Answer:
A) 12752.55 J
B) 12507.75 J
C) 12262.95 J
Explanation:
We are given;
Mass;m = 85 kg
Vertical distance; d = 15 m
From change in kinetic energy, the work done by the applied force to pull the spelunker is given by;
Change in kinetic energy = Wa + Wg
Where Wg = -mgd
A) In the first stage;the the work done is given by;
Wa = mgd + ½m(v_f)² - ½m(v_i)²
Since initially stationary, v_i = 0
So, we have;
Wa = mgd + ½m(v_f)²
v_f = 2.4 m/s
So,
Wa = (85 × 9.81 × 15) + ((1/2) × 85 × 2.4²)
Wa = 12752.55 J
B) for the second stage, there is a constant speed of 2.4 m/s
So, v_f = v_i
Thus; Wa = mgd
Wa = (85 × 9.81 × 15)
Wa = 12507.75 J
C) he finally decelerated to zero.
So v_f = 0 while v_i is 2.4 m/s
Thus;
Wa = mgd - ½m(v_i)²
Wa = (85 × 9.81 × 15) - (½ × 85 × 2.4²)
Wa = 12507.75 - 244.8
Wa = 12262.95 J
At a pressure of one atmosphere oxygen boils at −182.9°C and freezes at −218.3°C. Consider a temperature scale where the boiling point of oxygen is 100.0°O and the freezing point is 0°O. Determine the temperature on the Oxygen scale that corresponds to the absolute zero point on the Kelvin scale.
Answer: -254.51°O
Explanation:
Ok, in our scale, we have:
-182.9°C corresponds to 100° O
-218.3°C corresponds to 0°
Then we can find the slope of this relation as:
S = (100° - 0°)/(-182.9°C - (-218.3°C)) = 2.82°O/°C
So we can have the linear relationship between the scales is:
Y = (2.82°O/°C)*X + B
in this relation, X is the temperature in Celcius and Y is the temperature in the new scale.
And we know that when X = -182.9°C, we must have Y = 0°O
then:
0 = (2.82°O/°C)*(-182.9°C) + B
B = ( 2.82°O/°C*189.9°C) = 515.778°O.
now, we want to find the 0 K in this scale, and we know that:
0 K = -273.15°C
So we can use X = -273.15°C in our previous equation and get:
Y = (2.82°O/°C)*(-273.15°C) + 515.778°O = -254.51°O
A 50 g ice cube floats in 195 g of water in a 100 g copper cup; all are at a temperature of 0°C. A piece of lead at 96°C is dropped into the cup, and the final equilibrium temperature is 12°C. What is the mass of the lead?
Answer:
The mass of the lead will be "1.127 kg".
Explanation:
The given values are:
(Ice) m₁ = 50 g i.e.,
0.050 kg
(Water) m₂ = 195 g i.e.,
0.190 kg
(Copper cup) m₃ = 100 g i.e.,
0.100 kg
m₁, m₂ and m₃ at temperature,
t₁ = 0°C
Temperature of lead,
t₂ = 96°C
Temperature of Final equilibrium,
t₃ = 12°C
Let m₄ be the mass of the lead.
On applying formula, we get
⇒ [tex]m_{1}L+m_{1}s_{1} \Delta t+m_{2}s_{2} \Delta t+m_{2}s_{2} \Delta t=m_{4}s_{4} \Delta t[/tex]
On putting the estimated values, we get
⇒ [tex](0.050)(334)+(0.050)(4186)(12-0)+(0.190)(4186)(12-0)+(0.100)(387)(12-0)=m_{4} (128)(96-12)[/tex]
⇒ [tex]16.7+2511.6+9544.08+50.7=10752\times m_{4}[/tex]
⇒ [tex]12,123.08=10752\times m_{4}[/tex]
⇒ [tex]m_{4}=\frac{12,123.08}{10752}[/tex]
⇒ [tex]m_{4}=1.127 \ kg[/tex]
A 0.20 kg mass on a horizontal spring is pulled back 2.0 cm and released. If, instead, a 0.40 kg mass were used in this same experiment, the total energy of the system would A 0.20 mass on a horizontal spring is pulled back 2.0 and released. If, instead, a 0.40 mass were used in this same experiment, the total energy of the system would Be half as large. Double. Remain the same.
Answer:
The total mechanical energy does not change if the value of the mass is changed. That is, remain the same
Explanation:
The total mechanical energy of a spring-mass system is equal to the elastic potential energy where the object is at the amplitude of the motion. That is:
[tex]E=U=\frac{1}{2}kA^2[/tex] (1)
k: spring constant
A: amplitude of the motion = 2.0cm
As you can notice in the equation (1), the total mechanical energy of the system does not depend of the mass of the object. It only depends of the amplitude A and the spring constant.
Hence, if you use a mass of 0.40kg the total mechanical energy is the same as the obtained with a mas 0.20kg
Remain the same
After using a mass of 0.40 kg the total mechanical energy is the same as the obtained with a mass of 0.20 kg
If the object is at the amplitude of the motion, the total mechanical energy of a spring-mass system is equal to the elastic potential energy.
