Answer:
Explanation:
Let T be the tension in the string .
For hanging mass
net force = m₁g - T
m₁g - T = m₁a
For mass placed on horizontal plane
T = m₂a
m₁g - m₂a = m₁a
m₁g = a ( m₂ +m₁)
a = m₁g / ( m₂ +m₁)
= 5 x 9.8 / 19
= 2.579 m /s²
m₁g - T = m₁a
T = m₁g - m₁a
= m₁ ( g - a )
= 5 ( 9.8 - 2.579)
= 36.10 N
Suppose the truck that’s transporting the box In Example 6.10 (p. 150) is driving at a constant speed and then brakes and slows at a constant acceleration. While coming to a stop, the driver looks in the rear-view mirror and notices that the box is not slipping. In what direction is the frictional force acting on the box?
Answer:
Friction acts in the opposite direction to the motion of the truck and box.
Explanation:
Let's first review the problem.
A moving truck applies the brakes, and a box on it does not slip.
Now when the truck is applying brakes, only it itself is being slowed down. Since the box is slowing down with the truck, we can conclude that it is friction that slows it down.
The box in the question tries to maintains its velocity forward when the brakes are applied. We can think of this as the box exerting a positive force relative to the truck when the brakes are applied. When we imagine this, we can also figure out where the static friction will act to stop this positive force. Friction will act in the negative direction. Or in other words, friction will act in the opposite direction to the motion of the truck and box. This explains why the box slows down with the truck, as friction acts to stop its motion.
Final answer:
The frictional force acting on the box in a decelerating truck acts forward, towards the direction of the truck's original motion. This occurs as a result of trying to prevent the box from slipping backward by opposing its tendency to remain in motion, illustrating key concepts from Newton's laws of motion.
Explanation:
When the truck brakes and slows down at a constant acceleration, the box inside does not slip due to the frictional force acting on it. The direction of the frictional force on the box will be forward, towards the direction of the truck's original motion.
This might initially seem counterintuitive, but it's essential to understand that the frictional force is what keeps the box from sliding backward as the truck decelerates. According to Newton's first law of motion, an object in motion will stay in motion with the same speed and in the same direction unless acted upon by an unbalanced external force.
In this case, the frictional force acts as that external force, opposing the box's tendency to remain in motion while the truck slows down.
Friction, in physics, acts opposite to the direction of motion and is necessary for stopping the movement of objects. Since the truck is decelerating, the box tries to maintain its state of motion (Newton's first law), but the frictional force acts forward relative to the truck to prevent the box from slipping backward.
This example beautifully demonstrates the role of friction in everyday phenomena, aligning with concepts like Newton's laws of motion and the interaction between surfaces in contact.
Calculate the time required (in years) for water to penetrate a layer of clay that is 40 cm deep when exposed to a hydraulic gradient of 1 cm/cm. The permeability of clay is 1x10-8 cm/sec. If there is 30 cm static head of water on the clay layer, how long will it take for moisture to penetrate the 40 cm clay layer (in years)?
Answer:
The time required to penetrate the 40 cm clay layer is 126.82 years
Explanation:
Given:
Hydraulic gradient = 1 [tex]\frac{cm}{cm}[/tex]
Permeability of clay [tex]K = 1 \times 10^{-8}[/tex] [tex]\frac{cm}{sec}[/tex]
Time required to penetrate 40 cm clay layer,
[tex]= \frac{40}{K}[/tex]
[tex]= \frac{40}{1 \times 10^{-8} }[/tex]
[tex]= 40 \times 10^{8}[/tex] sec
But we have to find time in years,
[tex]= \frac{40 \times 10^{8} }{24 \times 60 \times 60 \times 365}[/tex] years
[tex]= 126.82[/tex] years
Therefore, the time required to penetrate the 40 cm clay layer is 126.82 years
Answer:
its a
Explanation:
3 In a tempering process, glass plate, which is initially at a uniform temperature Ti , is cooled by suddenly reducing the temperature of both surfaces to Ts. The plate is 20 mm thick, and the glass has a thermal diffusivity of 6 107 m2 /s.
Answer:
t=63s
tempersture gradient -2.36 x 10⁴
Explanation:
A girl of mass 55 kg throws a ball of mass 0.80 kg against a wall. The ball strikes the wall horizontally with a speed of 25 m/s, and it bounces back with this same speed. The ball is in contact with the wall 0.050 s. What is the magnitude of the average force exerted on the wall by the ball?
Answer:
[tex]F=800N[/tex]
Explanation:
The average force is defined as the mass of the body multiplied by its average velocity over the contact time. According to the third Newton's law, the magnitude of the average force exerted on the wall by the ball is equal to the magnitude of the average force exerted on the ball by the wall. Thus:
[tex]F=m\frac{v_f-v_i}{t}\\F=0.80kg\frac{25\frac{m}{s}-(-25\frac{m}{s})}{0.05s}\\F=800N[/tex]
Answer:
800 N
Explanation:
Force: This can be defined as the product of mass and acceleration. The S.I unit of force is Newton(N).
