Answer:
1.2 x 10⁻¹¹ N
Explanation:
B = magnitude of magnetic field at the equator = 5 x 10⁻⁵ T
q = magnitude of charge acquired by the head = 4 x 10⁻⁹ C
v = speed of driving at the equator = 60 m/s
Magnitude of magnetic force on the head at the equator is given as
F = q v B
Inserting the above values in the equation
F = (4 x 10⁻⁹) (60) (5 x 10⁻⁵)
F = 1.2 x 10⁻¹¹ N
Two wires A and B of equal length and circular cross section are made of the same metal, but the resistance of wire B is four times that of wire A. What is the ratio between their spokes ?: a) rB / rA = 2, b) rA / rB = 2, c) rB / rA = 4, d) rA / rB = 16
Answer:
Option B is the correct answer.
Explanation:
We have resistance [tex]R=\frac{\rho l}{A}[/tex]
The resistance of wire B is four times that of wire A
[tex]R_B=4R_A\\\\\frac{\rho_B l_B}{A_B}=4\times \frac{\rho_A l_A}{A_A}\\\\\frac{A_A}{A_B}=4\\\\\frac{\pi r_A^2}{\pi r_B^2}=4\\\\\frac{r_A}{r_B}=2[/tex]
Option B is the correct answer.
A car is rounding an unbanked circular turn with a speed of v = 35 m/s. The radius of the turn is r = 1500 m. What is the magnitude ac of the car’s centripetal acceleration?
Answer:
The magnitude of the car's centripetal acceleration is ac= 0.816 m/s² .
Explanation:
r= 1500m
Vt= 35 m/s
ac= Vt²/r
ac= (35m/s)² / 1500m
ac=0.816 m/s²
A proton moving to the right at 6.2 × 105 ms-1 enters a region where there is an electric field of 62 kNC-1 directed to the left. Describe qualitatively the motion of the proton in this filed. What is the time taken by the proton to come back to the point where it entered the field? (Use the standard values of the mass and charge of a proton)
Answer:
2.06 x 10^-7 s
Explanation:
For going opposite to the applied electric field
u = 6.2 x 10^5 m/s, v = 0 , q = 1.6 x 19^-19 C, E = 62000 N/C,
m = 1.67 x 106-27 Kg, t1 = ?
Let a be the acceleration.
a = q E / m = (1.6 x 10^-19 x 62000) / (1.67 x 10^-27) = 6 x 10^12 m/s^2
Use first equation of motion
v = u + a t1
0 = 6.2 x 10^5 - 6 x 10^12 x t1
t1 = 1.03 x 10^-7 s
Let the distance travelled is s.
Use third equation of motion
V^2 = u^2 - 2 a s
0 = (6.2 x 10^5)^2 - 2 x 6 x 10^12 x s
s = 0.032 m
Now when the proton moves in the same direction of electric field.
Let time taken be t2
use second equation of motion
S = u t + 1/2 a t^2
0.032 = 0 + 1/2 x 6 x 10^12 x t2^2
t2 = 1.03 x 10^-7 s
total time taken = t = t1 + t2
t = 1.03 x 10^-7 + 1.03 x 10^-7 = 2.06 x 10^-7 s
A vertical straight wire carrying an upward 24-A current exerts an attractive force per unit length of 88 X 104N/m on a second parallel wire 7.0 cm away. What current (magnitude and direction) flows in the second wire?
Answer:
The current flows in the second wire is [tex]1.3\times10^{10}\ A[/tex]
Explanation:
Given that,
Upward current = 24 A
Force per unit length[tex]\dfrac{F}{l} =88\times10^{4}\ N/m [/tex]
Distance = 7.0 cm
We need to calculate the current in second wire
Using formula of magnetic force
[tex]F=ILB[/tex]
[tex]\dfrac{F}{l}=\dfrac{\mu I_{1}I_{2}}{2\pi r}[/tex]
Where,
[tex]\dfrac{F}{l}[/tex]=force per unit length
I₁= current in first wire
I₂=current in second wire
r = distance between the wires
Put the value into the formula
[tex]88\times10^{4}=\dfrac{4\pi\times10^{-7}\times24\times I_{2}}{2\pi \times7\times10^{-2}}[/tex]
[tex]I_{2}=\dfrac{88\times10^{4}\times7\times10^{-2}}{2\times\times10^{-7}\times24}[/tex]
[tex]I_{2}=1.3\times10^{10}\ A[/tex]
Hence, The current flows in the second wire is [tex]1.3\times10^{10}\ A[/tex]
A major-league pitcher can throw a baseball in excess of 41.0 m/s. If a ball is thrown horizontally at this speed, how much will it drop by the time it reaches a catcher who is 17.0 m away from the point of release?
The ball thrown by a major-league pitcher to a catcher 17.0m away, with a speed of 41.0m/s, would drop approximately 0.84 meters due to the gravitational acceleration.
