Answer:
B. 0.50 meters/second^2
Explanation:
Given:
mass of table (m) = 10.0 kgEli's force (F1) = 120 NAndy's force (F2) = -125 NArranging Formulas and Substituting Values, Part 1:
F = F1 + F2F = 120 + (-125)F = -5 NThe -5 means that the table will move along the direction of the force applied by Andy.Arranging Formulas and Substituting Values, Part 2:
m * a = Fa = F / Ma = -5 / 10.0 kga = -0.5 m/s^2Answer:
B. 0.50 meters/second^2
Explanation:
Question 19 of 20
8 44%
Look at the diagram. To overcome the friction force, the man could use a slippery
substance like oil or graphite spray on the floor. These substances are examples of
what?
Enter your answer
Substances like oil or graphite spray are examples of - lubricants.
Lubrication is the control of friction and a friction-reducing film between moving surfaces such as the given diagram. The lubricant used can be a fluid, solid, or plastic substance.
Oil reduces the amount of friction by introducing a layer of a smooth and slippery substance that makes the surface smooth and slippery as well.Graphite Spray is a Lubricant that contains high purity graphite along with an organic binder that will adhere the coating to most surfaces.This leaves a dry, clean film of highly lubricous graphite.It lubricates and reduces friction in parts so they slide or rotate easily.Thus, oil or graphite spray are examples of - lubricants.
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A box with a mass of 100.0 kg slides down a ramp with
a 50 degree angle.
What is the weight of the box?
N
What is the value of the normal force? Round the
answer to the nearest whole number.
N
What is the acceleration of the box? (Disregard friction
and air resistance.) Round the answer to the nearest
tenth.
m/s2
Answer:
Weight of the box? 980 N
Value of the normal force? 630 N
Acceleration of the box? 7.5 m/s^2
Explanation:
I got the answer right
2. Look at the table of typical densities. You have a 1 cm volume of each of the materials.
a List them in order of smallest mass to biggest mass.
b Which is the odd one out? Why?
Answer:
um were is it
Explanation:
Answer:
i think its b
Explanation:
if the velocity is at 0 and the accelerationis at 0 what is the motion
why is it more difficult to revolve a stone by tieing it in a longer string then by tieing it to a shorter string
A 20.0 g rubber bullet traveling with a velocity of 60.0 m/s hits a stationary
3.00 kg wood block resting on a frictionless surface. The rubber bullet
ricochets off the wood block with a velocity of 40.0 m/s in the opposite
direction.
A. Find the velocity of the wood block.
Answer:
0.667 m/s
Explanation:
Given that,
Mass of a rubber bullet, m₁ = 20 g = 0.02 kg
Initial speed of rubber bullet, u₁ = 60 m/s
Mass of the wood block, m₂ = 3 kg
Initial speed of the wooden block, u₂ = 0
A rubber bullet hits the stationary wood block. The rubber bullet ricochets off the wood block with a velocity of 40.0 m/s in the opposite direction.
Final speed of rubber bullet, v₁ = -40 m/s
Using the conservation of momentum. It will remains conserved. Let v₂ is the velocity of the wood block.
[tex]m_1u_1+m_2u_2=m_1v_1+m_2v_2\\\\0.02 \times 60+3\times 0=0.02 \times (-40)+3\times v_2\\\\1.2=-0.8+3v_2\\\\2=3v_2\\\\v_2=0.667\ m/s[/tex]
So, the velocity of the wood block is 0.667 m/s.
A sound source produces 85 sound wave compressions in 5 seconds. the wave length of the sound is 20m. calculate the
a)frequency of the sound waves
b)the speed of the wave
Answer: it would be 14.1 mm
Explanation: Sound is a wave. More specifically, sound is defined to be a disturbance of matter that is transmitted from its source outward. A disturbance is anything that is moved from its state of equilibrium. Some sound waves can be characterized as periodic waves, which means that the atoms that make up the matter experience simple harmonic motion.
A vibrating string produces a sound wave as illustrated in Figure 14.2, Figure 14.3, and Figure 14.4. As the string oscillates back and forth, part of the string’s energy goes into compressing and expanding the surrounding air. This creates slightly higher and lower pressures. The higher pressure... regions are compressions, and the low pressure regions are rarefactions. The pressure disturbance moves through the air as longitudinal waves with the same frequency as the string. Some of the energy is lost in the form of thermal energy transferred to the air. You may recall from the chapter on waves that areas of compression and rarefaction in longitudinal waves (such as sound) are analogous to crests and troughs in transverse waves.
