Answer:
[tex]\Delta S=-0.11\frac{kJ}{kg*K}[/tex]
Explanation:
Hello,
In this case, we can compute the entropy change by using the following equation containing both pressure and temperature:
[tex]\Delta S=Cp\ ln(\frac{T_2}{T_1} )-R\ ln(\frac{p_2}{p_1} )[/tex]
Thus, we use the given data to obtain (2 MPa = 2000 kPa):
[tex]\Delta S=0.846\frac{kJ}{kg*K} \ ln(\frac{800K}{400K} )-0.1889\frac{kJ}{kg*K} \ ln(\frac{2000kPa}{50kPa} )\\\\\Delta S=-0.11\frac{kJ}{kg*K}[/tex]
Best regards.
When a spinning bike wheel is placed horizontally, hung from a pivot at one end, the axis of rotation of the wheel will swing in a horizontal circle. In which direction does it turn?a) upwardb) downwardc) horizontally, CWd) horizontally, CCW
Answer:
answer is D
Explanation:
horizontally, CCW
A 1.53-kg piece of iron is hung by a vertical ideal spring. When perturbed slightly, the system is moves up and down in simple harmonic oscillations with a frequency of 1.95 Hz and an amplitude of 7.50 cm. If we choose the total potential energy (elastic and gravitational) to be zero at the equilibrium position of the hanging iron, what is the total mechanical energy of the system
Answer:
E = 0.645J
Explanation:
In order to calculate the total mechanical energy of the system, you take into account that if the zero of energy is at the equilibrium position, then the total mechanical energy is only the elastic potential energy of the spring.
You use the following formula:
[tex]E=U_e=\frac{1}{2}kA^2[/tex] (1)
k: spring constant = ?
A: amplitude of the oscillation = 7.50cm = 0.075m
The spring constant is given by:
[tex]f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}[/tex]
[tex]k=4\pi^2f^2m[/tex] (2)
f: frequency of the oscillation = 1.95Hz
m: mass of the piece of iron = 1.53kg
You replace the expression (1) into the equation (2) and replace the values of all parameters:
[tex]E=\frac{1}{2}(4\pi^2f^2m)A^2=2\pi^2f^2mA^2\\\\E=2\pi^2(1.95Hz)^2(1.53kg)(0.075m)^2=0.645J[/tex]
The totoal mechanical energy of the system is 0.645J
Two 10-cm-diameter charged rings face each other, 18.0 cmcm apart. Both rings are charged to 30.0 nCnC . What is the electric field strength
Answer:
E=7453.99 V/m
Explanation:
The electric field on the charged is given by
E= Kqx/(r^2 +x^2)^3/2
Where;
K= constant of Coulomb's law
q= magnitude of charge= 30.0×10^-9 C
r= radius of the rings= 5 cm or 0.05m
x= distance between the rings = 18cm = 0.18 m
Substituting values;
E= 9.0×10^9 × 30.0×10^-9 × 0.18 / [(0.05^2 + (0.18)^2]^3/2
E= 48.6/(2.5×10^-3 + 0.0324)^3/2
E= 48.6/(0.0025 + 0.0324)^3/2
E= 48.6/6.52×10^-3
E=7453.99 V/m
Two moons orbit a planet in nearly circular orbits. Moon A has orbital radius r, and moon B has orbital radius 16r. Moon A takes 10 days to complete one orbit. How long does it take moon B to complete an orbit
Answer:
Kepler's Third Law: The square of the period of any planet about the sun is proportional to cube of its mean distance from the sun.
Mathematically: T^2 = K R^3
So (TA / TB)^2 = (RA / RB)^3
TB^2 = TA^2 * (RB / RA)^3
TB^2 = 10^2 * 16^3
TB = (409600)^1/2 = 640 days
How would the magnetic field lines appear for a bar magnet cut at the midpoint, with the two pieces placed end to end with a space in between such that the cut edges are closest to each other? What would the general shape of the field lines look like? What would the field lines look like in between the two pieces?
Answer:
Explanation:
check this out and rate me
a playground merry-go-round of radius r = 2.20 m has a moment of inertia i = 245 kg · m2 and is rotating at 11.0 rev/min about a frictionless vertical axle. facing the axle, a 26.0-kg child hops onto the merry-go-round and manages to sit down on the edge. what is the new angular speed of the merry-go-round?