[tex]\bold {E = U =\dfrac 1{2} kA^2}[/tex]
Where,
k: spring constant
A: amplitude of motion = 2.0 cm
In the equation, the total mechanical energy only depends of the amplitude and the spring constant. It does not depend on the mass of the object. It
Therefore, after using a mass of 0.40 kg the total mechanical energy is the same as the obtained with a mass of 0.20 kg
To know more about mechanical energy,
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A 50 gram meterstick is placed on a fulcrum at its 50 cm mark. A 20 gram mass is attached at the 12 cm mark. Where should a 40 gram mass be attached so that the meterstick will be balanced in rotational equilibrium
Answer:
The 40g mass will be attached at 69 cm
Explanation:
First, make a sketch of the meterstick with the masses placed on it;
--------------------------------------------------------------------------
↓ Δ ↓
20 g.................50 cm.................40g
38 cm y cm
Apply principle of moment;
sum of clockwise moment = sum of anticlockwise moment
40y = 20 (38)
40y = 760
y = 760 / 40
y = 19 cm
Therefore, the 40g mass will be attached at 50cm + 19cm = 69 cm
12cm 50 cm 69cm
--------------------------------------------------------------------------
↓ Δ ↓
20 g.................50 cm.................40g
38 cm 19 cm
A bowling ball traveling with constant speed hits pins at the end of a bowling lane 16.5m long. The bowler hears the sound of the ball hitting the pins 2.65s after the ball is release from her hand. What is the speed of the ball down the lane, assuming that the speed of sound is 340.0m/s
Answer: The speed of the ball is 7.64 m/s.
Explanation:
The distance between the player and the pins is 16.5m
if the velocity of the ball is V, then the time in which the ball reaches the pins is:
T = 16.5/V
Now, after this point, the sound needs
T' = 16,5/340 = 0.049 seconds to reach the player, this means that the time in that the ball needs to reach te pins is:
2.65 s - 0.49s = 2.16s
Then we have:
T = 2.16s = 16.5/V
V = 16.5/2.16 m/s = 7.64 m/s
Two conductors made of the same material are connected across the same potential difference. Conductor A has seven times the diameter and seven times the length of conductor B. What is the ratio of the power delivere
Complete question:
Two conductors made of the same material are connected across the same potential difference. Conductor A has seven times the diameter and seven times the length of conductor B. What is the ratio of the power delivered to A to power delivered to B.
Answer:
The ratio of the power delivered to A to power delivered to B is 7 : 1
Explanation:
Cross sectional area of a wire is calculated as;
[tex]A = \frac{\pi d^2}{4}[/tex]
Resistance of a wire is calculated as;
[tex]R = \frac{\rho L}{A} \\\\R = \frac{4\rho L}{\pi d^2} \\\\[/tex]
Resistance in wire A;
[tex]R = \frac{4\rho _AL_A}{\pi d_A^2}[/tex]
Resistance in wire B;
[tex]R = \frac{4\rho _BL_B}{\pi d_B^2}[/tex]
Power delivered in wire;
[tex]P = \frac{V^2}{R}[/tex]
Power delivered in wire A;
[tex]P = \frac{V^2_A}{R_A}[/tex]
Power delivered in wire B;
[tex]P = \frac{V^2_B}{R_B}[/tex]
Substitute in the value of R in Power delivered in wire A;
[tex]P_A = \frac{V^2_A}{R_A} = \frac{V^2_A \pi d^2_A}{4 \rho_A L_A}[/tex]
Substitute in the value of R in Power delivered in wire B;
[tex]P_B = \frac{V^2_B}{R_B} = \frac{V^2_B \pi d^2_B}{4 \rho_B L_B}[/tex]
Take the ratio of power delivered to A to power delivered to B;
[tex]\frac{P_A}{P_B} = (\frac{V^2_A \pi d^2_A}{4\rho_AL_A} ) *(\frac{4\rho_BL_B}{V^2_B \pi d^2_B})\\\\ \frac{P_A}{P_B} = (\frac{V^2_A d^2_A}{\rho_AL_A} )*(\frac{\rho_BL_B}{V^2_B d^2_B})\\\\[/tex]
The wires are made of the same material, [tex]\rho _A = \rho_B[/tex]
[tex]\frac{P_A}{P_B} = (\frac{V^2_A d^2_A}{L_A} )*(\frac{L_B}{V^2_B d^2_B})\\\\[/tex]
The wires are connected across the same potential; [tex]V_A = V_B[/tex]
[tex]\frac{P_A}{P_B} = (\frac{ d^2_A}{L_A} )* (\frac{L_B}{d^2_B} )[/tex]
wire A has seven times the diameter and seven times the length of wire B;
[tex]\frac{P_A}{P_B} = (\frac{ (7d_B)^2}{7L_B} )* (\frac{L_B}{d^2_B} )\\\\\frac{P_A}{P_B} = \frac{49d_B^2}{7L_B} *\frac{L_B}{d^2_B} \\\\\frac{P_A}{P_B} =\frac{49}{7} \\\\\frac{P_A}{P_B} = 7\\\\P_A : P_B = 7:1[/tex]
Therefore, the ratio of the power delivered to A to power delivered to B is
7 : 1
A nonuniform electric field is given by the expression = ay î + bz ĵ + cx , where a, b, and c are constants. Determine the electric flux (in the +z direction) through a rectangular surface in the xy plane, extending from x = 0 to x = w and from y = 0 to y = h. (Use any variable or symbol stated above as necessary.)