The frequency of sound waves is 144.1 m.
What is Sound waves?More specifically, sound is defined to be a disturbance of matter that is transmitted from its source outward. A disturbance is anything that is moved from its state of equilibrium.
Some sound waves can be characterized as periodic waves, which means that the atoms that make up the matter experience simple harmonic motion.
This creates slightly higher and lower pressures. The higher pressure... regions are compressions, and the low pressure regions are rarefactions. The pressure disturbance moves through the air as longitudinal waves with the same frequency as the string.
Therefore, The frequency of sound waves is 144.1 m.
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An artery with a 3 cm radius carries blood away from the heart. In a constricted region that is partially blocked with plaque, the radius shrinks to 2 cm and the blood's velocity is 5 m/s. What is the bloods velocity in the 3 cm radius unconstricted region?
Answer:
2.32m/sExplanation:
What is flow rate?
Flow rate is the volume of fluid per unit time flowing past a point through the area
the formula is Q=Av
where A= area
v= flow velocity
step one:
given data
radius r1= 3cm= 0.03m
radius r2=2cm= 0.02m
velocity v1=?
velocity v2=5m/s
The area A1
[tex]Area= \pi r^2\\\\Area- A1= 3.142*0.03^2\\\\A1= 0.0028m^2[/tex]
The area A2
[tex]Area= \pi r^2\\\\Area- A2= 3.142*0.02^2\\\\A1= 0.0013m^2[/tex]
Step two:
For the continuous flow
Q1=Q2
A1v1=A2v2
0.0028*v1=0.0013*5
v1=0.0065/0.0028
v1=2.32m/s
The bloods velocity in the 3 cm radius unconstricted region is 2.32m/s
A bowling ball with a mass of 7.0kg strikes a pin that had a mass of 2.0kg the pin flies forward with a velocity of 6.0m/s, and the ball continues forward at 4.0 m/s. What was the original velocity of the ball?
The conservation of momentum P states that the amount of momentum remains constant when there are not external forces.
We don't have external forces, so:
[tex]P_0 = P_1\\m_bv_{0b}+m_pv_{0p}=m_bv_{1b}+m_pv_{1p}\\[/tex]
Where:
mb is the mass of the bowling ball mp the mass of the pin[tex]v_{0b}\quad and\quad v_{0p}[/tex] the initial velocities of the bowling ball and the pin.[tex]v_{1b}\quad and\quad v_{1p}[/tex] the final velocities of the bowling ball and the pin.Solving for v0b:
[tex]v_{0b} =\dfrac{m_bv_{1b}+m_pv_{1p}- m_pv_{0p}}{m_{b}}\\\\v_{0b} =\dfrac{(7\;kg)(4\;m/s)+(2\;kg)(6\;m/s)- (2\;kg)(0 \;m/s)}{7\;kg}\\v_{0b}=\dfrac{40}{7}\;m/s\\\\\boxed{v_{0b}\approx5.71\;m/s}[/tex]
R/ The original velocity of the ball was 5.71 m/s.The original velocity of the ball is 5.71 m/s.
The principle of conservation of momentum: In a closed system, The total momentum before collision is equal to total momentum after collision.
From the principle of conservation of momentum,
MU+mu = MV+mv.................... Equation 1
Where M = mass of the bowling ball, m = mass of the pin, U = initial velocity of the bowling ball, u = initial velocity of the pin, V = final velocity of the bowling ball, v = final velocity of the pin.
From the question,
Given: M = 7 kg, m = 2 kg, u = 0 m/s (at rest), v = 6.0 m/s, V = 4 m/s.
Substitute these values into equation 1 and solve for U
7(U)+2(0) = 7(4)+2(6)
7U = 28+12
7U = 40
U = 40/7
U = 5.71 m/s.
Hence, The original velocity of the ball is 5.71 m/s.
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16.) A person walks 4 meters east, then walks 3 meters north. Determine distance and
displacement.
Answer:
Long question good luck:)
Explanation:
Answer:7
Explanation:
because its the right answer
compared to a maritime tropical air mass, a maritime polar air mass has?
A maritime polar air mass has relatively higher moisture and is relatively cooler than the tropical air mass.
