Answer:
-i - 7j
Explanation:
The computation of components of the force is shown below:-
torque T = r cross F
T = (4.00 i + 5.00 j + 0 k) X (1.00 i + 7.00 j)
Now we will cross multiplying vectorilly
T = 4 × (iXi) + 28 × (jXi) + 0 + 5 × (iXj)35 × (jXj) + 0
T = 4 × 0 + 28 × (-k) + 5 × (k) + 35 × 0
T = 28 × (-k) + 5 × (k) = -23k
net torque |T| = 23 N - m
direction >> negative k
Or Simply we can do by the below method
r × f
28 - 5 = 23 K
-i - 7j
In a high school swim competition, a student takes 1.6 s to complete 1.5 somersaults. Determine the average angular speed of the diver, in rad/s, during this time interval.
Answer:
The angular speed is [tex]w = 5.89 \ rad/s[/tex]
Explanation:
From the question we are told that
The time taken is [tex]t = 1.6 s[/tex]
The number of somersaults is n = 1.5
The total angular displacement during the somersault is mathematically represented as
[tex]\theta = n * 2 * \pi[/tex]
substituting values
[tex]\theta = 1.5 * 2 * 3.142[/tex]
[tex]\theta = 9.426 \ rad[/tex]
The angular speed is mathematically represented as
[tex]w = \frac{\theta }{t}[/tex]
substituting values
[tex]w = \frac{9.426}{1.6}[/tex]
[tex]w = 5.89 \ rad/s[/tex]
In the photoelectric effect, if the intensity of light shone on a metal increases, what will happen?
Answer:
C) There will be more electrons ejected
Explanation:
The number of electrons ejected whenever a photoelectric effect is identified it is proportional to the intensity of the incident light
Nevertheless, the photoelectrons' maximal kinetic energy is independent of their light intensity
Therefore, the maximum speed of the electron ejected doesn't really depend on the light intensity
So, if the intensity rises, only the number of electrons ejected will rised
Therefore the option c is correct
Answer:
C) There will be more electrons ejected
Explanation:
In the photoelectric effect, photons with an energy of E are shone upon a piece of metal, and if the energy of the photons overcome the work function ϕ of the metal, then electrons with will be ejected from the metal with a kinetic energy KE.
E_photon = Φ + KE
Each photon is capable of ejecting one electron from the metal. Therefore, increasing the intensity of the light (the number of photons shone on the metal) will increase the number of electrons ejected from the metal.
A projectile is fired at time t = 0.0 s from point o at the edge of a cliff, with initial velocity components of Vox = 30 m/s and Voy = 100 m/s. The projectile rises, and then falls into the sea
at point P. The time of flight of the projectile is 25 s. Assume air resistance is negligible.
t
What is the height of the cliff?
560 m
450 m
780 m
400 m
640 m
Answer:
It would be 450 or 640. My final answer would be about 450
Explanation: Because it would't be to high if it was shot Voy = 100
btw i think i know what i know what i am talking about.
The answer would be about 450 m.
What peak is considered a cliff?The top isn't the standard for a cliff to be reckoned as a cliff as such. Any steep rock face particularly at the edge of the sea can be specified as a cliff.
A 'clifftop' just refers to any pinnacle of a cliff. A 'plateau' is any flat extended geologic floor. An 'overhang' is a part of a structure or formation that protrudes from the primary frame and rests such that it is 'overhanging' the ground (striking above it).