Answer:
8.92 rpm
Explanation:
Given that
Radius of the merry go round, r = 2.2 m
Initial moment of inertia, I1 = 245 kgm²
Initial speed of rotation, w1 = 11 rpm
Mass of the child, m = 26 kg
To solve the problem, we use the law conservation of momentum
I1w1 = I2w2
I2 = mr² + I1
I2 = 245 + 26 * 2.2
I2 = 245 + 57.2
I2 = 302.2 kgm²
Now, applying the formula, we have
I1w1 = I2w2
245 * 11 = 302.2 * w2
w2 = 2695 / 302.2
w2 = 8.92 rpm
Thus, the new angular speed of the merry go round is 8.92 rpm
A 0.410 cm diameter plastic sphere, used in a static electricity demonstration, has a uniformly distributed 35.0 pC charge on its surface. What is the potential (in V) near its surface
Answer:
The potential is [tex]V = 153.659 \ V[/tex]
Explanation:
From the question we are told that
The diameter of the plastic sphere is [tex]d = 0.410 \ cm = 0.0041 \ m[/tex]
The magnitude of the charge is [tex]q = 35.0 pC = 35.0 *10^{-12} \ C[/tex]
The radius of the plastic sphere is mathematically evaluated as
[tex]r = \frac{d}{2}[/tex]
=> [tex]r = \frac{0.0041}{2}[/tex]
[tex]r = 0.00205 \ m[/tex]
The potential near the surface is mathematically represented as
[tex]V = \frac{k * q}{r }[/tex]
Where k is the Coulombs constant with value [tex]9 *10^{9} \ kg\cdot m^3\cdot s^{-4} \cdot A^{-2}.[/tex]
substituting values
[tex]V = \frac{9*10^9 * 35 *10^{-12}}{0.00205}[/tex]
[tex]V = 153.659 \ V[/tex]
Which of the following is not a benefit of improved cardiorespiratory fitness
Answer:
C - Arteries grow smaller
Explanation:
The option choices are:
A. Faster post-exercise recovery time
B. Lungs expand more easily
C. Arteries grow smaller
D. Diaphragm grows stronger
Explanation:
There are many advantages of cardiorespiratory fitness. It can decrease the risk of heart disease, lung cancer, type 2 diabetes, stroke, and other diseases. Cardiorespiratory health helps develop lung and heart conditions and enhances feelings of well-being.
A ball bouncing against the ground and rebounding is an example of an elastic collision. Describe two different methods of evaluating this interaction, one for which momentum is conserved, and one for which momentum is not conserved. Explain your answer.
Answer:
Momentum is conserved when there are no outside forced present and it has an equal and opposite reaction, also momentum is conserved the ball's momentum is transferred to the ground. This first instance is the case of a Closed system.
The second case where momentum is not conserved is when there is a variation or difference in the moment of the ball because of influence of external forces
What would be the Roche limit (in units of Earth radii) if the Earth had the same mass, but its radius was increased to 1.5 Earth radii?
First calculate the density of this new, larger, Earth. Now use this new density and the new radius in the calculator above to determine the Roche limit for this new larger 'Earth.
Answer:
Roche limit = 1.89 of earth radius
Explanation:
We know that,
Mass of earth = 5.972 × 10²⁷ g
New radius = 1.5(old radius) = 1.5(6.371 × 10⁸) = 9.5565 × 10⁸
Density of earth = 5.5132 g/cm³
New density of earth = Mass of earth / (4/3)πr³
New density of earth = 5.972 × 10²⁷ kg / (4/3)(22/7)( 9.5565 × 10⁸)³
New density of earth = 1.634 g/cm³
Roche limit = [2(Density of earth)/(New density of earth)]¹/³r
Roche limit = 1.89 of earth radius
hich muscle fibers are best suited for activities that involve lifting large, heavy objects for a short period of time? cardiac slow twitch intermediate fast twitch
Answer:
Dead lifting uses tho muscle fundamentals
Explanation:
Answer:
Fast twitch
Explanation:
Edmentum
A 0.140-kg baseball is dropped from rest. It has a speed of 1.20 m/s just before it hits the ground, and it rebounds with an upward speed of 1.00 m/s. The ball is in contact with the ground for 0.0140 s.
