The ball's position in the air at time t is given by the vector,
p(t) = (60 i + 64 k) t + 1/2 (-6 j - 32 k) t ²
and its velocity is given by
v(t) = (60 i + 64 k) + (-6 j - 32 k) t
The ball is in the air for as long as it takes for the vertical (k) component of the position vector to reach 0, so we solve,
64 t - 32/2 t ² = 0 ==> t = 0 OR t = 4
and so the ball is in the air for 4 s.
After this time, the ball has position vector
p(4) = (60 i + 64 k) t + 1/2 (-6 j - 32 k) t ² = 240 i - 48 j
which has magnitude
||p(4)|| = √(240² + (-48)²) = 48 √26 ≈ 244.8 ft
in a direction θ in the x,y plane from the positive x axis such that
tanθ = -48/240 = -1/5 ==> θ = -arctan(1/5) ≈ -11.3º
or 11.3º South of East.
The ball hits the ground with speed
||v(4)|| = ||60 i - 24 j - 64 k|| = √(60² + (-24)² + (-64)²) = 4 √517 ≈ 91.0 ft/s
kinematic relationships can be used in all dimensions and finding the position and velocity when reaching the ground is
r = 244.8 ft, tea = 21.8º from East to South
v = 91.0 ft / s
given parameters
Initial velocity v = (60 i + 64 k) ft / sBody acceleration a = (-6 j - 32 k) ft / s²
to find
where it reaches the groundground speed
Kinematics allows finding the position, velocity and acceleration of the body, in this case we have a problem in three dimensions, where they establish a Cartesian coordinate system, a method to solve this exercise is to solve each component independently
a) The acceleration of gravity acts on the z axis, so let's find the time it takes to reach the ground, if the initial vertical velocity is v_{oz} = 64 ft/s and the acceleration is a_z = g = -32 ft / s², we assume that the ball leaves the ground (z₀ = 0)
z = z₀ + v_{oz} t + ½ a_z t²
when reaching the ground its height of zero and
0 = 0 + v_{oz} t + ½ a_z t²
t (v_{oz) + ½ a_z t) = 0
t (64 - 16 t) = 0
the solusion of this squadron is
y = 0
t = 4 s
the first time is when it leaves and the second time is for when it reaches the ground, therefore the flight time is t = 4s
with this time we find the displacement is each exercise
X axis
in this axis there is no acceleration, so we use the uniform motion relationships
vₓ = x / t
x = vₓ t
x = 60 4
x = 240 ft
Y Axis
on this axis there is an acceleration of a_y = -6 ft/s and an initial velocity v_{oy} = 0
we use the kinematic relation
y = v_{oy} t + ½ a_y t²
y = 0 - ½ 6 4²
y = - 48 ft
let's use the Pythagoras theorem to find the position
r² = (x -x₀) ² + (y -y₀) ² + (z -z₀) ²
r² = (240-0) ² + (-48-0) ² + (0-0) ²
r = 244.8 ft
We use trigonometry to find the direction
tan θ = y / x
θ = tan⁻¹ [tex]\frac{y}{x}[/tex]
θ = tan⁻¹ [tex]\frac{96}{240}[/tex]
θ = -21.8º
This angle is measured clockwise from the x axis, it can also be read
θ = 21.8º from East to South
b) Let's look for the speed when we hit the ground
X axis
vₓ = v_{ox} + aₓ t
vₓ = 60 - 0
vₓ = 60 ft / s
Y Axis
v_y = v_{oy} + a_y t
v_y = 0 - 6 4
v_y = -24 ft / s²
Z axis
v_z = v_{oz} + a_z t
v_z = 64 -32 4
v_z = -64 ft / s
With the Pytagoras theorem find the modulus of this speed is
v² = vx² + vy² + vz²
v² = 60² + 24 ² + 64²
v = 91.0 ft / s
In conclusion, the kinematic relationships can be used in all dimensions and finding the position and velocity when reaching the ground is
a) r = 244.8 ft, θ = 21.8º from East to South
b) v = 91.0 ft / s
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Select the correct answer.
Which person is vulnerable to identity theft?
A.
Beverley opens a line of credit to purchase a household appliance.
B.
Deborah fills out her income tax form and includes her Social Security number.
C. Josiah misses three monthly car loan payments in a row.
D.
