Answer:
It can rotate in any direction. The sign of the angular acceleration depends on how you set the reference system, it can be both negative or positive.
A 2.0 kg block rests on a level surface. The coefficient of static friction is, and the coefficient of kinetic friction is A horizontal force, X, is applied to the block. As X is increased, the block begins moving. Describe how the force of friction varies as X increases from the moment the block is at rest to when it begins moving. Indicate how you could determine the force of friction at each value of X―before the block starts moving, at the point it starts moving, and after it is moving. Show your work.
ps. I had to change F to X because of brainly.
By Newton's second law, the net force acting on the block in the vertical direction is
∑ F [ver] = n - mg = 0
where n = magnitude of normal force and mg = weight of the block. It follows that n = mg.
When the block is at rest, the applied force X will not be enough to move the box until it can overcome the maximum mag. of static friction. If µ[s] is the coefficient of static friction, then the maximum mag. of the frictional force is
f = µ[s] n = µ[s] mg
The net horizontal force would be
∑ F [hor] = X - µ[s] mg = 0
so a minimum force of X = µ[s] mg is required to get the block moving. Any mag. smaller than this and the block stays at rest/in equilibrium.
Once the mag. of X exceeds µ[s] mg, the block will begin to move. At that point, if the coefficient of kinetic friction is µ[k], then the net force on the block is
∑ F [hor] = X - µ[k] mg = 0
so a minimum force of X = µ[k] mg would be needed to keep the block moving at constant speed, or otherwise X = µ[k] mg + ma if the block is accelerating with mag. a.
The principles here are captured in the attached plot.
A 2-column table with 5 rows. The first column has entries empty, time of trial number 1 (seconds), time of trial number 2 (seconds), time of trial number 3 (seconds), average time (seconds). The second column labeled one quarter checkpoint has entries 2. 15, 2. 05, 02. 02, 02. 7. Use the table to answer the questions. What is the fastest time trial for the first quarter checkpoint? seconds What is the slowest time trial for the first quarter checkpoint? seconds What is the range of times measured for this checkpoint? seconds.
The fastest time trial for the first quarter checkpoint is 2.02 s.
The slowest time trial for the first quarter checkpoint is 2.7 s.
The range of the times measured for the checkpoint is 0.68 s.
The given parameters;
Time for quarter checkpoint, = 2.15, 2.05, 2.02, 2.7The fastest time trial for the first quarter checkpoint is the least measured time value.
fastest time trial = least time measured
fastest time trial = 2.02 s
The slowest time trial for the first quarter checkpoint is the highest measured time value.
slowest time trial = 2.7 s
The range of the times measured for the checkpoint is difference between the fastest time and slowest time.
Range = fastest time - slowest time
Range = 2.7 - 2.02
Range = 0.68 s
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Answer:
2.02, 2.15, 0.13
Explanation:
A 2-column table with 5 rows. The first column has entries empty, time of trial number 1 (seconds), time of trial number 2 (seconds), time of trial number 3 (seconds), average time (seconds). The second column labeled one quarter checkpoint has entries 2.15, 2.05, 2.02, 2.07.
Use the table to answer the questions.
What is the fastest time trial for the first quarter checkpoint?
2.02
seconds
What is the slowest time trial for the first quarter checkpoint?
2.15 seconds
What is the range of times measured for this checkpoint?
0.13 seconds
A single covalent bond is stronger than a single hydrogen bond so why does a group of polar molecules tend to have a higher boiling point than a group of non polar molecules
Answer:
this question makes no sense
Explanation:
like how do you get this question
Answer:They require more energy to break intermolecular forces hence polar molecules have higher melting points and boiling points than non-polar molecules of similar size, shape and number of electrons.
Explanation:
what voltage is measured across the 15 ohm resistor
Which electromagnetic wave has a lowest frequency?
Group of answer choices
A) x-rays
B) ultraviolet light
C) microwaves
D) infrared light
E) visible light
Answer:
E.visible lights
Explanation:
hope its attachments
What is the mass of an object that has a weight of 110N ?
Answer:
The mass of the object is 11.2 Kg.
Hope you could get an idea from here.
Doubt clarification - use comment section.
The mass of an object that has a weight of 110 N is 11.2kg.