Required:
What is the average force exerted by the ground on the ball during this time? Also explain whether it's upwards or downwards.
Answer:
22 N upward
Explanation:
From the question,
Applying newton's second law of motion
F = m(v-u)/t....................... Equation 1
Where F = Average force exerted by the ground on the ball, m = mass of the baseball, v = final velocity, u = initial velocity, t = time of contact
Note: Let upward be negative and downward be positive
Given: m = 0.14 kg, v = -1.00 m/s, u = 1.2 m/s, t = 0.014 s
Substitute into equation 1
F = 0.14(-1-1.2)/0.014
F = 0.14(-2.2)/0.014
F = 10(-2.2)
F = -22 N
Note the negative sign shows that the force act upward
A satellite orbits the earth at a distance of 1.50 × 10^{7} m above the planetʹs surface and takes 8.65 hours for each revolution about the earth. The earthʹs radius is 6.38 × 10^{6} m. The acceleration of this satellite is closest to
Answer:
a = 0.43 m/s²
Explanation:
First, we need to find the velocity of the satellite:
Velocity = V = Distance Covered/Time Taken
here,
Distance = 1 revolution = 2π(1.5 x 10⁷ m) = 9.43 x 10⁷ m
Time = (8.65 hours)(3600 s/1 hour) = 31140 s
Therefore,
V = (9.43 x 10⁷ m)/(31140 s)
V = 3028.26 m/s
Now, the acceleration of the satellite will be equal to the centripetal acceleration, with the center of circular motion as the center of earth:
a = V²/R
where,
R = 1.5 x 10⁷ m + 0.638 x 10⁷ m
R = 2.138 x 10⁷ m
Therefore,
a = (3028.26 m/s)²/(2.138 x 10⁷ m)
a = 0.43 m/s²
A rod 16.0 cm long is uniformly charged and has a total charge of -25.0 µC. Determine the magnitude and direction of the electric field along the axis of the rod at a point 42.0 cm from its center.
Answer:
-1.4x10^6N/C
Explanation:
Pls see attached file
The magnitude of the electric field.
Magnitude is the size of the object in properties that is determines the size of the object. It also displays the result of the order of class of the object. The direction of the electric field tells us about the position of the field in four different directions. As per the question, the answer is 1.4x10^6N/C.
The rod of 16cm of total length is given. Has a charge of a total of -25.0uc. The rod's axis is pointed at 42.0 cm from its center and is given in the question. The rod Length will be then 0.16m and the total change will be 25x10 cm and point where the electricity will be calculated is shown by the axis of the rod at the distance of 42 cms.The magnitude and direction will be calculated based on the measure of the formula of E. This answer to the question will be 1.4x10^6N/C.Learn more about the uniformly charged.
brainly.com/question/12088419.
Calculate the change in internal energy of the following system: A balloon is cooled by removing 0.652 kJkJ of heat. It shrinks on cooling, and the atmosphere does 389 JJ of work on the balloon. Express your an
Question:
Calculate the change in internal energy of the following system: A balloon is cooled by removing 0.652 kJ of heat. It shrinks on cooling, and the atmosphere does 389 J of work on the balloon. Express your answer in Joules (J)
Answer:
-263J
Explanation:
Though its difficult and infact impossible to measure the internal energy of a system, the change in internal energy ΔE, can however be determined. This change when it is accompanied by work(W) and transfer of heat(Q) in or out of the system, can be calculated as follows;
ΔE = Q + W ----------------(i)
Q is negative if heat is lost. It is positive otherwise
W is negative if work is done by the system. It is positive otherwise.