Air masses:Maritime polar (mP) is cold but moist due to its origination over the oceans. The desert region air masses (hot and dry) are designated by ‘cT’ for ‘continental tropical’. As these air masses move around the earth they can begin to acquire additional attributes.
The continental Tropical (cT) air mass originates in arid or desert regions in the middle or lower latitudes, principally during the summer season. It is strongly heated in general, but its moisture content is very low.
Continental tropical air masses are extremely hot and dry. Arctic, Antarctic, and polar air masses are cold. The qualities of arctic air are developed over ice and snow-covered ground. Arctic air is deeply cold, colder than polar air masses
Maritime Polar (mP) air masses develop over the polar areas of both the Northern and the Southern hemispheres. They generally contain considerably more moisture than the cP air masses. As they move inland in middle and high latitudes, heavy precipitation may occur.
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when analyzing a position time graph the direction of the slope positive or negative tells us:
Answer:
Explanation:
Positive values for position indicate that the object is in front of the starting point and negative values tell us that the object is behind the starting point. (time = 9.5, position = 0) the object is at the starting point.
Answer:
Positive values for position indicate that the object is in front of the starting point and negative values tell us that the object is behind the starting point. (time = 9.5, position = 0) the object is at the starting point.
Explanation:
A +4.0 uC charge is placed on the x axis at x= +3.0 m, and a -2.0 uC is located on the y-axis at y= -1.0 m. Point A is on the y axis at y= +4.0 m. Determine the electric potential at point A (relative to zero at the origin).
Answer:
The potential is [tex]V_A = 9600 \ V[/tex]
Explanation:
From the question we are told that
The magnitude of the charge is [tex]q_1 = 4 \mu C = 4*10^{-6} \ C[/tex]
The position of the charge is [tex]x = + 3.0 \ m[/tex]
The magnitude of the second charge is [tex]q_2 = -2.0 \mu C = -2.0 *10^{-6} \ C[/tex]
The position is [tex]y_1 = - 1.0 \ m[/tex]
The position of point A is [tex]y_2 = + 4.0 \ m[/tex]
Generally the electric potential at A due to the first charge is mathematically represented as
[tex]V_a = \frac{k * q_1 }{r_1 }[/tex]
Here k is the coulombs constant with value [tex]k = 9*10^{9} \ \ kg\cdot m^3\cdot s^{-4} \cdot A^{-2}[/tex]
[tex]r_1[/tex] is the distance between first charge and a which is mathematically represented as
[tex]r_1 = \sqrt{x^2 + y_2 ^2 }[/tex]
=> [tex]r_1 = \sqrt{3^2 + 4 ^2 }[/tex]
=> [tex]r_1 = 5 \ m[/tex]
So
[tex]V_a = \frac{9*10^9 * 4*10^{-6} }{5 }[/tex]
[tex]V_a = 7200 \ V[/tex]
Generally the electric potential at A due to the second charge is mathematically represented as
[tex]V_b = \frac{k * q_2 }{r_2 }[/tex]
Here k is the coulombs constant with value [tex]k = 9*10^{9} \ \ kg\cdot m^3\cdot s^{-4} \cdot A^{-2}[/tex]
[tex]r_2[/tex] is the distance between second charge and a which is mathematically represented as
[tex]r_2 = y_2 - y[/tex]
=> [tex]r _2 = 4.0 - (-1.0)[/tex]
=> [tex]r = 5 \ m[/tex]
So
[tex]V_a = \frac{9*10^9 * -2*10^{-6} }{5 }[/tex]
[tex]V_a = -3600 \ V[/tex]
So the net potential difference at point A due to the charges is mathematically represented as
[tex]V_n = V_a + V_b[/tex]
=> [tex]V_n = 7200 - 3600[/tex]
=> [tex]V_n = 3600 V[/tex]
Generally the net potential difference at the origin due to both charges is mathematically represented as
[tex]V_N = V_c + V_d[/tex]
Here
[tex]V_c = \frac{k * q_1 }{x}[/tex]
=> [tex]V_c = \frac{9*10^9 * 4*10^{-6} }{3}[/tex]
=> [tex]V_c = 12000 V[/tex]
and
[tex]V_d= \frac{k * q_2 }{y}[/tex]
=> [tex]V_c = \frac{9*10^9 * -2*10^{-6} }{1}[/tex]
=> [tex]V_c =- 18000 V[/tex]
Generally the net potential difference at the origin is
[tex]V_N = 12000 - 18000[/tex]
=> [tex]V_N = -6000[/tex]
Generally the potential difference at A relative to zero at the origin is mathematically evaluated as
[tex]V_A = V_n - V_N[/tex]
=> [tex]V_A = 3600 - (-6000)[/tex]
=> [tex]V_A = 9600 \ V[/tex]
Ted Williams drops a ball from 12.7 meter to a desk that is 1.5 meter tall. What is the final speed of the ball right before it hits the desk?