Randell uses a computer at a public library to view his bank account online.
Reset
Next
Answer:
d
Explanation:
Randell uses a computer at a public library to view his bank account online represents one of the cases when a when person is vulnerable to identity theft, therefore the correct answer is option D.
What is identity theft?Identity theft occurs when criminals steal your confidential info and use it to create fresh accounts, rent or purchase property, file false financial records, or carry out other illegal activities.
This implies that a thief may steal sensitive data such as names, birthdates, Social Security numbers, information from driver's licenses, residences, and credit card or bank account numbers.
Once they have this information, they may use it to buy goods, apply for credit and debit cards, or even utilize it to pay for medical care with the assistance of health coverage.
Thus,the correct answer is option D.
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The carbon isotope C-14 is used for carbon dating. C-14 (mass 2.34 x 10-26 kg) decays by beta-decay, in which the nucleus emits an electron (the beta particle) and a subatomic particle called a neutrino. In one such decay, the electron and neutrino are emitted at right angles to each other. The electron (mass 9.11 x 10-31 kg) has a speed of 5.0 x 107 m/s and the neutrino has a momentum of 8.0 x 10-24 kg m/s. What is the recoil speed of the daughter nucleus?
Answer:
Explanation:
We shall apply the law of conservation of momentum to calculate the momentum and the speed of daughter nucleus .
Since the velocity of electron is very high we shall apply relativistic formula to calculate its momentum .
= [tex]\frac{mv}{\sqrt{1-(\frac{v}{c})^2 } }[/tex]
[tex]= \frac{9.11 \times 10^{-31}\times 5\times 10^7}{\sqrt{1-(\frac{5\times10^7}{3\times 10^8})^2 } }[/tex]
45.55 x 10⁻²⁴ x 1.176
= 53.58 x 10⁻²⁴ .
momentum of neutrino = 8 x 10⁻²⁴ . They are perpendicular to each other so total momentum
= √ [( 53.58 x 10⁻²⁴ )²+(8 x 10⁻²⁴)²]
= 54.17 x 10⁻²⁴
Hence the momentum of recoiled daughter nucleus will be same but in opposite direction
velocity of recoil = momentum / mass
= 54.17 x 10⁻²⁴ / 2.34 x 10⁻²⁶
= 23.15 x 10² m /s
An 80.0 kg man sits on a scale in his car. The car is driving at a speed of 11.0 m/s right as it passes over the top of a semicircular hill of radius 17.0 m. What does the scale read right when he is at the top
Answer:
F / g = 138 kg
Explanation:
For this exercise let's use Newton's second law
F- W = m a
the force is equal to the back of the balance
in this case the acceleration is centripetal
a = v² / r
we substitute
F - m g = m v² / r
F = m (g + v²/ r)
calculus
F = 80 (9.8 + 11²/17)
F = 1353 N
the balance reading is this value between gravity
F / g = 1353 / 9.8
F / g = 138 kg
A segment of wire of length L is along the x axis centered at x=0. Which of the following is a correct integral expression for the magnetic field at point P (centered on the wire segment at y=b) due to the current I flowing left to right in the segment of length L? In all answers below the limits of integration are from -L/2 to L/2.
a. μ0I/4π∫ dx b/(b2 + x2)3/2 kb. μ0I/4π∫ dx b/(b2 + x2)3/2 j c. μ0I/4π∫ dx /(b2 + x2) kd. -μ0I/4π∫ dx /(b2 + x2)1/2 ke. none of 1-5
Answer:
b. μ0I/4π∫ dx b/(b2 + x2)³/² j
Explanation:
The wire of length L centered at origin ( x =0 and y=0 ) carries current of I . We have to find magnetic field at point ( x = 0 , y = b ) .
First of all we shall consider magnetic field due to current element idx which is at x distance away from origin . magnetic field
dB = [tex]\frac{\mu _0idx}{4\pi(x^2+b^2)^2 }[/tex]
component of magnetic field along y- axis at point P
[tex]\frac{\mu _0idx}{4\pi(x^2+b^2)^2 }cos\theta[/tex]
where θ is angle between y - axis and dE .
component of magnetic field along y- axis at point P
[tex]\frac{\mu _0idx}{4\pi(x^2+b^2)^2 }\times \frac{b}{\sqrt{x^2+b^2} }[/tex]
[tex]\frac{\mu _0ibdx}{4\pi(x^2+b^2)^\frac{3}{2} }[/tex]
The same magnetic field will also exist due to current element dx at x distance away on negative x - axis
The perpendicular component will cancel out .