A helium nucleus (charge = 2e, mass = 6.63 10-27 kg) traveling at 6.20 105 m/s enters an electric field, traveling from point circled A, at a potential of 1.50 103 V, to point circled B, at 4.00 103 V. What is its speed at point circled B?
Answer:
[tex]v_B=3.78\times 10^5\ m/s[/tex]
Explanation:
It is given that,
Charge on helium nucleus is 2e and its mass is [tex]6.63\times 10^{-27}\ kg[/tex]
Speed of nucleus at A is [tex]v_A=6.2\times 10^5\ m/s[/tex]
Potential at point A, [tex]V_A=1.5\times 10^3\ V[/tex]
Potential at point B, [tex]V_B=4\times 10^3\ V[/tex]
We need to find the speed at point B on the circle. It is based on the concept of conservation of energy such that :
increase in kinetic energy = increase in potential×charge
[tex]\dfrac{1}{2}m(v_A^2-v_B^2)=(V_B-V_A)q\\\\\dfrac{1}{2}m(v_A^2-v_B^2)={(4\times 10^3-1.5\times 10^3)}\times 2\times 1.6\times 10^{-19}=8\times 10^{-16}\\\\v_A^2-v_B^2=\dfrac{2\times 8\times 10^{-16}}{6.63\times 10^{-27}}\\\\v_A^2-v_B^2=2.41\times 10^{11}\\\\v_B^2=(6.2\times 10^5)^2-2.41\times 10^{11}\\\\v_B=3.78\times 10^5\ m/s[/tex]
So, the speed at point B is [tex]3.78\times 10^5\ m/s[/tex].
An object spins in a horizontal circle with a radius of 15.0 cm. The rotations are timed and the amount of time it takes for it to go around once is 0.56 s. The centripetal force is measured to be 6.1 N.According to the experiment, the speed of the object is closest to:'
Answer:
1.7 m/s
Explanation:
Relevant Data provided as per the question below:-
Radius = 15.0 cm
Time = 0.56 s.
Based on the above information
The computation of the speed of the object is shown below:-
[tex]Velocity = \frac{2\times \pi \times Radius}{Time}[/tex]
[tex]Velocity = \frac{2\times \frac{22}{7} \times 0.15}{0.56}[/tex]
[tex]= \frac{0.942857}{0.56}[/tex]
= 1.683 m/s
or
= 1.7 m/s
Therefore for computing the speed of the object or velocity we simply applied the above formula by considering the pi and all other given data
The drawing shows a top view of a hockey puck as it slides across frictionless ice. Three forces act on the puck, and it is in equilibrium. The force F is applied at the center and has a magnitude of 32 N. The force F1 is applied at the top edge, and F2 is applied half way between the center and the bottom edge. Find the magnitude of F1 and F2.
Answer:
The values of the forces are
[tex]F_1 = 10.6 \ N[/tex] , [tex]F_2 = 21.33 \ N[/tex]
Explanation:
The diagram for this question is shown on the first uploaded image
From the question we are told that
The magnitude of F is [tex]F = 32 \ N[/tex]
Generally at equilibrium the torque is mathematically evaluated as
[tex]\sum \tau = 0[/tex]
From the diagram we have
[tex]r * F_1 - [\frac{r}{2} ] F_2 + 0 F = 0[/tex]
=> [tex]F_1 = 0.5 F_2[/tex]
Generally at equilibrium the Force is mathematically evaluated as
[tex]\sum F = 0[/tex]
From the diagram
[tex]F - F_ 1 - F_2 = 0[/tex]
substituting values
[tex]32 - (0.5F_2 ) - F_2 = 0[/tex]
[tex]F_2 = 21.33 \ N[/tex]
So
[tex]F_1 = 0.5 * 21.33[/tex]
[tex]F_1 = 10.6 \ N[/tex]
In a remote location near the North Pole, an iceberg floats in a lake. Next to the lake (assume it is not frozen) sits a comparably sized glacier sitting on land. If both chunks of ice should melt (and the melted ice all goes into the lake), which ice chunk would give the greatest increase in the level of the lake water, if any?
Answer:
The correct answer to the following question will be "Glacier".
Explanation:
A large mass of frozen ground that flows very steadily through both the valley as well as spreads out through the middle. Throughout several centuries, glaciers have formed from compacted snow in places whereby snow builds up quicker although this melts.The iceberg has been on the bay, so there's no adjustment in a size equivalent to the whole iceberg.So that the above seems to be the right answer.
Answer:
Glaciers not only transport material as they move, but they also sculpt and carve away the land beneath them. A glacier's weight, combined with its gradual movement, can drastically reshape the landscape over hundreds or even thousands of years. The ice erodes the land surface and carries the broken rocks and soil debris far from their original places, resulting in some interesting glacial landforms.
Explanation:
How much force is needed to cause a 15 kilogram bicycle to accelerate at a rate of 10
meters per second per second?
