Answer:
peak value of magnetic field is
[tex]B_o = 2.9 \times 10^{-6} T[/tex]
peak value of electric field is
[tex]E_o = 868 N/C[/tex]
Explanation:
As we know that the intensity of electromagnetic waves is given by the formula
[tex]I = \epsilon_0 E_{rms}^2 c[/tex]
here we have
[tex]I = 1000 W/m^2[/tex]
[tex]\epsilon_0 = 8.85 \times 10^{-12}[/tex]
[tex]c = 3\times 10^8[/tex]
now we have
[tex]1000 = (8.85 \times 10^{-12})E_{rms}^2(3 \times 10^8)[/tex]
[tex]E_{rms} = 614 N/C[/tex]
now peak value of electric field is given as
[tex]E_o = \sqrt2 E_{rms}[/tex]
So the peak value of electric field is given as
[tex]E_o = 868 N/C[/tex]
Now we have relation between electric field and magnetic field as
[tex]B_o = \frac{E_o}{c}[/tex]
[tex]B_o = \frac{868}{3\times 10^8}[/tex]
[tex]B_o = 2.9 \times 10^{-6} T[/tex]
A 170 kg astronaut (including space suit) acquires a speed of 2.25 m/s by pushing off with his legs from a 2600 kg space capsule. What is the average change in speed in meters/second of the space capsule?
Explanation:
Mass of the astronaut, m₁ = 170 kg
Speed of astronaut, v₁ = 2.25 m/s
mass of space capsule, m₂ = 2600 kg
Let v₂ is the speed of the space capsule. It can be calculated using the conservation of momentum as :
initial momentum = final momentum
Since, initial momentum is zero. So,
[tex]m_1v_1+m_2v_2=0[/tex]
[tex]170\ kg\times 2.25\ m/s+2600\ kg\times v_2=0[/tex]
[tex]v_2=-0.17\ m/s[/tex]
So, the change in speed of the space capsule is 0.17 m/s. Hence, this is the required solution.
Introduction Team Problem 2 You jump into the deep end of Legion Pool and swim to the bottom with a pressure gauge. The gauge at the bottom reads 39,240 N/m2. How deep is the pool if the density of water is 1000 kg/m3?
Answer:
Depth of the pool, h = 4.004 cm
Explanation:
Pressure at the bottom, P = 39240 N/m²
The density of water, d = 1000 kg/m³
The pressure at the bottom is given by :
P = dgh
We need to find the depth of pool. Let h is the depth of the pool. So,
[tex]h=\dfrac{P}{dg}[/tex]
[tex]h=\dfrac{39240\ N/m^2}{1000\ kg/m^3\times 9.8\ m/s^2}[/tex]
h = 4.004 m
So, the pool is 4.004 meters pool. Hence, this is the required solution.
Find the de Broglie wavelength of a ball of mass 0.20 kg just before it strikes the Earth after being |dropped from a building 50 m tall. IMPORTANT: Step by step *typed* format preferred. If hand written, please write clearly and legibly. Please be sure answer is CORRECT. Thanks!
Answer:
Wavelength of the ball is [tex]\lambda=1.05\times 10^{-34}\ m[/tex]
Explanation:
It is given that,
Mass of the ball, m = 0.2 kg
It strikes the Earth after being dropped from a building of 50 m tall, h = 50 m
In this time, the potential energy is converted to kinetic energy. On applying the conservation of energy as :
[tex]\dfrac{1}{2}mv^2=mgh[/tex]
[tex]v=\sqrt{2gh}[/tex].............(1)
The De-broglie wavelength of the ball is given by :
[tex]\lambda=\dfrac{h}{mv}[/tex]
[tex]\lambda=\dfrac{h}{m\sqrt{2gh}}[/tex]
[tex]\lambda=\dfrac{6.63\times 10^{-34}\ J-s}{0.2\ kg\times \sqrt{2\times 9.8\ m/s^2\times 50\ m}}[/tex]
[tex]\lambda=1.05\times 10^{-34}\ m[/tex]
Hence, this is the required solution.
A long jumper can jump a distance of 7.4 m when he takes off at an angle of 45° with respect to the horizontal. Assuming he can jump with the same initial speed at all angles, how much distance does he lose (in m) by taking off at 429?
Answer:
0.02 m
Explanation:
R₁ = initial distance jumped by jumper = 7.4 m
R₂ = final distance jumped by jumper = ?