What is Weight?The weight of an object is defined as the force that acts on the object due to gravity. It is a vector quantity in some cases such as when the force of gravity acts on an object but it is also a scalar quantity where the force of gravity has a magnitude.
Weight is the gravitational force of attraction on an object due to the presence of another heavy object, such as the Earth or the Moon. It is expressed by multiplying the mass by the gravitational acceleration.
In above case,
Weight (W) = 110 N
Acceleration due to gravity (g) = 9.8 m/s^2
Let the mass of the object be m.
By using the formula,
W = mg,
[tex]110 N = 9.8 m/s^2 *m\\m = 110 N/ 9.8 m/s^2[/tex]
so, m = 11.2 Kg
Thus, the mass of an object that has a weight of 110 N is 11.2kg.
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4 What type of circuit is described by each of the following statements?
Answer series or parallel
a All components are connected end-to-end.
b. The current in the circuit divides so that some flows through one component
and the rest through another component
Two lamps are connected side by side so that each lights brightly
d The current has the same valur everywhere in the circuit
C
Liquid X of volume 0.5m3 and density 900kgm-3 was mixed with liquid Y of volume 0.4m3 and density 800kgm-3. What was the density of the mixture?
Answer:
Density of the mixture = 855.56kgm-3
Explanation:
Density = Mass / Volume
Volume of Liquid X = 0.5m³
Density of Liquid X = 900kgm-3
Mass of Liquid X = Density × Volume
= 900kgm-3 × 0.5m³ = 450kg
Volume of Liquid Y = 0.4m³
Density of Liquid Y = 800kgm-3
Mass of Liquid Y = Density × Volume
= 800kgm-3 × 0.4m³= 320kg
As X and Y are mixed, we add their masses and volumes together:
Mass = 770kg
Volume = 0.9m³
Now we can find the density of the mixture:
Density = 770kg / 0.9m³ = 855.56kgm-3 (rounded to the 2nd decimal)
What is the final temperature if it requires 5000 J of heat to warm 2.38892 x10-2 kg of water that starts at 5oC? Remember Cp for water is 4186 J/kgC
The final temperature of water is equal to 50.9999°C
Given the following data:
Mass = [tex]2.38892 \times 10^{-2}\;kg[/tex]Quantity of heat = 5000 J Specific heat capacity of water = 4186 J/kg°CTo determine the final temperature of water:
Mathematically, quantity of heat is given by the formula;
[tex]Q=mc\theta[/tex]
Where:
Q represents the quantity of heat.m represents the mass of an object.c represents the specific heat capacity.∅ represents the change in temperature.Substituting the given parameters into the formula, we have;
[tex]5000=2.38892 \times 10^{-2}\times 4186 \times \theta\\\\5000=100.0001912 \theta\\\\ \theta=\frac{5000}{100.0001912} \\\\ \theta=49.9999^{\circ}C[/tex]
For the final temperature:
[tex]\theta = T_2 - T_1\\\\T_2 = \theta+T_1\\\\T_2 = 49.9999 + 50[/tex]
Final temperature = 50.9999°C
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29) A cheetah can accelerate from rest to 25
m/s in 6 s. Assuming that the cheetah moves
with constant acceleration, what distance does
it cover in the first 3 s
Answer:
[tex]\huge\boxed{\sf S = 18.75 \ meters}[/tex]
Explanation:
Given Data:
Initial Velocity = Vi = 0 m/s (rest)
Final Velocity for 6 seconds = Vf = 25 m/s
Time (1) = T1 = 6 seconds
Time (2) = T2 = 3 seconds
Required:
Distance for 3 seconds = S = ?
Solution:
For 6 seconds, the acceleration will be:
[tex]\displaystyle a = \frac{Vf-Vi}{t} \\\\a = \frac{25 - 0}{6} \\\\a = 25 / 6\\\\\boxed{a = 4.167 \ m/s^2}[/tex]
Since, acceleration is constant, it will be the same at 3 seconds as well.
Using second equation of motion to find Distance (S) with time being 3 seconds:
[tex]\displaystyle S= Vit+\frac{1}{2} at^2\\\\S = (0)(3)+ \frac{1}{2} (4.167)(3)^2\\\\S = \frac{1}{2} (4.167)(9)\\\\S = \frac{37.5}{2} \\\\\boxed{S = 18.75 \ meters}\\\\\rule[225]{225}{2}[/tex]
Hope this helped!