O A. 15 newtons
OB. 1.5 newtons
C. 150 newtons
OD. 10 newtons
Two red blood cells each have a mass of 9.0 x 10-14 kg and carry a negative charge spread uniformly over their surfaces. The repulsion from the excess charge prevents the cells from clumping together. One cell carries -2.5pC and the other -3.30 pC, and each cell can be modeled as a sphere 3.75 × 10-6 m in radius. If the red blood cells start very far apart and move directly toward each other with the same speed.
1. What initial speed would each need so that they get close enough to just barely touch?
2. What is the maximum acceleration of the cells as they move toward each other and just barely touch?
Answer:
Explanation:
Given that:
The mass of the cell is 9.0 x 10^-14 kg
The charges of the cell is -2.5pC and the other -3.30 pC
[tex]q_1=-2.5\times10^{-12}C \ \ and \ \ q_2=-3.75\times10^{-12}C[/tex]
Radius is 3.75 × 10-6 m
The final distance is twice the radius
i.e [tex]2*(3.75 \times 10^{-6}) = 7.5*10^{-6}m[/tex]
The formula for the velocity of the cell is
[tex]mv^2=\frac{q_1q_2}{4\pi \epsilon 2 r} \\[/tex]
[tex]v=\sqrt{\frac{q_1q_2}{4\pi \epsilon 2 r} }[/tex]
[tex]=\sqrt{\frac{(-2.5\times10^{-12})(-3.3\times10^{-12}}{4(3.14)(8.85\times10^{-112}(2\times3.75\times10^{-6})(9\times10^{-14})} } \\\\=\sqrt{\frac{(-8.25\times10^{-24})}{(7503.03\times10^{-32})} } \\\\=\sqrt{109955.5779} \\\\=331.60m/s[/tex]
The maximum acceleration of the cells as they move toward each other and just barely touch is
[tex]ma= \frac{q_1q_2}{4\pi \epsilon (2r)^2} \\\\a= \frac{q_1q_2}{4\pi \epsilon (2r)^2(m)}[/tex]
[tex]=\frac{(-2.5\times10^{-12})(-3.3\times10^{-12})}{4(3.14)(8.85\times10^{-12})(2\times3.75\times10^{-6})^2(9\times10^{-14})}[/tex]
[tex]=\frac{(-8.25\times10^{-24})}{(56272.725\times10^{-38})} \\\\=1.47\times10^{10}m/s^2[/tex]
The answers obtained are;
1. The initial speed of each of the red blood cells is [tex]v= 331.66\,m/s[/tex].
2. The maximum acceleration of the cells is [tex]a=1.47\times 10^{10}\,m/s^2[/tex].
The answer is explained as shown below.
We have, the mass of the red blood cell;
[tex]m=9\times 10^{-14}\,kg[/tex]Also, the charges of the cells are;
[tex]q_1=-2.5\times 10^{-12}\,C[/tex] and[tex]q_2=-3.30\times 10^{-12}\,C[/tex]The distance between the charges when they barely touch will be two times the radius of each charge.
[tex]r=2\times r\,'=2\times3.75\times10^{-6}\,m=7.5\times10^{-6}\,m[/tex]Kinetic Energy of moving charges1. As both the cells are negatively charged they will repel each other.
So, for the cells to come nearly close, their kinetic energies must be equal to the electric potential between them.[tex]\frac{1}{2}mv^2+ \frac{1}{2}mv^2=k\frac{q_1 q_2}{r^2}[/tex]Where, [tex]k=9\times10^9\,Nm^2/C^2[/tex] is the Coulomb's constant.Now, substituting all the known values in the equation, we get;
[tex](9\times 10^{-14}\,kg)\times v^2=9\times 10^9Nm^2/C^2\times\frac{(-2.5\times 10^{-12}\,C)\times(-3.30\times 10^{-12}\,C)}{7.5\times10^{-6}\,m}[/tex][tex]v^2=9\times 10^9Nm^2/C^2\times\frac{(-2.5\times 10^{-12}\,C)\times(-3.30\times 10^{-12}\,C)}{7.5\times10^{-6}\,m\times(9\times 10^{-14}\,kg)} =110000\,m^2/s^2[/tex]
[tex]\implies v=\sqrt{110000\,m^2/s^2}=331.66\,m/s[/tex]Electrostatic force between two charges2. Also as the force between them is repulsive, there must be an acceleration to make them barely touch each other.
[tex]ma=k\frac{q_1 q_2}{r^2}[/tex]Substituting the known values, we get;
[tex](9\times 10^{-14}\,kg)\times a=9\times 10^9Nm^2/C^2\times\frac{(-2.5\times 10^{-12}\,C)\times(-3.30\times 10^{-12}\,C)}{(7.5\times10^{-6}\,m)^2}[/tex]
[tex]\implies a=9\times 10^9Nm^2/C^2\times\frac{(-2.5\times 10^{-12}\,C)\times(-3.30\times 10^{-12}\,C)}{(7.5\times10^{-6}\,m)^2\times(9\times 10^{-14}\,kg) }[/tex]
[tex]a=1.47\times 10^{10}\,m/s^2[/tex]Find out more information about moving charges here:
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The coefficient of linear expansion of steel is 11 x 10 perc . A steel ball has a volume of
exactly 100 cm at 0 C. When heated to 100 C, its volume becomes:
Question: The coefficient of linear expansion of steel is 11 x 10⁻⁶ per °c . A steel ball has a volume of
exactly 100 cm³ at 0 C. When heated to 100 C, its volume becomes:
Answer:
100.11 cm³
Explanation:
From the question,
γ = (v₂-v₁)/(v₁Δt)...................... Equation 1
Where γ = coefficient of volume expansion, v₂ = final volume, v₁ = initial volume, Δt = change in temperature.