From the question;
Q = -0.652kJ = -652J {the negative sign shows heat loss}
W = +389J {the positive sign shows work done on the system(balloon)}
Substitute these values into equation (i) as follows;
ΔE = -652 + 389
ΔE = -263J
Therefore the change in internal energy is -263J
PS: The negative sign shows that the process is exothermic. This means that the system (balloon) lost some energy to the environment.
what is quantic fisic
Answer:
it is the physics that explains how everything works. The best description we have of the. nature of the particles that make up matters and the forces with which they interact. It underlines how atoms work, and so why chemistry and biology work as they do
If an ocean wave takes 2 minutes to generate one full cycle, from crest to crest, then we will call this the frequency of the wave. What would its period be
Answer:
Explanation:
The period of the wave will be 2 minutes
frequency = 1 / period
= 1 / 2
= .5 per minute
When a nucleus at rest spontaneously splits into fragments of mass m1 and m2, the ratio of the momentum of m1 to the momentum of m2 is
Answer:
p₁ = - p₂
the moment value of the two particles is the same, but its direction is opposite
Explanation:
When a nucleus divides spontaneously, the moment of the nucleic must be conserved, for this we form a system formed by the initial nucleus and the two fragments of the fission, in this case the forces during the division are internal and the moment is conserved
initial instant. Before fission
p₀ = 0
since they indicate that the nucleus is at rest
final moment. After fission
[tex]p_{f}[/tex] = m₁ v₁ + m₂ v₂
p₀ = p_{f}
0 = m₁ v₁ + m₂v₂
m₁ v₁ = -m₂ v₂
p₁ = - p₂
this indicates that the moment value of the two particles is the same, but its direction is opposite
What is the electric flux Φ3 through the annular ring, surface 3? Express your answer in terms of C , r1, r2, and any constants.
Answer:
Pls see attached file
Explanation:
The electric flux through annular ring is parallel to the surface everywhere hence the angle of filled lines will be π/2 and thus cosine of this angle is zero leads to the electric flux Φ3 = 0.
What is electrical flux?Electric flux the flow of electric field lines that passing over a given area in in unit time. This is actually the field line density in a surface. This physical quantity is dependent on the magnitude of field, radius of the object if it is a ring and the charge.
Let the area of an infinitesimal surface be dA and the field acting is E then flux is the dot product E(r) .dA.
The field respect to a position r for the radius r1 is written as follows:
E(r) = (C/r² )
where, c is a proportionality constant for r.
The integrand equation for the electric flux is written as follows:
Ф3= E(r).dA = E(r).dA cos ∅
Consider the surface 3 in the annular ring where dA is normal to the field E(r) and the electric field is parallel to everywhere in the surface so the angle will be π/2. Thus ,cos π/2 is zeo making Ф3.
To find more about electric flux, refer the link below:
https://brainly.com/question/14850656
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Your question is incomplete. But your complete question includes the image attached with the answer.
1. An unknown charged particle passes without deflection through crossed electric and magnetic fields of strengths 187,500 V/m and 0.1250 T, respectively. The particle passes out of the electric field, but the magnetic field continues, and the particle makes a semicircle of diameter 25.05 cm.
Part A. What is the particle's charge-to-mass ratio?
Part B. Can you identify the particle?
a. can't identify
b. proton
c. electron
d. neutron
2. A neutral lithium atom has three electrons. Two of these electrons form an "inner core," but the third-the valence electron-orbits at a much larger radius. From the valence electron's perspective, it is orbiting a spherical ball of charge having net charge +1e (i.e., the three protons in the nucleus and the two inner-core electrons). The energy required to ionize a lithium atom is 5.14 eV. Hint: Consider the energy needed to remove the electron and the force needed to give the electron a circular orbit.
Part A. According to Rutherford's nuclear model of the atom, what is the orbital radius of the valence electron? Express your answer with the appropriate units.
Part B. According to Rutherford's nuclear model of the atom, what is the speed of the valence electron? Express your answer with the appropriate units.
Answer:
a. q/m = 9.59 ×10⁷ C/kg
b. for proton = 9.58 ×10⁷ C/kg
for electron = 1.758 ×10¹¹ C/kg
for neutron, q= 0, q/m= 0
Explanation:
In his experiments on "cathode rays" during which he discovered the electron, J.J. Thomson showed that the same beam deflections resulted with tubes having cathodes made of different materials and containing various gases before evacuation.
A) Are these observations important? Explain.
B) When he applied various potential differences to the deflection plates and turned on the magnetic coils, alone or in combination the fluorescent screen continued to show a single small glowing patch. Argue whether his observation is important.
C) Do calculations to show that the charge-to-mass ratio Thomson obtained was huge compared with that of any macroscopic object or of any ionized atom or molecule.
