Answer:
0.1 s
Explanation:
The net force on the log is F - f = ma where F = force due to winch = 2850 N, f = kinetic frictional force = μmg where μ = coefficient of kinetic friction between log and ground = 0.45, m = mass of log = 300 kg and g = acceleration due to gravity = 9.8 m/s² and a = acceleration of log
So F - f = ma
F - μmg = ma
F/m - μg = a
So, substituting the values of the variables into the equation, we have
a = F/m - μg
a = 2850 N/300 kg - 0.45 × 9.8 m/s²
a = 9.5 m/s² - 4.41 m/s²
a = 5.09 m/s²
Since acceleration, a = (v - u)/t where u = initial velocity of log = 0 m/s (since it was a rest before being pulled out of the ditch), v = final velocity of log = 0.5 m/s and t = time taken for the log to reach a speed of 0.5 m/s.
So, making t subject of the formula, we have
t = (v - u)/a
substituting the values of the variables into the equation, we have
t = (v - u)/a
t = (0.5 m/s - 0 m/s)/5.09 m/s²
t = 0.5 m/s ÷ 5.09 m/s²
t = 0.098 s
t ≅ 0.1 s
What is the friction force on a box that has a mass of 15kg as it slides across the floor. The coefficient of friction of the not very clean floor is 0.25
please explain everything including formula used
Answer:
36.75 N
Explanation:
Applying
F = mgμ................. Equation 1
Where F = Friction force on the box, m = mass of the box, g = acceleration due to gravity of the box, μ = coefficient of static friction
From the question,
Given: m = 15 kg, μ = 0.25
Constant: g = 9.8 m/s²
Substitute these values into equation 1
F = 15(9.8)(0.25)
F = 36.75 N
Hence the friction force on the box is 36.75 N
According to the model, when was the universe at its most dense?
A) During the Dark Ages where matter increased in mass.
B) Just before the Big Bang where all matter existed in a singularity.
C) During the nuclear fusion events, as the atoms become more massive.
D) Current day, as the number of galaxies, solar systems, and planets have increased.
Answer:
The Answer is D
Explanation:
Hope this helps!!!!
Sunsets are a deep red because A) tiny particles in the air are more efficient at scattering short wavelength light than they are at scattering long wavelength light. Hence, long wavelength light ends up coming directly towards you. B) most polluting gases and dust particles in the air are reddish in color and lend their color to that of the sky. C) air molecules absorb red light more efficiently than they do blue light because of their electron orbitals. D) air molecules absorb blue light more efficiently than they do red light because of their electron orbitals.
Answer:
i think its A
Answer:tiny particles in the air are more efficient at scattering short wavelength light than they are at scattering long
Explanation:
Help!!
A table is pushed across the floor for a distance of 32 m with a force of 320 N in 150 seconds. How much power was used?
A.70.2W
B.68.3W
C.56.7W
D.49.8W
Compute the work done on the table:
W = Fd = (320 N) (32 m) = 10,240 J
Divide this by the given time duration to get the power output:
P = W/∆t = (10,240 J) / (150 s) ≈ 63.3 W
A hoop rolls with constant velocity and without sliding along level ground. Its rotational kinetic energy is:______a- half its translational kinetic energyb- the same as its translational kinetic energyc- twice its translational kinetic energyd- four times its translational kinetic energy
Answer:
The same as its translational KE.
The easy way to do this is to make up numbers and use them.
So, I'll say m=2 and r=3. I will also say v=3 .
Rot. Inertia of a hoop is mr^2. So the rot KE is: 1/2 (mr^2)(w^2)
note: (1/2*I*w^2)
Translational kinetic energy is basically normal KE, so 1/2(m)(v^2)
Now, lets plug our made up values in:
Rot Ke : 1/2 (9*2)(3/3) *note w = v/r
Tran Ke: 1/2(2)(9)
Rot Ke: 9
Tran Ke: 9
9=9, same.