Explanation:The subject of this question involves physics, specifically the concept of projectile motion. When a pitcher throws a ball with a horizontal speed, the ball also experiences a vertical motion due to gravity. Therefore, even though the ball is thrown horizontally, it will gradually drop as it travels towards the catcher. To figure out how far it drops, we can use the physics equation for motion under constant acceleration: distance = 0.5 * acceleration * time2.
Here, the acceleration is due to gravity, which is approximately 9.8m/s2. The time can be calculated by dividing the horizontal distance (17.0 m) by the horizontal speed (41.0 m/s), giving us approximately 0.4146 s. Substituting these into the equation: 0.5 * 9.8m/s2 * (0.4146s)2 = 0.84 meters, the drop of the ball is approximately 0.84 meters.
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A cat named SchrÖdinger walks along a uniform plank that is 4.00 m long and has a mass of 7.00 kg. The plank is supported by two sawhorses, one that is 0.44 m from the left end of the plank, and the other that is 1.50 m from the right end. When the cat reaches the very right end of the plank, the plank starts to tip. What is Schrodinger’s mass? Note: Just when the plank begins to tip, the normal force on the plank from the sawhorse on the left will go to zero.
Answer:
2.3 kg
Explanation:
L = length of the plank = 4 m
M = mass of the plank = 7 kg
m = mass of cat = ?
[tex]F_{c}[/tex] = Weight of the cat = mg
[tex]F_{p}[/tex] = Weight of the plank = Mg = 7 x 9.8 = 68.6 N
ED = 2 m
CD = 1.5 m
EC = ED - CD = 2 - 1.5 = 0.5 m
Using equilibrium of torque about sawhorse at C
[tex]F_{p}[/tex] (EC) = [tex]F_{c}[/tex] (CD)
(68.6) (0.5) = (mg) (1.5)
(68.6) (0.5) = (m) (1.5) (9.8)
m = 2.3 kg
Determine the length of a copper wire that has a resistance of 0.172 ? and cross-sectional area of 7.85 × 10-5 m2. The resistivity of copper is 1.72 × 10-8 ?·m.
Final answer:
To determine the length of a copper wire given its resistance, cross-sectional area, and the resistivity of copper, use the formula L = (R·A)/ρ. The length of the copper wire, based on the provided resistance of 0.172 Ω, cross-sectional area of 7.85 × [tex]10^-^5[/tex] m², and resistivity of 1.72 × [tex]10^-^8[/tex]Ω·m, is calculated to be 7.85 meters.
Explanation:
The question involves determining the length of a copper wire given its resistance, cross-sectional area, and the resistivity of copper. The formula to calculate the length of the wire is derived from Ohm's law, which in terms of resistivity (ρ) is expressed as R = ρL/A, where R is the resistance, L is the length of the wire, A is the cross-sectional area, and ρ is the resistivity of the material.
We are provided with the resistance (R) as 0.172 Ω, the cross-sectional area (A) as 7.85 × [tex]10^-^5[/tex] m², and the resistivity (ρ) of copper as 1.72 × [tex]10^-^8[/tex] Ω·m. By rearranging the formula to solve for L, we get L = (R·A)/ρ. Substituting the given values into the formula, we find the length of the copper wire.
Let's calculate: L = (0.172 Ω × 7.85 × [tex]10^-^5[/tex] m^2) / (1.72 × [tex]10^-^8[/tex] Ω·m) = 7.85 m. Therefore, the length of the copper wire is 7.85 meters.
Express 79 m in units of (a) centimeters
(b) feet
(c) inches and
(d) miles.
Answer: a) 7,00 centimeters
(b) 259. 19 feet
(c) 3110.28 inches
(d) 0.049 miles
Explanation:
(a) We know that 1 meter = 100 centimeters
Therefore,
[tex]79\ m= 7,900\text{ centimeters}[/tex]
(b)Since 1 meter = 3.28084 feet
Then, [tex]79\ m= 79\times3.28084=259.18636\approx 259.19\text{ feet}[/tex]
(c) Since, 1 feet = 12 inches.
[tex]79\ m=259.19\text{ feet}=259.19\times12=3110.28\text{ inches}[/tex]
(d) [tex]\text{Since 1 feet= }\dfrac{1}{5280}\text{ mile}[/tex]
[tex]79\ m=259.19\text{ feet}=\dfrac{259.19}{5280}= 0.0490890151515\approx0.049\text{ miles}[/tex]
Consider the mechanism. Step 1: 2A↽−−⇀B+C equilibrium Step 2: B+D⟶E slow Overall: 2A+D⟶C+E Determine the rate law for the overall reaction, where the overall rate constant is represented as ????.