From the question,
F = m(v-u)/t ....................... Equation 1
Where F = force exerted on the wall by the ball, m = mass of the ball, v = final velocity, u = initial velocity, t = time.
Note: Assuming the direction of the initial motion to be negative
Given: m = 0.80 kg, v = 25 m/s ( bounce back), u = -25 m/s, t = 0.05 s
Substitute into equation 1
F = 0.8[25-(25)]/0.05
F = 0.8(50)/0.05
F = 0.8(1000)
F = 800 N
Suppose you analyze standardized test results for a country and discover almost identical distributions of physics scores for female and male students. Which of the following would NOT be an explanation based upon what you read in the textbook? Select one:
O a. Physics ability is less likely than math to reflect real biological differences.
O b. There were no gender differences in the teaching and learning of physics in this country
c. Physics ability is not defined as gendered in this country
O d . The standardized testing took place in a relatively gender equal society.
The textbook would not likely suggest 'Physics ability is less likely than math to reflect real biological differences' as an explanation for identical physics scores between genders. This because research shows academic abilities aren't influenced by biological gender differences but rather other factors like sociocultural influences, teaching methods, and individual variance.
Explanation:The statement that would not be an explanation for finding the almost identical distribution of physics scores between male and female students, based on information provided in a textbook, is a. Physics ability is less likely than math to reflect real biological differences.
This claim implies there is a biological basis for differing abilities in physics between genders. However, research has largely debunked this concept, showing that abilities in academic subjects are not influenced by biological gender differences. Instead, other factors like sociocultural influences, teaching methods, and individual variance play a significant role.
Options b, c, and d all refer to sociocultural aspects such as the way physics is taught, the societal concept of gender roles, and the equality of the society. These, not inherent biological differences, could explain identical distributions of physics scores between genders.
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In a transpiration experiment, the air bubble has an initial volume of 4.33 mL, and an initial pressure of 101.9 kPa. What will be the pressure reading after the plant transpires 0.26 mL of water from the tubing
Answer:
Explanation:is in the picture below
A proton (being 1836 times heavier than an electron) gains how much energy when moving through a potential increase of one volt?
Answer:
1.6*10⁻¹⁹ J = 1 eV
Explanation:
As the potential is defined as energy per unit charge, the increase in the electrical potential energy of a proton, can be written as follows:[tex]\Delta U_{e} = e*\Delta V[/tex]
where e = 1.6*10⁻¹⁹ C, and ΔV = 1 V = 1 J/C⇒ [tex]\Delta U_{e} = e*\Delta V = 1.6e-19 C* 1.0 V = 1.6e-19 J = 1 eV[/tex]
As the electrical potential energy only depends on the charge and the voltage, the mass has no influence.Explanation:
Below is an attachment containing the solution.
Teams red and blue are having a tug-of-war. According to Newton's third law, the force with which the red team pulls on the blue team exactly equals the force with which the blue team pulls on the red team.
Answer: The statement is TRUE.
Explanation: The Newton's third law states that a body that experiments an external force responds with an opposite force with same magnitude. Therefore, the statement is true.
A 20-kg child starts at the center of a playground merry-go-round that has a radius of 1.5 mm and rotational inertia of 500kg⋅m2 and walks out to the edge. The merry-go-round has a rotational speed of 0.20 s−1 when she is at the center.
What is its rotational speed when she gets to the edge?
Answer:
w = 0.189 rad/ s
Explanation:
This exercise we work with the conservation of the moment, the system is made up of the merry-go-round and the child, for which we write the moment of two instants
Initial
L₀ = I₀ w₀
Final
[tex]L_{f}[/tex] = I w
L₀ = L_{f}
I₀ w₀ = I_{f} w
.w = I₀/I_{f} w₀
The initial moment of inertia is
I₀ = 500 kg. m2
The final moment of inertia
[tex]I_{f}[/tex] = 500 + m r²
I_{f} = 500 + 20 1.5
I_{f} = 530 kg m²
Initial angular velocity
w₀ = 0.20 rad / s
Let's calculate
w = 500/530 0.20
w = 0.189 rad / s
The atoms that constitute your body are mostly empty space, and structures such as the chair you're sitting on are composed of atoms that are also mostly empty space. So why don't you fall through the chair
Explanation:
As we all know that the atom mainly composed of two parts, Nucleus at the center and the electrons dancing around it. And, yes there is a vast distance between nucleus and the electrons and it's about ten billionth of a meter. So the question is if there is such an empty space then why not our body just move through the chair and fall? So the answer is, it's all due to the forces. Just like the cracking of lightning through the void, specks of electrons and nuclei are constantly interacting through the force called electromagnetic forces. During each interaction there is an exchange of energy particles called photons. So the each photon swapped is equals to the push or pull or the forces exerted across the emptiness. So these forces enable us to sit on chair or to do other things.