θ₁ = initial angle of jump = 45°
θ₂ = final angle of jump = 42.9°
[tex]v[/tex] = speed at which jumper jumps at all time
initial distance jumped is given as
[tex]R_{1}=\frac{v^{2}Sin2\theta _{1} }{g}[/tex]
final distance jumped is given as
[tex]R_{2}=\frac{v^{2}Sin2\theta _{2} }{g}[/tex]
Dividing final distance by initial distance
[tex]\frac{R_{2}}{R_{1}}=\frac{Sin2\theta _{1}}{Sin2\theta _{2}}[/tex]
[tex]\frac{R_{2}}{7.4}=\frac{Sin2(42.9)}{Sin2(45))}[/tex]
[tex]R_{2} =7.38[/tex]
distance lost is given as
d = [tex]R_{1} - R_{2} [/tex]
d = 7.4 - 7.38
d = 0.02 m
Use the work energy theorem to solve each of these problems and neglect air resistance in all cases. a) A branch falls from the top of a 98.0 m tall redwood tree, starting from rest. How fast is it moving when it reaches the ground? b) A volcano ejects a boulder directly upward 545 m into the air. How fast was the boulder moving just as it left the volcano?
Final answer:
The branch falls from 98.0 m with a final speed of 44 m/s, and the boulder must be ejected with a velocity of 103.4 m/s to reach a height of 545 m, based on the work-energy theorem and neglecting air resistance.
Explanation:
Work Energy Theorem Examples
For a branch falling from the height of a tree with no air resistance, the work-energy theorem tells us that the kinetic energy gained by the branch when it hits the ground must equal the potential energy it had at the top.
Using the formula for gravitational potential energy (PE = mgh, where m is mass, g is the acceleration due to gravity, and h is height) and kinetic energy (KE = 0.5 * m * v^2), we can solve for the velocity when the branch reaches the ground.
For instance, using a height of 98.0 m for the redwood tree and the acceleration due to gravity g = 9.8 m/s^2, we find that the final velocity (v) can be calculated as:
PE_top = KE_bottom
mgh = 0.5 * m * v^2
v = sqrt(2gh) = sqrt(2 * 9.8 m/s^2 * 98.0 m) = 44 m/s
For the boulder ejected from a volcano, we're looking for the initial velocity required for the boulder to reach a maximum height of 545 m. The potential energy at the top (PE_top) will be equal to the kinetic energy at the bottom (KE_bottom) since we're negating air resistance.
PE_top will be m * g * h. KE_bottom = PE_top = 0.5 * m * v^2, so solving for the initial velocity (v) gives us:
v = sqrt(2gh) = sqrt(2 * 9.8 m/s^2 * 545 m) = 103.4 m/s
An earthquake 45 km from a city produces P and S waves that travel outward at 5000 m/s and 3000 m/s, respectively. Once city residents feel the shaking of the P wave, how much time do they have before the S wave arrives in seconds?
Answer:
The S wave arrives 6 sec after the P wave.
Explanation:
Given that,
Distance of P = 45 km
Speed of p = 5000 m/s
Speed of S = 3000 m/s
We need to calculate the time by the P wave
Using formula of time
[tex]t = \dfrac{D}{v}[/tex]
Where, D = distance
v = speed
t = time
Put the value in to the formula
[tex]t_{p} =\dfrac{45\times1000}{5000}[/tex]
[tex]t_{p} = 9\ sec[/tex]
Now, time for s wave
[tex]t_{s}=\dfrac{45000}{3000}[/tex]
[tex]t =15\ sec[/tex]
The required time is
[tex]\Delta t=t_{s}-t_{p}[/tex]
[tex]\Delta t=15-9[/tex]
[tex]\Delta t =6\ sec[/tex]
Hence, The S wave arrives 6 sec after the P wave.
Consider a transformer used to recharge rechargeable flashlight batteries, that has 500 turns in its primary coil, 4 turns in its secondary coil, and an input voltage of 120 V. What is the voltage output in volts, of the transformer used for to charge the batteries?
Answer:
Output Voltage = 0.96 volts
Explanation:
As we know by the equation of transformer we have
[tex]\frac{V_1}{V_2} = \frac{N_1}{N_2}[/tex]
here we know that
[tex]V_1[/tex] = voltage of primary coil = 120 Volts
[tex]V_2[/tex] = voltage of secondary coil
[tex]N_1[/tex] = number of turns in primary coil = 500
[tex]N_2[/tex] = number of turns in secondary coil = 4
now we will have
[tex]\frac{120}{V} = \frac{500}{4}[/tex]
[tex]V = 0.96 Volts[/tex]
Three charged particles are positioned in the xy plane: a 50-nC charge at y = 6 m on the y axis, a −80-nC charge at x = −4 m on the x axis, and a 70-nc charge at y = −6 m on the y axis. What is the electric potential (relative to a zero at infinity) at the point x = 8 m on the x axis?