This is magnetic field dE due to small current element
Magnetic field due to whole wire
[tex]\int\limits^\frac{L}{2} _\frac{-L }{ 2 } }\frac{\mu _0ibdx}{4\pi(x^2+b^2)^\frac{3}{2} } \, dx[/tex]
What is the momentum of a 2 kg ball traveling at 2m/s
Answer:
4
Explanation:
p=m×v
m=2kg
v=2m/s
2×2=4
Answer:
4kgms/1
Explanation:
p=m×v
=2kg×2m/s
=4kgms/1
A planetary nebula is a(n) ____. Group of answer choices a shell of gas ejected by and expanding away from an extremely hot dying low-mass star a shell of gas ejected by and expanding away from an extremely hot dying high-mass star an expanding atmosphere of a low-mass star as it becomes a red giant a contracting shell of ionized interstellar medium absorbed by a dying low-mass star a contracting shell of dusty material from planets destroyed by a dying low-mass star
Answer: Planetary nebular is
A shell of gas ejected by and expanding away from an extremely hot dying high-mass star
An expanding atmosphere of a low-mass star as it becomes a red Giants.
Explanation:
Planetary nebular is a form of nebula emission that comprises an expanding and glowing shell of gas that is ejected from red giant stars.
Planetary nebulae play an important role in the chemical evolution of the Milky Way by expelling elements into the interstellar medium from stars where those elements were produced.
A ball is kicked horizontally with a speed of 5.0 ms-1 from the roof of a house 3 m high. When will the ball hit the ground?
Answer:
the time taken for the ball to hit the ground is 0.424 s
Explanation:
Given;
velocity of the ball, u = 5 m/s
height of the house which the ball was kicked, h = 3m
Apply kinematic equation;
h = ut + ¹/₂gt²
where;
h is height above ground
u is velocity
g is acceleration due to gravity
t is the time taken for the ball to hit the ground
Substitute the given values and solve for t
3 = 5t + ¹/₂(9.8)t²
3 = 5t + 4.9t²
4.9t² + 5t -3 = 0
a = 4.9, b = 5, c = -3
Solve for t using formula method
[tex]t = \frac{-5 +/-\sqrt{5^2-4(4.9*-3)}}{2(4.9)} = \frac{-5+/-(9.154)}{9.8} \\\\t = \frac{-5+9.154}{9.8} \ or \ \frac{-5-9.154}{9.8} \\\\t = \frac{4.154}{9.8} \ or \ \frac{-14.154}{9.8} \\\\t = 0.424 \ sec \ or -1.444 \ sec\\\\Thus, t = 0.424 \ sec[/tex]
What piece of equipment should be used to handle radioactive sources?
The brake in most cars makes use of a hydraulic system. This system consists of a fluid filled tube connected at each end to a piston. Assume that the piston attached to the brake pedal has a cross sectional area of 3 cm2 and the piston attached to the brake pad has a 2 cross section area of 15 cm . When you apply a force of 50 Newton to the piston attached to the brake pedal, how much will be the force at the brake pad
Answer:
The force at the brake pad = 250 N
Explanation:
The hydraulic brake system works on the Pascal's Principle for pressure transmission in fluids; the pressure applied to a fluid is transmitted undiminished in all directions.
For hydraulic systems, the pressure applied to the brake pedal is transmitted undiminished through the fluid filled tube, connected at each end to a piston, to the brake pad.
Hence, mathematically,
P(brake pedal) = P(break pad)
Pressure is given as the force applied divided by the cross sectional Area perpendicular to the direction of applied force.
P(brake pedal) = (Force applied on the brake pedal) ÷ (Cross Sectional Area of the brake pedal)
Force applied on the brake pedal = 50 N
Cross Sectional Area of the brake pedal = 3 cm²
P(brake pedal) = (50/3) = 16.67 N/cm²
P(brake pad) = P(brake pedal) = 16.67 N/cm²
P(brake pad) = (Force applied on the brake pad) ÷ (Cross Sectional Area of the brake pad)
Force applied on the brake pad = F = ?
