Answer:
Uhh 2 one
Explanation
The Great Sandini is a 60 kg circus performer who is shot from a cannon (actually a spring gun). You don't find many men of his caliber, so you help him design a new gun. This new gun has a very large spring with a very small mass and a force constant of 1300 N/m that he will compress with a force of 6500 N. The inside of the gun barrel is coated with Teflon, so the average friction force will be only 50 N during the 5.0 mm he moves in the barrel.
Required:
At what speed will he emerge from the end of the barrel, 2.5 mabove his initial rest position?
Answer:
22m/s
Explanation:
Mass, m=60 kg
Force constant, k=1300N/m
Restoring force, Fx=6500 N
Average friction force, f=50 N
Length of barrel, l=5m
y=2.5 m
Initial velocity, u=0
[tex]F_x=kx[/tex]
Substitute the values
[tex]6500=1300x[/tex]
[tex]x=\frac{6500}{1300}=5[/tex]m
Work done due to friction force
[tex]W_f=fscos\theta[/tex]
We have [tex]\theta=180^{\circ}[/tex]
Substitute the values
[tex]W_f=50\times 5cos180^{\circ}[/tex]
[tex]W_f=-250J[/tex]
Initial kinetic energy, Ki=0
Initial gravitational energy, [tex]U_{grav,1}=0[/tex]\
Initial elastic potential energy
[tex]U_{el,1}=\frac{1}{2}kx^2=\frac{1}{2}(1300)(5^2)[/tex]
[tex]U_{el,1}=16250J[/tex]
Final elastic energy,[tex]U_{el,2}=0[/tex]
Final kinetic energy, [tex]K_f=\frac{1}{2}(60)v^2=30v^2[/tex]
Final gravitational energy, [tex]U_{grav,2}=mgh=60\times 9.8\times 2.5[/tex]
Final gravitational energy, [tex]U_{grav,2}=1470J[/tex]
Using work-energy theorem
[tex]K_i+U_{grav,1}+U_{el,1}+W_f=K_f+U_{grav,2}+U_{el,2}[/tex]
Substitute the values
[tex]0+0+16250-250=30v^2+1470+0[/tex]
[tex]16000-1470=30v^2[/tex]
[tex]14530=30v^2[/tex]
[tex]v^2=\frac{14530}{30}[/tex]
[tex]v=\sqrt{\frac{14530}{30}}[/tex]
[tex]v=22m/s[/tex]
The resistivity of blood is related to its hematocrit, the volume fraction of red blood cells in the blood. A commonly used equation relating the hematocrit h to the blood resistivity rho (in Ω⋅m) is rho=1.32/(1−h)−0.79. In one experiment, blood filled a graduated cylinder with an inner diameter of 0.90 cm. The resistance of the blood between the 1.0 cm and 2.0 cm marks of the cylinder was measured to be 198 Ω.
Required:
What was the hematocrit for this blood?
Answer:
[tex]0.35598[/tex]
Explanation:
r = Radius = [tex]\dfrac{0.9}{2}=0.45\ \text{cm}[/tex]
R = Resistance = [tex]198\ \Omega[/tex]
A = Area = [tex]\pi r^2[/tex]
l = Length of blood in cylinder = 1 cm
h = Hematocrit of the blood
Resistivity is given by
[tex]\rho=\dfrac{1.32}{1-h}-0.79[/tex]
Resistance is given by
[tex]R=(\dfrac{1.32}{1-h}-0.79)\dfrac{l}{\pi r^2}\\\Rightarrow h=1-\dfrac{1.32}{\dfrac{R\pi r^2}{l}+0.79}\\\Rightarrow h=1-\dfrac{1.32}{\dfrac{198\times \pi\times (0.45\times 10^{-2})^2}{0.01}+0.79}\\\Rightarrow h=0.35598[/tex]
The hematocrit of the blood is [tex]0.35598[/tex].