Answer:
V = 48 Volts
Explanation:
Since we know that electric potential is a scalar quantity
So here total potential of a point is sum of potential due to each charge
It is given as
[tex]V = V_1 + V_2 + V_3[/tex]
here we have potential due to 50 nC placed at y = 6 m
[tex]V_1 = \frac{kQ}{r}[/tex]
[tex]V_1 = \frac{(9\times 10^9)(50 \times 10^{-9})}{\sqrt{6^2 + 8^2}}[/tex]
[tex]V_1 = 45 Volts[/tex]
Now potential due to -80 nC charge placed at x = -4
[tex]V_2 = \frac{kQ}{r}[/tex]
[tex]V_2 = \frac{(9\times 10^9)(-80 \times 10^{-9})}{12}[/tex]
[tex]V_2 = -60 Volts[/tex]
Now potential due to 70 nC placed at y = -6 m
[tex]V_3 = \frac{kQ}{r}[/tex]
[tex]V_3 = \frac{(9\times 10^9)(70 \times 10^{-9})}{\sqrt{6^2 + 8^2}}[/tex]
[tex]V_3 = 63 Volts[/tex]
Now total potential at this point is given as
[tex]V = 45 - 60 + 63 = 48 Volts[/tex]
Constants Part A wo charged dust particles exert a force of 7.5x102 N n each other What will be the force if they are moved so they are only one-eighth as far apart? Express your answer using two significant figures.
Final answer:
The force between two charged particles will increase by a factor of 64 if their separation distance is reduced to one-eighth of the original distance, according to Coulomb's Law.
Explanation:
The student is asking about how the force between two charged particles changes when their separation distance changes. Coulomb's Law, which describes the electrostatic interaction between electrically charged particles, states that the force (F) between two point charges is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance (r) between them. Coulomb's Law is mathematically expressed as F = k * (|q¹*q|²/r²), where k is Coulomb's constant, and q1 and q2 are the charges.
If the distance between the particles is reduced to one-eighth of the original distance, the new force F' can be calculated using the relation F' = F * (1/(1/8)²), since the force varies inversely with the square of the distance. After performing the calculations, we find that the new force is 64 times the original force of 7.5x10² N.
A block is moving at constant speed due to a horizontal force pulling to the right. The coefficient of kinetic friction, Hk, between the block and the surface is 0.20 and the mag- nitude of the frictional force is 100.0 N, what is the weight of the block? (a 400 N (b) 600 N (c) 500 N (d) 267 N
Answer:
c) 500 N
Explanation:
If the block moves with constant speed there is no acceleration. We draw a free body diagram and define the forces on the body.
[tex]F_{f}=FrictionForce\\F_{e}=ExternalForce\\N=NormalForce\\W=Weight\\[/tex]
The equation for the frictional force is:
[tex]F_{f}=N*H_{k}[/tex]
Where [tex]H_{k}[/tex] is the kinetic friction coefficient. We write the equilibrium equation in the y direcction:
[tex]\sum F_{y}\rightarrow N-W=0\\N=W[/tex]
We replace this result in the equation for the frictional force:
[tex]F_{f}=W*H_{k}[/tex]
We replace the data given by the exercise and find the weight
[tex]W=\frac{F_{f}}{H_{k}}\\W=\frac{100}{0.2}=500\: N[/tex]
Suppose that an experiment determines that the amount of work required for a Force field to move a particle from the point ( 1 , 2 ) to the point ( − 5 , 3 ) along the curve C 1 is 1.2 J and the work done by the field in moving the particle along another curve C 2 between the same two points is 1.4 J. What can you say about the field, and why? g
Answer:
From the given information we can infer that the field is not conservative.
Explanation:
For a conservative field the work done on an object in moving it from a position given by co-ordinates [tex](x_{1},y_{1},z_{1})[/tex] to another position with co-ordinates [tex](x_{2},y_{2},z_{2})[/tex] shall be independent of the path we take in between to reach our final position (by definition of a conservative field). But in the given case since the initial and the final position of both the curves [tex]C_{1},C_{2}[/tex] coincide but the work done along both the paths is different thus we conclude that the field is not conservative.
Car A uses tires for which the coefficient of static friction is 0.169 on a particular unbanked curve. The maximum speed at which the car can negotiate this curve is 23.7 m/s. Car B uses tires for which the coefficient of static friction is 0.826 on the same curve. What is the maximum speed at which car B can negotiate the curve?
Answer:
[tex]v_B = 52.4 m/s[/tex]
Explanation:
For unbanked road the maximum friction force will provide centripetal force to the car.
So here we will have
[tex]F_c = \frac{mv^2}{R}[/tex]
Since we know that centripetal force here is due to friction force
[tex]F_c = F_f[/tex]
[tex]\mu mg = \frac{mv^2}{R}[/tex]
now for two cars we will have
[tex]\mu_A m_A g = \frac{m_A v_A^2}{R}[/tex]
also we have
[tex]\mu_B m_B g = \frac{m_B v_B^2}{R}[/tex]
now by division of two equations
[tex]\frac{\mu_A}{\mu_B} = \frac{v_A^2}{v_B^2}[/tex]
[tex]\frac{0.169}{0.826} = \frac{23.7^2}{v_B^2}[/tex]
so we will have
[tex]v_B = 52.4 m/s[/tex]
Final answer:
The maximum speed at which car B can negotiate the curve is approximately 52.4 m/s, assuming the same curve and conditions as for car A.