A ball is thrown horizontally at a speed of 24 meters per second from the top of a cliff. If the ball hits the ground 6.0 seconds later, approximately how high is the cliff?
Answer:
144 meters
Explanation:
it takes 6 seconds to hit the ground right and the ball lays off 24 m per second .
so by the time the ball hits the ground 6 seconds passed. so that means the cliff is 6.0×24=144
An empty 12,954 kg railroad car, traveling at a speed of 28 m/s strikes a partially filled 17,616 kg railroad car moving in the same direction at a speed of 5 m/s. What is the total momentum of the two railroad cars AFTER the collision?
Answer:
450792 kgm/s
Explanation:
by conservation of momentum,
total momentum AFTER collision = total momentum BEFORE collision
=mv+m'v'
=12954×28+17616×5
=450792 kgm/s
A boy throws a ball straight up with a speed of 21.5 m/s. The ball has a mass of 0.19 kg. How much gravitational potential energy will the ball have at the top of its flight? (Assume there is no air resistance.) A. 43.9 J B. 37.5 J C. 48.5 J D. 41.2 J
Answer:
Explanation:
The equation fo potential energy is PE = mgh, where m is the mass of the ball, g is the pull of gravity (constant at 9.8), and h is the max height of the ball. What we do not have here is that height. We need to first solve for it using one-dimensional equations. What we have to know above all else, is that the final velocity of an object at its max height is always 0. That allows us to use the equation
[tex]v_f=v_0+at[/tex] where vf is the final velocity and v0 is the initial velocity. We will find out how long it takes for the object to reach that max height first and then use that time to find out what that max height is. Baby steps here...
0 = 21.5 + (-9.8)t and
-21.5 = -9.8t so
t = 2.19 seconds (Keep in mind that if I used the rules correctly for sig fig's, the answer you SHOULD get is not one shown, so I had to adjust the sig fig's and break the rules. But you know what they say about rules...)
Now we will use that time to find out the max height of the object in the equation
Δx = [tex]v_0t+\frac{1}{2}at^2[/tex] and filling in:
Δx = [tex]21.5(2.19)+\frac{1}{2}(-9.8)(2.19)^2[/tex] which simplifies down a bit to
Δx = 47.1 - 23.5 so
Δx = 23.6 meters.
Now we can plug that in to the PE equation to find the PE of the object:
PE = (.19)(9.8)(23.6) so
PE = 43.9 J
A string that is under 50.0N of tension has linear density 5.0g/m. A sinusoidal wave with amplitude 3.0cm and wavelength 2.0m travels along the string. What is the maximum speed of a particle on the string
Answer:
9.42 m/s
Explanation:
Applying,
V' = Aω.............. Equation 1
Where V' = maximum speed of the string, A = Amplitude of the wave, ω = angular velocity.
But,
ω = 2πf................. Equation 2
Where f = frequency, π = pie
And,
f = v/λ................ Equation 3
Where, λ = wave length, v = velocity
Also,
v = √(T/μ)................. Equation 4
Where T = Tension, μ = linear density.
From the question,
Given: T = 50.0 N, μ = 5.0 g/m = 0.005 kg/m
Substitute into equation 4
v = √(50/0.005)
v = √(10000)
v = 100 m/s
Also Given: λ = 2.0 m
Substitute into equation 3
f = 100/2
f = 50 Hz.
Substitute the value of f into equation 2
Where π = constant = 3.14
ω = 2(3.14)(50)
ω = 314 rad/s
Finally,
Given: A = 3.0 cm = 0.03 m
Substitute into equation 1
V' = 0.03(314)
V' = 9.42 m/s
1. Convert the following length into meters
a. 123.50mm
b. 560cm
c. 100dm
d. 125.89km
Two thin conducting plates, each 56.0 cm on a side, are situated parallel to one another and 7.0 mm apart. If 10^-10 electrons are moved from one plate to the other, what is the electric field between the plates?