make v₂ the subject of the equation
v₂ = v₁+γv₁Δt..................... Equation 2
Given: v₁ = 100 cm³, γ = 11×10⁻⁶/°C, Δt = 100 °C.
Substitute into equation 2
v₂ = 100+100(11×10⁻⁶)(100)
v₂ = 100+0.11
v₂ = 100.11 cm³
g Suppose a uniform slender rod has length L and mass m. The moment of inertia of the rod about about an axis that is perpendicular to the rod and that passes through its center of mass is given by Icm=112mL2. Find Iend, the moment of inertia of the rod with respect to a parallel axis through one end of the rod.
Answer:
[tex]I_e = \frac{1}{3}*m*L^2[/tex]
Explanation:
Solution:-
- Here we are given the moment of inertia of a uniform slender rod with mass ( m ) and length ( L ). The thickness / radius / diameter of the rod is considered to be insignificant.
- The moment of inertia ( Ir ) of a rod with an axis perpendicular to it at its center is given as:
[tex]I_r = \frac{1}{12}*m*L^2[/tex]
- We are to determine the moment of inertia of the rod at any one of its ends using the parallel axis theorem.
- The theorem is mostly used to translate the pivotal axis to any point on the mass or in space. With respect to that point the moment of inertia is determined using the parallel axis theorem. The moment of inertia of the object at its center of mass must be known to apply the theorem.
- The theorem is expressed as:
[tex]I_e = I_r + m*d^2[/tex]
Here,
d: Is the distance between the center of mass and the arbitrary point.
- Since we are asked to determine the moment of inertial at one of the rod's ends. We can evaluate the distance " d " from its center of mass to its end. The center lies at " L / 2 " distance from either of its ends. Hence, d = L / 2.
- We will plug in the parameters in the theorem and evaluate:
[tex]I_e = \frac{1}{12}*m*L^2 + m*[\frac{L}{2} ]^2 \\\\I_e = \frac{1}{12}*m*L^2 + m*\frac{L^2}{4} \\\\I_e = m*L^2 * [ \frac{1}{12}+ \frac{3}{12} ] = m*L^2 *\frac{4}{12} \\\\I_e = \frac{1}{3}*m*L^2[/tex]
A 4.5 kg ball swings from a string in a vertical circle such that it has constant sum of kinetic and gravitational potential energy. Ignore any friction forces from the air or in the string. What is the difference in the tension between the lowest and highest points on the circle
Answer:
88.29 N
Explanation:
mass of the ball = 4.5 kg
weight of the ball will be = mass x acceleration due to gravity(9.81 m/s^2)
weight W = 4.5 x 9.81 = 44.145 N
centrifugal forces Tc act on the ball as it swings.
At the top point of the vertical swing,
Tension on the rope = Tc - W.
At the bottom point of the vertical swing,
Tension on the rope = Tc + W
therefore,
difference in tension between these two points will be;
Net tension = tension at bottom minus tension at the top
= Tc + W - (Tc - W) = Tc + W -Tc + W
= 2W
imputing the value of the weight W, we have
2W = 2 x 44.145 = 88.29 N
A small, rigid object carries positive and negative 3.00 nC charges. It is oriented so that the positive charge has coordinates (−1.20 mm, 1.20 mm) and the negative charge is at the point (1.70 mm, −1.30 mm).
Required:
a. Find the electric dipole moment of the object.
b. The object is placed in an electric field E = (7.80 103 î − 4.90 103 ĵ). Find the torque acting on the object.
c. Find the potential energy of the object–field system when the object is in this orientation.
d. Assuming the orientation of the object can change, find the difference between the maximum and the minimum potential energies of the system,
Answer:
Umax = 105.8nJ
Umin =-105.8nJ
Umax-Umin = 211.6nJ
Explanation:
A mass of 50.00g hangs from a 7.00cm-long spring that is rigidly attached to a ceiling. The mass is pulled down slightly, let go, and is observed to make 8 round trips (up and back down) in 14.00s. What is the stiffness constant for this spring
Answer:
0.645 N/M
Explanation:
Given
Mass=50.00g
We have to convert into the kg
So Mass =0.050 Kg
[tex]Time\ = \frac{14}{8}\ = 1.75\ sec[/tex]
We know that
[tex]T\ =2\ PI\sqrt{\frac{M}{K} }[/tex]........................Eq(1)
Where T= time
and M= Mass
K= Stiffness constant
On squaring both side we get
[tex]K=\frac{4\pi^{2} M}{T^{2} }[/tex]....Eq(2)
Putting the value of M ,T and π in Eq(2) we get
K=0.645 N/M
If the magnitude of the magnetic field is 2.50 mT at a distance of 12.6 cm from a long straight current carrying wire, what is the magnitude of the magnetic field at a distance of 19.8 cm from the wire
Answer:
The magnetic field at a distance of 19.8 cm from the wire is 1.591 mT
Explanation:
Given;
first magnetic field at first distance, B₁ = 2.50 mT
first distance, r₁ = 12.6 cm = 0.126 m
Second magnetic field at Second distance, B₂ = ?