Explanation:
To determine the maximum speed at which car B can negotiate the curve, we need to use the relationship between the coefficient of static friction and the maximum speed a car can maintain without slipping on a curve.
The centripetal force required to keep a car moving in a circle of radius r is provided by the frictional force, which is the product of the coefficient of static friction μ and the normal force N, which for a flat curve is equal to the weight of the car.
Since car A, with a coefficient of static friction of 0.169, can negotiate the curve at a speed of 23.7 m/s, we can use proportionality to find the speed for car B, as the friction forces are proportional to the coefficients of static friction of their respective tires:
VB = VA ∙ √(μB / μA)
Substituting the given values:
VB = 23.7 m/s ∙ √(0.826 / 0.169)
VB = 23.7 m/s ∙ √(4.89) ≈ 23.7 m/s ∙ 2.21
VB ≈ 52.4 m/s
Therefore, the maximum speed at which car B can negotiate the curve is approximately 52.4 m/s, assuming the same curve and conditions as for car A.
A mass of 5 kg of saturated water vapor at 150 kPa is heated at constant pressure until the temperature reaches 200°C. Calculate the work done by the steam during this process.
Answer:
The work done by the steam is 213 kJ.
Explanation:
Given that,
Mass = 5 kg
Pressure = 150 kPa
Temperature = 200°C
We need to calculate the specific volume
Using formula of work done
[tex]W=Pm\DeltaV[/tex]
[tex]W=Pm(\dfrac{RT_{1}}{P_{atm}}-\dfrac{RT_{2}}{P_{atm}}[/tex]
[tex]W=\dfrac{PmR}{P_{atm}}(T_{2}-T_{1})[/tex]
Where,R = gas constant
T = temperature
P = pressure
[tex]P_{atm}[/tex]=Atmosphere pressure
m = mass
Put the value into the formula
[tex]W=\dfrac{150\times10^{3}\times5\times287.05}{1.01\times10^{5}}\times(473-373)[/tex]
[tex]W=213\ kJ[/tex]
Hence, The work done by the steam is 213 kJ.
The work done by 5 kg of saturated water heated at constant pressure to 200°C can be calculated by applying the first law of thermodynamics, W = PΔV, where W is the work done, P is the pressure, and ΔV is the change in volume of the gas. However, without information on the change in volume in this scenario, the actual calculations cannot be performed.
Explanation:The subject of this question is Physics, specifically thermodynamics, which deals with the transfer of heat and the work performed during this process. The given scenario involves a mass of 5 kg of saturated water vapor being heated at a constant pressure (150 kPa) until the temperature raises to 200°C.
We can apply the first law of thermodynamics to answer this question. The law states that the change in the internal energy of a system is equal to the heat supplied to the system minus the work done by the system. In this case, the work done by the system is the pressure times the change in volume (W = pΔV).
The scenario doesn't provide a direct change in volume, meaning the actual calculations can't be performed in this context. However, if the initial and final volumes were available, the calculations would follow the process described above. It is worth noting that for water vapor (or any other ideal gas), the volume would increase when heated if the pressure is held constant, so in theory, the system would perform positive work.
Learn more about Thermodynamics here:https://brainly.com/question/35546325
#SPJ3
Electric power is to be generated by installing a hydraulic turbine generator at a site 120 m below the free surface of a large water reservoir that can supply water at a rate of 2400 kg/s steadily. Determine the power generation potential.
Answer:
2.82 × 10^6 watt
Explanation:
H = 120 m,
Mass per second = 2400 kg/s
Power = work / time
Power = m g H / t
Power = 2400 × 9.8 × 120 / 1
Power = 2.82 × 10^6 Watt
When 23Na is bombarded with protons, the products are 20Ne and A. a neutron B. an alpha particle C. a deuteron D. a gamma ray particle E. two beta particles
Answer:
The correct option is B. an alpha particle.
Explanation:
When ²³Na is bombarded with protons [tex]_{1}^{1}\textrm{H}[/tex], ²⁰Ne and one alpha particle [tex]_{2}^{4}\textrm{He}[/tex] is released.
The reaction is as follows:
[tex]_{11}^{23}\textrm{Na} + _{1}^{1}\textrm{H} \rightarrow _{10}^{20}\textrm{Ne} + _{2}^{4}\textrm{He}[/tex]
Therefore, an alpha particle and ²⁰Ne are released when ²³Na is bombarded with protons.