State three factors affecting pressure in liquids
Answer:
Density of liquid
Depth of liquid
Acceleration due to gravity
A dog finds a toy at rest on the floor. The dog pushes the toy horizontally on a frictionless floor with a net force of 2.0 Newtons for 3.0 meters. How much kinetic energy does the toy gain? Round your answer to the nearest whole number.
Answer:
the kinetic energy gained by the toy is 6J.
Explanation:
Given;
net applied to the toy by dog, F = 2 N
distance moved by the toy, d = 3 m
Apply the principle of work-energy theorem to determine the kinetic energy gained by the toy.
ΔK.E = W
= F x d
= 2 x 3
= 6 J
Therefore, the kinetic energy gained by the toy is 6J.
A potter’s wheel moves from rest to an angular speed of 0.10 rev/s in 36.5 s.
Assuming constant angular acceleration,
what is its angular acceleration in rad/s2?
Answer in units of rad/s2
.
Answer:
please find attached pdf
Explanation:
the product of 2.03 and 0.05
Answer:
2.03 x 0.05= 0.1015
.........
Two identical conducting spheres, fixed in place, attract each other with an electrostatic force of 0.152 N when their center-to-center separation is 60.9 cm. The spheres are then connected by a thin conducting wire. When the wire is removed, the spheres repel each other with an electrostatic force of 0.0395 N. Of the initial charges on the spheres, with a positive net charge, what was (a) the negative charge on one of them and (b) the positive charge on the other
Answer:
a) lets take the positive charge to be
q1 = 3.13 * 10^-6 C
b) lets take the negative charge to be
q2 = 3.13 * 10^-6 C
Explanation:
Given data:
electrostatic force between Two identical conducting spheres = 0.152 N
Center-to-center separation = 60.9 cm = 60.9 * 10^-2 m
When wire is removed
electrostatic force between the spheres ( repelling force ) = 0.0395 N
Determine the negative and positive charges on the identical spheres
q1, q2 = 6.26 * 10^-12 C ( positive net charge )
lets take the positive charge to be
q1 = 3.13 * 10^-6 C
lets take the negative charge to be
q2 = 3.13 * 10^-6 C
since q1,q2 = 6.26 * 10^-12 C
attached below is detailed solution
what is costant error
Answer:
In a scientific experiment, a constant error -- also known as a systematic error -- is a source of error that causes measurements to deviate consistently from their true value.
Explanation:
Metals are amazing, and they are all around us. You can probably easily identify them by their shiny surfaces and tinny sound when you tap them.
a. True
b. False
Answer:
a. True
Explanation:
Metals are amazing, and they are all around us. You can probably easily identify them by their shiny surfaces and tinny sound when you tap them.
Generally, metals are mainly hard, shiny, and can be hammered into thin sheets. Also, another property of a metal is that, it is a good conductor of heat and even electricity because there valence (outer) electrons are delocalized. Some examples of chemical elements that are metals are iron, gold, sodium, silver, copper etc.
The resolution of a telescope is ultimately limited by the diameter of its objective lens or mirror. A typical amateur astronomer's telescope may have a 6.09 in diameter mirror. What is the minimum angular separation (in arc seconds) of two stars that can be resolved with a 6.09 in scope
Answer:
θ = 3.19 arc second
Explanation:
The resolution of a telescope is given by the rayleigh criterion, which establishes that two objects are separated if the principal maximum of diffraction of one of them coincides with the first minimum of diffraction of the second object, based on this the solution is given by the first diffraction minimum, the a slit is
a sin θ = m λ
with m = 1
in the case of circular apertures the equation must be found in polar coordinates, therefore a numerical constant is introduced
a sin θ = 1.22 λ
Angles are measured in radians and in these experiments they are small
sin θ = θ
θ= 1.22 λ / a
in this case a = 6.09 in, the wavelength is wrong = 550 10⁻⁹ m which is the maximum resolution of the human eye
l
et's reduce the magnitudes to the SI system
d = 6.09‘ 2.54 10⁻-2 m / 1 inch = 15.4686 10-2 m
let's calculate
θ = 1.22 550 10-9 / 15.468 10-2
θ = 15.5 10⁻⁶ rad
rad = 2.06 105 s
θ = 15.5 10⁻⁶ rad 2.06 105s/ 1 rad
θ = 3.19 s
What principle does a heat engine take advantage of in order to use heat to perform work?
