Answer: her rotational kinetic energy increases
Kinetic and
A brick is resting on a smooth wooden board that is at a 30° angle. What is one way to overcome the static friction that is holding the brick in place?
11 point)
O raise the board to a higher angle
O press down on the brick in a direction that is perpendicular to the board
O roughen up the texture of the wooden board
o lower the board so it's level with the ground
Answer:
raise the board to a higher angle
Explanation:
Static friction is the force opposite to the applied force.
Static friction is dependent on the angle of inclination, it means as the angle of incline increases, the force of friction will increases as normal force will decrease.
So, if the board will be raised to a higher angle, it will increase the angle of incline and will overcome the static friction and block will be able slide.
Hence, the correct option is "raise the board to a higher angle".
Vector has a magnitude of 6.0 m and points 30° north of east. Vector has a magnitude of 4.0 m and points 30° east of north. The resultant vector + is given by
Answer:
The resultant vector is [tex]\vec R = \vec A + \vec B = 7.196\,i + 6.464\,j[/tex].
Explanation:
First, each vector is determined in terms of absolute coordinates:
6-meter vector with direction: 30º north of east.
[tex]\vec A = (6\,m)\cdot (\cos30^{\circ} \,i + \sin 30^{\circ}\,j)[/tex]
[tex]\vec A = 5.196\,i + 3\,j[/tex]
4-meter vector with direction: 30º east of north.
[tex]\vec B = (4\,m)\cdot (\cos 60^{\circ}\,i + \sin 60^{\circ}\,j)[/tex]
[tex]\vec B = 2\,i + 3.464\,j[/tex]
The resultant vector is obtaining by sum of components:
[tex]\vec R = \vec A + \vec B = 7.196\,i + 6.464\,j[/tex]
The resultant vector is [tex]\vec R = \vec A + \vec B = 7.196\,i + 6.464\,j[/tex].
Exercise 2.4.5: Suppose we add possible friction to Exercise 2.4.4. Further, suppose you do not know the spring constant, but you have two reference weights 1 kg and 2 kg to calibrate your setup. You put each in motion on your spring and measure the frequency. For the 1 kg weight you measured 1.1 Hz, for the 2 kg weight you measured 0.8 Hz. a) Find k (spring constant) and c (damping constant). Find a formula for the mass in terms of the frequency in Hz. Note that there may be more than one possible mass for a given frequency. b) For an unknown object you measured 0.2 Hz, what is the mass of the object? Suppose that you know that the mass of the unknown object is more than a kilogram.
Answer:
a) k = 95.54 N / m, c = 19.55 , b) m₃ = 0.9078 kg
Explanation:
In a simple harmonic movement with friction, we can assume that this is provided by the speed
fr = -c v
when solving the system the angular value remains
w² = w₀² + (c / 2m)²
They give two conditions
1) m₁ = 1 kg
f₁ = 1.1 Hz
the angular velocity is related to frequency
w = 2π f₁
Let's find the angular velocity without friction is
w₂ = k / m₁
we substitute
(2π f₁)² = k / m₁ + (c / 2m₁)²
2) m₂ = 2 kg
f₂ = 0.8 Hz
(2π f₂)² = k / m₂ + (c / 2m₂)²
we have a system of two equations with two unknowns, so we can solve it
we solve (c / 2m)² is we equalize the expression
(2π f₁)² - k / m₁ = (2π f₂²) 2 - k / m₁
k (1 / m₂ - 1 / m₁) = 4π² (f₂² - f₁²)
k = 4π² (f₂² -f₁²) / (1 / m₂ - 1 / m₁)
a) Let's calculate
k = 4 π² (0.8² -1.1²) / (½ -1/1)
k = 39.4784 (1.21) / (-0.5)
k = 95.54 N / m
now we can find the constant of friction
(2π f₁) 2 = k / m₁ + (c / 2m₁)²
c2 = ((2π f₁)² - k / m₁) 4m₁²
c2 = (4ππ² f₁² - k / m₁) 4 m₁²
let's calculate
c² = (4π² 1,1² - 95,54 / 1) 4 1²
c² = (47.768885 - 95.54) 8
c² = -382.1689
c = 19.55
b) f₃ = 0.2 Hz
m₃ =?