Answer:
[tex]E=576.5V/m[/tex]
Explanation:
From the question we are told that:
Length [tex]l=56.0cm=0.56m[/tex]
Distance apart [tex]d=7.0mm=0.007m[/tex]
Electron Transferred [tex]n=10^{-10}[/tex]
Therefore
Total Charge
Since Charge on each electron is
[tex]e=1.602*10^{-19}[/tex]
Therefore
[tex]T=1.602*10^{-19} *10^{10}[/tex]
[tex]T=1.602*10^{-9}[/tex]
Generally the equation for Charge density is mathematically given by
[tex]\rho=T/A[/tex]
Where
Area
[tex]A=0.56*0.56[/tex]
[tex]A=0.3136[/tex]
Therefore
[tex]\rho=1.602*10^{-9}/0.3136[/tex]
[tex]\rho=5.10*10^{-9}[/tex]
Generally the equation for Electric Field in the capacitor is mathematically given by
[tex]E=\frac{\rho}{e_0}[/tex]
[tex]E=\frac{5.10*10^{-9}}{8.85x10{-12}}[/tex]
[tex]E=576.5V/m[/tex]
The lever of a car lift has an area of 0.2 meters squared, and the area of the lift under the car is 8
meters squared. If you push with a force of 3 newtons, how much force will be applied to the
car?
Answer:
THE ANSWER IS SOMETHING LIKE 55
A constant force moves an object along the line segment from to . Find the work done if the distance is measured in meters and the force in newtons.
This question is incomplete, the complete question is;
Flag
A constant force F = 6i+8j-6k moves an object along a straight line from point (6, 0, -10) to point (-6, 7, 2).
Find the work done if the distance is measured in meters and the magnitude of the force is measured in newtons.
Answer:
the work done is -88 J
Explanation:
Given the data in the question;
we know that;
Work done = F × S
where constant force F = ( 6i + 8j - 6k )
S = ( -6i + 7j + 2k ) - ( 6i + 0j - 10k )
S = ( (-6i - 6i) + (7j - 0j) + ( 2k - ( -10k) ) )
S = ( -12I + 7j + 12k )
so
Work force = ( 6i + 8j - 6k ) × ( -12I + 7j + 12k )
Work force = ( 6 × -12 ) + ( 8 × 7 ) + ( -6 × 12 )
Work force = -72 + 56 - 72
Work force = -88 J
Therefore, the work done is -88 J
instrument used in measurement Amount of substance
Answer:
For liquids: A measuring cylinder is used.
For solid: Over flow can is used
Answer:
i think a measuring cylinder
The cycle is a process that returns to its beginning, but it does not repeat
itself.
True
False
A swimmer heading directly across a river that is 200 m wide reaches the opposite bank in 6 min 40 s. During this swim, she is swept downstream 480 m. How fast can she swim in still water
Answer:
The speed of the swimmer in stil water is 0.5 m/s
Explanation:
Given;
total time taken to swim across = 6 mins 40 s = (6 x 60s) + 40 s = 400 s
width of the river, = 200 m
Please find the image attached for explanation.
Which of the following elements has the largest atomic radius?
Silicon
Aluminum
Sulfur
Phosphorous
Answer:
francium
Atomic radii vary in a predictable way across the periodic table. As can be seen in the figures below, the atomic radius increases from top to bottom in a group, and decreases from left to right across a period. Thus, helium is the smallest element, and francium is the largest.
Using only astronomical data from the Appendix E in the textbook, calculate the speed of the planet Venus in its essentially circular orbit around the sun.