There are three identical train locomotives on separate tracks. Each applies equal force to begin moving. The first train consists of a single empty freight car, the second train consists of 50 empty freight cars, and the third train consists of 50 equally loaded freight cars. Rank the accelerations of the trains. All of the freight cars, when empty, are identical.
Answer:
From lowest to highest acceleration:
3rd train
2nd train
1st train
Explanation:
The acceleration of an object can be found by using Newton's second law:
[tex]a=\frac{F}{m}[/tex]
where
a is the acceleration
F is the net force on the object
m is the mass of the object
We notice that for equal values of the forces F, the acceleration a is inversely proportional to the mass, m. Therefore, greater mass means lower acceleration, and viceversa.
So, the train with lowest acceleration is the one with largest mass, i.e. the 3rd train consisting of 50 equally loaded freight cars. Then, the 2nd train has larger acceleration, since it consists of 50 empty freight cars (so its mass is smaller). Finally, the 1st train (a single empty car) is the one with largest acceleration, since it is the train with smallest mass.
The acceleration of the three trains can be ranked as follows: the single empty freight car train, the 50 empty freight cars train, and the 50 equally loaded freight cars train.
Explanation:The accelerations of the three trains can be ranked as follows:
The single empty freight car trainThe 50 empty freight cars trainThe 50 equally loaded freight cars trainWhen an equal force is applied to each train, the train with fewer cars will experience a greater acceleration. This is because the force is distributed over fewer cars, resulting in a larger net force acting on each car.
For example, consider two cars:
Single empty freight car train: The force is applied to only one car, resulting in a larger net force per car50 empty freight cars train: The force is distributed over 50 cars, resulting in a smaller net force per carTherefore, the single empty freight car train will have the highest acceleration, followed by the 50 empty freight cars train, and finally the 50 equally loaded freight cars train.
A glass of cold water has a temperature of 42.0°E What is this in units of K?
Answer:
42°C is equivalent to 315.15 K.
Explanation:
In this question, we need to convert the temperature from one scale to another. A glass of cold water has a temperature of 42 °C. The conversion from degree Celsius to kelvin is given by :
[tex]T_k=T_c+273.15[/tex]
Where
[tex]T_c\ and\ T_k[/tex] are temperatures in degree Celsius and kelvin respectively.
So, [tex]T_k=42+273.15[/tex]
[tex]T_k=315.15\ K[/tex]
So, 42°C is equivalent to 315.15 K. Hence, this is the required solution.
Water from Old Faithful Geyser shoots upward to a height of about 50 m once every 90 minutes. What must be the pressure in the hot springs below the ground in order for the water to go that high?
Solution:
Given:
height of geyser, h = 50 m
speed of water at ground can be given by third eqn of motion with v = 0 m/s:
[tex]u^{2} = v^{2} + 2gh[/tex]
putting v = 0 m/s in the above eqn, we get:
u = [tex]\sqrt{2gh}[/tex] (1)
Now, pressure is given by:
[tex]\Delta p = \frac{1}{2}u^{2}\rho \\[/tex] (2)
where,
[tex]\Delta p[/tex] = density of water = 1000 kg/[tex]m^{3}[/tex]
g = 9.8 m/[tex]s^{2}[/tex]
Using eqn (1) and (2):
[tex]\Delta p = \frac{1}{2}(\sqrt{2gh})^{2}\rho[/tex]
⇒ [tex]\Delta p = \frac{1}{2}\rho gh[/tex]
[tex]\Delta p = \frac{1}{2}\times2\times 1000\times 9.8\times 50[/tex] = [tex]\Delta p[/tex] = 490000 Pa
[tex]p_{atm}[/tex] = 101325 Pa
For absolute pressure, p:
p = [tex]\Delta p[/tex] + [tex]p_{atm}[/tex]
p = 490000 - 101325
p = 388675 Pa
Therefore, pressure in hot springs for the water to attain the height of 50m is 388675 Pa
A 5-cm-diameter horizontal jet of water with a velocity of 40 m/s relative to the ground strikes a flat plate that is moving in the same direction as the jet at a velocity of 10 m/s. The water splatters in all directions in the plane of the plate. How much force does the water stream exert on the plate?
To calculate the force exerted by a stream of water on a moving plate, determine the change in momentum of the water as it hits the plate. By taking into account the relative velocity of the water to the plate, its cross-sectional area, and the density of water, we find the force exerted to be 17670 Newtons.
Explanation:The subject of this question is related to physics, specifically fluid dynamics. It involves calculating the force exerted by a jet of water on a moving plate. The plate and the water jet are moving in the same direction, but at different speeds. To calculate this, we need to understand the change in momentum of the water as it hits the plate.