A. Magnets can convert motion into electrical energy
B. The speed of atoms increases when heat is added
C. Fluids expand when heat is added to them
D. Energy can be created from nothing
A 6 kg box with initial speed 8 m/s slides across the floor and comes to a stop after 2.4 s. A) What is the coefficient of kinetic friction?B) How far does the box move? C) You put a 5 kg block in the box, so the total mass is now 11 kg, and you launch this heavier box with an initial speed of 7 m/s. How long does it take to stop?
Answer:
A. Coefficient of kinetic friction, μ = 0.34
B. The box moves a distance of 9.64 m before coming to a stop
C. The heavier box will stop after 2.1 seconds
Explanation:
a. The coefficient of kinetic or sliding friction is given as: μ = F/R
where applied force, F = m × (∆v)/t
∆v = v - u
∆v = 0 - 8 m/s = -8 m/s; t = 2.4 s
R = normal reaction = m×g
where g = 9.8 m/s²
Substituting in the kinetic friction formula; μ = m∆v/t ÷ 1/m×g
μ = ∆v/g×t
μ = 8 / 9.8 × 2.4
μ = 0.34
b. Using the equation v² = u² + 2as to calculate the distance travelled by the box
where v = 0 m/s; u = 8.0 m/s; a = ? s = ?
From F = ma = μR
a = μR/m = (μ × m × g)/m
a = μg
a = 0.34 × 9.8
a = 3.32 m/s²
This is negative acceleration or deceleration
Substituting in the equation of motion
8² + 2 × -3.32 × s = 0
-6.64s = -64
s = 9.64 m
Therefore, the box moves a distance of 9.64 m before coming to a stop
c. The coefficient of friction is independent of mass.
Using the formula in (a): μ = ∆v/g×t
t = ∆v/μg
t = 7/0.34 × 9.8
t = 2.10 s
Therefore, the heavier box will stop after 2.1 seconds
The coefficient of kinetic friction of the box is 0.34.
The distance traveled by the box is 9.6 m.
The time taken for the heavier box to stop is 2.1 s.
The coefficient of kinetic frictionThe coefficient of kinetic friction of the box is calculated as follows;
[tex]\mu mg = ma\\\\\mu g = a\\\\\mu g = \frac{v}{t} \\\\\mu = \frac{v}{gt} \\\\\mu = \frac{8}{9.8 \times 2.4} \\\\\mu = 0.34[/tex]
The distance traveled by the boxThe distance traveled by the box is calculated as follows;
[tex]v^2 = u^2 + 2as\\\\2as = -u^2\\\\s = \frac{-u^2}{2a} \\\\s = \frac{-u^2}{2\mu g} \\\\s = \frac{-(8)^2}{2\times 0.34 \times 9.8} \\\\s = -9.6 \ m\\\\|s| = 9.6 \ m[/tex]
The time taken for the heavier box to stop is calculated as;
[tex]\mu = \frac{v}{gt} \\\\t = \frac{v}{\mu g} \\\\t = \frac{7}{0.34 \times 9.8}\\\\ t = 2.1 \ s[/tex]
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Two long parallel wires 20 cm apart carry currents of 5.0 A and 8.0 A in the between the two wires where the magnetic field is zero?
a. yes, midway between the wires
b. yes, 12 cm from the 5-A wire
c. yes, 7.7 cm from the 5-A wire
d. no
Answer:
c. yes, 7.7 cm from the 5-A wire
Explanation:
Given;
distance between the two wires, r = 20 cm = 0.2 m
first current, I₁ = 5.0 A
second current, I₂ = 8.0 A
The magnetic field due to the two wires occurs in different directions and it can be zero at this region.
Let x be the distance from 5 A wire where the magnetic field is zero.