Venus = 4.87x10^24
Answer:
[tex]v=3.49\times 10^4\ m/s[/tex]
Explanation:
Given that,
Mass of Venus, [tex]M_V=4.87\times 10^{24}\ kg[/tex]
We know that,
Mass of Sun, [tex]M_s=1.98\times 10^{30}\ kg[/tex]
The distance between the center of Sun and the center of Venus is [tex]1.08\times 10^{11}\ m[/tex]
We need to find the peed of the planet Venus in its essentially circular orbit around the sun. using the formula,
[tex]v=\sqrt{\dfrac{GM_s}{r}}[/tex]
Put all the values,
[tex]v=\sqrt{\dfrac{6.67\times 10^{-11}\times 1.98\times 10^{30}}{1.08\times 10^{11}}}\\\\v=3.49\times 10^4\ m/s[/tex]
So, the speed of the planet venus is [tex]3.49\times 10^4\ m/s[/tex].
Problem
A charged particle is moving in the presence of uniform magnetic field. The mass of the particle
is m = 10−6 kg its charge is Q = 10−5 C and the magnetic field vector is B~ = (1T, 0, 0). At the
beginning the velocity vector of the particle is ~v0 = (12 m/s, 0, 5 m/s).
a.) How large will the x component of the velocity of the particle be in t = 2 s?
b.) Where will the particle be in t = 3.14 s?
c.) How large will the magnitude of the velocity be in t = 2.5 s?
Answer:
Answer is a I checked the work
What are the messing forces that would make the object be in equilibrium?
Answer:
A) 20 N, B) 20 N, & C) 8 N
Explanation:
For the object to be in equilibrium, the upward forces must be equal to the downward forces and the forward forces must be equal to the backward forces.
1. Determination of A and B.
Forward forces = Backward forces
A + 10 + B = 25 + 25
A + 10 + B = 50
Collect like terms
A + B = 50 – 10
A + B = 40
Assume A and B to be equal. Thus, A is 20 N and B is 20 N.
2. Determination of C
Upward forces = Downward forces
C + 112 = 20 + 100
C + 112 = 120
Collect like terms
C = 120 – 112
C = 8 N
Thus, for the object to be in equilibrium, A must be 20 N, B must be 20 N and C must be 8N.
what is the frequency of a wave related to
Answer:
Frequency is the number of complete oscillations or cycles or revolutions made in one second.
if p=2i+4j+3k and q=I+5j-2k,find P×q.
Answer:
[tex]p\times q=-23i+7j+6k[/tex]
Explanation:
We are given that
p=2i+4j+3k
q=i+5j-2k
We have to find pxq
We know that
[tex]p\times q=\begin{vmatrix} i&j &k\\ 2&4 & 3\\ 1& 5 & -2\end{vmatrix}[/tex]
[tex]p\times q=i(-8-15)-j(-4-3)+k(10-4)[/tex]
[tex]p\times q=-23i+7j+6k[/tex]
Hence,[tex]p\times q=-23i+7j+6k[/tex]
You swing a bat and hit a heavy box with a force of 1273 N. The force the box exerts on the bat is Group of answer choices less than 1273 N if the box moves. exactly 1273 N whether or not the box moves. None of the above choices are correct. exactly 1273 N only if the box does not move. greater than 1273 N if the bat bounces back. greater than 1273 N if the box moves.
Answer:
exactly 1273 N whether or not the box moves.
Explanation:
In the case when the bat is swing and it is hitted to a heavy box having a force of 1273 N so here the force of the box that exert on the box should be accurately 1273 N even if the box is moved or not. As the third law of the newton should be equivalent & the opposite reaction
Therefore as per the given situation, the above represent the answer
you are given a set of facts regarding a lens : object heigh, and dostance to objects. Given this jnformation, how can you tell if you're dealing with a concave or convex lens
Answer:
concave curves inward like an hourglass and convex is an outward curve like a football
Explanation:
hope this helps
A spring scale hung from the ceiling stretches by 6.1cm when a 2.0kg mass is hung from it. The 2.0kg mass is removed and replaced with a 2.8kg mass.What is the stretch of the spring?
If the pressure of a gas is really due to the random collisions of molecules with the walls of the container, why do pressure gauges – even very sensitive ones – give perfectly steady readings? Shouldn't the gauge be continually jiggling and fluctuating? Explain.