D) Could Thomson observe any deflection of the beam due to gravitation?
Answer:
A) his observation is of little importance ,
B) This observation is very important since the movement of the point of light depends on the relationship between the already magnetic electric force
C) see that in this second case it is 4 times less
D) the force of gravity is of the order of 10⁻⁴⁰
therefore it is 10²⁸ times less than the electric force,
Explanation:
A) This observation is of little importance since the cacodylate ray tube always emits electrons, regardless of the material of which it is made.
B) This observation is very important since the movement of the point of light depends on the relationship between the already magnetic electric force
C) the elect's load is 1.6 10⁻¹⁹ C its mass is 9.1 10⁻³¹ kg, let's look for its relation
e / m = 1.6 10⁻¹⁹ / 9.1 10⁻³¹
e / m = 0.1 758 10 10¹² N
look for this in the case of an atom, let's use the lightest atom hydrogen
the homogenize have an electron of charge 1.6 10⁻¹⁹ C
and a mass of 1.6735575 10⁻²⁷ ka
e / M = 1.6 10⁻⁻¹⁹ / 1.67 10⁻²⁷
e / M = 0.96 10⁸ N
We see that in this second case it is 4 times less
D) the force of gravity is of the order of 10⁻⁴⁰
therefore it is 10²⁸ times less than the electric force, therefore it should not contribute to the movement of the light beam
In a nuclear physics experiment, a proton (mass 1.67×10^(−27)kg, charge +e=+1.60×10^(−19)C) is fired directly at a target nucleus of unknown charge. (You can treat both objects as point charges, and assume that the nucleus remains at rest.) When it is far from its target, the proton has speed 2.50×10^6 m/s. The proton comes momentarily to rest at a distance 5.31×10^(−13)m from the center of the target nucleus, then flies back in the direction from which it came. What is the number of the protons the nucleus has? Assume no electron cloud is there, ε0=8.85x10^(-12) C^2/(Nm^2)
The given question is incomplete. The complete question is as follows.
In a nuclear physics experiment, a proton (mass [tex]1.67 \times 10^(-27)[/tex]kg, charge +e = [tex]+1.60 \times 10^(-19)[/tex] C) is fired directly at a target nucleus of unknown charge. (You can treat both objects as point charges, and assume that the nucleus remains at rest.) When it is far from its target, the proton has speed [tex]2.50 \times 10^6[/tex] m/s. The proton comes momentarily to rest at a distance [tex]5.31 \times 10^(-13)[/tex] m from the center of the target nucleus, then flies back in the direction from which it came. What is the electric potential energy of the proton and nucleus when they are [tex]5.31 \times 10^{-13}[/tex] m apart?
Explanation:
The given data is as follows.
Mass of proton = [tex]1.67 \times 10^{-27}[/tex] kg
Charge of proton = [tex]1.6 \times 10^{-19} C[/tex]
Speed of proton = [tex]2.50 \times 10^{6} m/s[/tex]
Distance traveled = [tex]5.31 \times 10^{-13} m[/tex]
We will calculate the electric potential energy of the proton and the nucleus by conservation of energy as follows.
[tex](K.E + P.E)_{initial}[/tex] = [tex](K.E + P.E)_{final}[/tex]
[tex](\frac{1}{2} m_{p}v^{2}_{p}) = (\frac{kq_{p}q_{t}}{r} + 0)[/tex]
where, [tex]\frac{kq_{p}q_{t}}{r} = U = Electric potential energy[/tex]
U = [tex](\frac{1}{2}m_{p}v^{2}_{p})[/tex]
Putting the given values into the above formula as follows.
U = [tex](\frac{1}{2}m_{p}v^{2}_{p})[/tex]
= [tex](\frac{1}{2} \times 1.67 \times 10^{-27} \times (2.5 \times 10^{6})^{2})[/tex]
= [tex]5.218 \times 10^{-15} J[/tex]
Therefore, we can conclude that the electric potential energy of the proton and nucleus is [tex]5.218 \times 10^{-15} J[/tex].
Two 2.0-cm-diameter insulating spheres have a 6.70 cm space between them. One sphere is charged to +70.0 nC, the other to -40.0 nC. What is the electric field strength at the midpoint between the two spheres?