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A standard 1 kilogram weight is a cylinder 48.5 mm in height and 49.0 mm in diameter. What is the density of the material? kg/m3
Answer:
Density = 10,933.93 kg/m^3
the density of the material is 10,933.93 kg/m^3
Explanation:
Density is the mass per unit volume
Density = mass/volume = m/V
Volume of a cylinder V = πr^2 h
Given;
Height h = 48.5mm = 0.0485 m
Radius r = diameter/2 = 49mm÷2 = 24.5mm = 0.0245m
Substituting the values;
Volume V = π×(0.0245^2)×0.0485
V = 0.000091458438030 m^3
V = 0.000091458 m^3
The mass is given as;
Mass = 1 kg
So, the density can be calculated as;
Density = 1/0.000091458
Density = 10933.92825785 kg/m^3
Density = 10,933.93 kg/m^3
the density of the material is 10,933.93 kg/m^3
A small cylinder made of a diamagnetic material is brought near a bar magnet. One end of the cylinder is brought near the North pole of the magnet and is repelled. What happens when the other end of the cylinder is brought near the North pole of the magnet? Group of answer choices
Answer:
it attracts
Explanation:
since in a magnetic body there are two poles
(north and south poles)if the first pole was repeled when brought near the North Pole therefore the other end is going to attarct because the first end was also a North Pole while the second end will be a south pole
A vector quantity has direction, a scalar quantity does not.
Explanation:
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Which compound is composed of oppositely charged ions?
Answer:
Option A. Li2O
Explanation:
To know which of the compound contains oppositely charged ions, let us determine the nature of each compound. This is illustrated below:
Li2O is an ionic compound as it contains a metal (Lithium, Li) and non metal (oxygen, O). Ionic compounds are charactized by the presence of aggregate positive and negative charge ions. This is true because they are formed by the transfer of electron(s) from the metallic atom to the non-metallic atom.
2Li —> 2Li^+ + 2e
O2 + 2e —> O^2-
2Li + O2 + 2e —> 2Li^+ + O^2- + 2e
2Li + O2 —> 2Li^+ O^2- —> Li2O
OF2 is a covalent compound as it contains non metals only (i.e oxygen, O and fluorine, F). Covalent compounds are characterised by the presence of molecules. This is true because they are formed from the sharing of electron(s) between the atoms involved.
PH3 is a covalent compound as it contains non metals only (i.e phosphorus, P and hydrogen, H).
SCl2 is a covalent compound as it contains non metals only (i.e sulphur, S and chlorine, Cl).
From the above information, we can see that only Li2O contains oppositely charged ions.
Answer:
A
Explanation:
Just took the test
Suppose the ring rotates once every 4.30 s . If a rider's mass is 53.0 kg , with how much force does the ring push on her at the top of the ride?
The complete question is;
In an amusement park ride called The Roundup, passengers stand inside a 16-m-diameter rotating ring. After the ring has acquired sufficient speed, it tilts into a vertical plane.
Suppose the ring rotates once every 4.30 s . If a rider's mass is 53.0 kg , with how much force does the ring push on her at the top of the ride?
Answer:
F_top = 385.36 N
Explanation:
We are given;
mass;m = 52 kg
Time;t = 4.3 s
Diameter;d = 16m
So,Radius;r = 16/2 = 8m
The formula for the centrifugal force is given as;
F_c = mω²R
Where;
R = radius
Angular velocity;ω = 2πf
f = frequency = 1/t = 1/4.3 Hz
F_c = 53 × (2π × 1/4.3)² × 8 = 905.29 N.
The force at top would be;
F_top = F_c - mg
F_top = 905.29 - (9.81 × 53) N
F_top = 385.36 N
The force at the top of ride will be "385.36 N".
Force and mass:According to the question,
Rider's mass, m = 52 kg
Time, t = 4.3 s
Diameter, d = 16 m
Radius, r = [tex]\frac{16}{2}[/tex] = 8 m
Frequency, f = [tex]\frac{1}{t}[/tex] = [tex]\frac{1}{4.3}[/tex] Hz
We know the formula,
Centrifugal force, [tex]F_c[/tex] = mω²R
or,
Angular velocity, ω = 2πf
By substituting the values in the above formula,
[tex]F_c = 53(2\pi \times (\frac{1}{4.3})^2\times 8 )[/tex]
[tex]= 905.29[/tex] N
hence,
The top force will be:
→ [tex]F_{top} = F_c[/tex] - mg
By substituting the values,
[tex]= 905.29-(9.81\times 53)[/tex]
[tex]= 385.36[/tex] N
Thus the above response is correct.