~AH1807
Which statement accurately describes how the acceleration of an object in free fall changes?
O A. It accelerates downward at a constant rate.
O
B. It accelerates downward at an increasing rate.
C. It accelerates downward at an irregular rate.
D. It accelerates downward at a decreasing rate.
Answer: A
Explanation: An object in free fall only is affected by gravity, so the acceleration is 9.8 m/s²
the purpose of many scientific investigations is to test a {n}
Answer:
Scientific investigation is a quest to find the answer to a question using the scientific method.
Explanation:
the scientific method is a systematic process that involves using measurable observations to formulate, test or modify a hypothesis.
One ball of mass 0.600kg travelling 9.00m/s to the right collides head on elastically with a second ball of mass 0.300kg travelling 8.00m/s to the left.what are their velocities after collisions?
Let m₁ and v₁ denote the mass and initial velocity of the first ball, and m₂ and v₂ the same quantities for the second ball. Momentum is conserved throughout the collision, so
m₁ v₁ + m₂ v₂ = m₁ v₁' + m₂ v₂'
where v₁' and v₂' are the balls' respective velocities after the collision.
Kinetic energy is also conserved, so
1/2 m₁ v₁² + 1/2 m₂ v₂² = 1/2 m₁ (v₁')² + 1/2 m₂ (v₂')²
or
m₁ v₁² + m₂ v₂² = m₁ (v₁')² + m₂ (v₂')²
From the momentum equation, we have
(0.600 kg) (9.00 m/s) + (0.300 kg) (-8.00 m/s) = (0.600 kg) v₁' + (0.300 kg) v₂'
which simplifies to
10.0 m/s = 2 v₁' + v₂'
so that
v₂' = 10.0 m/s - 2 v₁'
From the energy equation, we have
(0.600 kg) (9.00 m/s)² + (0.300 kg) (-8.00 m/s)² = (0.600 kg) (v₁')² + (0.300 kg) (v₂')²
which simplifies to
67.8 J = (0.600 kg) (v₁')² + (0.300 kg) (v₂')²
or
226 m²/s² = 2 (v₁')² + (v₂')²
Substituting v₂' yields
226 m²/s² = 2 (v₁')² + (10.0 m/s - 2 v₁')²
which simplifies to
3 (v₁')² - (20.0 m/s) v₁' - 63.0 m²/s² = 0
Solving for v₁' using the quadratic formula gives two solutions,
v₁' ≈ -2.33 m/s or v₁' = 9.00 m/s
but the second solution corresponds to the initial conditions, so we omit that one.
Then the second ball has velocity
v₂' = 10.0 m/s - 2 (-2.33 m/s)
v₂' ≈ 14.7 m/s
The magnetic field 0. 02 m from a wire is 0. 1 T. What is the magnitude of the magnetic field 0. 01 m from the same wire? 0. 01 T 0. 05 T 0. 1 T 0. 2 T.
The magnitude of the magnetic field 0.01 m from the same wire is 0.2 T.
Given to us:
Magnetic field, [tex]B_1 = 0.1\ T[/tex]
Radius of wire, [tex]R_1 = 0.02\ m[/tex]
To find out the magnitude of the magnetic field 0. 01 m from the same wire, we need to find out current first. we will use the formula,
[tex]B = \dfrac{\mu_oI }{2\pi R},\\\rn\\where,\\B= magnetic\ field\\\mu_o = 4\pi\times 10^{-7} m\cdot kg\cdot s^{-2} A^{-2}\ is\ the\ magnetic\ constant\\I= current\\R= radius\ of\ the\ wire[/tex]
Putting the values,
[tex]B_1 = \dfrac{\mu_oI }{2\pi R_1},\\\rn\\\\0.1= \dfrac{4\times \pi \times 10^{-7}\times I}{2\times \pi\times0.02}\\\\I=10,000\ A[/tex]
Now, for [tex]B_2[/tex]
[tex]B_2 = \dfrac{\mu_oI }{2\pi R_2},\\\rn\\\\B_2= \dfrac{4\times \pi \times 10^{-7}\times 10,000}{2\times \pi\times0.01}\\\\B_2= 0.2\ T[/tex]
Hence, the magnitude of the magnetic field 0. 01 m from the same wire is 0.2 T.
