) =.?
Next, we shall convert 8 hours to seconds. This can be obtained as follow:
1 h = 3600 s
Therefore,
8 h = 8 h × 3600 s / 1h
8 h = 28800 s
Next, we shall determine the energy. This can be obtained as follow:
Power (P) = 77 W
Time (t) = 28800 s
Energy (E) =.?
Energy (E) = Power (P) / time (t)
E = P/t
E = 77 /28800
E = 2.67×10¯³ J
Finally, we shall convert 2.67×10¯³ J to calorie. This can be obtained as follow:
4.184 J = 1 calorie
Therefore,
2.67×10¯³ J = 2.67×10¯³ J × 1 Cal / 4.184 J
2.67×10¯³ J = 6.38×10¯⁴ Calorie.
Therefore, 6.38×10¯⁴ Calorie of food is used during the 8 hours period.
A 1000kg boat is being towed by a rope that makes an angle of 20 degrees with respect to the x-axis. The boat accelerates horizontally from rest to 10 m/s in 5.0 seconds. During this time, a total drag force of 500N acts on the boat from the water. What is the tension in the rope during this time?
Answer:
Explanation:
acceleration of the boat a = (10 - 0) / 5 = 2 m /s²
Let tension in the rope be T
Horizontal component = T cos20
force acting on boat = T cos20 - 500
applying newton's law
T cos20 - 500 = mass x acceleration
T cos20 - 500 = 1000 x 2 = 2000
T cos20 = 2500 N .
T x .94 = 2500
T = 2660 N .
Can u help me thanks so
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6. A cat walks 1.5km South and then 2.4km East. What is the total displacement of the cat?
7. A boy scout troop hikes 10.0 km East and then hikes an additional 7km North. What is the total displacement of the boy scout troop?
8. A go-kart is moving 23 m/s on the track. The track is 152m long. How long will take for the go-kart to reach the end of the track?
9. If a snail moves at a pace of .01m/s and it travels for 1 hr. How far does the snail get?
10. It takes 3 hr and 10 min to ride the tram to the top of Pike’s Peak in Colorado. The tram will travel a total distance of 14.32 km. What is the speed of the tram?
Answer:
6. 3.9 km SE
7. 17 km NE
8. 3496 seconds
9. 36 m
10. 13.27 km/m
Explanation:
6. and 7. added both numbers
8. multiplied both numbers
9. 60 x 60 = 3600
3600 x 0.01 = 36
10. 60 x 3 = 180
180 + 10 =190
190 / 14.32 = 13.27
A(n)............ exerts a force on a system. It is symbolised as a subscript to the force.
1- Contact
2- Agent
3- Field
4- Push
How much force is needed to accelerate a 2,000 kg car at a rate of 5 m/s2? A. 2,500 N B. 15,000 N C. 2,000 N D. 10,000 N Reset
Answer: 10,000 N
Explanation: USE , f = ma
Which is true about the temperature of a gas? (2 points)
оа
At a higher temperature, the particles have a higher average kinetic energy.
Oь
At a higher temperature, the volume of the gas is smaller.
Ос
At a lower temperature, there are more particles in the gas.
Od
At a lower temperature, the particles in the gas move faster.