(2πf₃)² = k / m₃ + (c / 2m₃) 2
we substitute the values
(4π² 0.2²) = 95.54 / m₃ + 382.1689 2/4 m₃²
1.579 = 95.54 / m₃ + 95.542225 / m₃²
let's call
x = 1 / m₃
x² = 1 / m₃²
- 1.579 + 95.54 x + 95.542225 x² = 0
60.5080 x² + 60.5080 x -1 = 0
x² + x - 1.65 10⁻² = 0
x = [1 ±√ (1- 4 (-1.65 10⁻²)] / 2
x = [1 ± 1.03] / 2
x₁ = 1.015 kg
x₂ = -0.015 kg
Since the mass must be positive we eliminate the second results
x₁ = 1 / m₃
m₃ = 1 / x₁
m₃ = 1 / 1.1015
A solenoid is designed to produce a magnetic field of 3.50×10^−2 T at its center. It has a radius of 1.80 cm and a length of 46.0 cm , and the wire can carry a maximum current of 13.0 A.
Required:
a. What minimum number of turns per unit length must the solenoid have?
b. What total length of wire is required?
Answer:
a. 2143 turns/m
b. 111.5 m
Explanation:
a. The minimum number of turns per unit length (N/L) can be found using the following equation:
[tex] B = \frac{\mu_{0}NI}{L} [/tex]
[tex] \frac{N}{L} = \frac{B}{\mu_{0}I} = \frac{3.50 \cdot 10^{-2} T}{4\pi \cdot 10^{-7} Tm/A*13.0 A} = 2143 turns/m [/tex]
Hence, the minimum number of turns per unit length is 2143 turns/m.
b. The total length of wire is the following:
[tex] N = 2143 turns/m*L = 2143 turns/m*46.0 \cdot 10^{-2} m = 986 turns [/tex]
Since each turn has length 2πr of wire, the total length is:
[tex] L_{T} = N*2\pi r = 986 turn*2*\pi*1.80 \cdot 10^{-2} m = 111.5 m [/tex]
Therefore, the total length of wire required is 111.5 m.
I hope it helps you!
Calculate the density of the following material.
1 kg helium with a volume of 5.587 m³
700 kg/m³
5.587 kg/m³
0.179 kg/m³
Answer:
[tex]density \: = \frac{mass}{volume} [/tex]
1 / 5.587 is equal to 0.179 kg/m³
Hope it helps:)
Answer:
The answer is
0.179 kg/m³Explanation:
Density of a substance is given by
[tex]Density \: = \frac{mass}{volume} [/tex]
From the
mass = 1 kg
volume = 5.583 m³
Substitute the values into the above formula
We have
[tex]Density \: = \frac{1 \: kg}{5.583 \: {m}^{3} } [/tex]
We have the final answer as
Density = 0.179 kg/m³Hope this helps you
Explain how blood circulation takes place in humans?
Blood comes into the right atrium from the body, moves into the right ventricle and is pushed into the pulmonary arteries in the lungs. After picking up oxygen, the blood travels back to the heart through the pulmonary veins into the left atrium, to the left ventricle and out to the body's tissues through the aorta.
Hope it helps you
Mark my answer as brainlist
have a nice day
What characteristic makes Biology a science, but not Art History?
Using a process of testing ideas and gathering evidence
o Writing books about the subject
O Having a college degree to study it
Discussing and sharing ideas
Answer:
Using a process of testing ideas and gathering evidence
Explanation:
Open the sash half way up, take the beaker containing the dry ice / water out of the hood, and slowly move it from right in front of the hood all the way down to the floor. At what point do the fumes stop getting sucked up by the fume hood?
Answer:
The fumes stop getting sucked up by the fume hood once the beaker is pulled out of the hood.
how much is need to lift a load of 100n placed at a distance of 29 cm from fulcrum if effort is applied at 60cm from the fulcrum on opposite side of the load? calculate mechanical advantage and velocity ratio of the lever
Answer:
206.8965517 n
Explanation:
First, we need to see that 60:29 is 2.078965517:1. Then we need to multiply the energy put 29 cm from the fulcrum by 2.078965517, giving us the end result of our answer.
A charge of uniform density (0.74 nC/m) is distributed along the x axis from the origin to the point x = 10 cm. What is the electric potential (relative to zero at infinity) at a point, x = 23 cm, on the x axis? Hint: Use Calculus to solve this problem.
Answer:
V = - 3.85 V
Explanation:
The electric potential of a continuous charge distribution is
V = k ∫ dq / r
to find charge differential let's use the concept of linear density
λ = dq / dx
dq = λ dx
the distance from a load element to the point of interest
x₀ = 23 cm = 0.23 m
r = √ (x-x₀)² = x - x₀
we substitute
v = k ∫ λ dx / (x-x₀)
we integrate and evaluate between x = 0 and x = l = 0.10 cm
V = k λ [ln (x-x₀) - ln (-x₀)]
V = k λ ln ((x-x₀) / x₀)
let's calculate
V = 9 10⁹ 0.74 10⁻⁹ ln ((0.23 - 0.10) / 0.23)
V = - 3.85 V
A third wire of the same material has the same length and twice the diameter as the first. How far will it be stretched by the same force
Complete question is;
A force stretches a wire by 0.60 mm. A second wire of the same material has the same cross section and twice the length.
a) How far will it be stretched by the same force?
b) A third wire of the same material has the same length and twice the diameter as the first. How far will it be stretched by the same force?