Second distance, r₂ = ?
Magnetic field for a straight wire is given as;
[tex]B = \frac{\mu I}{2 \pi r}[/tex]
Where:
μ is permeability
B is magnetic field
I is current flowing in the wire
r distance to the wire
[tex]Let \ \frac{\mu I}{2\pi} \ be \ constant; = K\\\\B = \frac{K}{r} \\\\K = Br\\\\B_1r_1 = B_2r_2\\\\B_2 =\frac{B_1r_1}{r_2} \\\\B_2 = \frac{2.5*10^{-3} *0.126}{0.198} \\\\B_2 = 1.591 *10^{-3}\ T\\\\B_2 = 1.591 \ mT[/tex]
Therefore, the magnetic field at a distance of 19.8 cm from the wire is 1.591 mT
A glass tube 1 mm in diameter is dipped into glycerin. The density of the glycerin is 1260 kg/m3, surface tension is 6.3x10-2 N/m, and the contact angle is zero. The capillary rise of the glycerin is most nearly:
Answer:
The capillary rise of the glycerin is most nearly [tex]y = 0.0204 \ m[/tex]
Explanation:
From the question we are told that
The diameter of the glass tube is [tex]d = 1 \ mm = 0.001 \ m[/tex]
The density of glycerin is [tex]\rho = 1260 \ kg /m^3[/tex]
The surface tension of the glycerin is [tex]\sigma = 6.3 *10^{-2} \ N /m[/tex]
The capillary rise of the glycerin is mathematically represented as
[tex]y = \frac{4 * \sigma * cos (\theta )}{ \rho * g * d}[/tex]
substituting value
[tex]y = \frac{4 * 6.3 *10^{-2} * cos (0 )}{ 1260 * 9.8 * 0.001}[/tex]
[tex]y = 0.0204 \ m[/tex]
Therefore the height of the glass tube the glycerin was able to cover is
[tex]y = 0.0204 \ m[/tex]
A 25 kg box is 220 N pulled at constant speed up a frictionless inclined plane by a force that is parallel to the incline. If the plane is inclined at an angle of 25o above the horizontal, the magnitude of the applied force is
Answer:
F = 103.54N
Explanation:
In order to calculate the magnitude of the applied force, you take into account that the forces on the box are the applied force F and the weight of the box W.
The box moves with a constant velocity. By the Newton second law you have that the sum of forces must be equal to zero.
Furthermore, you have that the sum of forces are given by:
[tex]F-Wsin\theta=0[/tex] (1)
F: applied force = ?
W: weight of the box = Mg = (25kg)(9.8m/s^2) = 245N
θ: degree of the incline = 25°
You solve the equation (1) for F:
[tex]F=Wsin\theta=(245N)sin(25\°)=103.54N[/tex] (2)
The applied force on the box is 103.54N
A device for acclimating military pilots to the high accelerations they must experience consists of a horizontal beam that rotates horizontally about one end while the pilot is seated at the other end. In order to achieve a radial acceleration of 27.9 m/s2 with a beam of length 5.21 m , what rotation frequency is required
Answer: the angular frequency is 2.31 rad/s
Explanation:
The data we have is:
Radial acceleration A = 27.9 m/s^2
Beam length r = 5.21m
The radial acceleration is equal to the velocity square divided the radius of the circle (the lenght of the beam in this case)
And we can write the velocity as:
v = w*r where r is the radius of the circle, and w is the angular frequency.
w = 2pi*f
where f is the "normal" frequency.
So we have:
A = (v^2)/r = (r*w)^2/r = r*w^2
We can replace the values and find w.
27.9m/s^2 = 5.21m*w^2
√(27.9/5.21) = w = 2.31 rad/s
Four fixed point charges are at the corners of a square with sides of length L. Q1 is positive and at (OL) Q2 is positive and at (LL) Q3 is positive and at (4,0) Q4 is negative and at (0,0) A) Draw and label a diagram of the described arrangement described above (include a coordinate system). B) Determine the force that charge Q1 exerts on charge Qz. C) Determine the force that charge Q3 exerts on charge Q2. D) Determine the force that charge Q4 exerts on charge Q2. E) Now assume that all the charges have the same magnitude (Q) and determine the net force on charge Q2 due to the other three charges. Reduce this to the simplest form (but don't put in the numerical value for the force constant).