Answer:
Explanation:
The distance of middle point from centres of spheres will be as follows
From each of 2 cm diameter sphere
R = 1 + 6.7 / 2 = 4.35 cm = 4.35 x 10⁻² m
Expression for electric field = Q / 4πε R²
Electric field due to positive charge
E₁ = 70 x 10⁻⁹ x 9 x 10⁹ / 4.35² x 10⁻⁴
= 33.3 x 10⁴ N/C
Electric field due to negative charge
E₂ = 40 x 10⁻⁹ x 9 x 10⁹ / 4.35² x 10⁻⁴
= 19.02 x 10⁴ N/C
E₁ and E₂ act in the same direction so
Total field = (33.3 + 19.02 ) x 10⁴
= 52.32 x 10⁴ N/C .
A wet shirt is put on a clothesline to dry on a sunny day. Do water molecules lose heat and condense, gain heat and condense or gain heat and evaporate
For a wet shirt is put on a clothesline to dry on a sunny day, water molecules gain heat and evaporate.
When a clothe is placed on a line to dry, the idea is to ensure that the water molecules should evaporate.
For the water molecules to evaporate, they must gain more energy that will enable them to transit from liquid to gaseous state.
Recall that he change from liquid to vapor requires energy, this is why water molecules gain energy when they evaporate.
Learn more: https://brainly.com/question/5019199
As an ice skater begins a spin, his angular speed is 3.14 rad/s. After pulling in his arms, his angular speed increases to 5.94 rad/s. Find the ratio of teh skater's final momentum of inertia to his initial momentum of inertia.
Answer:
I₂/I₁ = 0.53
Explanation:
During the motion the angular momentum of the skater remains conserved. Therefore:
Angular Momentum of Skater Before Pulling Arms = Angular Momentum of Skater After Pulling Arms
L₁ = L₂
but, the formula for angular momentum is:
L = Iω
Therefore,
I₁ω₁ = I₂ω₂
I₂/I₁ = ω₁/ω₂
where,
I₁ = Initial Moment of Inertia
I₂ = Final Moment of Inertia
ω₁ = Initial Angular Velocity = 3.14 rad/s
ω₂ = Final Angular velocity = 5.94 rad/s
Therefore,
I₂/I₁ = (3.14 rad/s)/(5.94 rad/s)
I₂/I₁ = 0.53
A tube of water is open on one end to the environment while the other end is closed. The height of the water relative to the base is 100 cm on the open end and 40 cm on the closed end. What is the absolute pressure of the water at the top of the closed end in units of atm
Answer:
1.06 atm
Explanation:
On the open end of the tube, the pressure will be the sum of atmospheric pressure and the pressure due to the height of water
The pressure due to a height of water = ρgh
where ρ is the density of water = 1000 kg/m^3
g is the acceleration due to gravity = 9.81 m/s^2
h is the height of the water column
The height of water column on the open end = 100 cm = 1 m
pressure on this end = ρgh = 1000 x 9.81 x 1 = 9810 Pa
Atmospheric pressure = 101325 Pa
The total pressure on the open end = 101325 Pa + 9810 Pa = 111135 Pa
The pressure due to the water column on the closed end = ρgh
The height of the water in the closed end = 40 cm = 0.4 m
The pressure due to this column of water = 1000 x 9.81 x 0.4 = 3924 Pa
The resultant pressure on the water on the top of the closed end of the tube = 111135 Pa - 3924 Pa = 107211 Pa
In atm unit, this pressure = 107211/101325 = 1.06 atm
A crate resting on a horizontal floor (\muμs = 0.75, \muμk = 0.24 ) has a horizontal force F = 93 Newtons applied to the right. This applied force is the maximum possible force for which the crate does not begin to slide. If you applied this same force after the crate is already sliding, what would be the resulting acceleration (in meters/second2) ?