Answer:
0.15 mm
Explanation:
According to Hooke's Law,
E = Stress(σ)/Strain(ε)
Where E is youngs modulus
Formula for stress is;
Stress(σ) = Force(F)/Area(A)
Formula for strain is;
Strain(ε) = Change in length/original length = (Lf - Li)/Li
We are also told that a second wire of the same material has the same cross section and twice the length.
Thus;
Rearranging Hooke's Law to get the constants on one side, we have;
F/(AE) = ε
Thus from the conditions given;
ε1 = 0.6/Li
ε2 = (Change in length)/(2*Li)
And ε1 = ε2
Thus;
0.6/Li = Change in length/(2*Li)
Li will cancel out and we now have;
Change in length = 2 × 0.6 = 1.2 mm
Finally, we are told A third wire of the same material has the same length and twice the diameter as the first.
Area of a circle;A1 = πd²/4
Now, we are told d is doubled.
Thus, new area of the new circle is;
A2 = π(2d)²/4 = πd²
Rearranging Hooke's Law,we have;
F/A = εE
Since F and E are now constants, we have;
F/E = constant = Aε
Thus;
A1(ε1) = A2(ε2)
A1 = πd²/4
e1 = 0.60/Li
A2 = πd²
e2 = Change in length/Li
Thus;
((πd²/4) × 0.6)/Li = (πd² × Change in length)/ Li
Rearranging, Li and πd² will cancel out to give;
0.6/4 = Change in length
Change in length = 0.15 mm
Positive charge Q is placed on a conducting spherical shell with inner radius R1 and outer radius R2. The electric field at a point r < R1 is:
Answer:
E = 0 r <R₁
Explanation:
If we use Gauss's law
Ф = ∫ E. dA = [tex]q_{int}[/tex] / ε₀
in this case the charge is distributed throughout the spherical shell and as we are asked for the field for a radius smaller than the radius of the spherical shell, therefore, THERE ARE NO CHARGES INSIDE this surface.
Consequently by Gauss's law the electric field is ZERO
E = 0 r <R₁
Which columns are mislabeled?
Answer:
first order date and most recent order date
Explanation:
it was switched. column 5 should be most recent order date because it's 2020 while column 6 should be first order date because it was in 2019
The ability of a water strider to move along the surface of water without breaking the surface is due to:
Answer:
The ability of a water strider to move along the surface of water without breaking the surface is due to:
SURFACE TENSION
Explanation:
this is because Water molecules are more attracted to each other than they are to other materials, so they generate a force to stay together called surface tension. Which allows the strider to move without breaking the surface
An emf is induced in response to a change in magnetic field inside a loop of wire. Which of the following changes would increase the magnitude of the induced emf? A. Straighten the wire out to be flat B. Reduce the resistance of the wire of which the loop is made C. Turning the plane of the loop to be parallel to the magnetic field D. Reducing the diameter of the loop
Answer:
changing the magnetic field more rapidly
Explanation:
According to Faraday's law, whenever there is a change in the magnetic lines of force, it leads the production of induced emf. The magnitude of induced emf is proportional to to the rate of change of flux.
Hence if the magnetic field inside a loop of wire is changed rapidly, the magnitude of induced emf increases in accordance with Faraday's law of electromagnetic induction stated above when the magnetic field is changed more rapidly, hence the answer.
Light of wavelength 550 nm is incident on a slit having a width of 0.200 mm. The viewing screen is 1.90 m from the slit. Find the width of the central bright fringe
Answer:
The width of Center bright fringe is 10.2mm
Explanation:
Given that if
Y/ L << 1 then
Sin theta will be approx Y/L
So sin theta approx Y/L = lamda/a
Y= a x lambda/a
By substituting
1.9x 10^ -3m x 550*10^-9/ 0.2 x 10^-3m
= 5.2mm
But
Change in y = 2y = 10.4mm
In a double‑slit interference experiment, the wavelength is lambda=487 nm , the slit separation is d=0.200 mm , and the screen is D=48.0 cm away from the slits. What is the linear distance Δx between the eighth order maximum and the fourth order maximum on the screen?