Answer:
A) See Annex
B) Fq₁₂ = K * Q₁*Q₂ /16 [N] (repulsion force)
C) Fq₃₂ = K * Q₃*Q₂ /16 [N] (repulsion force)
D) Fq₄₂ = K * Q₄*Q₂ /32 [N] (attraction force)
E) Net force (its components)
Fnx = (2,59/64 )* K*Q² [N] in direction of original Fq₃₂
Fny =(2,59/64 )* K*Q² [N] in direction of original Fq₁₂
Explanation:
For calculation of d (diagonal of the square, we apply Pythagoras Theorem)
d² = L² + L² ⇒ d² = 2*L² ⇒ d = √2*L² ⇒ d= (√2 )*L
d = 4√2 units of length (we will assume meters, to work with MKS system of units)
B) Force of Q₁ exerts on charge Q₂
Fq₁₂ = K * Q₁*Q₂ /(L)² Fq₁₂ = K * Q₁*Q₂ /16 (repulsion force in the direction indicated in annex)
C) Force of Q₃ exerts on charge Q₂
Fq₃₂ = K * Q₃*Q₂ /(L)² Fq₃₂ = K * Q₃*Q₂ /16 (repulsion force in the direction indicated in annex)
D) Force of -Q₄ exerts on charge Q₂
Fq₄₂ = K * Q₄*Q₂ / (d)² Fq₄₂ = K * Q₄*Q₂ /32 (Attraction force in the direction indicated in annex)
E) Net force in the case all charges have the same magnitude Q (keeping the negative sign in Q₄)
Let´s take the force that Q₄ exerts on Q₂ and Q₂ = Q ( magnitude) and
Q₄ = -Q
Then the force is:
F₄₂ = K * Q*Q / 32 F₄₂ = K* Q²/32 [N]
We should get its components
F₄₂(x) = [K*Q²/32 ]* √2/2 and so is F₄₂(y) = [K*Q²/32 ]* √2/2
Note that this components have opposite direction than forces Fq₁₂ and
Fq₃₂ respectively, and that Fq₁₂ and Fq₃₂ are bigger than F₄₂(x) and F₄₂(y) respectively
In new conditions
Fq₁₂ = K * Q₁*Q₂ /16 becomes Fq₁₂ = K * Q²/ 16 [N] and
Fq₃₂ = K* Q₃*Q₂ /16 becomes Fq₃₂ = K* Q² /16 [N]
Note that Fq₁₂ and Fq₃₂ are bigger than F₄₂(x) and F₄₂(y) respectively
Then over x-axis we subtract Fq₃₂ - F₄₂(x) = Fnx
and over y-axis, we subtract Fq₁₂ - F₄₂(y) = Fny
And we get:
Fnx = K* Q² /16 - [K*Q²/32 ]* √2/2 ⇒ Fnx = K*Q² [1/16 - √2/64]
Fnx = (2,59/64 )* K*Q²
Fny has the same magnitude then
Fny =(2,59/64 )* K*Q²
The fact that Fq₁₂ and Fq₃₂ are bigger than F₄₂(x) and F₄₂(y) respectively, means that Fnx and Fny remains as repulsion forces
The robot HooRU is lost in space, floating around aimlessly, and radiates heat into the depths of the cosmos at the rate of 13.1 W. HooRU's surface area is 1.55 m2 and the emissivity of its surface is 0.287. Ignoring the radiation that HooRU absorbs from the cold universe, what is HooRU's temperature T?
Answer:
The temperature is [tex]T = 168.44 \ K[/tex]
Explanation:
From the question ewe are told that
The rate of heat transferred is [tex]P = 13.1 \ W[/tex]
The surface area is [tex]A = 1.55 \ m^2[/tex]
The emissivity of its surface is [tex]e = 0.287[/tex]
Generally, the rate of heat transfer is mathematically represented as
[tex]H = A e \sigma T^{4}[/tex]
=> [tex]T = \sqrt[4]{\frac{P}{e* \sigma } }[/tex]
where [tex]\sigma[/tex] is the Boltzmann constant with value [tex]\sigma = 5.67*10^{-8} \ W\cdot m^{-2} \cdot K^{-4}.[/tex]
substituting value
[tex]T = \sqrt[4]{\frac{13.1}{ 0.287* 5.67 *10^{-8} } }[/tex]
[tex]T = 168.44 \ K[/tex]
Cart A, with a mass of 0.20 kg, travels on a horizontal air trackat 3.0m/s and hits cart B, which has a mass of 0.40 kg and is initially traveling away from Aat 2.0m/s. After the collision the center of mass of the two cart system has a speed of:____________.A. zeroB. 0.33m/sC. 2.3m/sD. 2.5m/sE. 5.0m/s
Answer:
[tex]\large \boxed{\text{C. 2.3 m/s}}[/tex]
Explanation:
Data:
[tex]m_{\text{A} } = \text{0.20 kg};\,v_{\text{Ai}} = \text{3.0 m/s}\\m_{\text{B} } = \text{0.40 kg};\,v_{\text{Bi}} = \text{2.0 m/s}\\[/tex]
Calculation:
This is a perfectly inelastic collision. The two carts stick together after the collision and move with a common final velocity.