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A 90.0-kg ice hockey player hits a 0.150-kg puck, giving the puck a velocity of 45.0 m/s. If both are initially at rest and if the ice is frictionless, how far does the player recoil in the time it takes the puck to reach the goal 15.0 m away
Answer:
0.0241 m
Explanation:
mass of the hockey player m1 = 90 kg
mass of puck m2 = 0.150 kg
puck velocity v1= 45 m/s
distance traveled by puck to reach the goal =15.0 m.
now accoding to momentum conservation law
90×45+0.15×v2 = 0 [ since, If both are initially at rest and if the ice is frictionless,]
therefore, v2= -0.0725 m/s.
Now time taken by the puck to reach the goal
t= 15/45 = 1/3 sec.
therefore, how far does the player recoil in the time
=0.0725×1/3= 0.0241 m.
the distance travelled by the player( recoil ) in the time the puck reach the goal is 0.025m.
Given the data in the question
Mass of the player; [tex]m_1 = 90.0kg[/tex]Mass of puck; [tex]m = 0.150kg[/tex]Since they were both at rest initially
Initial velocity of player; [tex]u_1 = 0[/tex]Initial velocity of puck; [tex]u = 0[/tex]Velocity of player after the hit; [tex]v_1 = \ ?[/tex]Velocity of puck after the hit; [tex]v = 45.0m/s[/tex]Distance to the goal; [tex]s = 15.0m[/tex]Using conservation of liner momentum:
[tex]mu + m_1u_1 = mv+ m_1v_1[/tex]
Now, Since they were both at rest initially
[tex]0 = mv+ m_1v_1[/tex]
We substitute in our values to find the velocity of the player after the hit ( recoil velocity )
[tex]0 =[ 0.150kg * 45.0m/s ] + [ 90.0kg * v_1 ]\\\\0 = 6.75kg.m/s + [ 90.0kg * v_1 ]\\\\90.0kg * v_1 = -6.75kg.m/s \\\\v_1 = -\frac{6.75kg.m/s}{90.0kg} \\\\v_1 =- 0.075m/s[/tex]
{ The negative sign shows that the velocity of both the player and the puck are in opposite direction }
Hence, recoil velocity of the player is 0.075m/s
Now, we determine the time taken for the puck to trach the goal using the relation between distance, velocity and time .
Time = Distance / Velocity
We substitute our values into the expression
[tex]t = \frac{s}{v} \\\\t = \frac{15.0m}{45m/s} \\\\t = 0.3333s[/tex]
Hence, the time taken for the puck to reach the goal is 0.3333 seconds.
Next, we determine the distance travelled by the player( recoil ) in the time the puck reach the goal using the relation between distance, velocity and time .
Time = Distance / Velocity
We substitute in our values
[tex]t = \frac{s}{v}\\\\0.3333s = \frac{s}{0.075m/s} \\\\s = 0.3333s * 0.075m/s\\\\s = 0.025m[/tex]
Therefore, the distance travelled by the player( recoil ) in the time the puck reach the goal is 0.025m.
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If a negative point charge is placed at P without moving the original charges, the net electrical force the charges ±Q will exert on it is
Answer:
The particle P moves directly upwards
Explanation:
Lets designate the negative point charge at point P as particle P
The +Q charge will exert an attractive force on the particle P.
The -Q charge will exert a repulsive force on the particle P
The +Q charge exerts an upwards and leftward force on particle P
The -Q charge exerts an upwards and rightward force on particle P
Since the charges are equidistant from the particle P, and are of equal magnitude, the rightward force and the leftward force will cancel out, leaving just the upward force on the particle P.
The effect of the upward force is that the particle P moves directly upwards
Identify the term used to describe the ability of a liquid to flow against gravity up a narrow tube.
Answer:
This would be capillary action.
Explanation:
The physics behind it is gravity adhesion. The forces that attract between dissimilar molecules or atoms, in our case the contact area between the particles of the liquid and the particles forming the tube.
Faraday's Law states that the negative of the time rate of change of the flux of the magnetic field through a surface is equal to which of the following quantities?
a. The flux of the magnetic field through a surface which has the loop as its boundary.
b. The negative of the time rate of change of the flux of the magnetic field through a surface which has the loop as its boundary.
c. The line integral of the magnetic field around the closed loop.
d. The flux of the electric field through a surface which has the loop as its boundary.