First, we calculate the velocity of water relative to the plate, which is 40 m/s - 10 m/s = 30 m/s. The water's diameter is 5 cm, which gives us a cross-sectional area (A) of 19.63 cm2. We can obtain the volume flow rate (Q) of the water from Q = Av, which gives 0.589 m3/s. Multiplying this by the density of water (ρ=1000 kg/m3) gives a mass flow rate (ṁ) of 589 kg/s.
The force exerted by the water on the plate is equal to the rate of change of momentum of the water. Since the water is coming to rest on the plate, the initial momentum of the water relative to the plate is -ṁv, and the final momentum is zero. Therefore, the force exerted on the plate is Δp/Δt = ṁv = 589 kg/s * 30 m/s = 17670 N.
Learn more about Fluid Dynamics here:https://brainly.com/question/30578986
#SPJ3
The force exerted by the water stream on the plate is [tex]\( 2356.2 \, \text{N} \).[/tex]
To determine the force exerted by the water stream on the moving plate, we need to apply the principles of fluid dynamics and the conservation of momentum.
Given Data:
Diameter of the water jet, [tex]\( D = 5 \, \text{cm} = 0.05 \, \text{m} \)[/tex]Velocity of the water jet relative to the ground, [tex]\( V_{\text{jet}} = 40 \, \text{m/s} \)[/tex]Velocity of the plate, [tex]\( V_{\text{plate}} = 10 \, \text{m/s} \)[/tex]Step-by-Step Solution:
1. Calculate the cross-sectional area of the jet:
[tex]\[ A = \pi \left(\frac{D}{2}\right)^2 = \pi \left(\frac{0.05}{2}\right)^2 = \pi \left(0.025\right)^2 = \pi \times 0.000625 = 0.001963495 \, \text{m}^2 \][/tex]
2. Determine the velocity of the water relative to the plate:
[tex]\[ V_{\text{rel}} = V_{\text{jet}} - V_{\text{plate}} = 40 \, \text{m/s} - 10 \, \text{m/s} = 30 \, \text{m/s} \][/tex]
3. Calculate the mass flow rate of the water jet:
[tex]\[ \dot{m} = \rho \cdot A \cdot V_{\text{jet}} \][/tex]
Assuming the density of water [tex]\( \rho \) is \( 1000 \, \text{kg/m}^3 \):[/tex]
[tex]\[ \dot{m} = 1000 \, \text{kg/m}^3 \times 0.001963495 \, \text{m}^2 \times 40 \, \text{m/s} \][/tex]
[tex]\[ \dot{m} = 78.54 \, \text{kg/s} \][/tex]
4. Apply the conservation of momentum to find the force:
The change in momentum of the water relative to the plate will result in the force exerted on the plate. Since the water splatters in all directions in the plane of the plate, we assume the relative velocity of the water to the plate goes to zero after hitting the plate (as it splatters and no longer moves in a single direction).
The force exerted by the water on the plate is given by the rate of change of momentum:
[tex]\[ F = \dot{m} \cdot V_{\text{rel}} \][/tex]
Substituting the values:
[tex]\[ F = 78.54 \, \text{kg/s} \times 30 \, \text{m/s} \][/tex]
[tex]\[ F = 2356.2 \, \text{N} \][/tex]
Saturated steam coming off the turbine of a steam power plant at 40°C condenses on the outside of a 3-cm-outerdiameter, 35-m-long tube at a rate of 63 kg/h. Determine the rate of heat transfer from the steam to the cooling water flowing through the pipe. The properties of water at the saturation temperature of 40°C are hfg
Answer:
Q = 30.07 kJ/sec
Explanation:
GIVEN DATA:
temperature of steam = 40 degree celcius
mass flow rate = 63 kg/h
from saturated water tables,
from temperature 40 °, enthalpy of evaporation[tex]h_f[/tex] value is 2406 kj/kg
rate of heat transfer (Q) can be determine by using following relation
[tex]Q = \dot{m} h_f[/tex]
putting all value to get Q value
Q = 45 *2406
Q = 108270 kJ/h
[tex]Q = 108270 *\frac{1}{3600} kJ/s[/tex]
Q = 30.07 kJ/sec
Explanation:
According to the water table, the value of enthalpy of evaporation at a temperature of [tex]40^{o}C[/tex] is 2406.0 kJ/kg.
Hence, we will calculate the rate of heat transfer by using the formula as follows.
Q = [tex]m \times h_{fg}[/tex]
where, m = mass
[tex]h_{fg}[/tex] = enthalpy of evaporation
Putting the given values into the above formula as follows.
Q = [tex]m \times h_{fg}[/tex]
= [tex]63 kg/h \times 2406.0 kJ/kg[/tex]
= 151578 kJ/h
or, = [tex]151578 \times \frac{1}{3600} kJ/s[/tex]
= 42.105 kW
Thus, we can conclude that rate of heat transfer from the steam to the cooling water flowing through the pipe is 42.105 kW.