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please help quickly,the picture and choices are above.
Answer:
C because it make sens
C the light wave traveled through ice and then through a Dimond.
The normal force of a parked car on a level surface is 15,000 Newtons. What is the force of the car?
Answer:
The force of the car is 15000N.
Explanation:
The unit of force is Newtons (N), so based on the question, the force is 15000 Newtons.
[tex] \huge{ \mathrm{question \hookleftarrow}}[/tex]
Calculate equivalent resistance of two resistors R1 and R2 in parallel where,
[tex] \sf R_ 1 = (6 \pm0.2 )\: \: ohms[/tex]
[tex] \sf R_ 2 = (3 \pm0.1 )\: \: ohms[/tex]
[tex]it \: was \: helpful \: to \: you[/tex]
# be careful#
Suppose that the Sun started shrinking in size, without losing any mass. What would be the effect of the Sun's change on the orbits of the planets
Answer:
F = G M m / R^2 gravitational force on planet of mass m.
None of these quantities change in the given hypothesis so
there will be no change in the orbit of mass m
if the speed of a car is increased by 50%, bywhat factor will its minimum braking distance be increased, assuming all else is the same?
When the speed of a car increases by a factor of 50%, the minimum braking distance will also be increased by a factor.
Braking distance will increase by a factor 2.25
Solution
From Newton's equation of motion, we can say that;
v² = u² + 2as
Where initial velocity is zero, we have;
v² = 2as
s = v²/2a
s is the distance and v is the final speed.
50% increase in speed means it has increased by a factor of 1.50
Hence we have
1.50²v² is directly proportional to the 1.50²d
Distance = 1.50²v² = 2.25d
Hence, breaking distance will increase by a factor 3.0625
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2.
A swimmer swims 1 m/s due north against a current of 3 m/s due south. What is the resultant velocity of
the boat?
The resultant velocity of the boat is equal to 2 m/s due south.
Given the following data:
Swimmer's speed = 1 m/s due northOcean current speed = 3 m/s due south.To determine the resultant velocity of the boat:
Resultant velocity can be defined as the sum (addition) of each of the vector velocity acting on a physical or an object. Thus, it is simply a combination of two or more single vector velocity.
Note: When the velocities are acting in the same direction, you will add them up while you will subtract when the velocities are acting in opposite directions.
In this scenario, the velocities are acting in opposite directions. Thus, we would subtract as follows:
[tex]V = 3 - 1[/tex]
V = 2 m/s due south.
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what role does gravity have in the motion of planets around the sun?
Answer:
The gravity helps the planets stay together near the Sun, without it Earth would be floating away with the planet eventually becoming frozen up, thus the role gravity have in the motion of planets around the sun is by keeping them together.
Explanation:
Hope this helps!
if a 5kg ball is traveling at 20 m/s and it’s stopped in 4s, what’s the impulse on the ball?
why does polishing the surface of a metal extend fatigue life
Answer:
they are machined with shape characteristics which maximize the fatigue life of a metal.
Explanation:
they are highly polished to provide the surface characteristics which enable the best fatigue life.
How much energy has 4x 1010 m³ of water collected in a reservoir at a 2. 3. height of 100 m from the power house? What kind of energy is that? (Given, mass of 1 m³ of water = 1000 kg)
Explanation:
[tex] \rule{999pt}{66646pt}[/tex]
Physics Practice Help just need an explanation really not looking for an answer
What variables would be required for a computational model that predicts change in the energy of a falling object?(1 point)
A.)Momentum, speed, mass
B.)Height, speed, mass
C.)Acceleration, speed, mass
D,) x, y, z
This question involves the concepts of the law of conservation of energy and kinetic energy.
The variables required for a computational model that predicts change in the energy of a falling object are "B.)Height, speed, mass".
According to the law of conservation of energy the change in energy can be given in the following form for a falling object:
Loss in Potential Energy = Gain in Kinetic Energy
[tex]mgh = \frac{1}{2}mv^2[/tex]
It is clear from the formula that the variables involved in the computational model are mass, height, and velocity.