Answer:
(C). The line integral of the magnetic field around a closed loop
Explanation:
Faraday's law states that induced emf is directly proportional to the time rate of change of magnetic flux.
This can be written mathematically as;
[tex]EMF = -\frac{\delta \phi _B}{\delta t}[/tex]
[tex](\frac{\delta \phi _B}{\delta t} )[/tex] is the rate of change of the magnetic flux through a surface bounded by the loop.
ΔФ = BA
where;
ΔФ is change in flux
B is the magnetic field
A is the area of the loop
Thus, according to Faraday's law of electric generators
∫BdL = [tex]\frac{\delta \phi _B}{\delta t}[/tex] = EMF
Therefore, the line integral of the magnetic field around a closed loop is equal to the negative of the rate of change of the magnetic flux through the area enclosed by the loop.
The correct option is "C"
(C). The line integral of the magnetic field around a closed loop
Faraday's Law states that the negative of the time rate of change of the flux of the magnetic field through a surface is equal to: D. The flux of the electric field through a surface which has the loop as its boundary.
In Physics, the surface integral with respect to the normal component of a magnetic field over a surface is the magnetic flux through that surface and it is typically denoted by the symbol [tex]\phi[/tex].
Faraday's Law states that the negative of the time rate of change ([tex]\Delta t)[/tex] of the flux of the magnetic field ([tex]\phi[/tex]) through a surface is directly proportional to the flux ([tex]\phi[/tex]) of the electric field through a surface which has the loop as its boundary.
Mathematically, Faraday's Law is given by the formula:
[tex]E.m.f = -N\frac{\Delta \phi}{\Delta t}[/tex]
Where:
N is the number of turns.Read more: https://brainly.com/question/15121836
Find the ratio of the gravitational force between two planets if the masses of both planets are quadrupled but the distance between them stays the same.
Answer:
The ratio of the new force over the original force is 16
Explanation:
Recall the formula for the gravitational force between two masses M1 and M2 separated a distance D:
[tex]F_G=G\,\frac{M_1\,\,M_2}{D^2}[/tex]
So now, if the masses M1 and M2 are quadrupled and the distance stays the same, the new force becomes:
[tex]F'_G=G\,\frac{4M_1\,\,4M_2}{D^2}=G\,\frac{16\,\,M_1\,\,M_2}{D^2}=16\,\,G\,\frac{M_1\,\,M_2}{D^2}= 16\,\,F_G[/tex]
which is 16 times the original force.
So the ratio of the new force over the original force is 16
The ratio of the gravitational force between two planets if the masses of both planets are quadrupled but the distance between them stays the same is 16:1.
What does Newton's law of gravitation state?Newton's law of gravitation states that any particle of matter in the universe attracts any other with a force varying directly as the product of the masses and inversely as the square of the distance between them.
The formula for Newton's law of gravitation is:
[tex]F = G \frac{m_1m_2}{r^{2} }[/tex]
where,
F is the gravitational force.G is the gravitational constant.m₁ and m₂ are the masses of both objects.r is the distance between the objects.The initial force between the planets is:
[tex]F_1 = G \frac{m_1m_2}{r^{2} }[/tex]
The force between the planets if the masses of both planets are quadrupled but the distance between them stays the same is:
[tex]F_2 = G \frac{4m_14m_2}{r^{2} } = 16 G \frac{m_1m_2}{r^{2} }[/tex]
The ratio of F₂ to F₁ is:
[tex]\frac{F_2}{F_1} =\frac{16 G \frac{m_1m_2}{r^{2} }}{G \frac{m_1m_2}{r^{2} }} = \frac{16}{1}[/tex]
The ratio of the gravitational force between two planets if the masses of both planets are quadrupled but the distance between them stays the same is 16:1.
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A single-turn circular loop of radius 6 cm is to produce a field at its center that will just cancel the earth's field of magnitude 0.7 G directed at 70 below the horizontal north direction. Find the current in the loop.