Answer : The rate law for the overall reaction is, [tex]R=\frac{K[A]^2[D]}{[C]}[/tex]
Explanation :
As we are given the mechanism for the reaction :
Step 1 : [tex]2A\rightleftharpoons B+C[/tex] (equilibrium)
Step 2 : [tex]B+D\rightarrow E[/tex] (slow)
Overall reaction : [tex]2A+D\rightarrow C+E[/tex]
First we have to determine the equilibrium constant from step 1.
The expression for equilibrium constant will be,
[tex]K'=\frac{[B][C]}{[A]^2}[/tex]
Form this, the value of [B] is,
[tex][B]=\frac{K'[A]^2}{[C]}[/tex] ............(1)
Now we have to determine the rate law from the slow step 2.
The expression for law will be,
[tex]Rate=K''[B][D][/tex] .............(2)
Now put equation 1 in 2, we get:
[tex]Rate=K''\frac{K'[A]^2}{[C]}[D][/tex]
[tex]Rate=\frac{K[A]^2[D]}{[C]}[/tex]
Therefore, the rate law for the overall reaction is, [tex]R=\frac{K[A]^2[D]}{[C]}[/tex]
The rate law for the reaction is obtained from the slow step (Step 2) and modified with information from the fast equilibrium step (Step 1), considering B as an intermediate reactant. The overall rate law becomes: Rate = k [ ((k1/k-1)[A]^2/[C]) ][D]
Explanation:The rate law for a reaction is established based upon the rate-determining (slowest) step of the reaction. So, first, we identify the slow step of the reaction. From the information given, we can see that the slow step is Step 2, with reactants B and D forming product E.
The rate law for this step of the reaction is written as: Rate = k [B][D], where k is the rate constant for this reaction and [B] and [D] are the concentrations of reactants B and D respectively.
However, reactant B is an intermediate since it doesn’t appear in the overall reaction. So, the rate of formation of B in the fast equilibrium Step 1 becomes important. Here, B is produced from reactant A. The forward and backward rates are equal at equilibrium. So, the rate law for Step 1 could be written as: Rate = k1[A]^2 = k-1[B][C], giving us [B]= (k1/k-1)[A]^2/[C],
Substituting [B] in the rate law for step 2, we obtain the rate law for the overall reaction as: Rate = k [ ((k1/k-1)[A]^2/[C]) ][D], where k is the rate constant for the slow step and k1 and k-1 are the rate constants for the forward and reverse reactions of Step 1.
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A uniform non-conducting ring of radius 2.2 cm and total charge 6.08 µC rotates with a constant angular speed of 2.01 rad/s around an axis perpendicular to the plane of the ring that passes through its center. What is the magnitude of the magnetic moment of the rotating ring?
Answer:
The magnitude of the magnetic moment of the rotating ring is [tex]2.96\times10^{-9}\ Am^2[/tex]
Explanation:
Given that,
Radius = 2.2 cm
Charge [tex]q = 6.08\times10^{-6}\ C[/tex]
Angular speed = 2.01 rad/s
We need to calculate the time period
[tex]T = \dfrac{2\pi}{\omega}[/tex]
Now, The spinning produced the current
Using formula for current
[tex] I = \dfrac{q}{T}[/tex]
We need to calculate the magnetic moment
Using formula of magnetic moment
[tex]M = I A[/tex]
Put the value of I and A into the formula
[tex]M=\dfrac{q}{T}\times A[/tex]
[tex]M=\dfrac{q\times\omega}{2\pi}\times \pi\times r^2[/tex]
Put the value into the formula
[tex]M=\dfrac{6.08\times10^{-6}\times2.01\times(2.2\times10^{-2})^2}{2}[/tex]
[tex]M=2.96\times10^{-9}\ Am^2[/tex]
Hence, The magnitude of the magnetic moment of the rotating ring is [tex]2.96\times10^{-9}\ Am^2[/tex]
The magnetic moment of the rotating ring with a radius of 2.2 cm, a total charge of 6.08 µC, and an angular velocity of 2.01 rad/s is approximately 1.176 x 10⁻⁵ A⋅m².
Explanation:The magnitude of the magnetic moment (μ) of a rotating charged ring can be found using the formula given by the product of the charge (Q) and the angular velocity (ω) and the radius (r) squared, as μ = Qωr²/2. For a uniform non-conducting ring rotating about an axis perpendicular to its plane, this relationship can be used to determine the magnetic moment generated by its rotation.
To calculate the magnetic moment for the given scenario:
Carrying out the multiplication gives the magnetic moment:
μ = (6.08 x 10-6 C)(2.01 rad/s)(0.000484 m2)/2 = 0.01176 x 10-6 C⋅m2/s
This results in a magnetic moment of approximately μ ≈ 1.176 x 10-5 A⋅m2.
The overall energy involved in the formation of CsCl from Cs(s) and Cl2(g) is −443 kJ/mol. Given the following information: heat of sublimation for Cs is +76 kJ/mol, bond dissociation energy for 12Cl2 is +121 kJ/mol, Ei1 for Cs is +376 kJ/mol, and Eea for Cl(g) is −349 kJ/mol. What is the magnitude of the lattice energy for CsCl?