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Answer:
b) height, speed, mass
Explanation:
just took the test
What are the three different social perspectives on sport
Answer:
functionalist theory,feminist theory. discipline of sociology
Which best describes how plant and animal cells carry out the basic functions of life?
A.
Plant and animal cells do not need food molecules for growth.
B.
Plant and animal cells do not make molecules for their own growth and repair.
C.
Plant and animal cells break down food for growth and repair.
D.
Plant and animal cells perform life functions such as making food molecules for growth.
how much force would be needed to push a box weighing 30 N up a ramp that ahas an ideal mechanical advantage of 3
Answer:
60n would be the needed force
The force needed to push a box weighing 30 N up a ramp that has an ideal mechanical advantage of 3 is equal to 10 N.
What is the mechanical advantage?The mechanical advantage can be described as the ratio of the input force to the output force. The mechanical advantage of any machine can be determined by the ratio of the forces involved to do the work.
The ratio of the resistance force to the effort is called the actual mechanical advantage which will be comparatively less. The efficiency of a machine is always determined by equating the ratio of its output to its input.
The efficiency of the machine is equal to the ratio of the actual mechanical advantage (M.A.) and theoretical mechanical advantage. Mechanical advantage can be defined as the force produced by a machine to the force applied to it.
Given the load = 30 N and the ideal mechanical advantage = 3
Mechanical advantage = Load/ Effort
Input force or effort = Load/ M.A.
Force = 30/3
Input Force = 10 N
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Mars is 7.83x10^10m [^10 is an exponent] from planet earth. The planet Earth is 5.98x10^24kg [^24 is an exponent] while Mars has a mass of 6.42x10^23kg [^23 is an exponent]. What is the gravitational attraction between the two planets? G=6.67×10^-11 (-11 is an exponent)
Answer:
Approximately [tex]4.18 \times 10^{16}\; \rm N[/tex].
Explanation:
Consider two objects of mass [tex]m_{1}[/tex] and [tex]m_{2}[/tex]. Let [tex]r[/tex] denote the distance between the center of mass of each object. Let [tex]G[/tex] denote the gravitational constant. ([tex]G \approx 6.67 \times 10^{-11}\; {\rm m^{3}\cdot kg^{-1}\cdot s^{-2}}[/tex].)
By Newton's Law of Universal Gravitation, the size of gravitational attraction between these two objects would be:
[tex]\begin{aligned}F &= \frac{G\, m_{1}\, m_{2}}{r^{2}}\end{aligned}[/tex].
In this question, [tex]m_{1} = 5.98\times 10^{24}\; {\rm kg}[/tex] and [tex]m_{2} = 6.24 \times 10^{23}\; {\rm kg}[/tex] are the mass of the two planets. The distance between the two planets is [tex]r = 7.83 \times 10^{10}\; \rm m[/tex] (approximately the same as the distance between the center of mass of planet Earth and the center of mass of Mars.)
Apply Newton's Law of Universal Gravitation to find the size of gravitational attraction between the two planets:
[tex]\begin{aligned}F &= \frac{G\, m_{1}\, m_{2}}{r^{2}} \\ &= \frac{1}{(7.83 \times 10^{10}\; {\rm m})^{2}} \\ &\quad \times (6.67 \times 10^{11}\; {\rm m^{3} \cdot kg^{-1} \cdot s^{-2}}) \\ &\quad \times (5.98 \times 10^{24}\; {\rm kg}) \\ &\quad \times (6.42 \times 10^{23}\; {\rm kg}) \\ &\approx 4.18 \times 10^{16}\; {\rm kg \cdot m \cdot s^{-2}} \end{aligned}[/tex].
Since [tex]1\; {\rm kg \cdot m \cdot s^{-2}} = 1\; {\rm N}[/tex], the size of gravitational attraction between the two planets would be approximately [tex]4.18 \times 10^{16}\; {\rm N}[/tex].
A ball falling from a height of 5 m was caught at some height after being reflected off the floor. Find the magnitude of the movement of the ball. If the distance traveled is equal to 7 meters.
Answer:
3m
Explanation:
I can't make a drawing right know, but it's all around vectors. First of all the ball travels down 5 meters, so we have a vector pointing down, then it is reflected and travels 2 meters up. You have to sum the vectors so you obtain that the distance traveled is: 5-2 meters = 3m