[tex]B = \frac{\mu_o}{2\pi} [\frac{I_1}{x} -\frac{I_2}{r-x} ] = 0\\\\ \frac{\mu_o}{2\pi} [\frac{I_1}{x} -\frac{I_2}{r-x} ] = 0\\\\ \frac{\mu_o}{2\pi} [\frac{5}{x} -\frac{8}{0.2-x} ] = 0\\\\\frac{5}{x} -\frac{8}{0.2-x} = 0\\\\\frac{5}{x} = \frac{8}{0.2-x}\\\\5(0.2-x) = 8x\\\\1 -5x = 8x\\\\1 = 5x \ + \ 8x\\\\1 = 13x\\\\x = \frac{1}{13} \\\\x = 0.077 \ m\\\\x = 7.7 \ cm[/tex]
Therefore, the correct option is c. yes, 7.7 cm from the 5-A wire
If a biker rides west for 50 miles from his starting position, then turns and bikes back east for 80 miles. What is his displacement?
Answer:
Displacement = 30 miles due east.
Explanation:
Let the distance due west be A
Let the distance due east be B
Given the following data;
A = 50 miles
B = 80 miles
To find the displacement;
Displacement can be defined as the change in the position of a body or an object. It is a vector quantity because it has both magnitude and direction.
Thus, the displacement would be calculated by subtracting distance A from distance B because the rider rode in opposite directions.
Displacement = B - A
Displacement = 80 - 50
Displacement = 30 miles due east.
Messages from the Perseverance Rover on Mars reach Earth in 11 min. The speed of light is 3.00 x 108 m/s. Using this information, how far is Mars from Earth?
Answer:
[tex]d=1.98\times 10^{11}\ m[/tex]
Explanation:
Messages from the Perseverance Rover on Mars reach Earth in 11 min i.e. time is 660 s
The speed of light is [tex]3\times 10^8\ m/s[/tex]
We need to find the distance between Mars and Earth. Let the distance be d.
We know that,
Distance = speed × time
So,
[tex]d=660\times 3\times 10^8\\\\d=1.98\times 10^{11}\ m[/tex]
So, Mars is [tex]1.98\times 10^{11}\ m[/tex] from the Earth.
Candice is examining a cell under a microscope. She has identified a cell wall, a nucleus, and a chloroplast. What type of organism does this most likely belong to?
A. A plant B. An animal C. A fungus D. A bacterium
Answer:
A plant
Explanation:
because animals don't have cell walls, and fungus and bacteria dont have chloroplasts
In the absence of friction, if a force acting on a moving object stops acting, the object will
Answer:
Keep on moving
Explanation:
Newton's first law states "that, if a body is at rest or moving at a constant speed in a straight line, it will remain at rest or keep moving in a straight line at constant speed unless it is acted upon by a force."
help plz! what vibrates in following types of wave motion 1)light wave 2)sound waves 3)x-rays 4)water waves
Answer:
I believe it's 2) sound waves
Explanation:
With sound waves, the energy travels along in the same direction as the particles vibrate. This type of wave is known as a longitudinal wave, so named because the energy travels along the direction of vibration of the particles.
An object is placed in material a at point P, as shown in the diagram. The light is refracted when it strikes the interface with material b. When viewed from material b, at which point will the image appear?
Captain Jack Sparrow has been marooned on an island in the Atlantic by his crew, and decides to builda raft to escape. The wind seems quite steady, and first blows him due east for 11km, and then 6km ina direction 6degrees north of east. Confident that he will eventually find himself in safety, he fallsasleep. When he wakes up, he notices the wind is now blowing him gently 11degrees south of east -but after traveling for 21km, he finds himself back on the island.
Variable Name Min Max Step Sample Value
thetab 5 10 1 6
a 10 20 11 1
b 5 15 1 6
c 20 30 1 21
thetac 10 15 11 1
Required:
How far (in km) did the wind blow him while he was sleeping?