Answer:
The current is [tex]I = 6.68 \ A[/tex]
Explanation:
From the question we are told that
The radius of the loop is [tex]r = 6 \ cm = 0.06 \ m[/tex]
The earth's magnetic field is [tex]B_e = 0.7G= 0.7 G * \frac{1*10^{-4} T}{1 G} = 0.7 *10^{-4} T[/tex]
The number of turns is [tex]N =1[/tex]
Generally the magnetic field generated by the current in the loop is mathematically represented as
[tex]B = \frac{\mu_o * N * I}{2 r }[/tex]
Now for the earth's magnetic field to be canceled out the magnetic field generated by the loop must be equal to the magnetic field out the earth
[tex]B = B_e[/tex]
=> [tex]B_e = \frac{\mu_o * N * I }{ 2 * r}[/tex]
Where [tex]\mu[/tex] is the permeability of free space with value [tex]\mu _o = 4\pi * 10^{-7} N/A^2[/tex]
[tex]0.7 *10^{-4}= \frac{ 4\pi * 10^{-7} * 1 * I}{2 * 0.06}[/tex]
=> [tex]I = \frac{2 * 0.06 * 0.7 *10^{-4}}{ 4\pi * 10^{-7} * 1}[/tex]
[tex]I = 6.68 \ A[/tex]
The current in the loop will be "6.68 A".
Magnetic fieldAccording to the question,
Radius of loop, r = 6 cm or,
= 0.06 m
Earth's magnetic field, [tex]B_e[/tex] = 0.7 G or,
= 0.7 × [tex]\frac{1\times 10^{-4}}{1 G}[/tex]
= 0.7 × 10⁻⁴ T
Number of turns, N = 1
We know the relation,
→ B = [tex]\frac{\mu_0\times N\times I}{2r}[/tex]
or,
B = [tex]B_e[/tex]
then,
→ [tex]B_e[/tex] = [tex]\frac{\mu_0\times N\times I}{2r}[/tex]
By substituting the values,
0.7 × 10⁻⁴ = [tex]\frac{4 \pi\times 10^{-7}\times 1\times I}{2\times 0.06}[/tex]
hence,
The current will be:
I = [tex]\frac{2\times 0.06\times 0.7\times 10^{-4}}{4 \pi\times 10^{-7}\times 1}[/tex]
= 6.68 A
Thus the above approach is correct.
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Two cylindrical resistors are made from the same material. The shorter one has length L, diameter D, and resistance R1. The longer one has length 8L, diameter 4D, and resistance R2. How do the resistances of these two resistors compare
Answer:
the resistance of the longer one is twice as big as the resistance of the shorter one.
Explanation:
Given that :
For the shorter cylindrical resistor
Length = L
Diameter = D
Resistance = R1
For the longer cylindrical resistor
Length = 8L
Diameter = 4D
Resistance = R2
So;
We all know that the resistance of a given material can be determined by using the formula :
[tex]R = \dfrac{\rho L }{A}[/tex]
where;
A = πr²
[tex]R = \dfrac{\rho L }{\pi r ^2}[/tex]
For the shorter cylindrical resistor ; we have:
[tex]R = \dfrac{\rho L }{\pi r ^2}[/tex]
since 2 r = D
[tex]R = \dfrac{\rho L }{\pi (\frac{2}{2 \ r}) ^2}[/tex]
[tex]R = \dfrac{ 4 \rho L }{\pi \ D ^2}[/tex]
For the longer cylindrical resistor ; we have:
[tex]R = \dfrac{\rho L }{\pi r ^2}[/tex]
since 2 r = D
[tex]R = \dfrac{ \rho (8 ) L }{\pi (\frac{2}{2 \ r}) ^2}[/tex]
[tex]R = \dfrac{32\rho L }{\pi \ (4 D) ^2}[/tex]
[tex]R = \dfrac{2\rho L }{\pi \ (D) ^2}[/tex]
Sp;we can equate the shorter cylindrical resistor to the longer cylindrical resistor as shown below :
[tex]\dfrac{R_s}{R_L} = \dfrac{ \dfrac{ 4 \rho L }{\pi \ D ^2}}{ \dfrac{2\rho L }{\pi \ (D) ^2}}[/tex]
[tex]\dfrac{R_s}{R_L} ={ \dfrac{ 4 \rho L }{\pi \ D ^2}}* { \dfrac {\pi \ (D) ^2} {2\rho L}}[/tex]
[tex]\dfrac{R_s}{R_L} =2[/tex]
[tex]{R_s}=2{R_L}[/tex]
Thus; the resistance of the longer one is twice as big as the resistance of the shorter one.