Answer : The magnitude of the lattice energy for CsCl is, 667 KJ/mole
Explanation :
The steps involved in the born-Haber cycle for the formation of [tex]CsCl[/tex] :
(1) Conversion of solid calcium into gaseous cesium atoms.
[tex]Cs(s)\overset{\Delta H_s}\rightarrow Cs(g)[/tex]
[tex]\Delta H_s[/tex] = sublimation energy of calcium
(2) Conversion of gaseous cesium atoms into gaseous cesium ions.
[tex]Ca(g)\overset{\Delta H_I}\rightarrow Ca^{+1}(g)[/tex]
[tex]\Delta H_I[/tex] = ionization energy of calcium
(3) Conversion of molecular gaseous chlorine into gaseous chlorine atoms.
[tex]Cl_2(g)\overset{\frac{1}{2}\Delta H_D}\rightarrow Cl(g)[/tex]
[tex]\Delta H_D[/tex] = dissociation energy of chlorine
(4) Conversion of gaseous chlorine atoms into gaseous chlorine ions.
[tex]Cl(g)\overset{\Delta H_E}\rightarrow Cl^-(g)[/tex]
[tex]\Delta H_E[/tex] = electron affinity energy of chlorine
(5) Conversion of gaseous cations and gaseous anion into solid cesium chloride.
[tex]Cs^{1+}(g)+Cl^-(g)\overset{\Delta H_L}\rightarrow CsCl(s)[/tex]
[tex]\Delta H_L[/tex] = lattice energy of calcium chloride
To calculate the overall energy from the born-Haber cycle, the equation used will be:
[tex]\Delta H_f^o=\Delta H_s+\Delta H_I+\Delta H_D+\Delta H_E+\Delta H_L[/tex]
Now put all the given values in this equation, we get:
[tex]-443KJ/mole=76KJ/mole+376KJ/mole+121KJ/mole+(-349KJ/mole)+\Delta H_L[/tex]
[tex]\Delta H_L=-667KJ/mole[/tex]
The negative sign indicates that for exothermic reaction, the lattice energy will be negative.
Therefore, the magnitude of the lattice energy for CsCl is, 667 KJ/mole
The magnitude of Lattice energy of CsCl is -676 kJ/mol.
Given Here,
Enthalpy of sublimation of Cs [tex]\rm \bold{ \Delta H(sub ) }[/tex] = +76 kJ/mo
Ionization Energy for Potassium IE(Cs) = +376 kJ/mol
Electron affinity for Chlorine is EA(Cl) = −349 kJ/mol.
Bond dissociation energy of Chlorine, BE(Cl) = +121 kJ/mol
Enthalpy of formation for CsCl, [tex]\rm \bold{ \Delta H(f) }[/tex] = −436.5 kj/mol .
The lattice energy of KCl can be calculated from the formula,
[tex]\rm \bold {U(CsCl) = \Delta Hf(CsCl) - [ \Delta H(sub) + IE(Cs) +BE(Cl_2) + EA(Cl)]}[/tex]
U( CsCl) = -436 - [+76 +376 +121 kJ/mol -349]
U( CsCl) = -676 kJ/mol
Hence we can calculate that the magnitude of Lattice energy of CsCl is -676 kJ/mol.
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A gray kangaroo can bound across level ground with each jump carrying it 9.6 m from the takeoff point. Typically the kangaroo leaves the ground at a 28º angle. If this is so: Part A) What is its takeoff speed? Express your answer to two significant figures and include the appropriate units. Part B) What is its maximum height above the ground? Express your answer to two significant figures and include the appropriate units
Answer:
(A) 11 m/s
(B) 1.3 m
Explanation:
Horizontal range, R = 9.6 m
Angle of projection, theta = 28 degree
(A)
Use the formula of horizontal range
R = u^2 Sin 2 theta / g
u^2 = R g / Sin 2 theta
u^2 = 9.6 × 9.8 / Sin ( 2 × 28)
u = 10.65 m/s
u = 11 m/s
(B)
Use the formula for maximum height
H = u^2 Sin ^2 theta / 2g
H =
10.65 × 10.65 × Sin^2 (28) / ( 2 × 9.8)
H = 1.275 m
H = 1 .3 m
(a)The take-off speed is the speed at the start of takeoff. The take-off speed of the kangaroo will be 11 m/sec.
(b)The height achieved during takeoff is the maximum height. The maximum height above the ground will be 1.3 meters.
what is the maximum height achieved in projectile motion?It is the height achieved by the body when a body is thrown at the same angle and the body is attaining the projectile motion.The maximum height of motion is given by
[tex]H = \frac{u^{2}sin^2\theta }{2g}[/tex]
What is a range of projectile?The horizontal distance is covered by the body when the body is thrown at some angle is known as the range of the projectile. It is given by the formula
[tex]R = \frac{u^{2}sin2\theta}{g}[/tex]
(a)Take-of velocity =?
given
Horizontal range = 9.6m.