The conservation of momentum equation is
[tex]\begin{array}{rcl}m_{\text{A}}v_{\text{Ai}} +m_{\text{B}} v_{\text{Bi}}&=&(m_{\text{A}} + m_{\text{B}})v_{\text{f}}\\0.20\times 3.0 + 0.40\times 2.0 & = & (0.20 + 0.40)v_{\text{f}}\\0.60 + 0.80 & = & 0.60v_{\text{f}}\\1.40 & = & 0.60v_{\text{f}}\\v_{\text{f}}&=& \dfrac{1.40}{0.60}\\\\& = & \textbf{2.3 m/s}\\\end{array}\\\text{The centre of mass has a velocity of $\large \boxed{\textbf{2.3 m/s}}$}[/tex]
A 1.0-kg ball on the end of a string is whirled at a constant speed of 2.0 m/s in a horizontal circle of radius 1.5 m. What is the work done by the centripetal force during one revolution
Answer:
The work done by the centripetal force is always, zero.
Explanation:
The formula for the work done by a force on an object is given as follows:
W = F d Cos θ
where,
W = Magnitude of the Work Done
F = Force applied to the body
θ = Angle between the direction of force and direction of motion of the object
In case of the circular motion, the force is the centripetal force. The centripetal force is always directed towards the center of the circle. While, the object moves in a direction, which is tangential to the circle. Hence, the angle between them is always 90°. Therefore,
W = F d Cos 90°
W = F d (0)
W = 0 J
Hence, the work done by the centripetal force is always, zero.
Question 4
3 pts
I am approaching a traffic light at a speed of 135 km/h when I suddenly notice that
the light is red. I slam on my brakes and come to a stop in 4.29 seconds. What is the
acceleration of the car as I screech to a complete stop? (Note that an object that slows down
simply has a negative acceleration.)
& show work please I want to also understand
Answer:
The deceleration of the car is [tex]\approx -0.065m/s^{2}[/tex]
Explanation:
to solve this, we will have to apply the knowledge that will be got from the equations of motion.
There are several equations of motion, and depending on the parameters given in the problem, we can choose the perfect equation that can best be used to solve the problem.
In this case, since we are given the velocity and time, and we are solving for the acceleration, we will use this formula
[tex]v = u +at[/tex]
where v= final velocity = 0
u = initial velocity = 135Km/h [tex]\approx 0.278 m/s[/tex]
t= time = 4.29 seconds.
[tex]a = \frac{v - u}{t}[/tex]
[tex]a =\frac{0-0.278}{4.29} \approx 0.065m/s^{2}[/tex]
Hence, the deceleration of the car is [tex]\approx -0.065m/s^{2}[/tex]
Consider two identical containers. Container A is filled with water to the top. Container B has a block of wood floating in it, but the level of the water is also at the top. Which container weighs more
Answer:
Container A will weigh more
Explanation:
Both containers are identical, so we assume that they weigh the same.
They both have the same volume, and will contain an equal volume of a material.
Since they both contain water to the top, this means that their volume is fully occupied. But container B contain a block of wood floating in it.
The fact that the block of wood floats in the water in container B shows that it is less dense than the water around it, and in the container A, this same space is completely filled with water.
What we derive from this is that the portion of space contained by the block of wood in container B is occupied by water in container A, but, in container B, the density of this space is lesser now, since the wood block floats.
Since density is mass per unit volume, and weight is proportional to mass, then we can see that the weight of this volume portion in container B is lesser than that of container A. The consequence is that container A will weigh more than container B because of this extra weight.
Calculate the speed of a proton that is accelerated from rest through an electric potential difference of 114 V. m/s (b) Calculate the speed of an electron that is accelerated through the same potential difference. m/s
Answer:
A) v = 148,242.72 m/s
B) v = 6,328,025.58 m/s
Explanation:
To solve this, we will equate electric potential to kinetic energy.
Formula for Electric potential is qV where q is charge and V is potential difference.
While formula for kinetic energy is ½mv² where m is mass and v is velocity
Thus;
qV = ½mv²
Let us make the velocity the formula;
v = √(2qV/m)
A) PROTON
Charge of proton has a constant value of 1.6 × 10^(-19) C
Mass of proton has a constant value of 1.66 × 10^(-27) kg
We are given that potential difference = 114 V.
So, v = √(2qV/m)
Thus; v = √(2*1.6 × 10^(-19)*114/(1.66 × 10^(-27)))
v = 148,242.72 m/s
B) ELECTRON
Charge of electron has a constant value of 1.6 × 10^(-19) C
Mass of electron has a constant value of 9.11 × 10^(-31) kg
v = √(2qV/m)
Thus;
v = √(2*1.6*10^(-19)*114)/(9.11 × 10^(-31)))
v = 6,328,025.58 m/s
what is the orbital speed for a satellite 3.5 x 10^8m from the center of mars? Mars mass is 6.4 x 10^23 kg
Answer:
v = 349.23 m/s
Explanation:
It is required to find the orbital speed for a satellite [tex]3.5\times 10^8\ m[/tex] from the center of mass.
Mass of Mars, [tex]M=6.4\times 10^{23}\ kg[/tex]
The orbital speed for a satellite is given by the formula as follows :
[tex]v=\sqrt{\dfrac{GM}{r}} \\\\v=\sqrt{\dfrac{6.67\times 10^{-11}\times 6.4\times 10^{23}}{3.5\times 10^8}} \\\\v=349.23\ m/s[/tex]
So, the orbital speed for a satellite is 349.23 m/s.