Answer:
d₃ = 37,729 km, θ= 5.1º North of West
Explanation:
This is a velocity addition problem, the easiest way to solve it is to decompose the velocities in a Cartesian system, the x-axis coincides with the West-East direction and the y-axis with the South-North direction
* first displacement is
d₁ₓ = 11 km
* second offset is
cos 6 = d₂ₓ / d₂
sin 6 = d_{2y} / d₂
d₂ₓ = d₂ cos 6
d_{2y} = d₂ sin 6
d₂ₓ = 6 cos 6 = 5.967 km
d_{2y} = 6 sin 6 = 0.6272 km
* third displacement is unknown
* fourth and last displacement
cos (-11) = d₄ₓ / d₄
sin (-11) = d_{4y} / d₄
d₄ₓ = d₄ cos (-11)
d_{4y} = d₄ sin (-11)
d₄ₓ = 21 cos (-11) = 20.61 km
d_{4y} = 21 sin (-11) = -4.007 km
They tell us that at the end of the tour you are back on the island, so the displacement must be zero
X axis
x = d₁ₓ + d₂ₓ + d₃ₓ + d₄ₓ
0 = 11 +5.967 + d₃ₓ + 20.61
d₃ₓ = -11 - 5.967 - 20.61
d₃ₓ = -37.577 km
Y axis
y = d_{1y} + d_{2y} + d_{3y} + d_{4y}
0 = 0 + 0.6272 + d_{3y} -4.007
d_{3y} = 4.007 - 0.6272
d_{3y} = 3.3798 km
This distance can be given in the form of module and angle
Let's use the Pythagorean theorem for the module
d₃ = [tex]\sqrt{d_{3x}^2 + d_{3y}^2}[/tex]
d₃ = [tex]\sqrt{37.577^2 + 3.3798^2}[/tex]
d₃ = 37,729 km
Let's use trigonometry for the angle
tan θ = d_{3y} / d₃ₓ
θ = tan⁻¹ [tex]\frac{d_{3y}}{d_{3x}}[/tex]
θ = tan-1 (-3.3798 / 37.577)
θ = 5.1º
Since the y coordinate is positive and the x coordinate is negative, this angle is in the second quadrant, so the direction given in the form of cardinal coordinates is
θ= 5.1º North of West
A 60 kg student weighs 600 N.
He does a bungee jump.
Calculate the change in gravitational potential energy as the student falls 50 m.
Give the unit.
Will give brainliest!
Answer:
30 000 J/Nm
Explanation:
60 x 10 x 50
=600 x 50
=30000 J/Nm
The change in gravitational potential energy of the student is 18000 Joule.
What is gravitational potential energy?
The energy that an item has or acquires when its location changes as a result of being in a gravitational field is known as gravitational potential energy. Gravitational potential energy can be defined as an energy that has a connection to gravitational force or gravity.
Given parameters:
Mass of the student: m = 60 kg.
Weighs of the student: W = 600 N.
the student falls h =50 m.
Hence, change in gravitational potential energy of the student =
Weighs of the student × change in height
= 600 N × 30 m
= 18000 Joule.
So, the change in gravitational potential energy of the student is 18000 Joule.
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The work is calculated by replacing pressure with nRT/V then pulling nRT outside the integral. We are only allowed to pull T outside the integral in that example because:
a. we are always allowed to pull T outside a volume integral
b. T is constant for an isothermal process
c. it's an expansion not a compression
d. T has a positive value in that problem
Answer: b. T is constant for an isothermal process
Explanation:
In a general case for an ideal gas, we have the relation:
dW = -pdV.
If we integrate in both sides, we get:
W = ∫-pdV
Now, as the problem says, we can replace p by n*R*T/V
This is because of the equation for ideal gases.
p = n*R*T/V
replacing that in the equation we get
W = -∫pdV = -∫(n*R*T/V)dV
Now we want to take the n*R*T part outside of the integral.
But we only could do this if T does not depend on V.
If T is cosntant, then T does not depend on V, and we know that T is constant when we are in an isothermal process.
Where an isothermal process is a process where the temperature does not change, then T = constant.