[tex]\theta = 28^0[/tex]
[tex]g = 9.81 \frac{m}{sec^{2} }[/tex]
[tex]R = \frac{u^{2}sin2\theta}{g}[/tex]
[tex]u = \sqrt{\frac{Rg}{sin2\theta} }[/tex]
[tex]u = \sqrt{\frac{9.6\times9.81}{sin56^0} }[/tex]
[tex]u = 11 m /sec[/tex]
Hence the take-off speed of the kangaroo will be 11 m/sec.
(b) Maximum height =?
given,
[tex]u = 11 m /sec[/tex]
[tex]H = \frac{u^{2}sin^2\theta }{2g}[/tex]
[tex]H = \frac{(11)^{2}sin^2 58^0 }{2\times 9.81}[/tex]
[tex]\rm { H = 1.3 meter }[/tex]
Hence the maximum height above the ground will be 1.3 meters.
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The orbit of a certain a satellite has a semimajor axis of 1.7 x 107 m and an eccentricity of 0.25. What are the satellite's perigee and apogee (in km)?
The perigee and apogee of the satellite are 1.275 x 10^4 km and 2.125 x 10^4 km, respectively.
Explanation:The perigee and apogee of a satellite can be determined using the equations:
Perigee = Semimajor Axis - Eccentricity x Semimajor Axis
Apogee = Semimajor Axis + Eccentricity x Semimajor Axis
Given that the semimajor axis is 1.7 x 10^7 m and the eccentricity is 0.25, we can substitute these values into the equations to find:
Perigee = (1.7 x 10^7 m) - (0.25 x 1.7 x 10^7 m) = 1.275 x 10^7 m
Apogee = (1.7 x 10^7 m) + (0.25 x 1.7 x 10^7 m) = 2.125 x 10^7 m
Converting these values to km:
Perigee = 1.275 x 10^4 km
Apogee = 2.125 x 10^4 km
A charged capacitor is connected to an inductor to form an LC circuit with a frequency of oscillation f = 1.6 Hz. At time t = 0 the capacitor is fully charged. At a given instant later the charge on the capacitor is measured to be Q = 3 μC and the current in the circuit is equal to 75 μA. What is the maximum charge of the capacitor?
Answer:
[tex]Q = 8.61 \times 10^{-4} C[/tex]
Explanation:
Since in LC oscillation there is no energy loss
so here we can say that
initial total energy of capacitor = energy stored in capacitor + energy stored in inductor at any instant of time
so we can say
[tex]\frac{Q^2}{2C} = \frac{q^2}{2C} + \frac{1}{2}Li^2[/tex]
now we have
[tex]q = 3\mu C[/tex]
[tex]i = 75 \mu A[/tex]
now we have
[tex]Q^2 = q^2 + (LC) i^2[/tex]
we also know that
[tex]2\pi f = \frac{1}{\sqrt{LC}}[/tex]
[tex]2\pi(1.6) = \frac{1}{\sqrt{LC}}[/tex]
[tex]LC = 9.89 \times 10^{-3}[/tex]
now from above equation
[tex]Q^2 = (3\mu C)^2 + (9.89 \times 10^{-3})(75 \mu A)[/tex]
[tex]Q = 8.61 \times 10^{-4} C[/tex]
To find the maximum charge of the capacitor in the LC circuit, set up a cosine equation using the initial charge and current value.
Explanation:To find the maximum charge of the capacitor in the LC circuit, we need to utilize the relationship between the charge on the capacitor and the current in the circuit. At time t = 0, the capacitor is fully charged, so the initial charge is Q = 3 µC. The current in the circuit is equal to 75 µA. Knowing these values, we can set up a cosine equation to find q(t). By solving the equation, we can find the maximum charge of the capacitor.
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The gravitational acceleration is 9.81 m/s2 here on Earth at sea level. What is the gravitational acceleration at a height of 350 km above the surface of the Earth, where the International Space Station (ISS) flies? (The mass of the Earth is 5.97×1024 kg, and the radius of the Earth is 6370 km.) It is somewhat greater than 9.81 m/s2. It is zero, since the ISS is in the state of weightlessness. It is half of 9.81 m/s2. It is somewhat less than 9.81 m/s2. It is twice of 9.81 m/s2. It is 9.81 m/s2, the same.
Answer:
g = 8.82 m/s/s
It is somewhat less than 9.81 m/s2.
