Answer:
Specific gravity of other fluid = .854 (Approx)
Explanation:
Given:
Mass of water = 35 g
Mass of filled bottle with water = 98.44 g
Mass of filled bottle with fluid = 89.22 g
Computation:
Mass of water = 98.44g - 35g = 63.44g
Density of water = 1000 g/L
Volume of bottle = 63.44/1000 = 0.06344L
Mass of other liquid = 89.22g - 35g = 54.22g
Density of other liquid = 54.22g/0.06344L = 854.665826 g/L
Water has a specific gravity = 1
So , specific gravity of other fluid
1000 / 854.665826 = 1 / specific gravity of other fluid
Specific gravity of other fluid = .854 (Approx)
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]
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
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 bar slides vertically between two conducting rails without friction in a magnetic field. The rails are connected via a resistor. The bar reaches a constant velocity and slides for 2m. How much energy is dissipated in the resistor during these 2m's
Answer:
The energy dissipated is [tex]E = 9.8 J[/tex]
Explanation:
From the question we are told that
The distance covered at constant velocity d = 2 m
The velocity is [tex]v = 1.5 \ m/s[/tex]
Generally at constant velocity the magnetic force on the bar is mathematically represented as
[tex]F = m * g[/tex]
substituting values
[tex]F = 0.5 * 9.8[/tex]
[tex]F = 4.9 \ N[/tex]
The energy dissipated is mathematically evaluate as
[tex]E = F * d[/tex]
substituting the value
[tex]E = 4.9 * 2[/tex]
[tex]E = 9.8 J[/tex]
Water vapor is less dense than ice because:
a. molecules in the gas phase are in constant motion.
b. molecules in the gas phase have more potential energy than in solids.
c. molecules in the gas phase have more kinetic energy than in solids.
d. gaseous molecules have less mass.
e. molecules in the gas phase have more space between them than in solids,
Answer:
The correct answer is option E
Explanation:
Relative density of the different phases of the same compound like water are basically determined by their number of molecules per volume when each of the molecules have the same mass in each of their phases.
Now, for the water vapor phase, it's molecules have very little interaction with themselves and so they are at large distance apart, whereas in ice(solid), molecules are in continuous contact with each other because they are at close distance between each other. Therefore, it's obvious that there are less molecules per liter in water vapour than in ice, and thus the density is smaller.
The correct answer is option E
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
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
That 85 kg paratrooper from the 50's was moving at constant speed of 56 m/s because the air was applying a frictional drag force to him that matched his weight. If he fell this way for 40 m, how much heat was generated by this frictional drag force in J
Answer:
46648 J
Explanation:
mass m= 85 Kg
velocity v = 56 m/s
distance covered s =40 m
According to Question,
frictional drag force to him that matched his weight
[tex]\Rightarrow F_d =mg\\=85\times9.81=833 N[/tex]
Therefore, work done by practometer against the drag force = heat was generated by this frictional drag force in J
W=Q= F_d×s
=833×56 = 46648 J
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:
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 you go to the beach on a hot summer day, the temperature of the sand is much higher than the temperature of the water. If we assume the same amount of energy was supplied by the sun to both the sand and the water, does sand or water require more energy to raise its temperature?
Water requires more energy to raise its temperature than sand does. In fact, of all the common substances that we see around us every day, water is one of the BEST at storing heat energy.
This is a big part of the reason why we use frozen water to cool our soda, instead of cold wood or cold steel balls.
It's also a big part of the reason why we warm up the bed in the Winter with a hot water bag, instead of a bag of hot rocks or hot BBs.
On a hot summer day, the temperature of the sand is much higher than the temperature of the water. The same amount of energy was supplied by the sun to both the sand and the water, but the water required more energy to raise its temperature than the sand.
What is "specific heat"?The specific heat of any substance is explained by the amount of heat required to increase the temperature by 1 degree; here, the specific heat of water is much higher than that of sand. The sand needs 670 joules of energy to raise the temperature, while the water needs nearly 3800 joules of energy to raise one degree of temperature.