Cross Sectional Area of the brake pad = 15 cm²
16.67 = (F/15)
F = 16.67 × 15 = 250 N
Hence, the force at the brake pad = 250 N
Hope this Helps!!!
If the circuit in the motorcycle runs off a standard 12 V battery, and one of the headlights is switched on, current flowing through the headlight is measured at 3.75 A. What is the power usage of the headlamp? Report the answer to two significant digits
Answer:
45 W
Explanation:
Power: This can be defined as the rate at which electrical energy is consumed in a circuit. The unit of power is Watt (W).
From the question,
The expression for power is given as,
P = VI.................. Equation 1
Where V = Voltage, I = current.
Given: V = 12 V, I = 3.75 A.
Substitute into equation 1.
P = 12(3.75)
P = 45 W.
Answer:
P = 45 Watt
Explanation:
The electrical power used by an electrical device or the electrical circuit is given by the following formulae:
P = VI
P = I²R
P = V²/R
where,
P = Electrical Power Consumed by the Device
V = The Voltage applied to the circuit or device
R = Resistance of device or circuit
I = Current passing through the device or circuit
We have the following data for our circuit:
V = 12 volts
I = 3.75 A
Therefore, it is clear from the data that we will use the first formula:
P = VI
P = (12 volts)(3.75 A)
P = 45 Watt
Two long, parallel, current-carrying wires lie in an xy-plane. The first wire lies on the line y = 0.340 m and carries a current of 27.5 A in the +x direction. The second wire lies along the x-axis. The wires exert attractive forces on each other, and the force per unit length on each wire is 295 µN/m. What is the y-value (in m) of the line in the xy-plane where the total magnetic field is zero?
Answer:
The y-value of the line in the xy-plane where the total magnetic field is zero [tex]U = 0.1355 \ m[/tex]
Explanation:
From the question we are told that
The distance of wire one from two along the y-axis is y = 0.340 m
The current on the first wire is [tex]I_1 = (27.5i) A[/tex]
The force per unit length on each wire is [tex]Z = 295 \mu N/m = 295*10^{-6} N/m[/tex]
Generally the force per unit length is mathematically represented as
[tex]Z = \frac{F}{l} = \frac{\mu_o I_1I_2}{2\pi y}[/tex]
=> [tex]\frac{\mu_o I_1I_2}{2\pi y} = 295[/tex]
Where [tex]\mu_o[/tex] is the permeability of free space with a constant value of [tex]\mu_o = 4\pi *10^{-7} \ N/A2[/tex]
substituting values
[tex]\frac{ 4\pi *10^{-7} 27.5 * I_2}{2\pi * 0.340} = 295 *10^{-6}[/tex]
=> [tex]I_2 = 18.23 \ A[/tex]
Let U denote the line in the xy-plane where the total magnetic field is zero
So
So the force per unit length of wire 2 from line U is equal to the force per unit length of wire 1 from line (y - U)
So
[tex]\frac{\mu_o I_2 }{2 \pi U} = \frac{\mu_o I_1 }{2 \pi(y - U) }[/tex]
substituting values
[tex]\frac{ 18.23 }{ U} = \frac{ 27.5 }{(0.34 - U) }[/tex]
[tex]6.198 -18.23U = 27.5U[/tex]
[tex]6.198=45.73U[/tex]
[tex]U = 0.1355 \ m[/tex]
A tank holds a 2.38-m thick layer of oil that floats on a 1.24-m thick layer of brine. Both liquids are clear and do not intermix. Point O is at the bottom of the tank, on a vertical axis. The indices of refraction of the oil and the brine are 1.27 and 1.81, respectively. A ray originating at O reaches the brine-oil interface at the critical angle. What is the distance of this point from the axis?
Answer:
1.22m
Explanation:
Since
sinθ = refraction-of-the-oil/refraction-of-the-brine =1.27/1.81 = 0.702
θ = [tex]sin^-^1[/tex](0.702)
Hence
Critical angle = θ = 44.58°
tan(θ) = d/1.24
tan(44.58°) = d/1.24
Hence, 0.98 = d/1.24
The distance d = 0.98 x 1.24 = 1.22m
Give 2 examples for Newton’s first law of motion.
please answer this question!!!!!
dose sound travel faster in a warm room or a cold room? explain your answer
Answer:Sound travel faster in warm room.