Then n*R*T = constant = k
Because this is a constant, we could take it out of the integral so we get:
W = -∫(n*R*T/V)dV ) = -(n*R*T)∫(1/V)*dV
Then the correct option is option b.
b. T is constant for an isothermal process
Ocean waves crash on the beach at a velocity of 3.5 m/s. If the distance between the crests of each wave is 4 m, find the frequency of the waves.
a. 0.0088 Hz
b. 14.0 Hz
c. 1.14 Hz
d. 0.88 Hz
Answer:
d
Explanation:
velocity=frequency × wavelength
frequency=speed/wavelength
frequency=3.5÷4
=0.875~0.88
The frequency of the waves is (d) 0.88 Hz. So, correct answer is option (d).
What is frequency of wave?The frequency of a sinusoidal wave is the number of full oscillations performed by any wave constituent in a unit of time. According to the definition of frequency, if a body is moving periodically, it has completed one cycle after going through a number of situations or postures and then returning to its initial position. Therefore, frequency is a quantity that describes the rate of oscillation and vibration.
Given parameter,
Velocity of the waves = 3.5 m/s.
distance between the crests of each wave, that is, wavelength of the waves = 4 m.
We know that, for a wave transmission,
velocity of wave =frequency of wave × wavelength of wave
⇒ frequency of wave=speed of wave/wavelength of wave
⇒ frequency of wave =3.5 m/s ÷4m
=0.875 Hz
≈ 0.88 Hz
Hence, the frequency of the waves is 0.88 Hz.
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The spring is unstretched at the position x = 0. under the action of a force p, the cart moves from the initial position x1 = -8 in. to the final position x2 = 5 in. determine (a) the work done on the cart by the spring and (b) the work done on the cart by its weight.
This question is incomplete, the missing diagram is uploaded along this Answer below.
Answer:
a) the work done on the cart by the spring is 4.875 lb-ft
b) the work done on the cart by its weight is - 3.935 lb-ft
Explanation:
Given the data in the question;
(a) determine the work done on the cart by the spring
we calculate the work done on the cart by the spring as follows;
[tex]W_{spring}[/tex] = 1/2×k( [tex]x^{2} _{1}[/tex] - [tex]x^{2} _{2}[/tex] )
where k is spring constant ( 3 lb/in )
we substitute
[tex]W_{spring}[/tex] = 1/2 × 3( (-8)² - (5)² )
[tex]W_{spring}[/tex] = 1/2 × 3( 64 - 25 )
[tex]W_{spring}[/tex] = 1/2 × 3( 39 )
[tex]W_{spring}[/tex] = 58.5 lb-in
we convert to pound force-foot
[tex]W_{spring}[/tex] = 58.5 × 0.0833333 lb-ft
[tex]W_{spring}[/tex] = 4.875 lb-ft
Therefore, the work done on the cart by the spring is 4.875 lb-ft
b) the work done on the cart by its weight
work done by its weight;
[tex]W_{gravity}[/tex] = -mgsin∅( x₂ - x₁ )
we substitute in of values from the image below;
[tex]W_{gravity}[/tex] = -14 × sin(15°)( 5 - (-8) )
[tex]W_{gravity}[/tex] = -14 × 0.2588 × 13
[tex]W_{gravity}[/tex] = -47.1 lb-in
we convert to pound force-foot
[tex]W_{gravity}[/tex] = -47.1 × 0.0833333 lb-ft
[tex]W_{gravity}[/tex] = - 3.935 lb-ft
Therefore, the work done on the cart by its weight is - 3.935 lb-ft
a) the work done on the cart by the spring is 4.875 lb-ft.
b) the work done on the cart by its weight is - 3.935 lb-ft.
Calculation of the work done:a. The work done on the cart by the spring is
= 1/2 × 3( (-8)² - (5)² )
= 1/2 × 3( 64 - 25 )
= 1/2 × 3( 39 )
= 58.5 lb-in
Now we have to convert to pound force-foot
So,
= 58.5 × 0.0833333 lb-ft
= 4.875 lb-ft
b) Now
work done by its weight;
= -mgsin∅( x₂ - x₁ )
So,
= -14 × sin(15°)( 5 - (-8) )
= -14 × 0.2588 × 13
= -47.1 lb-in
Now we convert to pound force-foot
= -47.1 × 0.0833333 lb-ft
= - 3.935 lb-ft
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How does Physics help you as a student?