Despite the fact that the sun cast the same amount of light on both water and sand, sand heated up faster than water. The water has a high latent heat of vaporization, which means it needs more energy to vaporize. The animal body maintains homeostasis as a result of this water.
Hence, water requires more energy to raise the temperature due to its high specific heat.
Learn more about the specific heat, here
https://brainly.com/question/11297584
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In a contest, two tractors pull two identical blocks of stone thesame distance over identical surfaces. However, block A is moving twice as fast as block B when it crosses the finish line. Which statement is correct?a) Block A has twiceas much kinetic energy as block B.b) Block B has losttwice as much kinetic energy to friction as block A.c) Block B has losttwice as much kinetic energy as block A.d) Both blocks havehad equal losses of energy to friction.e) No energy is lostto friction because the ground has no displacement.
Answer:
d) Both blocks have had equal losses of energy to friction
Explanation:
As it is mentioned in the question that two tractors pull two same stone blocks having the identical distance over the same surfaces
Moreover, the block A is twice as fast than block B at the time of crossing the finish line
So based on the above information, it contains the losses of identical friction
And we also know that
Friction energy loss is
[tex]= \mu \times m \times g \times D[/tex]
It would be the same for both the blocks
hence, the option d is correct
The correct answer will be both blocks have had equal losses of energy to friction.
What is friction?Friction is defined as when any object is slides on a surface by means of any external force then the force in the opposite direction generated between the surface and the body restrict the motion of the body this force is called as the friction.
As it is mentioned in the question that two tractors pull two same stone blocks having the identical distance over the same surfaces.
Moreover, the block A is twice as fast as block B at the time of crossing the finish line.
So based on the above information, it contains the losses of identical friction.
And we also know that
Friction energy loss is
[tex]E_f=\mu m g D[/tex]
It would be the same for both the blocks
Hence both blocks have had equal losses of energy to friction.
To know more about friction, follow
https://brainly.com/question/24386803
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
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]
Which formula can be used to find velocity if kinetic energy and mass are known? v = The square root of one half times K E and mass. v = Square Root of StartFraction 2 times m over K E EndFraction. v = Square Root of K E times m v = Square Root of StartFraction 2 times K E over m EndFraction.
Answer:
v = Square Root of StartFraction 2 times K E over m EndFraction.
Explanation:
Kinetic Energy ,K.E = 1/2 × mass × velocity squared
K.E = 1/2 × mv2
Hence v2 = (2× K.E)/m
v = √ 2K.E/m
Expressing it in computer language we have :
v = Square Root of StartFraction 2 times K E over m EndFraction.
Answer:
v = Square Root of StartFraction 2 times K E over m EndFraction
Explanation:
also d on edgenuity:)
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]
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.
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
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
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.
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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which of the following terms refers to the amount of thermal energy need to change 1 kg of a substance from a liquid to a gas at its boiling point
Answer:
See the answer below.
Explanation:
"Latent Heat", also called the "Heat of Vaporization", is the amount of energy necessary to change a liquid to a vapour at constant temperature and pressure.
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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
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 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]
The larger the push, the larger the change in velocity. This is an example of Newton's Second Law of Motion which states that the acceleration an object experiences is
Answer:
According to Newtons 2nd law of motion ;
The acceleration an object experiences is as a result of the net force which is directly proportional to the magnitude of the net force, in the same direction as the net force, and inversely proportional to the mass of the object.
Explanation:
This law is simply saying ;
Force = Mass ×Acceleration
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An object of mass 3.07 kg, moving with an initial velocity of 5.07 m/s, collides with and sticks to an object of mass 2.52 kg with an initial velocity of -3.11 m/s. Find the final velocity of the composite objec
Answer:
This is an inelastic collision. This means, unfortunately, that KE cannot save you, at least in the problem's current form.