7–14 Engine oil at 80°C flows over a 12-m-long flat plate whose temperature is 30°C with a velocity of 3 m/s. Determine the total drag force and the rate of heat transfer over the entire plate per unit width.
Answer:
Total drag force is 88.74 N
Rate of heat transfer over the entire plate per unit width is 26451.6 W
Explanation:
The rate of heat transfer over the entire plate per unit width can be obtained by applying the Newton's law of cooling. The pictures attached are step by step solution to the question;
A hockey puck is sliding across a frozen pond with an initial speed of 9.3 m/s. It comes to rest after sliding a distance of 42.0 m. What is the coefficient of kinetic friction between the puck and the ice?
Answer:
The coefficient of kinetic friction between the puck and the ice is 0.11
Explanation:
Given;
initial speed, u = 9.3 m/s
sliding distance, S = 42 m
From equation of motion we determine the acceleration;
v² = u² + 2as
0 = (9.3)² + (2x42)a
- 84a = 86.49
a = -86.49/84
|a| = 1.0296
[tex]F_k = \mu_k N[/tex] = ma
where;
Fk is the frictional force
μk is the coefficient of kinetic friction
N is the normal reaction = mg
μkmg = ma
μkg = a
μk = a/g
where;
g is the gravitational constant = 9.8 m/s²
μk = a/g
μk = 1.0296/9.8
μk = 0.11
Therefore, the coefficient of kinetic friction between the puck and the ice is 0.11
A client with a head injury is being monitored for increased intracranial pressure (ICP). The client's blood pressure is 90/60 mm Hg and the ICP is 18 mm Hg; therefore their cerebral perfusion pressure (CPP) is
Answer:
[tex]CPP=52mm \ Hg[/tex]
Explanation:
Cerebral Perfusion Pressure is obtained by subtracting IntraCranial Pressure(ICP) from the Mean Arterial Pressure(MAP). Adequate cerebral perfusion requires a minimum goal of [tex]70mm \ Hg[/tex]. MAP is obtained using the formula:-
[tex]MAP=\frac{(diastolic \ blood \ pressure\times2)+\ systolic \ blood \ pressure)}{3}\\\\MAP=\frac{2\times60+90}{3}\\\\MAP=70mm \ Hg\\CPP=MAP-ICP\\CPP=70-18\\CPP=52mm \ Hg[/tex]
A battery has a terminal voltage of 12.0 V when no current flows. Its internal resistance is 2.0 Ω. If a 7.2 Ω resistor is connected across the battery terminals, what is the terminal voltage and what is the current through the 7.2 Ω resistor?
Answer:
Voltage= 9.4 V
Current= 1.3 A
Explanation:
Equivalent resistance will be the sum of two resistors hence
[tex]R_{equ}[/tex]=2+7.2=9.2Ω
From Ohm's law, V=IR where I is current, R is equivalent resistance and V is voltage
Terminal voltage will be given by V=e-Ir where r is internal resistance, I is current and e is electromotive force.
From Ohm's law, current is given by dividing emf of the battery by the equivalent resistance hence
[tex]I=\frac {e}{R_{equ}}\\I=\frac {12}{9.2}=1.304347826A\approx 1.3A[/tex]
This is the current through external resistor
Terminal voltage will be given by V=e-Ir hence V=12-(1.3*2)=9.4 V
Final answer:
The terminal voltage across the 7.2
Ω resistor when connected to a 12.0 V battery with 2.0
Ω internal resistance is approximately 9.392 V, and the current flowing through the resistor is approximately 1.304 A.
Explanation:
The student is asking about the calculation of terminal voltage and current through a resistor when connected to a battery that has internal resistance. This is a problem related to the topic of electrical circuits in Physics.
To find the current through the 7.2
Ω resistor, use Ohm's law which states that V = IR, where V is voltage, I is current, and R is resistance. Since the battery has an internal resistance, we need to consider the total resistance in the circuit which is the sum of the internal resistance and the resistance of the 7.2
Ω resistor.
Total resistance (Rtotal) = Internal resistance (r) + External resistance (R) = 2.0
Ω + 7.2
Ω = 9.2
Ω
Using Ohm's Law, the current (I) through the circuit can be calculated as follows:
I = V / Rtotal = 12.0 V / 9.2
Ω = 1.304
A (approximately).
To find the terminal voltage (Vterminal) across the 7.2
Ω resistor when the current is flowing, we account for the voltage drop across the internal resistance:
Vterminal = emf - I × r = 12.0 V - (1.304
A × 2.0
Ω) = 12.0 V - 2.608 V = 9.392 V (approximately).
The terminal voltage across the 7.2
Ω resistor when it is connected to the battery is 9.392 V, and the current flowing through the resistor is 1.304
A.