Explanation:The speed of sound depends on the temperature of the medium. Mathematically, the relation between the speed of the sound and the temperature is give by:v=
is the ratio of the specific heats
R is the gas constant
T is the temperature of the medium
We know that the temperature of the warm room is more as compared to the cold room.
So, it is clear that the sound travel faster in a warm room. The particles move faster when the temperature is high.
Two circular coils are concentric and lie in the same plane. The inner coil contains 110 turns of wire, has a radius of 0.014 m, and carries a current of 9.0 A. The outer coil contains 160 turns and has a radius of 0.022 m. What must be the magnitude of the current in the outer coil, such that the net magnetic field at the common center of the two coils is zero?
Answer:
The current flowing through the outer coils is
Explanation:
From the question we are told that
The number of turn of inner coil is [tex]N _i = 110 \ turns[/tex]
The radius of inner coil is [tex]r_i = 0.014 \ m[/tex]
The current flowing through the inner coil is [tex]I_i = 9.0 \ A[/tex]
The number of turn of outer coil is [tex]N_o = 160 \ turns[/tex]
The radius of outer coil is [tex]r_o = 0.022\ m[/tex]
For net magnetic field at the common center of the two coils to be zero the current flowing in the outer coil must be opposite to current flowing inner coil
The magnetic field due to inner coils is mathematically represented as
[tex]B_i = \frac{N_i \mu I}{2 r_i}[/tex]
The magnetic field due to inner coils is mathematically represented as
[tex]B_o = \frac{N_o \mu I_o}{2 r_o}[/tex]
Now for magnetic field at center to be zero
[tex]B_o = B_i[/tex]
So
[tex]\frac{N_i \mu I_i}{2 r_i} = \frac{N_o \mu I_o}{2 r_o}[/tex]
=> [tex]\frac{110 * 9}{2 * 0.014} = \frac{160 *I_o}{2 0.022}[/tex]
[tex]I_o = 9.72 \ A[/tex]
Two blocks can collide in a one-dimensional collision. The block on the left hass a mass of 0.30 kg and is initially moving to the right at 2.4 m/s toward a second block of mass 0.80 kg that is initially at rest. When the blocks collide, a cocked spring releases 1.2 J of energy into the system. (For velocities, use to mean to the right, - to mean to the left).A) What is the velocity of the first block after the collision?
B) What is the velocity of the second block after the collision?
Answer:
a) 3.632 m/s
b) 0.462 m/s
Explanation:
Using the law of conservation of momentum:
[tex]m_{1} u_{1} + m_{2} u_{2}= m_{1} V_{1} + m_{2} V_{2}[/tex]..........(1)
[tex]m_{1} = 0.30 kg\\u_{1} = 2.4 m/s\\m_{2} = 0.80 kg\\u_{2} = 0 m/s[/tex]
Substituting the above values into equation (1) and make V2 the subject of the formula:
[tex]0.3(2.4) + 0.80(0)= 0.3 V_{1} + 0.8 V_{2}\\[/tex]
[tex]V_{2} = \frac{0.72 - 0.3 V_{1}}{0.8}[/tex]..................(2)
Using the law of conservation of kinetic energy:
[tex]0.5m_{1} u_{1} ^{2} + 1.2 = 0.5m_{1} V_{1} ^{2} + 0.5m_{2} V_{2} ^{2}\\0.5(0.3) (2.4) ^{2} + 1.2 = 0.5(0.3) V_{1} ^{2} + 0.5(0.8)V_{2} ^{2}\\[/tex]
[tex]2.064 = 0.15 V_{1} ^{2} + 0.4V_{2} ^{2}[/tex].......(3)
Substitute equation (2) into equation (3)
[tex]2.064 = 0.15 V_{1} ^{2} + 0.4(\frac{0.72 - 0.3V_{1} }{0.8}) ^{2}\\2.064 = 0.15 V_{1} ^{2} + 0.4(\frac{0.5184 - 0.432V_{1} + 0.09V_{1} ^{2} }{0.64}) \\1.32096 = 0.096 V_{1} ^{2} + 0.20736 - 0.1728V_{1} + 0.036V_{1} ^{2} \\0.132 V_{1} ^{2} - 0.1728V_{1} - 1.1136 = 0\\V_{1} = 3.632 m/s[/tex]
Substituting [tex]V_{1}[/tex] into equation(2)
[tex]V_{2} = \frac{0.72 - 0.3 *3.632}{0.8}\\V_{2} = \frac{0.72 - 0.3 *(3.632)}{0.8}\\V_{2} = 0.462 m/s[/tex]
A machinist needs to remove a tight-fitting pin of material A from a hole in a block made of material B. The machinist heats both the pin and the block to the same high temperature and removes the pin easily. What statement relates the coefficient of thermal expansion of material A to that of material B?