Answer:
The acceleration is [tex]a = 5 \ m/s^2[/tex]
Explanation:
From the question we are told that
The coefficient of kinetic friction is [tex]\mu_k = 0.24[/tex]
The coefficient of static friction is [tex]\mu_s = 0.75[/tex]
The horizontal force is [tex]F_h = 93 \ N[/tex]
Generally the static frictional force is mathematically represented as
[tex]F_F = \mu_s * (m * g )[/tex]
The static frictional force is the equivalent to the maximum possible force for which the crate does not begin to slide So
[tex]F_h = F_F = \mu_s * (m * g )[/tex]
=> [tex]93 = \mu_s * (m * g )[/tex]
=> [tex]m = \frac{93}{\mu_s * g }[/tex]
substituting values
[tex]m = \frac{93}{0.75 * 9.8 }[/tex]
[tex]m = 12.65 \ kg[/tex]
When the crate is already sliding the frictional force is
[tex]F_s = \mu_k *(m * g )[/tex]
substituting values
[tex]F_s = 0.24 * 12.65 * 9.8[/tex]
[tex]F_s = 29.82 \ N[/tex]
Now the net force when the horizontal force is applied during sliding is
[tex]F_{net} = F_h - F_s[/tex]
substituting values
[tex]F_{net} = 93 - 29.8[/tex]
[tex]F_{net} = 63.2 \ N[/tex]
This net force is mathematically represented as
[tex]F_{net } = m * a[/tex]
Where a is the acceleration of the crate
So
[tex]a = \frac{F_{net}}{m }[/tex]
[tex]a = \frac{ 63.2}{12.65 }[/tex]
[tex]a = 5 \ m/s^2[/tex]
Let us suppose the magnitude of the original Coulomb force between the two charged spheres is FF. In this scenario, a third sphere touches the grey sphere and the red sphere multiple times, being grounded each touch. If the grey sphere is touched twice, and the red sphere is touched three times, what is the magnitude of the Coulomb force between the spheres now
Answer:
F ’= 1/32 F
We see that the value of the force is the initial force over 32
Explanation:
In this problem the sphere that is touching the others is connected to ground, after each touch,
Let's analyze the charge of the gray sphere, when you touch it for the first time, the charge is divided between the two spheres each having Q / 2, when the sphere separates and touches ground, its charge passes zero. When I touch the gray dial again, its charge is reduced by half
½ (Q / 2) = ¼ Q
For the red dial repeat the same scheme
with the first touch the charge is reduced to Q / 2
with the second touch e reduce to ½ (Q / 2) = ¼ Q
with the third toce it is reduced to ½ (¼ Q) = ⅛ Q
Now let's analyze what happens to the electric force
if the force is F for when the charge of each sphere is Q
F = k Q Q / r²
with the remaining charge strength is
F ’= k (¼ Q) (⅛ Q) / r²
F ’= 1/32 k Q Q / r²
F ’= 1/32 F
We see that the value of the force is the initial force over 32
A juggler throws two balls up to the same height so that they pass each other halfway up when A is rising and B is descending. Ignore air resistance and buoyant forces. Which statement is true of the two balls at that point?a. The only force acting on each ball is the gravitational force. b. There is an residual upward force from the hand on each ball. c. Only gravity acts on B but there is an additional residual force from the hand on A. d. There is a greater residual force from the hand on A than there is on B. e. There is an additional downwards force besides gravity on each ball.
Answer:
Only gravity acts on B but there is an additional residual force from the hand on A.
Explanation:
All bodies are constantly under the effect of gravity. Gravity is what gives us weight here on earth. Gravity acts downwards, and helps to decrease the deceleration of a body moving up and accelerating a body that travels downwards. For the ball A, traveling upwards, the upwards movement is due to the force on it impacted on it from the hand. As A tries to go up, gravity tries to decelerate it until it will come to a stop and then fall downwards under gravity. For body B, descending down means that only gravity forces acts on it at that point, if we ignore buoyant forces and air resistance. And B accelerates as it falls down towards the juggler.
You push very hard on a heavy desk, trying to move it. You do work on the desk: You push very hard on a heavy desk, trying to move it. You do work on the desk: whether or not it moves, as long as you are exerting a force. only if it starts moving. never-it does work on you. only if it doesn't move. None of the above.
Answer:
Only if it starts moving
Explanation:
Work done is defines as [tex]W=F.d=Fdcos\Theta[/tex]
In two case work done will be zero
First case is that when force and displacement are perpendicular to each other
What's more, other case is that when there is no displacement
So for work to be done there must have displacement, in the event that there is no displacement then there is no work done