You are given three iron rods. Two of the rods are magnets but the third one is not. How could you use the three rods to definitively identify the unmagnetized rod?
a. Test the rods a pair at a time in all possible orientations to determine which two rods have the strongest force between them; the leftover rod is unmagnetized.
b. Find one pair of rods that attract when placed end‑to‑end, regardless of orientation; the leftover rod is unmagnetized.
c. Test different pairs of rods in all possible orientations to find a rod that is not attracted to either of the other two rods; that rod is unmagnetized.
d. Test the rods one pair at a time in each possible end‑to‑end orientation until finding a pair that repels; the leftover rod is unmagnetized.
Answer:
d. Test the rods one pair at a time in each possible end‑to‑end orientation until finding a pair that repels; the leftover rod is unmagnetized.
Explanation:
Repulsion is the surest test of magnetisation . If two objects repel each other , they must be magnets . If two objects attract each other , they may not be magnets. In later case , one may be non - magnet. So the last test will help us identify the unmagnetised rod.
A carbon resistor is to be used as a thermometer. On a winter day when the temperature is 4.00 ∘C, the resistance of the carbon resistor is 217.0 Ω . What is the temperature on a spring day when the resistance is 215.1 Ω ? Take the temperature coefficient of resistivity for carbon to be α = −5.00×10−4 C−1 .
Answer:
21.5 °C.
Explanation:
Given:
α = −5.00 × 10−4 °C−1
To = 4°C
Ro = 217 Ω
Rt = 215.1 Ω
Rt/Ro = 1 + α(T - To)
215.1/217 = 1 + (-5 × 10^−4) × (T - 4)
-0.00876 = -5 × 10^−4 × (T - 4)
17.5 = (T - 4)
T = 21.5 °C.
The temperature on a spring day can be calculated by using the relationship between temperature and resistance. Given the initial resistance, the change in resistance and the temperature coefficient of resistivity for carbon, the spring day temperature can be deduced to be 21.51 °C.
Explanation:The change in temperature can be determined by using the equation ΔR = R0αΔT, where ΔR is the change in resistance, R0 is the initial resistance, α is the temperature coefficient of resistivity and ΔT is the change in temperature. The initial resistance of the carbon resistor is given as 217.0 Ω, the resistance on a spring day was 215.1 Ω, so the change in resistance ΔR is 215.1 - 217.0 = -1.9 Ω. The temperature coefficient for carbon is given as -5.00×10-4 C-1, and α is -5.00×10-4 C-1. We can rearrange the above equation to solve for ΔT: ΔT = ΔR / (R0 * α) = -1.9 Ω / (217.0 Ω * -5.00×10-4 C-1) = 17.51 C.
Thus, the temperature on the spring day would be the winter day temperature plus the calculated change in temperature: T_spring = T_winter + ΔT = 4.00 °C + 17.51 °C = 21.51 °C.
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Two long parallel wires are a center-to-center distance of 4.90 cm apart and carry equal anti-parallel currents of 4.10 A. Find the magnetic field intensity at the point P which is equidistant from the wires. (R = 6.00 cm).
Answer:
The magnetic field intensity at the point P, equidistant from the wires is
9.566 x [tex]10^{-5}[/tex] T
Explanation:
The magnetic field intensity can be obtained by resolving the vertical and horizontal components of the magnetic field. The step by step calculation is contained in the pictures attached;
A dielectric material is inserted between the charged plates of a parallel-plate capacitor. Do the following quantities increase, decrease, or remain the same as equilibrium is reestablished?
1. Charge on plates (plates remain connected to battery)
2. Electric potential energy (plates isolated from battery before inserting dielectric)
3.Capacitance (plates isolated from battery before inserting dielectric)
4. Voltage between plates (plates remain connected to battery)
5. Charge on plates (plates isolated from battery before inserting dielectric)
6. Capacitance (plates remain connected to battery)
7. Electric potential energy (plates remain connected to battery)
8. Voltage between plates (plates isolated from battery before inserting dielectric)
Answer:
1. Charge on plates (plates remain connected to battery) increases.
2. Electric potential energy (plates isolated from battery before inserting dielectric) decreases.
3.Capacitance (plates isolated from battery before inserting dielectric) increases.
4. Voltage between plates (plates remain connected to battery) remains the same.
5. Charge on plates (plates isolated from battery before inserting dielectric) remains the same.
6. Capacitance (plates remain connected to battery) increases.
7. Electric potential energy (plates remain connected to battery) increases.
8. Voltage between plates (plates isolated from battery before inserting dielectric) decreases.
Explanation:
When a dielectric material is inserted between the plates of a capacitor, the capacitance is increase by a factor K, the dielectric constant.
[tex]C = KC_0[/tex]
By the capacitance formula, the other factors change accordingly.
1. Charge on plates (plates remain connected to battery) increases, because charge and capacitance are directly proportional.
2. Electric potential energy (plates isolated from battery before inserting dielectric) decreases, because potential is inversely proportional to capacitance, and potential energy is given by the following formula
[tex]U = \frac{1}{2}CV^2[/tex]
3.Capacitance (plates isolated from battery before inserting dielectric) increases.