a. The situation is not possible, heating block B will shrink the hole in the material as the material expands with increasing temperature
b. Material B has the same coefficient of expansion as does material A
c. Material B has a negative coefficient of expansion and A has a positive coefficient of expansion
d. Material A has a greater coefficient of expansion than does material B
e. Material B has a greater coefficient of expansion than does material A
Answer:
C. Material B has a negative coefficient of expansion and A has a positive coefficient of expansion
Explanation:
If both material A and material B have the same coefficient of thermal expansion and they were heated to same temperature, both will expand making it impossible to remove the pin from the hole.
Also, if the coefficient of thermal expansion of any of the materials is higher than the other and they were subjected to the same temperature, the material with lower coefficient of thermal expansion will expand, making it impossible to remove the pin.
However, materials with negative coefficient of thermal expansion will contract on heating instead of expanding, while materials with positive coefficient of expansion will expand on heating. This makes it possible to remove the pin from the hole in the block.
When a voltage difference is applied to a piece of metal wire, a 8-mA current flows through it. If this metal wire is now replaced with a silver wire having twice the diameter of the original wire, how much current will flow through the silver wire
Answer:
Explanation:
The resistance of a wire can be given by the following expression
[tex]R=\frac{\rho\times L}{A }[/tex]
where R is resistance , ρ is specific resistance , L is length of wire and A is cross sectional area
specific resistance of metals are almost the same . So in the present case ρ and l are same . Hence the formula becomes
R = k / A where k is a constant .
The diameter of wire becomes two times hence area of cross section becomes 4 times or 4A .
Resistance becomes 1/4 times . Hence if resistance of metal wire is R , resistance of silver wire will be R / 4 .
current = voltage / resistance
In case of metal wire
8 x 10⁻³ = V / R
In case of silver wire
I = V / (R / 4 ) , I is current , V is potential difference .
I = 4 x V/R
= 4 x 8 x 10⁻³ A
= 32 mA.
Write the second law of motion’s formula and its unit.
please answer this question, for 13 points!!!!!!!!!
Answer:
The Formula Is F = m * a And It's Units Is (kg)(m/s2)
Explanation:
On the moon, what would be the force of gravity acting on an object that has a mass of 7kg?
Answer:
Force of gravity, F = 70 N
Explanation:
It is required to find the force of gravity acting on an object that has a mass of 7 kg. Force of gravity always acts in downward direction.
The force of gravity is equal to the weight of an object. It is given by :
[tex]F=mg[/tex]
g = acceleration due to gravity, for Earth, g = 10 m/s²
So,
[tex]F=7\times 10\\\\F=70\ N[/tex]
So, 70 N of force of acting on an object.
Light requires 4.5 years to travel from the nearest star to earth. If we could travel there in a spaceship going 90% of the speed of light, the trip would require 5.0 years according to clocks on earth. How much time would pass for the passengers in the ship
Answer:
Time according to earth clock (T0) = 0.22 years (Approx)
Explanation:
Given:
Time taken by light = 4.5 years
Time taken by ship = 5 years
Speed of light = c
Speed of ship (v) = 90% of c = 0.9c
Find:
Time according to earth clock (T0) = ?
Computation:
Time dilation is ,
[tex]T(Difference) = \frac{T0}{\sqrt{1-\frac{v^2}{c^2} } }\\\\ (5-4.5)= \frac{T0}{\sqrt{1-\frac{(0.9c)^2}{c^2} } }\\\\ 0.5=\frac{T0}{\sqrt{1-0.81} }\\\\T0 =0.2179[/tex]
Time according to earth clock (T0) = 0.22 years (Approx)
Three balls, with masses of 3m,2m and m, are fastened to a massless rod of length L. The rotational inertias about the ledt
I = MR^2
The Attempt at a Solution:::
I total = (3M)(0)^2 + (2M)(L/2)^2 + (M)(L)^2
I total = 3ML^2/2
It says the answer is 3ML^2/4 though.