(5 pt) You tie a cord to a pail of water, and your swing the pail in a vertical circular 0.700 m. What is the minimum speed must you give the pail at the highest point of the circle if no water is to spill from it
Answer:
The minimum speed required is 2.62m/s
Explanation:
The value of gravitational acceleration = g = 9.81 m/s^2
Radius of the vertical circle = R = 0.7 m
Given the mass of the pail of water = m
The speed at the highest point of the circle = V
The centripetal force will be needed must be more than the weight of the pail of water in order to not spill water.
Below is the calculation:
[tex]\frac{mV^{2}}{R} = mg[/tex]
[tex]V = \sqrt{gR}[/tex]
[tex]V = \sqrt{9.81 \times 0.7}[/tex]
[tex]V = 2.62 m/s[/tex]
How many excess electrons must be distributed uniformly within the volume of an isolated plastic sphere 20.0 cm in diameter to produce an electric field of 1450 N/C just outside the surface of the sphere
Answer:
1.007 × 10^(10) electron
Explanation:
We are given;
Electric Field;E = 1450 N/C
Diameter;d = 20 cm = 0.2 m
So, Radius: r = 0.2/2 = 0.1 m
Formula for Electric field just outside the sphere is given by the formula;
E = kq/r²
Where;
E is the magnitude of the electric field. q is the magnitude of the point charge r is distance from the point charge
k is a constant with a value of 9 x 10^(9) N.m²/C²
Making q the subject, we have;
q = Er²/k
Thus,
q = 1450 × 0.1²/(9 × 10^(9))
q = 1.61 × 10^(-9) C
Now, total charge q is also given by the formula;
q = Ne
Where;
e is charge on electron which is 1.6 × 10^(-19)
N is number of excess electrons
Making N the formula, we have;
N = q/e
N = (1.61 × 10^(-9))/(1.6 × 10^(-19))
N = 1.007 × 10^(10) electron
A small ferryboat is 4.70 m wide and 6.10 m long. When a loaded truck pulls onto it, the boat sinks an additional 5.00 cm into the river. What is the weight of the truck
Answer:
M = 1433.5 kg
Explanation:
This exercise is solved using the Archimedean principle, which states that the hydrostatic thrust is equal to the weight of the desalinated liquid,
B = ρ g V
with the weight of the truck it is in equilibrium with the push, we use Newton's equilibrium condition
Σ F = 0
B-W = 0
B = W
body weight
W = M g
the volume is
V = l to h
rho_liquid g (l to h) = M g
M = rho_liquid l a h
we calculate
M = 1000 4.7 6.10 0.05
M = 1433.5 kg
Which of the following is not considered a behavior?
A. eating
B. anxiety
C. sleeping
D. crying
━━━━━━━☆☆━━━━━━━
▹ Answer
B. Anxiety
▹ Step-by-Step Explanation
Anxiety isn't a behavior since it's a feeling. Behavior and feeling are different things therefore, anxiety is the correct answer.
Hope this helps!
- CloutAnswers ❁
Brainliest is greatly appreciated!
━━━━━━━☆☆━━━━━━━
Eating, sleeping, and crying all are considered as behaviors. However, anxiety cannot be considered as a behavior because it is a feeling. Thus, the correct option is B.
What is Anxiety?Anxiety is an intense feeling of excessive, and persistent worry and the fear about everyday situations. This includes fast heart rate, rapid breathing, sweating, and feeling tired constantly may occur.
Behavior is the range of actions and mannerisms which are made by individuals, organisms, systems or the artificial entities in some environment. These systems can include other systems or organisms as well as the inanimate physical environment. Behaviors include eating, sleeping, and crying. Anxiety is not a behavior, it is a feeling.