4. Voltage between plates (plates remain connected to battery) stays the same, because the voltage is applied by the battery.
5. Charge on plates (plates isolated from battery before inserting dielectric) remains constant. If the plates isolated from the battery, then the total charge is conserved.
6. Capacitance (plates remain connected to battery) increases when a dielectric is inserted.
7. Electric potential energy (plates remain connected to battery) increases, because when plates remain connected to battery the voltage remains the same. But the capacitance increases. Therefore, electric potential energy increases.
8. Voltage between plates (plates isolated from battery before inserting dielectric) decreases, because voltage is inversely proportional to capacitance.
Final answer:
When a dielectric is inserted into a capacitor, the capacitance increases regardless of whether the plates are connected to a battery or isolated. If connected to a battery, charge on the plates increases while voltage remains the same; if isolated, charge remains the same while voltage decreases.
Explanation:
When inserting a dielectric material into a parallel-plate capacitor, the effect on different parameters depends on whether the capacitor is connected to a battery or isolated:
Charge on plates (plates remain connected to battery): The charge on the plates will increase.
Electric potential energy (plates isolated from the battery before inserting the dielectric): The electric potential energy will decrease due to the polarization of the dielectric.
Capacitance (plates isolated from the battery before inserting the dielectric): The capacitance will increase.
Voltage between plates (plates remain connected to battery): The voltage will remain the same, as it is maintained by the battery.
Charge on plates (plates isolated from battery before inserting dielectric): The charge will remain the same as isolation prevents any charge exchange.
Capacitance (plates remain connected to battery): The capacitance will increase.
Electric potential energy (plates remain connected to the battery): The electric potential energy will increase, as more charge is accumulated due to the dielectric.
Voltage between plates (plates isolated from battery before inserting dielectric): The voltage will decrease because the same charge is now across a larger capacitance.
It is important to note that the introduction of a dielectric material affects the capacitance of a parallel-plate capacitor by reducing the effective electric field and permitting greater charge storage capability for a given voltage.
Consider an electron on the surface of a uniformly charged sphere of radius 1.2 cm and total charge 1.4 10-15 C. What is the "escape speed" for this electron
Answer:
Explanation:
electric potential of electron
= k Qq / r
= - 9 x 10⁹ x 1.4 x 10⁻¹⁵ x 1.6 x 10⁻¹⁹ / ( 1.2 x 10⁻² )
= - 16.8 x 10⁻²³ J
If v be the escape velocity
1/2 m v² = potential energy of electron
= 1/2 x 9.1 x 10⁻³¹ x v² = 16.8 x 10⁻²³
v² = 3.69 x 10⁸
v = 1.92 x 10⁴ m /s
Suppose that you wish to fabricate a uniform wire out of1.15 g of copper. Assume the wire has aresistance R = 0.300 , and all ofthe copper is used.
(a) What will be the length of the wire?
(b) What will be the diameter of the wire?
Answer:
(a) L = 149.2 cm
(b) d = 0.033 cm
Explanation:
Note that the resistivity of copper is
[tex]\rho = 1.72\times 10^{-8}~{\rm \Omega . m} = 1.72\times 10^{-6}~{\rm \Omega.cm}[/tex]
and the mass density of copper is
[tex]d = 8.96~{\rm g/cm^3}[/tex]
We will use the following formula to relate the resistance to other parameters
[tex]R = \frac{\rho L}{A} = \frac{\rho L}{\pi r^2}[/tex]
If all the copper with 1.15 g is used, then
[tex]m = dV\\1.15 = 8.96 \times (L\pi r^2)\\L\pi r^2 = 0.128[/tex]
Back to the resistance:
[tex]0.3 = \frac{1.72\times 10^{-6} L}{\pi r^2}\\L = \pi r^2 (1.74\times 10^5)\\L = \frac{0.128}{L}(1.74\times 10^5)\\L = 149.2~{\rm cm}[/tex]
Then, the diameter is
[tex]149.2\times \pi r^2 = 0.128\\r = 0.0165~{\rm cm}\\d = 2r = 0.033~{\rm cm}[/tex]
The deflection plates in an oscilloscope are 10 cm by 2 cm with a gap distance of 1 mm. A 100 volt potential difference is suddenly applied to the initially uncharged plates through a 975 ohm resistor in series with the deflection plates. How long does it take for the potential difference between the deflection plates to reach 75 volts
Answer: 23.92 ns
Explanation:
The capacitance, C = (e(0) * A) / d
C = [8.85*10^-12 * 2*10^-2 * 10*10^-2] / 1*10^-3
C = (1.77*10^-14) / 1*10^-3
C = 1.77*10^-11
C = 17.7 pF
Vc = V * [1 - e^-(t/CR)]
75 = 100 * [1 - e^-(t/CR)]
75/100 = 1 - e^-(t/CR)
e^-(t/CR) = 1 - 0.75
e^-(t/CR) = 0.25
If we take the log of both sides, we then have
Log e^-(t/CR) = Log 0.25
-t/CR = In 0.25
t = -CR In 0.25
t = - 17.7*10^-12 * 975 * -(1.386)
t = 2.392*10^-8 s
t = 23.92 ns
In a mattress test, you drop a 7.0 kg bowling ball from a height of 1.5 m above a mattress, which as a result compresses 15 cm as the ball comes to a stop.