Answer:
Δx = 4.68 x 10⁻³ m = 4.68 mm
Explanation:
The distance between the consecutive maxima, in Young's Double Slit Experiment is given bu the following formula:
Δx = λD/d
So, the distance between the eighth order maximum and the fourth order maximum on the screen will be given as:
Δx = 4λD/d
where,
Δx = distance between eighth order maximum and fourth order maximum=?
λ = wavelength = 487 nm = 4.87 x 10⁻⁷ m
d = slit separation = 0.2 mm = 2 x 10⁻⁴ m
D = Distance between slits and screen = 48 cm = 0.48 m
Therefore,
Δx = (4)(4.87 x 10⁻⁷ m)(0.48 m)/(2 x 10⁻⁴ m)
Δx = 4.68 x 10⁻³ m = 4.68 mm
An electron and a proton both moving at nonrelativistic speeds have the same de Broglie wavelength. Which of the following are also the same for the two particles?
(A) speed
(B) kinetic energy
(C) frequency
(D) momentum
Explanation:
The De-Broglie wavelength is given by :
[tex]\lambda=\dfrac{h}{p}[/tex]
h is Planck's constant
p is momentum
In this case, an electron and a proton both moving at nonrelativistic speeds have the same de Broglie wavelength. Mass of electron and proton is different. It means their velocity and energy are different.
Only momentum is the factor that remains same for both particles i.e. momentum.
In the Bohr model of the hydrogen atom, an electron in the 3rd excited state moves at a speed of 2.43 105 m/s in a circular path of radius 4.76 10-10 m. What is the effective current associated with this orbiting electron
Answer:
Current =,charge / time
Charge = e = 1.6E-19 coulombs
t = T time for 1 revolution (period)
v = S / T = distance traveled in 1 revolution / time for 1 revolution
T = S / v = 2 pi * 4.76E-10 / 2.43E5 = 1.23E-14
I = Q / T = 1.6E-19 / 1.23E-14 = 1.30E-5
An elderly sailor is shipwrecked on a desert island but manages to save his eyeglasses. The lens for one eye has a power of 1.28 diopters, and the other lens has a power of 8.50 diopters. What is the magnifying power of the telescope he can construct with these lenses
"How many wavelengths wide must a single slit be if, at a point 8o from the central maximum, there is a 72 rad phase difference between the top and bottom rays?"
Answer
82.3 wavelengths
A person of 70 kg standing on an un-deformable horizontal surface. She bends her knees and jumps up from rest, achieving a launching speed of 1.7 m/s. The launching process lasts 0.1 second. Calculate the average force exerted by the surface on the person during the launch.
Answer:
1190 N
Explanation:
Force: This can be defined as the product of mass and velocity. The unit of force is Newton(N).
From the question,
F = ma................. Equation 1
Where F = average force, m = mass, a = acceleration.
But,
a = (v-u)/t................ Equation 2
Where v = final velocity, u = initial velocity, t = time.
Substitute equation 2 into equation 1
F = m(v-u)/t.............. Equation 3
Given: m = 70 kg, v = 1.7 m/s, u = 0 m/s (from rest), t = 0.1 s.
Substitute into equation 3
F = 70(1.7-0)/0.1
F = 1190 N.
A bungee cord with a spring constant of 800 StartFraction N over m EndFraction stretches 6 meters at its greatest displacement. How much elastic potential energy does the bungee cord have? The bungee cord has J of elastic potential energy.
Explanation:
EE = ½ kx²
EE = ½ (800 N/m) (6 m)²
EE = 14,400 J
Answer:
14,400 J
Explanation:
Its the answer
) Calculate current passing in an electrical circuit if you know that the voltage is 8 volts and the resistance is 10 ohms
Explanation:
Hey, there!
Here, In question given that,
potential difference (V)= 8V
resistance (R)= 10 ohm
Now,
According to the Ohm's law,
V= R×I { where I = current}
or, I = V/R
or, I = 8/10
Therefore, current is 4/5 A or 0.8 A.
(A= ampere = unit of current).
Hope it helps...
A navigational beacon in deep space broadcasts at a radio frequency of 50 MHz. A spaceship approaches the beacon with a relative velocity of 0.40c. What is the frequency of the beacon radio signal that is detected on the ship? Answer in MHz with no decimal places.
A) 55 MHz
B) 60 MHz
C) 66 MHz
D) 71 MHz
E) 76 MHz
Answer:
To first order f' = f (1 - u/c) where observer and source are separating
f' = (1 + .4) = 70 MHz so (d) would be correct
The next term would be 1/2 (u/c)^2 = 1/2 * ,4^2 = .08 and the correction would be .08 * 50 = 4 Mhz
Most approximations would use (d).