Explanation:
As we know that the formula to find the gravitational acceleration is given as
[tex]g = \frac{GM}{r^2}[/tex]
here we know that
r = distance from center of earth
here we have
[tex]r = R + h[/tex]
[tex]r = (6370 + 350) km = 6720 km [/tex]
now we have
[tex]g = \frac{(6.67 \times 10^{-11})(5.97 \times 10^{24})}{(6720 \times 10^3)^2}[/tex]
[tex]g = 8.82 m/s^2[/tex]
In a particular case of Compton scattering, a photon collides with a free electron and scatters backwards. The wavelength after the collision is exactly double the wavelength before the collision. What is the wavelength of the incident photon?
Answer:
hence initial wavelength is [tex]\lambda =4.86\times10^{-12}m[/tex]
Explanation:
shift in wavelength due to compton effect is given by
[tex]\lambda ^{'}-\lambda =\frac{h}{m_{e}c}\times(1-cos\theta )[/tex]
λ' = the wavelength after scattering
λ= initial wave length
h= planks constant
m_{e}= electron rest mass
c= speed of light
θ= scattering angle = 180°
compton wavelength is
[tex]\frac{h}{m_{e}c}= 2.43\times10^{-12}m[/tex]
[tex]\lambda '-\lambda =2.43\times10^{-12}\times(1-cos\theta )[/tex]
[tex]\lambda '-\lambda =2.43\times10^{-12}\times(1+1 )[/tex] ( put cos 180°=-1)
also given λ'=2λ
putting values and solving we get
[tex]\lambda =4.86\times10^{-12}m[/tex]
hence initial wavelength is [tex]\lambda =4.86\times10^{-12}m[/tex]
Because of your knowledge of physics and interest in weapons, you've got a summer job with the FBI, your job is to determine if the weapon that was found at the scene of a crime was precisely the same with which the crime. For this your boss has asked you to determine the speed of exit of the weapon that was in the scene. Design an experiment in detail where you explain the results.
Answer:
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Some criminal justice careers require a college degree; colleges and universities offer a variety of degrees related to the criminal justice sector, including online degrees. Criminal justice degree jobs cover a wide range of areas.
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Explanation:
A 3.0 \Omega Ω resistor is connected in parallel to a 6.0 \Omega Ω resistor, and the combination is connected in series with a 4.0 \Omega Ω resistor. If this combination is connected across a 12.0 V battery, what is the power dissipated in the 3.0 \Omega Ω resistor?
Answer:
The power dissipated in the 3 Ω resistor is P= 5.3watts.
Explanation:
After combine the 3 and 6 Ω resistor in parallel, we have an 2 Ω and a 4 Ω resistor in series.
The resultating resistor is of Req=6Ω.
I= V/Req
I= 2A
the parallel resistors have a potential drop of Vparallel=4 volts.
I(3Ω) = Vparallel/R(3Ω)
I(3Ω)= 1.33A
P= I(3Ω)² * R(3Ω)
P= 5.3 Watts
Which of the following are true (choose all that apply)? Sound can travel through a vacuum. -Sound can travel through water. Light can travel through a vacuum Sound can travel through air. Light can travel through air Light can travel through water
Answer:
option (B) - true
option (C) - true
option (D) - true
option (E) - true
option (F) - true
Explanation:
Batman (mass = 88.8 kg) jumps straight down from a bridge into a boat (mass = 530 kg) in which a criminal is fleeing. The velocity of the boat is initially +14.5 m/s. What is the velocity of the boat after Batman lands in it?
Answer:
The velocity of the boat after Batman lands in it is 2.08 m/s.
Explanation:
Given that,
Mass of batman [tex]m_{1}= 88.8\ kg[/tex]
Mass of boat [tex]m_{2}=530\ kg[/tex]
Initial velocity = 14.5 m/s
We need to calculate the velocity of boat
Using conservation of momentum
[tex]m_{1}u=(m_{1}+m_{2})v[/tex]...(I)
Where, [tex]m_{1}[/tex]=mass of batman
[tex]m_{2}[/tex] =mass of boat
u=initial velocity
v = velocity of boat
Put the value in the equation
[tex]88.8\times14.5=530+88.8\times v[/tex]
[tex]v=\dfrac{88.8\times14.5}{530+88.8}[/tex]
[tex]v=2.08\ m/s[/tex]
Hence, The velocity of the boat after Batman lands in it is 2.08 m/s.
A car is traveling at 50 mi/h when the brakes are fully applied, producing a constant deceleration of 48 ft/s2. What is the distance covered before the car comes to a stop
Answer:
Distance covered by the car is 56.01 feet.
Explanation:
It is given that,
Initial velocity, u = 50 mi/h = 73.33 ft/s
Constant deceleration of the car, [tex]a=-48\ ft/s^2[/tex]
Final velocity, v = 0
Let s is the distance covered before the car comes to rest. It can be calculated using third equation of motion as :
[tex]v^2-u^2=2as[/tex]
[tex]s=\dfrac{v^2-u^2}{2a}[/tex]
[tex]s=\dfrac{0^2-(73.33\ ft/s^2)^2}{2\times -48\ ft/s^2}[/tex]
s = 56.01 ft
So, the distance covered before the car comes to a stop is 56.01 feet. Hence, this is the required solution.
The distance covered by the car as it comes to rest is 17.1 m.
To calculate the distance covered before the car come to stop, we use the formula below.
Formula:
v² = u²+2as............... Equation 1Where:
s = distance covered by the carv = final velocityu = initial velocitya = acceleration of the carmake s the subject of the equation
s = (v²-u²)/2a........................Equation 2From the question,
Given:
v = 50 mi/h = (50×0.447) = 22.35 m/su = 0 m/sa = 48 ft/s² = (48×0.3048) = 14.63 m/s²Substitute these values into equation 2
s = (22.35²-0²)/(2×14.63)s = 17.1 mHence, the distance covered by the car as it comes to rest is 17.1 m.
Learn more about distance here: https://brainly.com/question/17273444
Consider two cubes, one of aluminum and one of copper. If each cube measures 8 cm along an edge, calculate the mass of each cube.
Answer:
≈1,39 and 4.56 kg.
Explanation:
for more details see the attachment. Note, the ρ(Cu) means 'Density of copper', ρ(Al) means 'Density of aluminium'.
Young's Modulus of elasticity is a) Shear stress/Shear strain b) Tensile stress/Shear strain 9. c) Shear stress /Tensile strairn d) Tensile stress/Tensile strain e) None of these
Answer:
Option C is the correct answer.
Explanation:
Young's modulus is the ratio of tensile stress and tensile strain.
Bulk modulus is the ratio of pressure and volume strain.
Rigidity modulus is the ratio of shear stress and shear strain.
Here we are asked about Young's modulus which is the ratio of tensile stress and tensile strain.
Option C is the correct answer.
Answer:
C.
It is the force per unit area acting on the material’s surface.
Explanation:
At a certain instant after jumping from the airplane A, a skydiver B is in the position shown and has reached a terminal (constant) speed vB = 52 m/s. The airplane has the same constant speed vA = 52 m/s, and after a period of level flight is just beginning to follow the circular path shown of radius ρA = 2330 m. (a) Determine the velocity and acceleration of the airplane relative to the skydiver. (b) Determine the time rate of change of the speed vr of the airplane and the radius of curvature ρr of its path, both as observed by the nonrotating skydiver.
(a) The velocity of the airplane relative to the skydiver is 52 m/s, and the acceleration is directed radially inward. (b) The time rate of change of the speed [tex]\(v_r\)[/tex] of the airplane, as observed by the skydiver, is 0 m/s², and the radius of curvature [tex]\(\rho_r\)[/tex] of its path is 2330 m.
Explanation:(a) The velocity of the airplane relative to the skydiver is the vector difference of their individual velocities. Since both have the same constant speed of 52 m/s, the relative velocity is 52 m/s. The acceleration of the airplane, as observed by the skydiver, is directed radially inward due to the circular motion.
(b) The time rate of change of speed [tex](\(v_r\))[/tex] is the radial component of acceleration, which is 0 m/s² since the airplane is moving at a constant speed. The radius of curvature [tex](\(\rho_r\))[/tex] of its path is given as 2330 m, representing the circular path's curvature.
These results are derived from principles of relative motion and circular motion, providing insights into the dynamics of the airplane-skydiver system.
A 1200 kg car traveling north at 14 m/s is rear-ended by a 2000 kg truck traveling at 30 m/s. What is the total momentum before and after the collision?
Final answer:
The Physics question involves calculating the total momentum before and after a traffic collision. Total momentum before the collision is 76800 kg·m/s northward. Assuming a perfectly inelastic collision where the vehicles stick together, the total momentum remains unchanged after the collision.
Explanation:
To calculate the total momentum before the collision, we use the formula: momentum = mass × velocity. For the 1200 kg car traveling north at 14 m/s, its momentum is 1200 kg × 14 m/s = 16800 kg·m/s northward.
For the 2000 kg truck traveling at 30 m/s, its momentum is 2000 kg × 30 m/s = 60000 kg·m/s northward. The total momentum before the collision is the sum of these two momenta, so 76800 kg·m/s northward.
Assuming the collision is perfectly inelastic and the two vehicles stick together, the law of conservation of momentum states that the total momentum before the collision must be equal to the total momentum after the collision.
Therefore, the total momentum after the collision will also be 76800 kg·m/s northward.
A flywheel in the form of a uniformly thick disk of radius 1.93 m has a mass of 92.1 kg and spins counterclockwise at 419 rpm. Calculate the constant torque required to stop it in 1.25 min.