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The rotational inertia about the left is [tex]3ML^{2} /2[/tex].
What is meant by inertia?Inertia, property of a body by virtue of which it opposes any agency that attempts to put it in motion or, if it is moving, to change the magnitude or direction of its velocity. Inertia is a passive property and does not enable a body to do anything except oppose such active agents as forces and torques.To calculate the rotational inertia about the left[tex]I = I1 + I2 + I3\\ I= 3M(0^{2}) + 2M(L/2 )^{2} + M(L)^{2} \\I = 3ML^{2} /2[/tex]
The rotational inertia about the left is [tex]3ML^{2} /2[/tex]
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An antiproton is identical to a proton except it has the opposite charge, −e. To study antiprotons, they must be confined in an ultrahigh vacuum because they will annihilate−producing gamma rays−if they come into contact with the protons of ordinary matter. One way of confining antiprotons is to keep them in a magnetic field. Suppose that antiprotons are created with a speed of 1.5 × 10^4 m/s and then trapped in a 4.5 mT magnetic field.What minimum diameter must the vacuum chamber have to allow these antiprotons to circulate without touching the walls?Express your answer with the appropriate units.d = _________
Answer:
Explanation:
We know that, the force responsible for circulating in circular path is the centripetal force given by the force on charged particle due to magnetic field.
Here the charge is antiproton is
p = -1.6 * 10⁻¹⁹C
the speed of proton is given by 1.5 * 10 ⁴ m/s
the magnetic field is B = 4.5 * 10⁻³T
we have force due to magnetic field is equal to centripetal force
Bqv = mv² / r
Bq = mv / r
[tex]r = \frac{mv}{Bq} \\\\r=\frac{mv}{Bq} \\\\r=\frac{1.67 \times 10^-^2^7\times 1.5 \times 10^4}{4.5 \times 10^-^3\times 11.6\times 10^-^1^9} \\\\r=347.9\times 10^-^4\\\\r=3.479cm[/tex]
The diameter d of the vacuum chamber have to allow these antiprotons to circulate without touching the walls is
d = 2r
d = 2 * 3.479
d = 6.958
d ≅ 7cm
Say you want to make a sling by swinging a mass M of 2.3 kg in a horizontal circle of radius 0.034 m, using a string of length 0.034 m. You wish the mass to have a kinetic energy of 13.0 Joules when released. How strong will the string need to be
Answer:
T = 764.41 N
Explanation:
In this case the tension of the string is determined by the centripetal force. The formula to calculate the centripetal force is given by:
[tex]F_c=m\frac{v^2}{r}[/tex] (1)
m: mass object = 2.3 kg
r: radius of the circular orbit = 0.034 m
v: tangential speed of the object
However, it is necessary to calculate the velocity v first. To find v you use the formula for the kinetic energy:
[tex]K=\frac{1}{2}mv^2[/tex]
You have the value of the kinetic energy (13.0 J), then, you replace the values of K and m, and solve for v^2:
[tex]v^2=\frac{2K}{m}=\frac{2(13.0J)}{2.3kg}=11.3\frac{m^2}{s^2}[/tex]
you replace this value of v in the equation (1). Also, you replace the values of r and m:
[tex]F_c=(2.3kg)(\frac{11.3m^2/s^2}{0.034})=764.41N[/tex]
hence, the tension in the string must be T = Fc = 764.41 N
R1=3 ohms
R2=6 ohms
R4=18 ohms
R5= 15 ohms
R5=9 ohms
90 volts
What is the current running through the entire circuit?
Answer: current I = 1.875A
Explanation:
If the resistors are connected in series,
Then the equivalent resistance will be
R = 6 + 18 + 15 + 9
R = 48 ohms
Using ohms law
V = IR
Make current I the subject of formula
I = V/R
I = 90/48
I = 1.875A
And if the resistors are connected in parallel, the equivalent resistance will be
1/R = 1/6 + 1/18 + 1/15 + 1/9
1/R = 0.166 + 0.055 + 0.066 + 0.111
R = 1/0.3999
R = 2.5 ohms
Using ohms law
V = IR
I = 90/2.5
Current I = 35.99A
What is meant by a charge carrier hole in a semiconductor? Can it be created in a conductor?