Answer:
there is no fluctuation in the measurement because the quantity of molecule is too large and a quantity of some molecules is imperceptible.
Explanation:
The pressure measurement is carried out by calibrating the force exerted by the air on a surface of known area, suppose a small area 1 mm² = 0.01 cm²
To find out if the random movement of air molecules affects the pressure reading, let's calculate the number of molecules that reaches the pressure gauge.
In a system at atmospheric pressure and in a volume of 1 m³ (walls of 1 m each) there is one mole of air molecules, this mole is evenly distributed, so how many molecules fall on our surface
# _molecule = 6.02 10²³ 0.01 10⁻⁴ / 1
#_molecular = 6.02 10¹⁷ molecules per second
therefore the variation of the number of molecules is not very important
Consequently there is no fluctuation in the measurement because the quantity of molecule is too large and a quantity of some molecules is imperceptible.
assuming a filament in a 120W light bulb acts like a prefect blackbody, what is the temperature of the hottest portion of the filament if it has a surface area of 6.4×10^_5m^2. The stefan- boltzmann constant is 5.67×10^-8W/(m2.k2) A. 12OOk B. 2400K C. 2100K
Answer:
T = 2398 K
Explanation:
To calculate the emission of the light bulb we use the law is Stefan
P = σ A e T⁴
as they indicate that the filament is a black body, the emissivity is equal to 1 (e = 1)
T = [tex]\sqrt[4]{\frac{P}{ \sigma A} }[/tex]
let's calculate
T =[tex]\sqrt[4]{\frac{120}{5.67 \ 10^{-8} \ 6.4 \ 10^{-5}} }[/tex]
T = [tex]\sqrt[4]{33.06878 \ 10^{12} }[/tex]
T = 2,398 10³ K
T = 2398 K
a. A horse pulls a cart along a flat road. Consider the following four forces that arise in this situation.
1. the force of the horse pulling on the cart
2. the force of the cart pulling on the horse
3. the force of the horse pushing on the road
4. the force of the road pushing on the horse
b. Suppose that the horse and cart have started from rest; and as time goes on, their speed increases in the same direction. Which one of the following conclusions is correct concerning the magnitudes of the forces mentioned above?
1. Force 1 exceeds Force 2.
2. Force 2 is less than Force 3.
3. Force 2 exceeds Force 4.
4. Force 3 exceeds Force 4.
5. Forces 1 and 2 cannot have equal magnitudes.
Answer:
a) F₁ = F₂, F₃ = F₄, b) the correct answer is 3
Explanation:
a) In this exercise we have several action and reaction forces, which are characterized by having the same magnitude, but different direction and being applied to different bodies
Forces 1 and 2 are action and reaction forces F₁ = F₂
Forces 3 and 4 are action and reaction forces F₃ = F₄
as it indicates that the
b) how the car increases if speed implies that force 1> force3
F₁ > F₃
therefore the correct answer is 3
A block of mass m is moved over a distance d. An applied force F is directed perpendicularly to the block’s displacement. How much work is done on the block by the force F?
zero
Explanation:
Work W is defined as
W = F•d = Fdcos(theta)
and it is a dot product of the force and displacement and theta is angle between F and d Since the force is perpendicular to d, angle is 90° thus cos90 = 0. Hence work is zero.
The image shows the right-hand rule being used for a current-carrying wire.
An illustration with a right hand with fingers curled and thumb pointed up.
Which statement describes what the hand shows?
When the current flows down the wire, the magnetic field flows out on the left side of the wire and in on the right side of the wire.
When the current flows up the wire, the magnetic field flows out on the left side of the wire and in on the right side of the wire.
When the current flows down the wire, the magnetic field flows in on the left side of the wire and out on the right side of the wire.
When the current flows up the wire, the magnetic field flows in on the left side of the wire and out on the right side of the wire.
Answer:
The answer is (D): When the current flows up the wire, the magnetic field flows in on the left side of the wire and out on the right side of the wire.
Explanation:
