Answer:
67.5
Explanation:
[tex]F= BILsin\theta[/tex] where F is the magnitude of the force, B is the magnetic field, I is current, [tex]\theta[/tex] is the angle of inclination, L is the length of the wire.
Substituting 0.9 T for magnetic field B, 10 m for length, 15 A for current I and 30 degrees for [tex]\theta[/tex] then
[tex]F=0.9\times 15\times 10\times sin 30^{\circ}=67.5[/tex]
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.
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
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;
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.
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.
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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.
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
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.
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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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.
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.
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.
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;
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]
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
Consider a rigid steel beam of length L = 13 m and mass mb = 388 kg resting on two supports, one at each end. A worker of mass mw = 74 kg sits on the beam at a distance x from support A. Refer to the figure.
When the worker sits at a distance x = 7.5 m from support A, calculate the force, in newtons, that support B must exert on the beam in order for it to remain at rest. Use g with three significant figures.
Answer:
2321 N
Explanation:
Let g = 9.807 m/s2
Assume this is a uniform beam and the center of mass it at the geometric center, which is half way between the 2 ends, or h = 13/2 = 6.5 m from the left end (support A).
For the system to remain at rest, then the total moments around a point (let pick support A) must be 0. Moments created by each force is the product of the force magnitude and the moment arm, aka distance from the force to support A
[tex]\sum M_A = 0[/tex]
[tex]M_w + M_b + M_B = 0[/tex]
[tex]F_wx + F_bh + F_BL = 0[/tex]
[tex]m_wgx + m_bgh = -F_BL[/tex]
If we take upward direction be the positive direction, that means all the gravity acting downward are negative
[tex]74*(-9.807)*7.5 + 388(-9.807)*6.5 = -F_B13[/tex]
[tex]-13F_B = -30176.139[/tex]
[tex]F_B = -30176.139/-13 = 2321 N[/tex]
So the force at support B has a magnitude of 2321 N acting upward
The force that support B must apply to keep the beam balanced when the worker sits at a distance of 7.5m from support A is approximately 1597.92 N.
Explanation:This problem involves understanding of statics, specifically the principle of moments or torques. Assuming the system is in equilibrium (i.e., the beam does not rotate or translate), the sum of the forces and the sum of the torques must both be zero.
We start by calculating the total weight of the beam, which is simply its mass multiplied by the acceleration due to gravity, g. This gives mb * g = 388 kg * 9.81 m/s^2 = 3806.68 N, this force acts in the middle of the beam, that is at a distance of L/2 = 13m/2 = 6.5m from each support.
Similarly, the weight of the worker is given by mw * g = 74 kg * 9.81 m/s^2 = 725.94 N, this force acts at a distance of x = 7.5m from the left support (A) and L - x = 13m - 7.5m = 5.5m from the right support (B).
Now we can calculate the net force (Fn) and the net torque (Tn). We choose the left support (A) as the pivot point. The total force acting on the beam is the sum of the weights of the beam and the worker, and this must be supported by the two supports. Since sum of the forces must be zero: Fn = F_A + F_B = mass_beam * g + mass_worker * g.
The net torque is calculated from the forces acting at a distance from the pivot point. So we get, Tn = beam_weight * distance_beam - worker_weight * distance_worker + F_B * L = 0. From this, we can solve for F_B (force at support B), we find: F_B = (beam_weight * distance_beam - worker_weight * distance_worker) / L.
Plugging into these formulas, we get the force that support B must apply to keep the system in balance: F_B = (3806.68 N * 6.5m - 725.94 N * 7.5m) / 13m = 1597.92 N.
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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
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:
Air exits a turbine at 200 kPa and 1508C with a volumetric flow rate of 7000 liters/s. Modeling air as an ideal gas, determine the mass flow rate, in kg/s.
Answer:
mass flow rate = 11.464 kg/sec
Explanation:
given data
P = 200 kPa = 2 × [tex]10^{5}[/tex] Pa
temperature = 150°C = 423 k
volumetric flow rate = 7000 liters/s
solution
first we get here molar flow rate that is express as
molar flow rate = [tex]\frac{Pv}{RT}[/tex] .................1
put here value
molar flow rate = [tex]\frac{2 \times 10^5 \times 7}{8.314\times 423}[/tex]
molar flow rate = 398.06 mol/sec
and
now we get mass flow rate
mass flow rate = molar flow rate × average mol weight ..............2
mass flow rate = 398.06 × 28.8
mass flow rate = 11464.3 g/sec
mass flow rate = 11.464 kg/sec
Final answer:
The mass flow rate of air exiting the turbine is calculated using the ideal gas equation and the given conditions of pressure, temperature, and volumetric flow rate. By substituting the values into the equation, the mass flow rate is found to be 11.6 kg/s.