(c)How much work does the gravitational force do on the ball while it is compressing the mattress?
(d)How much work does the mattress do on the ball?
(e)Now model the mattress as a single spring with an unknown spring constant k, and consider the whole system formed by the ball, the earth and the mattress. By how much does the potential energy of the mattress increase as it compresses?
(f)What is the value of the spring constant k?
Answer:
(c) 10.29 J
(d) 113.19 J
(e) 113.19 J
(f) 10061 N/m
Explanation:
15 cm = 0.15 m
Let g = 9.8 m/s2
(c) The work done by gravitational force is the product of gravity force and the distance compressed
[tex]E_p = mgx = 7*9.8*0.15 = 10.29 J[/tex]
(d) By using law of energy conservation with potential energy reference being 0 at the maximum compression point. As the ball falls and come to a stop at the compression point, its potential energy is transferred to elastic energy, which is the work that the mattress does on the ball:
[tex]E_p = E_e[/tex]
[tex]E_e = mgh[/tex]
where h = 1.5 + 0.15 = 1.65 m is the vertical distance that it falls.
[tex]E_e = 7*9.8*1.65 = 113.19 J[/tex]
(e) Before the compression, the potential energy of the mattress is 0. After the compression, the potential energy is 113.19J. So it has increased by 113.19J due to the potential energy transferred from the falling ball.
(f) [tex]E_e = 113.19 = kx^2/2[/tex]
[tex]k0.15^2/2 = 113.19[/tex]
[tex]k = 10061 N/m[/tex]
Answer:
(C) Wg = 113.2J
(D) Wm = 10.3J
(E) E = 1/2kx² + mgh where h is the height above the mattress and x is the compressed distance in the mattress.
(F) k = 457N/m.
Explanation:
See attachment below.
A 2-kg disk is constrained horizontally but is free to move vertically. The disk is struck from below by a vertical jet of water. The speed and diameter of the water jet are 10 m/s and 25 mm at the nozzle exit. Obtain a general expression for the speed of the water jet as a function of height, h. Find the height to which the disk will rise and remain stationary
Answer:
h = 4.281 m
Explanation
Given Data,
Weight of disk = 2kg
Speed of water jet (V) = 10 m/s
diameter of water jet = 25 mm
Cal. the velocity of the water jet as function height by applying Bernoulli's eqtn of water surface to the jet
[tex]\frac{P}{\rho } +\frac{V^{2}}{2} + gz[/tex] = Constant
[tex]\frac{V_{0}^{2}}{2} + g(0) = \frac{V^{2}}{2} + g[/tex]
[tex]V=\sqrt{V_{0}^{2}-2gh}[/tex]
Relation between [tex]V_{0}[/tex] & V
[tex]m=\rho V_{0}A_{0}[/tex]
[tex]\rho VA=\rho V_{0}A_{0}[/tex]
[tex]VA= V_{0}A_{0}[/tex]
Momentum
[tex]F_{w}+F_{d}= \frac{\partial }{\partial t}\int_{cv} w\rho dA + \int_{cs} w\rho V dA[/tex]
[tex]-mg = w_{1}[-\rho VA] +w_{2}[\rho VA][/tex]
[tex]w_{1}[/tex] = V
[tex]w_{2}[/tex] = 0
[tex]mg = \rho V^{2}A[/tex]
[tex]mg = \rho VV_{0} A[/tex]
[tex]mg = \rho VV_{0} A_{0}[/tex]
[tex]mg = \rho V_{0} A_{0} \sqrt{V_{0} ^{2}-2gh }[/tex]
Solving for h
[tex]h = \frac{1}{2g}[V_{0}^{2}-\frac{mg}{\rho V_{0}A_{0}}][/tex]
g is gravitational acc.
[tex]= \frac{1}{2\times 9.81}[10^{2}-(\frac{2\times 9.81}{999\times10\times\frac{\Pi }{4}\times(0.025)^{2}})^{2} ][/tex]
[tex]= \frac{1}{19.62}[100-(\frac{19.68}{4.9038})^{2}][/tex]
[tex]= \frac{100-16.0078}{19.62}[/tex]
h = 4.281 m
h of disk on which it remains stationary.