Therefore, the correct option is B.
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A cylindrical shell of radius 7.00 cm and length 2.21 m has its charge uniformly distributed on its curved surface. The magnitude of the electric field at a point 15.2 cm radially outward from its axis (measured from the midpoint of the shell) is 36.0 kN/C. (a) Find the net charge on the shell.
Answer:
The net charge on the shell is 30x10^-9C
Explanation:
Pls see attached file
Refer the attached photo
Answer:
A
Explanation:
since the wooden bat is an opaque object placed after a translucent object, light will come through the plastic sheet but will be unable to go through the bat. hence the dark shadow of the bat on a lit sheet
Find the displacement. Will give brainliest!
Answer:
1000 m upwards
Explanation:
Displacement Formula: Average Velocity = Displacement/Total Time
Simply plug in our known variables and solve:
100 m/s = x m/10 seconds
100 m/s(10 s) = x m
m = 1000
Angular velocity in the z direction of a flywheel is w(t)=A + Bt2 The numerical values of the constants are A=2.75 and B=1.50. What is the angular acceleration α(t) when t=0s and t=5.00s?
Answer:
α(0) = 0 rad/s²
α(5) = 15 rad/s²
Explanation:
The angular velocity of the flywheel is given as follows:
w(t) = A + B t²
where, A and B are constants.
Now, for the angular acceleration, we must take derivative of angular velocity with respect to time:
Angular Acceleration = α (t) = dw/dt
α(t) = (d/dt)(A + B t²)
α(t) = 2 B t
where,
B = 1.5
AT t = 0 s
α(0) = 2(1.5)(0)
α(0) = 0 rad/s²
AT t = 5 s
α(5) = 2(1.5)(5)
α(5) = 15 rad/s²
A piano tuner hears a beat every 2.20 s when listening to a 266.0 Hz tuning fork and a single piano string. What are the two possible frequencies (in Hz) of the string? (Give your answers to at least one decimal place.)
Answer:
The lower frequency is [tex]f_1 = 265.55 \ Hz[/tex]
The higher frequency is [tex]f_2 = 266.4546 \ Hz[/tex]
Explanation:
From the question we are told that
The period is [tex]T = 2.20 \ s[/tex]
The frequency of the tuning fork is [tex]f = 266.0 \ Hz[/tex]
Generally the beat frequency is mathematically represented as
[tex]f_b = \frac{1}{T}[/tex]
substituting values
[tex]f_b = \frac{1}{2.20}[/tex]
[tex]f_b = 0.4546 \ Hz[/tex]
Since the beat frequency is gotten from the beat produced by the tuning fork and and the string then
The possible frequency of the string ranges from
[tex]f_1 = f- f _b[/tex]
to
[tex]f_2 = f + f_b[/tex]
Now substituting values
[tex]f_1 = 266.0 - 0.4546[/tex]
[tex]f_1 = 265.55 \ Hz[/tex]
For [tex]f_2[/tex]
[tex]f_2 = 266 + 0.4546[/tex]
[tex]f_2 = 266.4546 \ Hz[/tex]
Four identical charges particles of charge 1Uc, 2Uc,
3Uc and 4Uc
are placed at x = lm, x=2m,
x=3m and
x=5m. The electric field intensity
at origin is?
Answer:
17.94 kN/C is the electric field intensity at the origin due to the charges.
Explanation:
From the question, we are told that
The distance of 1 μC from origin = 1 m
The distance of 2 μC from origin = 2 m
The distance of 3 μC from origin = 3 m
The distance of 4 μC from origin = 5 m
Therefore, for us to find the electric field intensity, we'll solve below:
The formula for Electric field intensity = ( k * q ) / ( r * r )
where , r is distance ,
k = 9 * 10^9 ,
and , q is charge .