Let's see what conservation of momentum in both directions does ya:
Conservation in the x direction:
Only 1 object here has a momentum in the x direction initally.
m1v1i + 0 = (m1 + m2)(vx)
3.09(5.10) = (3.09 + 2.52)Vx
Vx = 2.81 m/s
Explanation:
Conservation in the y direction:
Again, only 1 object here has initial velocity in the y:
0 + m2v2i = (m1 +m2)Vy
(2.52)(-3.36) = (2.52 + 3.09)Vy
Vy = -1.51 m/s
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Now that you have Vx and Vy of the composite object, you can find the final velocity by doing Vf = √Vx^2 + Vy^2)
Vf = √(2.81)^2 + (-1.51)^2
Vf = 3.19 m/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
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
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Two astronauts, of masses 60 kg and 80 kg, are initially right next to each other and at rest in outer space. They suddenly push each other apart. What is their separation after the heavier astronaut has moved 12m
Answer:
The astronauts are separated by 28 m.
Explanation:
The separation of the astronauts can be found by conservation of linear momentum:
[tex] p_{i} = p_{f} [/tex]
[tex] m_{1}v_{1i} + m_{2}v_{2i} = m_{1}v_{1f} + m_{2}v_{2f} [/tex]
[tex] m_{1}*0 + m_{2}*0 = m_{1}v_{1f} + m_{2}v_{2f} [/tex]
[tex] m_{1}v_{1f} = -m_{2}v_{2f} [/tex]
[tex] v_{1f} = -\frac{m_{2}v_{2f}}{m_{1}} = -\frac{80v_{2f}}{60} [/tex]
Now, the distance (x) is:
[tex] x = \frac{v}{t} [/tex]
The distance traveled by the astronaut 1 is:
[tex] x_{1} = v_{1f}*t = -\frac{80v_{2f}}{60}*t [/tex] (1)
And, the distance traveled by the astronaut 2 is:
[tex] x_{2} = v_{2f}*t [/tex] (2)
From the above equation we have:
[tex] t = \frac{x_{2}}{v_{2f}} [/tex] (3)
By entering equation (3) into (1) we have:
[tex] x_{1} = -\frac{80v_{2f}}{60}*(\frac{x_{2}}{v_{2f}}) [/tex]
[tex] x_{1} = -\frac{4*12}{3} = -16 m [/tex]
The minus sign is because astronaut 1 is moving in the opposite direction of the astronaut 2.
Finally, the separation of the astronauts is:
[tex] x_{T} = |x_{1}| + x_{2} = (16 + 12)m = 28 m [/tex]
Therefore, the astronauts are separated by 28 m.
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The total separation between the two astronauts is 28m.
The given parameters:
masses of the astronauts, = 60 kg and 80 kgApply the principle of conservation of momentum to determine the final velocity of each astronauts as follows;
[tex]m_1u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2\\\\60(0) + 80(0) = 60(v_1) + 80(v_2)\\\\0 = 60v_1 + 80v_2\\\\-60v_1 = 80v_2\\\\v_1 = \frac{-80v_2}{60} \\\\v_1 = -1.333v_2[/tex]
Let the time when astronaut 2 moved 12 m = t
The distance traveled by astronaut 1 is calculated as;
[tex]x_1 = v_1 t\\\\x_1 = -1.333v_2t[/tex]
The distance traveled by astronaut 2 is calculated as;
[tex]x_2 = v_2 t\\\\12 = v_2t\\\\t = \frac{12}{v_2}[/tex]
Now solve for the distance of astronaut 1
[tex]x_1 = - 1.333v_2 \times t\\\\x_1 = -1.333 v_2 \times \frac{12}{v_2} \\\\x_1 = -16 \ m[/tex]
The total separation between the two astronauts is calculated as follows;
[tex]d = |x_1| + x_2\\\\d = 16 + 12\\\\d = 28 \ m[/tex]
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