To understand the terms in Faraday's law and to be able to identify the magnitude and direction of induced emf. Faraday's law states that induced emf is directly proportional to the time rate of change of magnetic flux. Mathematically, it can be written as
Answer: [tex]V_{\epsilon}\propto \frac{d\phi_{B}}{dt}[/tex]
Explanation:
A direct proportionality means a linear relationship between two variables and rate of change means an application of derivatives. Hence, the mathematical model is:
[tex]V_{\epsilon}\propto \frac{d\phi_{B}}{dt}[/tex]
Faraday's Law of Induction, primarily focused on Physics, reveals that the induced electromotive force in a coil is proportional to the rate of change of magnetic flux through the coil and is also dependent on the number of coil turns.
Explanation:Understanding Faraday's Law of Induction
Faraday's Law of Induction is a critical concept in electromagnetism, underlying the working principles of various electrical devices. The induced electromotive force (emf) is directly proportional to the time rate of change of magnetic flux. In simpler terms, when the magnetic environment of a circuit changes, an emf is induced. Faraday's Law is given by the equation EMF = -N(ΔΦ/Δt), where 'N' represents the number of turns in a coil, and (ΔΦ/Δt) is the rate of change of magnetic flux over time. This equation highlights three main factors influencing the induced EMF:
The induced EMF is proportional to the number of loops in the coil.It is proportional to the change in magnetic flux, ΔΦ.It is inversely proportional to the time interval, Δt, over which the change occurs.The negative sign in Faraday's Law is a reflection of Lenz's Law, which indicates that the induced emf always works to oppose the change in magnetic flux that produces it.
The intensity of a linearly polarized electromagnetic wave is directly proportional to the square of the electric field (e.g., I = k*E2). If the Receiver’s meter reading was directly proportional to the incident microwave’s intensity, the meter would read the relationship M = Mocos2 θ. Plot this relationship. Based on your graphs, discuss the relationship between the meter reading of the Receiver and the polarization and magnitude of the incident microwave.
Answer:
Please find attached file for complete answer solution and explanation of same question.
Explanation:
person is 1.9 m tall. Where should he place the top of a mirror on the wall so he can see the top of his head? Assume his eyes are 5.1 cm below the top of his head. Find a minimal vertical distance from the level of the person’s eye
Answer:
The minimal vertical distance from the level of the person’s eye is 2.55 cm.
Explanation:
Given;
height of the person = 1.9 m
his eyes are 5.1 cm below the top of his head
If top of his head is the same distance from the mirror as his eyes is from the mirror, then the top of the mirror must be half of his eyes level.
Therefore, the minimal vertical distance from the level of the person’s eye is ¹/₂ x 5.1 cm = 2.55 cm
The minimal vertical distance of the mirror from the level of the person’s eye is 2.55 cm.
Approximately 1.000 g each of four gasses H2, Ne, Ar, and Kr are placed in a sealed container all under1.5 atm of pressure. Assuming ideal behavior, determine the partial pressure of the H2 and Ne
Answer:
The partial pressure of H2 is 0.375 atm
The partial pressure of Ne is also 0.375 atm
Explanation:
Mass of H2 = 1 g
Mass of Ne = 1 g
Mass of Ar = 1 g
Mass of Kr = 1 g
Total mass of gas mixture = 1 + 1 + 1 + 1 = 4 g
Pressure of sealed container = 1.5 atm
Partial pressure of H2 = (mass of H2/total mass of gas mixture) × pressure of sealed container = 1/4 × 1.5 = 0.375 atm
Partial pressure of Ne = (mass of Ne/total mass of gas mixture) × pressure of sealed container = 1/4 × 1.5 = 0.375 atm
The partial pressures of H2 and Ne in the mixture under ideal gas behavior are approximately 1.278 atm and 0.128 atm respectively. This is calculated using their mole fractions and the total pressure.
Explanation:In this scenario, we're essentially dealing with ideal gas behavior. The partial pressure of a gas in a mixture can be determined by the mole fraction of the gas times the total pressure. The moles of each gas can be calcuated using the equation: moles = mass/molar mass. The molar masses of H2, Ne, Ar, and Kr are approximately 2 g/mol, 20 g/mol, 40 g/mol, and 84 g/mol, respectively.
For example, moles of H2 ≈ 1.000 g / 2 g/mol = 0.5 mol. Likewise, moles of Ne = 1.000 g / 20 g/mol = 0.05 mol, moles of Ar = 1.000 g / 40 g/mol = 0.025 mol, and moles of Kr = 1.000 g / 84 g/mol = 0.012 mol. The total moles of gas = 0.5 mol + 0.05 mol + 0.025 mol + 0.012 mol = 0.587 mol.
The mole fraction of H2 = 0.5 mol / 0.587 mol = 0.852. Therefore, the partial pressure of H2 = 0.852 * 1.5 atm = 1.278 atm. Applying the same calculations for Ne, the mole fraction of Ne = 0.05 mol / 0.587 mol = 0.085. Therefore, the partial pressure of Ne = 0.085 * 1.5 atm = 0.128 atm.