A force of 500 N is applied to a hydraulic jack piston that is 0.2 m in diameter. If the piston that supports the load has a diameter of 2.0 m, approximately how much mass can be lifted by the jack? Ignore any difference in height between the pistons.
Answer:
50,000 N
Explanation:
d = 0.2 m, r = 0.1 m, f = 500 N,
D = 2 m, R = 1 m
Let F be the weight lifted by the jack.
By use of Pascal's law
f/a = F / A
F = f x A / a
F = 500 x 3.14 x 1 x 1 / (3.14 x 0.1 x 0.1)
F = 50,000 N
A skier is pulled up a slope at a constant velocity by a tow bar. The slope is inclined at 21.5 ° with respect to the horizontal. The force applied to the skier by the tow bar is parallel to the slope. The skier's mass is 55.0 kg, and the coefficient of kinetic friction between the skis and the snow is 0.212. Find the magnitude of the force that the tow bar exerts on the skier.
Mass = 55.0 kg
Coefficient of friction = 0.212
Slope = 21.5 degrees
Gravity = 9.81
Magnitude of force = 55*9.81*(0.212*cos21.5 + sin21.5)
Magnitude of force = 304.17 Newtons
Round the answer as needed.
Answer:
Hope it can help you and mark me as a brainlist..
Wings on race cars push them into the track. The increased normal force makes large friction forces possible. At one Formula One racetrack, cars turn around a half-circle with diameter 190 m at 68 m/s. For a 610 kg vehicle, the approximate minimum static friction force to complete this turn is________.
Answer:
The minium static friction force to complete this turn is F= 29,690.94 N.
Explanation:
R= 95m
m= 610 kg
V= 68 m/s
an= V²/R
an= 48.67 m/s²
F= m * an
F= 29690.94 N
The approximate minimum static friction force to complete the turn is approximately 18806 N.
Explanation:When a race car turns a corner, the wings on the car push it into the track, increasing the normal force. This increased normal force allows for larger friction forces, which help the car maintain traction and complete the turn.
For the given Formula One racetrack with a half-circle turn of diameter 190 m and a speed of 68 m/s, we can find the approximate minimum static friction force using the centripetal force equation. The centripetal force in this situation is equal to the product of the mass (610 kg), the speed squared (68 m/s) squared, and the reciprocal of the radius (half of the diameter = 95 m).
The approximate minimum static friction force to complete this turn is approximately 18806 N.
Learn more about Static friction force here:https://brainly.com/question/17237604
#SPJ3
Rectangular Loop A current I flows in a rectangular loop of wire placed on the y plane and centered about the origin. The length of the two sides are a and b (a) Find the magnetic field at any location (0,0, 2) on the z-axis. (b) For za, b, what does your expression simplify to? (c) Compare your answer to (b) with the on-axis field for a circular loop at distances much greater than the loop radius Z. Three Parallel Wires Consider three co-planar equally-spaced (by distance d) parallel wires of negligible radius. Each wire carries current I, a in the same direction (a) Where does the magnetic field vanish? (b) Sketch the magnetic field nes in the plane perpendicular to the current flow e) Suppose you displace the middle wire a sinall distance δ in a direction perpendicular to the plane containing the wires and then release it. Analyze and describe the resulting motion of the middle wire
Answer:
mathy math
Explanation:
do you eat at baskin-robbins?
btw, answer is probably 78 (6-9)x 67,587r Xy
Suppose a woman does 463 J of work and dissipates 8,854 J of heat in the process. (a) What is the decrease in her internal energy, assuming no change in temperature or consumption of food? (That is, there is no other energy transfer.) kcal (b) What is her efficiency? %
Answer:
Part a)
decrease in internal energy is 2.23 k Cal
Part b)
Efficiency will be 4.9 %
Explanation:
Part A)
As per first law of thermodynamics we know that
[tex]Q = W + \Delta U[/tex]
here we have
Q = -8854 J
W = 463 J
now we have
[tex]-8854 J = 463 J + \Delta U[/tex]
so we have
[tex]\Delta U = -9317 J = - 2.23 kCal[/tex]
So decrease in internal energy is 2.23 k Cal
Part B)
Efficiency of the woman is given as
[tex]\eta = \frac{W}{\Delta U}[/tex]
here we have
[tex]\eta = \frac{463}{9317} = 0.049 [/tex]
So efficiency will be 4.9 %
What is the reactance of a capacitor at a 8.0 μF frequency of 56.0 Hz?
Answer:
35.55 ohm
Explanation:
C = 80 micro farad
F = 56 hertz
Reactance
Xc = 1 / (2 × pi × F × C)
Xc = 1 / ( 2 × 3.14 × 56 × 80 × 10^-6)
Xc = 35.55 ohm
What is the minimum number of 100 Ohm resistors that you need to design an effective resistor with 275 Ohm resistance.