Determine the value of the current in the solenoid so that the magnetic field at the center of the loop is zero tesla. Justify your answer.
Answer:
I will explain the concept of magnetic field and how it can be calculated.
Explanation:
The formula for magnetic field at the center of a loop is given as
B = μ[tex]_{o}[/tex]I / 2R
where B is the magnetic field
R is the radius of the loop
I is the current
and μ[tex]_{o}[/tex] is the magnetic permeability of free space which is a constant 4π × [tex]10^{-7}[/tex] newtons/ampere²
If the magnetic field at the center of the loop is 0, then μ[tex]_{o}[/tex]I = 0
I = 0 which means there will be no current flow in the loop.
Two wires carry current I1 = 73 A and I2 = 31 A in the opposite directions parallel to the x-axis at y1 = 3 cm and y2 = 13 cm. Where on the y-axis (in cm) is the magnetic field zero?
Answer:
The position on the y-axis where the magnetic field is zero is at y = 10 cm
Explanation:
The magnetic field B due to a long straight wire carrying a current, i at a distance R from the wire is given by
B = μ₀i/2πR
Now, let y be the point where the magnetic fields of both wires are equal.
So, the magnetic field due to wire 1 carrying current i₁ = 73 A is
B₁ = μ₀i₁/2π(y - 3) and
the magnetic field due to wire 2 carrying current i₂ = 31 A is
B₂ = μ₀i₂/2π(13 - y)
At the point where the magnetic field is zero, B₁ = B₂. So,
μ₀i₁/2π(y - 3) = μ₀i₂/2π(13 - y)
cancelling out μ₀ and 2π, we have
i₁/(x - y) = i₂/(13 - y)
cross-multiplying, we have
(13 - y)i₁ = (y - 3)i₂
Substituting the values of i₁ and i₂, we have
(13 - y)73 = (y - 3)31
949 - 73y = 31y - 93
Collecting like terms, we have
949 + 93 = 73y + 31y
1042 = 104y
dividing through by 104, we have
y = 1042/104
y = 10.02 cm
y ≅ 10 cm
So, the position on the y-axis where the magnetic field is zero is at y = 10 cm
Suppose a 58-turn coil lies in the plane of the page in a uniform magnetic field that is directed into the page. The coil originally has an area of 0.150 m2. It is stretched to have no area in 0.100 s. What is the magnitude (in V) and direction (as seen from above) of the average induced emf if the uniform magnetic field has a strength of 1.10 T? magnitude V direction ---Select--- †\
Answer:
95.7v
Explanation
Using Faraday's law of electromagnetic induction we know that rate of change in magnetic flux will induce EMF in closed loop
So it is given as
E= Ndစ/dt
E= N BA-0/ deta t
Given that
N = 58turns
B = 1.10T
A = 0.150m^²
Deta t= 0.1s
now we have
E = 58(1.10x0.150)/0.1
= 95.7v
Magnetic flux is decreasing, so the direction of the current will be to aid the decreasing flux $decrease= CLOCKWISE
Explanation:
A charged particle enters a magnetic field with an angle theta If theta equals 90 degrees what bath it will follow - If theta larger than zero and less than 90 degrees what path will it follow?
Given that,
A charged particle enters a magnetic field with an angle theta .
If [tex]\theta=90^{\circ}[/tex]
We know that,
If the angle is 90° then the charged particle enters perpendicular to the B.
B is magnetic field.
The charged particle will be follow of the circular path.
If the angle is greater than 0 and less than 90° then the charged particle will be show the helical path.
Hence, This is required answer.
A plastic balloon that has been rubbed with wool will stick to a wall.
a. Can you conclude that the wall is charged? If not, why not? If so, where does the charge come from?
b. Draw a series of charge diagrams showing how the balloon is held to the wall.
Answer:
Explanation:
When plastic balloon is rubbed with wool , charges are created on both balloon and silk in equal amount . Rubber balloon will acquire negative charge and silk will acquire positive charge .
Now when balloon is brought near a wall , there is induction of charge on the wall due to charge on the balloon . On the near surface of wall positive charge is produced and on the surface deep inside the wall negative charge is produced . The charge deep inside goes inside the earth but the positive charge near the surface of wall can not escape . It remains trapped by negative charge on the balloon .
hence there is mutual attraction between balloon and surface of wall is just like attraction between opposite charges . But once the ballon due to mutual attraction comes in contact with the wall , the charge on balloon and on wall neutralises each other and hence after some time the balloon falls off from the wall on the ground . It does not remain attracted to wall for ever . It happens due to neutralisation of charges on balloon and wall .