Answer:
Explanation:
Using the equation of motion
V² = u²+2g∆H
v is the final velocity
u is the initial velocity = 0m/s
g = 9.8m/s²
∆H = 12.7-1.5 = 11.2m
Substitute
v² =0²+2(9.8)(11.2)
v² = 219.52
v = √219.52
v = 14.82m/s
Hence the final.speed of the ball is 14.82m/s
For Valentine’s Day Sally received a helium-filled balloon at a party. On returning home she accidently left the balloon in the car. Later she went to get the balloon and found it was partially deflated. After being in the house for an hour she noticed it was fully inflated again. Explain why this happened.
Answer:
The helium molecules move closer together when it's cold causing the balloon to deflate
Explanation:
The possible reason behind it is: the temperature of outside was low, so the Helium gas inside the balloon compressed and it was partially deflated at the car. When she returned to the house, temperature of the gas of the balloon rises again and it was fully inflated again.
What is Charles' law of gas?The experimental gas law known as Charles' law, commonly referred to as the law of volumes, illustrates how gases tend to expand when heated. Charles's law can now be expressed as follows:
The Kelvin temperature and the volume will be directly proportional when the pressure on a sample of a dry gas is kept constant.
This direct proportional relationship can be expressed as follows:
V∝T.
Hence, as the temperature is low outside the house, the balloon was partially deflated at the car and the temperature rises again when it was inside the house, so it was fully inflated again inside the house.
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i don’t know what to put here but i just wanna leave
Create a flexibility workout using the 'Principle of Progression' over a four week period.
Answer:
You can go to youtub and search 4 week flexibility programs and use that and add the principle of progression.
Answer:
r
Explanation:
Differences in water temperature in the ocean create movement because
how (and why) does a planet change its speed while in orbit?
please help asap!!!
Answer:
A planet's orbital speed changes, depending on how far it is from the Sun. The closer a planet is to the Sun, the stronger the Sun's gravitational pull on it, and the faster the planet moves.
Explanation: hope that helps :)
Which term best explains why you hear an echo?
Answer:
reflection
Explanation:
Answer:
reflection
Explanation:
A p e x
mr.easton is 80 , and he thinks that all young people with longer hair use drugs .this is an example of:
Why is gravity so weird? .
Answer:
Why is gravity so weird? No force is more familiar than gravity — it's what keeps our feet on the ground, after all. And Einstein's theory of general relativity gives a mathematical formulation for gravity, describing it as a “warping” of space.
Answer:
Because it pulls us down to the centre the earth
What fonts do you use to create color coded lyrics
A certain sound wave has a wavelength of 2 m and a frequency of 170 hertz. calculate the velocity of this wave.
Answer:
340m/s
Explanation:
Given parameters:
Wavelength = 2m
Frequency = 170Hz
Unknown:
Velocity of the wave = ?
Solution:
The velocity of the wave is the product of frequency and wavelength of the wave;
V = F λ
V is the velocity
F is the frequency
λ is the wavelength
Insert the parameters and solve;
V = 2 x 170 = 340m/s
The velocity of the wave with the given wavelength and frequency is 340m/s.
Given the data in the question;
Wavelength; [tex]\lambda = 2m[/tex]Frequency; [tex]f = 170Hz = 170 s^{-1}[/tex]Velocity; [tex]v =\ ?[/tex]
To determine the velocity of the wave, we use the expression for the relations between wavelength, frequency and speed.
[tex]\lambda = \frac{v}{f}[/tex]
Where [tex]\lambda[/tex] is wavelength, f is frequency and v is speed.
We substitute our given values into the equation
[tex]2m = \frac{v}{170s^{-1}}\\\\v = 2m * 170s^{-1}\\\\v = 340m/s[/tex]
Therefore, the velocity of the wave with the given wavelength and frequency is 340m/s.