Answer:
The goal of physics is to understand how things work from first principles. ... Courses in physics reveal the mathematical beauty of the universe at scales ranging from subatomic to cosmological. Studying physics strengthens quantitative reasoning and problem solving skills that are valuable in areas beyond physics
Answer:
you get to understand why things happen this way
Explanation:
for example, are you not curious about why when standing in the bus and when the bus stops, you will might feel like you are going to fall ,
why does this happen because....
newton's laws explains it,
inertia causes you to be reluctant to change your initial state of motion due to your mass so you fall because you are still moving at the 'speed of the bus ' , something in like that
hope this helps,
please mark also
A book is moved once around the perimeter of a tabletop with dimensions 1.2 m x 1.8 m.
(a) If the book ends up at its initial position, what is its displacement
(b) What is the distance traveled?
Answer:
a) 0m
b) 6m
Explanation:
First, we need to remember:
Displacement: Difference between final and initial position.
Distance traveled: Total distance traveled.
a) If the final position is the same as the initial position, then:
final position = initial position
And we know that:
displacement = final position - initial position = 0
Then the displacement of the book is zero.
b)
We can assume that the book traveled along the perimeter of the table.
The table is a rectangle of width 1.2m and length 1.8m
Remember that for a rectangle of width W and length L, the perimeter is:
P = 2*L + 2*W
Then the perimeter of the table is:
P = 2*1.2m + 2*1.8m = 6m
This means that the distance traveled by the book is 6 meters.
If an object has a mass of 210g and the net force acting upon it is 5.0N, what is the acceleration of that object?
Answer:
24ms^-2 (2 sig figs)
Explanation:
F = ma
This means that the sum of all forces (or the net force) acting upon an object is equal to its mass x accelleration.
Its important to convert all parts of this equations in to SI units such that Force (N), Mass (kg) and Accelleration (ms^-2) to make sure that your answer is in the correct units.
F = ma so 5 = 0.21a
a = 5/0.21
a = 23.810 ms^-2 (5 sig figs)
If the magnitude of vector A⃗ is less than the magnitude of vectorB⃗ , then the x component of A⃗ is less than the x component ofB⃗ . If the magnitude of vector is less than the magnitude of vector, then the component of is less than the component of.
a. True
b. False
A bar magnet is attached solidly to a frictionless surface and its length is aligned with the x axis.To the right of the first magnet a short distance away is a second bar magnet with its center placed on the x axis and its length perpendicular to the x axis.The second magnet is free to move.Once placed in position at rest,which best describes the initial motion of the second magnet?
a. The magnet will move away from the fixed magnet.
b. The magnet will not move.
c. The magnet will move toward the fixed magnet.
d. The magnet will start to rotate.
Answer:
the correct answer is D
Explanation:
For this exercise we must remember that the poles of the same if not repel each other and the poles of different signs attract.
With this, let's analyze the situation presented.
The two magnets are perpendicular, with the second magnet to the right of the first.
We have two cases:
* first magnet with the north pole to the right
If the north of the second magnet is upwards there is a repulsion and the south pole of the second magnet there is an attraction with the north pole of the first magnet, so it would have a force that has to rotate the second magnet.
The force with the south pole of the first magnet that is at a greater distance is less, so the resultant of force is determined by the nearest poles.
If the poles of the second magnet are reversed, that is, the South pole up and the North pole down, the same result is obtained, but with a twist in the opposite direction.
* first magnet with the south pole to the right
Repels with the south pole of the second magnet and attracts with the north of the second magnet
therefore in both possibilities the second magnet acquires a rotational movement
Consequently the correct answer is D
What type of bond is CO2?||
Answer:
Lol
Explanation: