A person is standing a distance D = 5.8 m in front of a flat, vertical mirror. The distance from the ground to his eyes is H = 1.6 m. An object is placed on the ground a distance d = D/2 = 2.9 m in front of the mirror. At what height h should the bottom of the mirror be so that the person can see the bottom of the object?
Answers
Answer:
it is at height of y = 0.533 m from ground
Explanation:
As per law of reflection we know that angle of incidence = angle of reflection
so here we have
[tex]tan\theta_i = tan\theta_r[/tex]
here we know that
[tex]tan\theta_i = \frac{y}{d}[/tex]
also we have
[tex]tan\theta_r = \frac{H - y}{D}[/tex]
now we have
[tex]\frac{H - y}{D} = \frac{y}{d}[/tex]
here we have
[tex]\frac{1.6 - y}{5.8} = \frac{y}{2.9}[/tex]
[tex]3y = 1.6[/tex]
[tex]y = 0.533 m[/tex]
To see the bottom of the object in the mirror, the bottom of the mirror should be at a height of half the height of the person.
Explanation:To see the bottom of the object in the mirror, the person should be able to see an image of the bottom of the object reflected in the mirror.
Using the law of reflection, we can determine the height h at which the bottom of the mirror should be. The angle of incidence for the light from the bottom of the object is equal to the angle of reflection, so the height of the mirror should be equal to the height of the person's eyes, H, plus the height of the bottom of the object, h. We can calculate h using similar triangles:
h/H = d/D
where d is the distance from the object to the mirror and D is the distance from the person to the mirror. Substituting the given values, we have:
h = (d/D) * H
Substituting d = D/2, we get:
h = (D/2D) * H = H/2
Therefore, the bottom of the mirror should be at a height of half the height of the person.
Related Questions
At the same instant that a 0.50-kg ball is dropped from 25 m above Earth, a second ball, with a mass of 0.25 kg, is thrown straight upward from Earth's surface with an initial speed of 15 m/s. They move along nearby lines and pass each other without colliding. At the end of 2.0 s, the height above Earth's surface of the center of mass of the two-ball system is:
Answers
Answer:
The center of mass of the two-ball system is 7.05 m above ground.
Explanation:
Motion of 0.50 kg ball:
Initial speed, u = 0 m/s
Time = 2 s
Acceleration = 9.81 m/s²
Initial height = 25 m
Substituting in equation s = ut + 0.5 at²
s = 0 x 2 + 0.5 x 9.81 x 2² = 19.62 m
Height above ground = 25 - 19.62 = 5.38 m
Motion of 0.25 kg ball:
Initial speed, u = 15 m/s
Time = 2 s
Acceleration = -9.81 m/s²
Substituting in equation s = ut + 0.5 at²
s = 15 x 2 - 0.5 x 9.81 x 2² = 10.38 m
Height above ground = 10.38 m
We have equation for center of gravity
[tex]\bar{x}=\frac{m_1x_1+m_2x_2}{m_1+m_2}[/tex]
m₁ = 0.50 kg
x₁ = 5.38 m
m₂ = 0.25 kg
x₂ = 10.38 m
Substituting
[tex]\bar{x}=\frac{0.50\times 5.38+0.25\times 10.38}{0.50+0.25}=7.05m[/tex]
The center of mass of the two-ball system is 7.05 m above ground.
The resulting velocity of the center of mass is (A)[tex]11 m/s[/tex]downward.
To find the velocity of the center of mass of the two-ball system, we need to calculate the velocities of each ball after [tex]2.0[/tex] seconds and then use the center of mass formula.
Ball 1: This ball is dropped from [tex]25 m[/tex], so its initial velocity ([tex]u_1[/tex]) is[tex]0 m/s[/tex]. Using the equation for velocity under constant acceleration ([tex]v = u + at[/tex]), we get:[tex]v_1 = 0 + (9.8 \, \text{m/s}^2 \cdot 2.0 \, \text{s}) \\= 19.6 \, \text{m/s}[/tex](downward).
Ball 2: This ball is thrown upward with an initial velocity ([tex]u_2[/tex]) of [tex]15 m/s[/tex]. After [tex]2.0[/tex]seconds, using the same velocity equation:[tex]v_2 = 15 \, \text{m/s} - (9.8 \, \text{m/s}^2 \cdot 2.0 \, \text{s}) \\= 15 \, \text{m/s} - 19.6 \, \text{m/s} \\= -4.6 \, \text{m/s}[/tex]
(negative since it's moving downward).
Now, we calculate the velocity of the center of mass:Total mass ([tex]M[/tex]): [tex]0.50 kg + 0.25 kg = 0.75 kg[/tex]
Using the center of mass velocity formula: [tex]V_{\text{cm}} = \frac{m_1 \cdot v_1 + m_2 \cdot v_2}{M}[/tex]We get:
[tex]V_{\text{cm}} = \frac{0.50 \, \text{kg} \cdot 19.6 \, \text{m/s} + 0.25 \, \text{kg} \cdot (-4.6 \, \text{m/s})}{0.75 \, \text{kg}} \\= \frac{9.8 - 1.15}{0.75} \\= 11.0 \, \text{m/s}[/tex] (downward)
Complete question:-
At the same instant that a [tex]0.50kg[/tex] ball is dropped from [tex]25 m[/tex] above Earth, a second ball, with a mass of [tex]0.25 kg[/tex], is thrown straight upward from Earth's surface with an initial speed of [tex]15 m/s.\\[/tex]. They move along nearby lines and pass without colliding. At the end of [tex]2.0 s[/tex] the velocity of the center of mass of the two-ball system is:
A) [tex]11 m/s[/tex], down
B) [tex]11 m/s[/tex], up
C) [tex]15 m/s[/tex], down
D) [tex]15 m/s[/tex], up
E) [tex]20 m/s[/tex], down
A vector in the xy plane has a magnitude of 25 m and x component of 12 m. What is its angle with the positive x axis?
Answers
Final answer:
The angle a vector makes with the positive x-axis can be found using the cosine ratio; the angle is approximately 58.2 degrees for the vector with a magnitude of 25 m and an x-component of 12 m.
Explanation:
To determine the angle a vector makes with the positive x-axis, we can use trigonometry since we have the magnitude of the vector and the length of the x-component. Specifically, the cosine of the angle (θ) that the vector makes with the x-axis is equal to the x-component divided by the magnitude of the vector.
Here is the formula:
cos(θ) = (x-component) / (magnitude)
By substituting the given values:
cos(θ) = 12 m / 25 m
Now calculate the angle:
θ = cos⁻¹(12/25)
θ = 58.2° (approx.)
Therefore, the vector makes an angle of approximately 58.2 degrees with the positive x-axis.
A pressure cylinder has a diameter of 150-mm and has a 6-mm wall thickness. What pressure can this vessel carry if the maximum shear stress is not to exceed 25 MPa?
Answers
Answer:
p = 8N/mm2
Explanation:
given data ;
diameter of cylinder = 150 mm
thickness of cylinder = 6 mm
maximum shear stress = 25 MPa
we know that
hoop stress is given as =[tex]\frac{pd}{2t}[/tex]
axial stress is given as =[tex]\frac{pd}{4t}[/tex]
maximum shear stress = (hoop stress - axial stress)/2
putting both stress value to get required pressure
[tex]25 = \frac{ \frac{pd}{2t} -\frac{pd}{4t}}{2}[/tex]
[tex]25 = \frac{pd}{8t}[/tex]
t = 6 mm
d = 150 mm
therefore we have pressure
p = 8N/mm2
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.
Answers
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
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)?
Answers
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
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?
Answers
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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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.
Answers
Answer:
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Explanation:
A rocket sled accelerates at a rate of 49.0 m/s2 . Its passenger has a mass of 75.0 kg. (a) Calculate the horizontal component of the force the seat exerts against his body. Compare this with his weight using a ratio. (b) Calculate the direction and magnitude of the total force the seat exerts against his body.
Answers
Explanation:
It is given that,
Mass of the passenger, m = 75 kg
Acceleration of the rocket, [tex]a=49\ m/s^2[/tex]
(a) The horizontal component of the force the seat exerts against his body is given by using Newton's second law of motion as :
F = m a
[tex]F=75\ kg\times 49\ m/s^2[/tex]
F = 3675 N
Ratio, [tex]R=\dfrac{F}{W}[/tex]
[tex]R=\dfrac{3675}{75\times 9.8}=5[/tex]
So, the ratio between the horizontal force and the weight is 5 : 1.
(b) The magnitude of total force the seat exerts against his body is F' i.e.
[tex]F'=\sqrt{F^2+W^2}[/tex]
[tex]F'=\sqrt{(3675)^2+(75\times 9.8)^2}[/tex]
F' = 3747.7 N
The direction of force is calculated as :
[tex]\theta=tan^{-1}(\dfrac{W}{F})[/tex]
[tex]\theta=tan^{-1}(\dfrac{1}{5})[/tex]
[tex]\theta=11.3^{\circ}[/tex]
Hence, this is the required solution.
The horizontal component of the force the seat exerts against the passenger's body is 3675 N. The ratio of this force to the passenger's weight is 5. The total force the seat exerts has a magnitude of 3793 N.
Explanation:(a) To calculate the horizontal component of the force the seat exerts against the passenger's body, we can use Newton's second law, which states that force is equal to mass times acceleration. In this case, the mass of the passenger is 75.0 kg and the acceleration of the rocket sled is 49.0 m/s2. So the force exerted by the seat is:
Force = mass * acceleration
Force = 75.0 kg * 49.0 m/s2
Force = 3675 N
Now let's compare this force to the passenger's weight. The weight of an object is given by the formula:
Weight = mass * gravitational acceleration
Weight = 75.0 kg * 9.8 m/s2
Weight = 735 N
To find the ratio, we divide the force exerted by the seat by the weight of the passenger:
Ratio = Force / Weight
Ratio = 3675 N / 735 N
Ratio = 5
(b) The total force the seat exerts against the passenger's body has both a horizontal and vertical component. The direction of the total force is the same as the direction of the acceleration of the rocket sled. The magnitude of the total force can be found using the Pythagorean theorem:
Total Force = √(horizontal component2 + vertical component2)
Total Force = √(36752 + 7352)
Total Force = 3793 N
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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
Answers
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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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
Answers
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 dog, with a mass of 10.0 kg, is standing on a flatboat so that he is 22.5 m from the shore. He walks 7.8 m on the boat toward the shore and then stops. The boat has a mass of 46.0 kg. Assuming there is no friction between the boat and the water, how far is the dog from the shore now?
Answers
Answer:16.096
Explanation:
Given
mass of dog[tex]\left ( m_d\right )=10kg[/tex]
mass of boat[tex]\left ( m_b\right )=46kg[/tex]
distance moved by dog relative to ground=[tex]x_d[/tex]
distance moved by boat relative to ground=[tex]x_b[/tex]
Distance moved by dog relative to boat=7.8m
There no net force on the system therefore centre of mass of system remains at its position
0=[tex]m_d\times x_d+m_b\dot x_b[/tex]
0=[tex]10\times x_d+46\dot x_b[/tex]
[tex]x_d=-4.6x_b[/tex]
i.e. boat will move opposite to the direction of dog
Now
[tex]|x_d|+|x_b|=7.8[/tex]
substituting[tex] x_d [/tex]value
[tex]5.6|x_b|=7.8[/tex]
[tex]|x_b|=1.392m[/tex]
[tex]|x_d|=6.4032m[/tex]
now the dog is 22.5-6.403=16.096m from shore
The distance from the shore after the dog walks on the boat is calculated using the conservation of momentum. Since there are no external forces, the center of mass of the dog-boat system must stay the same. After the dog walks towards the shore, we determine the boat's movement in the opposite direction and calculate the dog's final distance from the shore.
Explanation:To solve how far the dog is from the shore after walking on the boat, we need to apply the law of conservation of momentum. Since there's no external force acting on the system (the dog plus the boat), the center of mass of the system will not move. Initially, the center of mass is stationary, and it should remain so as the dog walks.
We can calculate the initial position of the center of mass (CM) using the formula:
CM = (m × dog_position + M × boat_center_position) / (m + M)
As the dog walks 7.8 m towards the shore, the boat moves in the opposite direction to keep the center of mass of the system in the same place. Let's denote the distance that the boat moves as x. Using the conservation of momentum:
m × dog_walk = M × x
The distance from the shore after the dog walks = initial dog's distance from shore - dog's walk + boat's movement towards shore (which is x).
However, since we're keeping the center of mass stationary, boat's movement towards shore (x) is equal to the dog's walk distance (7.8 m) multiplied by the ratio of the dog's mass to the boat's mass.
Therefore:
x = (m/M) × dog_walk
And so, the final distance from the shore = 22.5m - 7.8m + x
A 64.9 kg sprinter starts a race with an acceleration of 3.89 m/s2. She keeps this acceleration for 17 m and then maintains the velocity for the remainder of the 100-m dash, what will be her time for the race?
Answers
Answer:
Time of race = 10.18 s
Explanation:
She keeps this acceleration for 17 m and then maintains the velocity for the remainder of the 100-m dash
Time to travel 17 m can be calculated
s = ut + 0.5at²
17 = 0 x t + 0.5 x 3.89 x t²
t = 2.96 s
Velocity after 2.96 seconds
v = 3.89 x 2.96 = 11.50 m/s
Remaining distance = 100 - 17 = 83 m
Time required to cover 83 m with a speed of 11.50 m/s
[tex]t=\frac{83}{11.50}=7.22s[/tex]
Time of race = 2.96 + 7.22 = 10.18 s
The distance between the lenses in a compound microscope is 18 cm. The focal length of the objective is 1.5 cm. If the microscope is to provide an angular magnification of -58 when used by a person with a normal near point (25 cm from the eye), what must be the focal length of the eyepiece?
Answers
Answer:
The focal length of the eyepiece is 4.012 cm.
Explanation:
Given that,
Focal length = 1.5 cm
angular magnification = -58
Distance of image = 18 cm
We need to calculate the focal length of the eyepiece
Using formula of angular magnification
[tex]m=-(\dfrac{i-f_{e}}{f_{0}})\dfrac{N}{f_{e}}[/tex]
Where,
[tex]i[/tex] = distance between the lenses in a compound microscope
[tex]f_{e}[/tex]=focal length of eyepiece
[tex]f_{0}[/tex]=focal length of the object
N = normal point
Put the value into the formula
[tex]-58=-(\dfrac{18-f_{e}\times25}{1.5\timesf_{e}})[/tex]
[tex]87f_{e}=450-25f_{e}[/tex]
[tex]f_{e}=\dfrac{450}{112}[/tex]
[tex]f_{e}=4.012\ cm[/tex]
Hence, The focal length of the eyepiece is 4.012 cm.
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.
Answers
(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 45 g bullet strikes and becomes embedded in a 1.55 kg block of wood placed on a horizontal surface just in front of the gun. If the coefficient of kinetic friction between the block and the surface is 0.28, and the impact drives the block a distance of 13.0 m before it comes to rest, what was the muzzle speed of the bullet in meters/second?
Answers
Answer:
299.51 m/s
Explanation:
m = mass of the bullet = 45 g = 0.045 kg
M = mass of the block = 1.55 kg
v = muzzle speed of the bullet
V = speed of bullet-block combination after the collision
μ = Coefficient of friction between the block and the surface = 0.28
d = distance traveled by the block = 13 m
V' = final speed of the bullet-block combination = 0 m/s
acceleration of the bullet-block combination due to frictional force is given as
a = - μg
using the kinematics equation
V'² = V² + 2 a d
0² = V² + 2 (- μg) d
0 = V² - 2 (μg) d
0 = V² - 2 (0.28) (9.8) (13)
V = 8.45 m/s
Using conservation of momentum for collision between bullet and block
mv = (M + m) V
(0.045) v = (1.55 + 0.045) (8.45)
v = 299.51 m/s
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?
Answers
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.
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
Answers
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:
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?
Answers
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
An astronaut leaves Earth in a spaceship at a speed of 0.96 c relative to an observer on Earth. The astronaut's destination is a star system 14.4 light-years away (one light-year is the distance light travels in one year.) According to the astronaut, how long does the trip take?
Answers
Answer:
t=4.2 years
Explanation:
velocity v= 0.96c
destination star distance = 14.4 light year
According to the theory of relativity length contraction
[tex]l= \frac{l_0}{\sqrt{1-\frac{v^2}{c^2}}}[/tex]
[tex]l_0= l{\sqrt{1-\frac{v^2}{c^2}}}[/tex]
putting values we get
[tex]l_0= 14.4{\sqrt{1-\frac{(0.96c)^2}{c^2}}}[/tex]
[tex]l_0= 4.032 light years[/tex]
now distance the trip covers
D= vt
[tex]4.032\times c= 0.96c\times t[/tex]
[tex]t= \frac{4.032c}{0.96c}[/tex]
t= 4.2 years
so the trip will take 4.2 years
Calculate the approximate radius of the 31P nucleus (assume r0 = 1.3x10-15 m).
Answers
Answer:
4.08 x 10⁻¹⁵ m
Explanation:
r₀ = constant of proportionality = 1.3 x 10⁻¹⁵ m
r = radius of the nucleus 31P = ?
A = mass number of the nucleus 31P = 31
Radius of the nucleus is given as
[tex]r=r_{o}A^{\frac{1}{3}}[/tex]
Inserting the values
[tex]r=(1.3\times 10^{-15})(31)^{\frac{1}{3}}[/tex]
[tex]r=(1.3\times 10^{-15})(3.14)[/tex]
r = 4.08 x 10⁻¹⁵ m
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?
Answers
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
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)
Answers
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 1m length of wire carrying a current of 7A lies on a horizontal table with a rectangular top of dimensions 0.6m x 0.8m. The ends of the wire are attached to opposite ends of a diagonal of the rectangle. A vertical magnetic field of 0.IT is present. What magnetic force acts on this segment of wire? a) 0.7N b) 0.98N c) 7N d) zero
Answers
Answer:
F = 0.7 N
Explanation:
As we know that the wire touch the ends of the diagonal of the rectangle
so here the length of the wire is given as
[tex]L = \sqrt{x^2 + y^2}[/tex]
[tex]L = \sqrt{0.6^2 + 0.8^2}[/tex]
[tex]L = 1 m[/tex]
now force due to magnetic field on current carrying wire is given as
[tex]F = iLB[/tex]
now we have
[tex]F = (7)(1)(0.1)[/tex]
[tex]F = 0.7 N[/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 ????.
Answers
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 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.
Answers
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
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.
Answers
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]
Electric Field of Ring of Charge A ring shaped charge has a radius of 0.250 meters and bears a total charge of +5.00 C What is the magnitude of the electric field along the axis of symmetry of the ring a distance of 0.500 meters from the center of the ring? ? A. 1.29 × 105 N/C 0 B. 1.67 × 105 N/C O C. 2.59 x 105 N/C ? D. 3.95 × 105 N/C
Answers
Answer:
6.44 × 10^10 N/C
Explanation:
Electric field due to the ring on its axis is given by
E = K q r / (r^2 + x^2)^3/2
Where r be the radius of ring and x be the distance of point from the centre of ring and q be the charge on ring.
r = 0.25 m, x = 0.5 m, q = 5 C
K = 9 × 10^9 Nm^2/C^2
E = 9 × 10^9 × 5 × 0.25 / (0.0625 + 0.25)^3/2
E = 6.44 × 10^10 N/C
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.
Answers
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.
During strengthening heat treatment, the _______ step traps the material in an unstable crystalline structure. a)-Quenching, b)-Tempering c)-Solution treatment
Answers
Answer: A) Quenching
Hope this helps
Light shines through a single slit whose width is 5.6 × 10-4 m. A diffraction pattern is formed on a flat screen located 4.0 m away. The distance between the middle of the central bright fringe and the first dark fringe is 3.3 mm. What is the wavelength of the light?
Answers
Answer:
462 nm
Explanation:
Given: width of the slit, d = 5.6 × 10⁻⁴ m
Distance of the screen, D = 4.0 m
Fringe width, β = 3.3 mm = 3.3 × 10⁻³ m
First dark fringe means n =1
Wavelength of the light, λ = ?
[tex] \beta = \frac{\lambda D}{d}\\ \Rightarrow \lambda = \frac{d \beta}{D} =\frac{5.6\times 10^{-4} \times 3.3 \times 10^{-3}}{4.0} = 4.62 \times 10^{-7}m = 462 nm[/tex]