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what can cause a magnetic domain to form?
A. iron atoms being closed together
B. magnetic material being placed close to a magnet
C. electric current flowing through iron
D. magnetic fields of atoms being aligned with each other
Answer:
D
Explanation:
Magnetic domains are always present in ferromagnetic materials due to the way the atoms bond to form the material. However, when a ferromagnetic material is in the unmagnetized condition, the magnetic domains are randomly oriented so that the magnetic field strength in the piece of material is zero.
Answer:
D. magnetic fields of atoms being aligned with each other
Explanation:
For a given radius, what happens to the angular acceleration as linear acceleration increases?
Linear acceleration and angular acceleration are directly proportional. The greater the linear acceleration, the larger the angular acceleration.
For example, the greater the angular acceleration of a bike's wheels, the greater the acceleration of the bike.
An object is swung in a horizontal circle on a length of string that is 0.93 m long. Its acceleration is 26.36 m/s2. What is the time it takes the object to complete one horizontal circle?
P.S. : I know the answer is 1.18 seconds but I just don’t know how to get there. Can someone explain to me how to do it? Thanks
Answer:
The object takes approximately 1.180 seconds to complete one horizontal circle.
Explanation:
From statement we know that the object is experimenting an Uniform Circular Motion, in which acceleration ([tex]a[/tex]), measured in meters per square second, is entirely centripetal and is expressed as:
[tex]a = \frac{4\pi^{2}\cdot R}{T^{2}}[/tex] (1)
Where:
[tex]T[/tex] - Period of rotation, measured in seconds.
[tex]R[/tex] - Radius of rotation, measured in meters.
If we know that [tex]a = 26.36\,\frac{m}{s^{2}}[/tex] and [tex]R = 0.93\,m[/tex], then the time taken by the object to complete one revolution is:
[tex]T^{2} = \frac{4\pi^{2}\cdot R}{a}[/tex]
[tex]T = 2\pi\cdot \sqrt{\frac{R}{a} }[/tex]
[tex]T = 2\pi\cdot \sqrt{\frac{0.93\,m}{26.36\,\frac{m}{s^{2}} } }[/tex]
[tex]T \approx 1.180\,s[/tex]
The object takes approximately 1.180 seconds to complete one horizontal circle.
Is block b speeding up or slowing down once it is set into downward motion and the cat is on block a?
Answer:
The answer is "Slowing down ".
Explanation:
please find the complete question in the attached file.
In this question, if the block B weight were accounted for by kinetic the friction of frame A, because the blocks pushed at a consistent speed throughout the beginning.
Afterward, on block A, the resistance intensity rises, which allows frames to also be negative, which is defined in the graph, that's why the answer Slowing down is correct.
An illustration to show the respective wavelengths of electromagnetic waves. There is a transverse wave with really small wavelengths to really long and are listed from smallest to largest are gamma ray, x-rays, ultraviolet, visible, infrared, microwave and radio. Which has the longest wavelength? Which has the highest frequency?
The wave which has the longest frequency if, listed from smallest to largest are gamma ray, x-rays, ultraviolet, visible, infrared, microwave, and radio is radio waves and gamma rays have the highest frequency.
What is an electromagnetic wave?The oscillations between an electric field and a magnetic field produce waves known as electromagnetic waves or EM waves. In other words, magnetic and electric fields oscillate to form electromagnetic (EM) waves.
The term "electromagnetic spectrum" (EMS) refers to the entire range of electromagnetic radiation that is now understood. Wavelengths rise from roughly 10⁻¹⁸ m to 100 km, and frequencies fall from 3 × 10²⁶ Hz to 3 × 10³ Hz in this range.
The following list of the seven electromagnetic waves is organized by decreasing wavelength and increasing frequency and energy: X-rays, gamma rays, UV rays, visible light, radio waves, microwaves, and infrared waves.
Thus, radio waves have the longest frequency and gamma rays have the highest frequency.
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Cold air is more dense than hot air. When I filled a 9 centimeter diameter balloon with cold air the mass was 1 gram and when I measured the mass of the same size balloon with hot air it was 0.5 grams. When molecules are cooled they move closer together and when they are heated up they move farther apart. Because of this more molecules can fit into a balloon when the air going in is cold than when the air going in is warm. (Please explain this to me)