Answer:
The materials used to make electronic components like transistor and integrated, circuit behave as if effective particles known as electron through them, causing electrical properties
Name and draw the devices that can convert the analog signal to digital
Answer:
Analog to digital converters
Explanation:
An analog-to-digital converter (ADC) is a device that converts analog signals such as sound into digital signals. Analog information is transmitted by modulating the signal and then amplifying the signal's strength. The conversion from analog to digital involves quantization of the input thereby reducing error or noise. The ADC produces the output data as a series of digital values (0's and 1's) with fixed precision.
Answer:
Analog to digital converter.
Explanation:
Three important type that covert analog signal to digital
1 flash ADC
2 Digital Ramp ADC
3 successive Approximation
A ladder 7.90 m long leans against the side of a building. If the ladder is inclined at an angle of 66.0° to the horizontal, what is the horizontal distance from the bottom of the ladder to the building? 10.3 Incorrect: Your answer is incorrect.
Answer:3.21 m
Explanation:
Given
Length of ladder [tex]L=7.9\ m[/tex]
inclination of ladder [tex]\theta =66^{\circ}[/tex]
If x is the horizontal distance between building and ladder then,
Using trigonometric relation
we get
[tex]\cos \theta =\frac{x}{L}[/tex]
[tex]x=L\cos \theta [/tex]
[tex]x=7.9\times \cos (66)[/tex]
[tex]x=3.21\ m[/tex]
Fill in the blanks for the following:
Arigid container of volume 100.0 Liters contains Oxygen gas. It is at room temperature (293 Kelvin), and is at atmospheric pressure (absolute pressure, meaning the gauge pressure is zero). Therefore, the number of moles of Oxygen molecules . Also, the rms-velocity inside is ________of these Oxygen molecules is most nearly ______, which is______ the rms-speed of the Nitrogen molecules just outside the container (the rigid container and its surroundings are in thermal equilibrium).
Answer:
a. 4.21 moles
b. 478.6 m/s
c. 1.5 times the root mean square velocity of the nitrogen gas outside the tank
Explanation:
Volume of container = 100.0 L
Temperature = 293 K
pressure = 1 atm = 1.01325 bar
number of moles n = ?
using the gas equation PV = nRT
n = PV/RT
R = 0.08206 L-atm-[tex]mol^{-1}[/tex][tex]K^{-1}[/tex]
Therefore,
n = (1.01325 x 100)/(0.08206 x 293)
n = 101.325/24.04 = 4.21 moles
The equation for root mean square velocity is
Vrms = [tex]\sqrt{\frac{3RT}{M} }[/tex]
R = 8.314 J/mol-K
where M is the molar mass of oxygen gas = 31.9 g/mol = 0.0319 kg/mol
Vrms = [tex]\sqrt{\frac{3*8.314*293}{0.0319} }[/tex]= 478.6 m/s
For Nitrogen in thermal equilibrium with the oxygen, the root mean square velocity of the nitrogen will be proportional to the root mean square velocity of the oxygen by the relationship
[tex]\frac{Voxy}{Vnit}[/tex] = [tex]\sqrt{\frac{Mnit}{Moxy} }[/tex]
where
Voxy = root mean square velocity of oxygen = 478.6 m/s
Vnit = root mean square velocity of nitrogen = ?
Moxy = Molar mass of oxygen = 31.9 g/mol
Mnit = Molar mass of nitrogen = 14.00 g/mol
[tex]\frac{478.6}{Vnit}[/tex] = [tex]\sqrt{\frac{14.0}{31.9} }[/tex]
[tex]\frac{478.6}{Vnit}[/tex] = 0.66
Vnit = 0.66 x 478.6 = 315.876 m/s
the root mean square velocity of the oxygen gas is
478.6/315.876 = 1.5 times the root mean square velocity of the nitrogen gas outside the tank
Popping popcorn is a thermodynamic process. Assume the pot remains covered while popcorn is being popped and the contents of the pot are the system. Which of the following correctly describes some feature of the system or what happens to the system undergoing this thermodynamic process?A. W > 0.B. Q>0.C. Tincreases.D. AU < 0.E. Pincreases.F. V-constant.
Answer:
the answer is C
Explanation:
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