now ,
electric field intensity at the origin = [ k * 10^(-6) / 1 * 1 ] +[ k * 2 * 10^(-6) / 2 * 2 ] + [ k * 3 * 10^(-6) / 3 * 3 ] + [ k * 4 * 10^(-6) / 5 * 5 ]
=> electric field intensity at the origin = k * 10^(-6) [ 1 + 1/2 + 1/3 + 4/25 ] N/C
=> electric field intensity at the origin = 9 * 10^9 * 10^(-6) * 1.99 N/C
=> electric field intensity at the origin = 17.94 kN/C
I attach a 4.1 kg block to a spring that obeys Hooke's law and supply 3.8 J of energy to stretch the spring. I release the block and it oscillates with a period of 0.13 s. What is the amplitude of oscillation
Answer:
The amplitude of the oscillation is 2.82 cm
Explanation:
Given;
mass of attached block, m = 4.1 kg
energy of the stretched spring, E = 3.8 J
period of oscillation, T = 0.13 s
First, determine the spring constant, k;
[tex]T = 2\pi \sqrt{\frac{m}{k} }[/tex]
where;
T is the period oscillation
m is mass of the spring
k is the spring constant
[tex]T = 2\pi \sqrt{\frac{m}{k} } \\\\k = \frac{m*4\pi ^2}{T^2} \\\\k = \frac{4.1*4*(3.142^2)}{(0.13^2)} \\\\k = 9580.088 \ N/m\\\\[/tex]
Now, determine the amplitude of oscillation, A;
[tex]E = \frac{1}{2} kA^2[/tex]
where;
E is the energy of the spring
k is the spring constant
A is the amplitude of the oscillation
[tex]E = \frac{1}{2} kA^2\\\\2E = kA^2\\\\A^2 = \frac{2E}{k} \\\\A = \sqrt{\frac{2E}{k} } \\\\A = \sqrt{\frac{2*3.8}{9580.088} }\\\\A = 0.0282 \ m\\\\A = 2.82 \ cm[/tex]
Therefore, the amplitude of the oscillation is 2.82 cm
A woman is standing at the rim of a nonuniform cylindrical horizontal platform initially at rest. The platform is free to rotate about frictionless orthogonal axle that goes through its center and has 4 m in diameter and moment of inertia of 500 kgm2. The woman then starts walking along the rim in clockwise direction at a constant speed of 1.50 m/s relative to the Earth. If a woman has 60 kg, how much work does she do to set herself and the platform into motion?
Answer:
e
Explanation:
i took it myself and got it right
Match each term to the best description. ::
1. Coherent
2. Diffraction
3. Grating
4. Interference
5. Specular dot
a. Composed of numerous narrowly spaced parallel slits or grooves
b. Having the same wavelength, frequency, and in-phase
c. Interaction of waves where they meet in space
d. The bending of waves near a boundary or as a wave passes through an opening
e. The zeroth order direct reflection fringe
is tantalum least reactive or more
Answer:
it is more reactive in high temperature than in low temperature.
A rod has length 0.900 mm and mass 0.500 kgkg and is pivoted at one end. The rod is not uniform; the center of mass of the rod is not at its center but is 0.500 mm from the pivot. The period of the rod's motion as a pendulum is 1.49 ss. What is the moment of inertia of the rod around the pivot
Answer:
The moment of inertia is [tex]I =0.14 \ kg \cdot m^2[/tex]
Explanation:
From the question we are told that
The length of the rod is [tex]l = 0.900 \ m[/tex]
The mass of the rod is [tex]m = 0.500 \ kg[/tex]
The distance of the center of mass from the pivot is [tex]d = 0.500 \ m[/tex]
The period of the rod's motion is [tex]T = 1.49 \ s[/tex]
Generally the period of the motion is mathematically represented as
[tex]T = 2 \pi * \sqrt{\frac{I}{m* g * d} }[/tex]
Where [tex]I[/tex] is the moment of inertia about the pivot so making [tex]I[/tex] the subject of formula
[tex]I = [\frac{T}{2\pi } ]^2 * m * g * d[/tex]
substituting values
[tex]I = [\frac{1.49}{2* 3.142 } ]^2 * 0.5 * 9.8 * 0.5[/tex]
[tex]I =0.14 \ kg \cdot m^2[/tex]