Explanation:
The problem involves finding the mass flow rate of air using the ideal gas model, given the exiting conditions from a turbine. The known variables are the pressure (200 kPa), temperature (150°C which is 423.15 K after conversion from Celsius to Kelvin), and volumetric flow rate (7000 liters/s or 7 m³/s). To find the mass flow rate, we can use the ideal gas equation PV = mRT, where P is the absolute pressure, V is the volume flow rate per unit time, m is the mass flow rate, R is the specific gas constant for air, and T is the absolute temperature.
The specific gas constant for air (R) is 287 J/kg·K, and it can be solved for m as:
m = PV / (RT)
Substituting the given values:
m = (200,000 Pa)(7 m³/s) / (287 J/kg·K)(423.15 K)
m = 11.599 kg/s (rounded to three significant figures)
Therefore, the mass flow rate of the air exiting the turbine is 11.6 kg/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.
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 15.0 g bullet is moving to the right with speed 270 m/s when it hits a target and travels an additional 25.0 cm into the target. What are the magnitude (in N) and direction of the stopping force acting on the bullet
Answer:
[tex]F=2187\ N[/tex]
Explanation:
Given:
mass of bullet,[tex]m=15\ g=0.015\ kg[/tex] initial velocity of bullet, [tex]u=270\ m.s^{-1}[/tex]displacement of the bullet in the target, [tex]s=25\ cm=0.25\ m[/tex]Here as given in the question the bullet penetrates the target by the given displacement of the bullet into it. During this process it faces deceleration and hence it comes to rest.
so, final velocity of the bullet, [tex]v=0\ m.s^{-1}[/tex]Now using the equation of motion:
[tex]v^2=u^2+2a.s[/tex]
where:
[tex]a=[/tex] acceleration of the bullet
[tex]0^2=270^2+2a\times 0.25[/tex]
[tex]a=145800\ m.s^{-2}[/tex]
Now the force of resistance offered by the target in stopping it:
[tex]F=m.a[/tex]
[tex]F=0.015\times145800[/tex]
[tex]F=2187\ N[/tex]
The temperature within a thin plate with thermal conductivity of 10 W/m/K depends on position as given by the following expression: TT=(100 K)????????−xx2/????????xx 2cos�yy/????????yy�+300 K Where, Lx = 1 m, and Ly = 2 m. At the point (0.4 m, 1 m), find: a. The magnitude of the heat flux b. The direction of the heat flux
Answer:
Heat flux = (598.3î + 204.3j) W/m²
a) Magnitude of the heat flux = 632.22 W/m²
b) Direction of the heat flux = 18.85°
Explanation:
- The correct question is the first image attached to this solution.
- The solution to this question is contained on the second and third images attached to this solution respectively.
Hope this Helps!!!
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:
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:
An uncharged metal sphere, A, is on an insulating base. A second sphere, B, of the same size, shape, and material carrying charge Q is brought close to, but not touching, sphere A. Describe what happens to the charges on A and B as they are brought close but not touching. If we now remove sphere B and place it far away, what is the charge on sphere A
Answer:
0
Explanation:
If we bring the charged sphere B close to, but not touching it , to the uncharged sphere A, as charges can move freely on the conductor, a charge -Q will be built on the outer surface of the sphere A, facing to sphere B.As the sphere A must remain neutral, a charge Q will be built on the surface, on the side farther to the sphere B, as the following condition must be met:Q +(-Q) =0.
If we now remove sphere B, and place it far away, there will be a charge redistribution within sphere A, making to disappear the separation between Q and -Q.The total charge on sphere A must be 0, as there is no charge transfer from sphere B to sphere A.Final answer:
When identical conducting spheres A and B with charges of –5 nC and –3 nC respectively are brought into contact, the total initial charge of –8 nC is shared equally. Upon separation, each sphere holds a final charge of –4 nC.