Learn more about Partial Pressure here:https://brainly.com/question/35280352
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When an object slides down an inclined plane, the angle between the displacement and the force of gravity is the same as the angle that the surface makes with the horizontal. Suppose you have a 2-kg cart rolling a distance of 1.5 m down a 30° incline. According to Eq. (7.8), what is the work done on it by the force of gravity? Select One of the Following:
(a) 3J
(b) 25.5 J
(c) 2.6J
(d) 14.7 J
Answer:
(d) 14.7 J.
Explanation:
Using the equation,
W = mghsin∅................. equation 1
Where W = work done on the cart by the force of gravity, g = acceleration due to gravity, h = height, ∅ = angle between the displacement and the force of gravity.
Given: m = 2 kg, g = 9.8 m/s², h = 1.5 m, ∅ = 30°
Substitute into equation 1
W = 2(9.8)(1.5)sin30
W = 2×9.8×1.5×0.5
W = 14.7 J.
The right option is (d) 14.7 J
The work done by gravity on a 2-kg cart rolling down a 1.5 m, 30° incline is 14.7 J, corresponding to option (d).
Explanation:When an object slides down an inclined plane, the work done by gravity can be calculated using the formula W = mgh, where W is the work done, m is the mass of the object, g is the acceleration due to gravity (9.8 m/s2), and h is the vertical height. The vertical height can be determined from the incline's length and the angle of inclination using the sine function (h = l · sin(θ)). For a 2-kg cart rolling a distance of 1.5 m down a 30° incline, the work done by gravity would be W = m · g · l · sin(θ) = 2 kg · 9.8 m/s2 · 1.5 m · sin(30°) = 2 · 9.8 · 1.5 · 0.5 = 14.7 J, which corresponds to option (d).
Calculate the force required to pull a copper ball of radius 1.69 cm upward through a viscous fluid at a constant speed of 9.3 cm/s. Take the damping constant of the fluid to be 0.884 kg/s\.\*
Answer:
[tex] m = \rho V[/tex]
Since we have an ball we can consider this like a sphere and the volume is given by [tex] V = \frac{4}{3} \pi r^3 = \frac{4}{3} \pi (1.69cm)^3 = 20.218 cm^3 = 0.00002022 m^3[/tex]
The density for the copper is approximately [tex] \rho = 8940 kg/m^3[/tex]
So then the mass is :
[tex] m =8940 kg/m^3 * 0.00002022m^3 = 0.1808 Kg[/tex]
And now we have everything in order to replace into the formula for F, like this:[tex] F = 0.1808 Kg *9.8 m/s^2 + 0.884 kg/s * 0.093 m/s= 1.772 +0.975 N = 2.747 N[/tex]And that would be the final answer for this case.
Explanation:
For this case if we assume that we have a damping motion the force action on the vertical direction would be:
[tex] F = mg + bv[/tex]
Where F represent the upward force on the copper ball
m represent the mass
g = 9.8 m/s^2 represent the gravity
b = 0.884 kg/s represent the proportionality constant
v = 9.3 cm/s = 0.093 m/s represent the velocity
We can solve for the mass from the following expression:
[tex] m = \rho V[/tex]
Since we have an ball we can consider this like a sphere and the volume is given by [tex] V = \frac{4}{3} \pi r^3 = \frac{4}{3} \pi (1.69cm)^3 = 20.218 cm^3 = 0.00002022 m^3[/tex]
The density for the copper is approximately [tex] \rho = 8940 kg/m^3[/tex]
So then the mass is :
[tex] m =8940 kg/m^3 * 0.00002022m^3 = 0.1808 Kg[/tex]
And now we have everything in order to replace into the formula for F, like this:[tex] F = 0.1808 Kg *9.8 m/s^2 + 0.884 kg/s * 0.093 m/s= 1.772 +0.975 N = 2.747 N[/tex]And that would be the final answer for this case.
A Rankine oval is formed by combining a source-sink pair, each having a strength of 36 ft2/s and separated by a distance of 15 ft along the x axis, with a uniform velocity of 12 ft/s (in the positive x direction). Determine the length of the oval.
Answer:
0.28 ft
Explanation:
We are given that
Strength=m=[tex]36ft^2/s[/tex]
Distance between source and sink=15 ft
Distance between the sink of the source and origin=[tex]a=\frac{15}{2}[/tex] ft
Uniform velocity, U=12 ft/s
We have to find the length of the oval.
Formula to find the half length of the body
[tex]\frac{l}{a}=(\frac{m}{\pi Ua}+1)^{\frac{1}{2}}[/tex]
Where a=Distance between sink of source and origin
U=Uniform velocity
m=Strength
l=Half length
Using the formula
[tex]\frac{l}{\frac{15}{2}}=(\frac{36}{\pi\times 12\times \frac{15}{2}}+1)^{\frac{1}{2}}[/tex]
[tex]l=\frac{2}{15}(\frac{36}{\pi\times 12\times \frac{15}{2}}+1)^{\frac{1}{2}}[/tex]
[tex]l=0.14[/tex]
Length of oval=[tex]2l=2(0.14)=0.28 ft[/tex]