Answer:
Torque = 99.48 N-m²
Explanation:
It is given that,
Radius of the flywheel, r = 1.93 m
Mass of the disk, m = 92.1 kg
Initial angular velocity, [tex]\omega_i=419\ rpm=43.87\ rad/s[/tex]
Final angular speed, [tex]\omega_f=0[/tex]
We need to find the constant torque required to stop it in 1.25 min, t = 1.25 minutes = 75 seconds
Torque is given by :
[tex]\tau=I\times \alpha[/tex]...........(1)
I is moment of inertia, for a solid disk, [tex]I=\dfrac{mr^2}{2}[/tex]
[tex]\alpha[/tex] is angular acceleration
[tex]I=\dfrac{92.1\ kg\times (1.93\ m)^2}{2}=171.53\ kgm^2[/tex]..............(2)
Now finding the value of angular acceleration as :
[tex]\omega_f=\omega_i+\alpha t[/tex]
[tex]0=43.87+\alpha \times 75[/tex]
[tex]\alpha =-0.58\ m/s^2[/tex]..........(3)
Using equation (2) and (3), solve equation (1) as :
[tex]\tau=171.53\ kgm^2\times -0.58\ m/s^2[/tex]
[tex]\tau=-99.48\ N-m^2[/tex]
So, the torque require to stop the flywheel is 99.48 N-m². Hence, this is the required solution.
A spacecraft of mass 1500 kg orbits the earth at an altitude of approximately 450 km above the surface of the earth. Assuming a circular orbit, what is the attractive force that the earth exerts to keep the spacecraft+ in orbit? Answer: (a) 1.28x 10^7 (N) (b) 2.99 x 10^7 (N) (c) 3.56 x 10^7 (N). (d) 4.11 x 10^7 (N) (e) 5.06x 10^7 (N)
Answer:
1.28 x 10^4 N
Explanation:
m = 1500 kg, h = 450 km, radius of earth, R = 6400 km
Let the acceleration due to gravity at this height is g'
g' / g = {R / (R + h)}^2
g' / g = {6400 / (6850)}^2
g' = 8.55 m/s^2
The force between the spacecraft and teh earth is teh weight of teh spacecraft
W = m x g' = 1500 x 8.55 = 1.28 x 10^4 N
A criminal is escaping across a rooftop and runs off the roof horizontally at a speed of 4.2 m/s, hoping to land on the roof of an adjacent bullding Air resistance is negligible. The horlzontal distance between the two buildings is D, and the roof of the adjacent building is 2.1 m below the jumping off polint Find the maximum value for D. (g 9.80ms
Answer:
Maximum separation between buildings, D = 1.806 m
Explanation:
Vertical motion of criminal
Initial speed = 0 m/s
Acceleration = 9.81 m/s²
Displacement = 2.1 m
We need to find time when he moves vertically 2.1 m.
We have
s = ut + 0.5at²
2.1 = 0 x t + 0.5 x 9.81 x t²
t = 0.43 s.
Horizontal motion of criminal
Initial speed = 4.2 m/s
Acceleration =0 m/s²
Time = 0.43 s
We need to find displacement.
We have
s = ut + 0.5at²
s = 4.2 x 0.43 + 0.5 x 0 x 0.43²
s = 1.806 m
So maximum separation between buildings, D = 1.806 m
Consider a grain of table salt which is made of positive
and negative ions (Na+and Cl−). Suppose each of these ions carries a charge of 1.60x10^−19 C and are 5.29x10^−11mapart. What is the magnitude of the electrostatic force
between them?
Answer:
Force, [tex]F=8.23\times 10^{-8}\ N[/tex]
Explanation:
It is given that,
Each ion in Na⁺ and Cl⁻ has a charge of, [tex]q=1.6\times 10^{-19}\ C[/tex]
Distance between two ions, [tex]d=5.29\times 10^{-11}\ m[/tex]
We need to find the electrostatic force. It is given by :
[tex]F=k\dfrac{q_1q_2}{d^2}[/tex]
[tex]F=9\times 10^9\times \dfrac{(1.6\times 10^{-19})^2}{(5.29\times 10^{-11})^2}[/tex]
[tex]F=8.23\times 10^{-8}\ N[/tex]
So, the magnitude of electrostatic force between them is [tex]F=8.23\times 10^{-8}\ N[/tex]. Hence, this is the required solution.
A person fires a 38 gram bullet straight up into the air. It rises, then falls straight back down, striking the ground with a speed of 345 m/s. The bullet embeds itself into the ground a distance of 8.9 cm before coming to a stop. What force does the ground exert on the bullet? Express your answer in newtons.
Answer:
Force exerted = 25.41 kN
Explanation:
We have equation of motion
v² = u²+2as
u = 345 m/s, s = 8.9 cm = 0.089 m, v = 0 m/s
0² = 345²+2 x a x 0.089
a = -668679.78 m/s²
Force exerted = Mass x Acceleration
Mass of bullet = 38 g = 0.038 kg
Acceleration = 668679.78 m/s²
Force exerted = 25409.83 N = 25.41 kN