Explanation:
When two conducting spheres, A and B, with different charges are brought into contact, the total charge is shared equally between them. Given sphere A has a charge of –5 nC and sphere B has a charge of –3 nC, the total initial charge is –8 nC. This charge will be evenly distributed between both spheres when they touch.
Therefore, after spheres A and B are brought into contact and then separated, the final charge on each sphere will be half of the total charge. This final charge will be calculated as follows:
Total charge before contact: –5 nC + –3 nC = –8 nCCharge on each sphere after contact and separation: –8 nC / 2 = –4 nCThus, after separation, both sphere A and B will have a charge of –4 nC each.
A copper rod of length 27.5 m has its temperature increases by 35.9 degrees celsius. how much does its length increase?(unit=m)
Explanation:
When copper rod is heated , its length increases
The increase in length can be found by the relation
L = L₀ ( 1 + α ΔT )
here L is the increased length and L₀ is the original length
α is the coefficient of linear expansion and ΔT is the increase in temperature .
The increase in length = L - L₀ = L₀ x α ΔT
Substituting all these value
Increase in length = 27.5 x 1.7 x 10⁻⁵ x 35.9
= 1.87 x 10⁻² m
Answer:0.01678325m
Explanation:
Original length(L1)=27.5m
Temperature rise(@)=35.9°C
Linear expansivity(α)=0.000017
Length increase(L)=?
L=α x L1 x @
L=0.000017 x 27.5 x 35.9
L=0.01678325m
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).
The slotted link is pinned at O, and as a result of the constant angular velocity u # = 3 rad>s it drives the peg P for a short distance along the spiral guide r = (0.4 u) m, where u is in radians. Determine the velocity and acceleration of the particle at the
Answer:
A) vᵣ = 1.2 m/s
B) vₜ = 0.4π m/s = 1.257 m/s
C) aᵣ = - 1.2π m/s² = - 3.771 m/s²
D) aₜ = 7.2 m/s²
Explanation:
Given,
θ' = 3 rad/s
r = 0.4 θ
Note that
θ' = (dθ/dt) = 3 rad/s
θ" = (d²θ/dt²) = (d/dt) (3) = 0 rad/s²
r' = (dr/dt) = (d/dt) (0.4θ) = 0.4 (dθ/dt) = 0.4 × θ' = 0.4 × 3 = 1.2 m/s
r" = (d²r/dt²) = 0.4 (d²θ/dt²) = 0 m/s²
A) Radial component of the velocity of P at the instant θ=π/3rad.
From the kinematics of a particle in a plane,
vᵣ = r' = 1.2 m/s
B) Transverse component of the velocity and acceleration of P at the instant θ=π/3rad.
vₜ = rθ' = (0.4 θ) (3) = 1.2 θ = 1.2 (π/3) = 0.4π m/s = 1.257 m/s
C) Radial component of the acceleration of P at the instant θ=π/3rad.
aᵣ = r" - r(θ'²) = 0 - (0.4θ)(3²) = - 3.6θ = - 3.6 (π/3) = - 1.2π m/s² = - 3.771 m/s²
D) Transverse component of the acceleration of P at the instant θ=π/3rad
aₜ = rθ" + 2r'θ' = (0.4θ)(0) + (2)(1.2)(3) = 0 + 7.2 = 7.2 m/s²
Hope this Helps!!!
The response requires understanding rotational kinematics and dynamics to determine the velocity and acceleration of a particle moving along a spiral guide, involving calculus and principles of rotational motion.
Explanation:The question pertains to kinematics in the context of rotational and spiral motion. In particular, it involves the motion of a particle along a spiral guide with a given radial dependency on the angle turned through (in radians). To find the velocity and acceleration of the particle, one must apply the equations of motion that relate linear and angular quantities and consider any given geometric relationships and constraints, such as the provided relationship between radial distance and angular position. To solve the tasks posed, one would typically employ differential calculus and the principles of rotational dynamics including tangential and radial components of velocity and acceleration.
For example, the velocity of the particle can be found by taking the time derivative of the radial position r, which in turn depends on the angular velocity ω. The acceleration would then be found by differentiating the velocity with respect to time or by explicitly using the tangential and radial acceleration components for rotational motion, which are related to ω and its time derivative (angular acceleration).
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.