Answer:
The minimum number of 100 Ω resistors that i need to design an effective resistor with 275Ω resistance are 8 resistors.
Explanation:
(2 Resistors of 100Ω in parallel) in series with (4 resistors of 100Ω in parallel) in series with 2 resistors of 100Ω.
2 resistors in parallel of 100Ω = 50Ω
+
4 resistors in parallel of 100Ω = 25Ω
+
2 resistors in series of 100 2Ω = 200Ω
=
275Ω
A 2,000-kg car moving east at 10.0 m/s collides with a 3,000-kg car moving north. The cars stick together and move as a unit after the collision, at an angle of 48.0° north of east and at a speed of 5.98 m/s. Find the speed of the 3,000-kg car before the collision.
To find the speed of the 3,000-kg car before the collision, use the principle of conservation of momentum. Calculate the total momentum of the combined cars after the collision based on the mass and speed. Apply the principle of conservation of momentum again to determine the initial speed of the 3,000-kg car.
Explanation:To find the speed of the 3,000-kg car before the collision, we can use the principle of conservation of momentum. In this case, the momentum before the collision is equal to the momentum after the collision. So, we can equate the momentum of the 2,000-kg car moving east to the momentum of the combined cars moving at an angle of 48.0° north of east.
Using trigonometry, we can determine the components of the velocity of the cars after the collision. The horizontal component will be the speed of the cars moving east, and the vertical component will be the speed of the cars moving north. From the given information, we can calculate the total momentum after the collision based on the mass and speed of the combined cars.
Finally, we can use the principle of conservation of momentum again to determine the speed of the 3,000-kg car before the collision. Rearranging the equation and substituting the known values, we can solve for the initial speed of the 3,000-kg car.
Learn more about Conservation of momentum here:https://brainly.com/question/32309314
#SPJ12
The speed of the 3,000-kg car before the collision is 7.93 m/s.
Explanation:To find the speed of the 3,000-kg car before the collision, we can use the principle of conservation of momentum. Before the collision, the 2,000-kg car is moving east at 10.0 m/s and the 3,000-kg car is moving north. After the collision, the two cars stick together and move as a unit. Using the given information, we can set up an equation
(2,000 kg)(10.0 m/s) + (3,000 kg)(v) = (5,000 kg)(5.98 m/s
Solving for v, we find that the speed of the 3,000-kg car before the collision is 7.93 m/s.
Learn more about speed of car before collision here:https://brainly.com/question/2100626
#SPJ11
Consider a generator that rotates its 200 turn, 0.18 m diameter coil at 3250 rpm in a 0.65 T field. Calculate the peak voltage of the generator.
Answer:
1124.8 Volt
Explanation:
N = 200, Diameter = 0.18 m, Radius, r = 0.09 m, f = 3250 rpm = 54.17 rps
B = 0.65 T
the peak value of induced emf is given by
e0 = N x B x A x ω
e0 = 200 x 0.65 x 3.14 x 0.09 x 0.09 x 2 x 3.14 x 54.17
e0 = 1124.8 Volt
The peak voltage of the generator can be calculated using Faraday's law of electromagnetic induction, taking into consideration the number of turns in the coil, the strength of the magnetic field, the area of the coil, and the speed at which the coil is rotating. In this case, it's 1100 volts.
Explanation:The subject of this question is a practical application of Faraday's law of electromagnetic induction, which is a topic typically covered in high school or college-level physics. We can find the peak voltage of a rotating generator using the formula for Induced electromotive force (emf), also known as voltage, in a rotating coil, which is given by ε_max = NBAω. Here, N is the number of turns in the coil, B is the magnetic field strength, A is the area of the coil, and ω is the angular velocity.
Let's calculate each parameter. The area A of the coil can be calculated using the formula for the area A = πr^2, where r is the radius of the coil. Given the diameter d=0.18 m, we get r = d/2 = 0.09 m, which gives A = π * (0.09 m)^2 = 0.025 m^2. The angular velocity ω in rad/s is calculated using the relation ω = 2πf, where f is the frequency in Hz. Given the frequency in rpm (revolutions per minute), we convert it to Hz by dividing by 60, hence f = 3250/60 = 54.16 Hz. Then ω = 2π * 54.16 Hz = 340.53 rad/s. Now, we can substitute the values into the emf equation: ε_max = (200 turns) * (0.65 T) * (0.025 m^2) * (340.53 rad/s) = 1100 volts. Hence, the peak voltage of the generator is 1100 volts.
Learn more about Peak Voltage Calculation here:https://brainly.com/question/21288598
#SPJ11
Suppose the sediment density (g/cm) of a randomly selected specimen from a certain region is normally distributed with mean 2.61 and standard deviation 0.82. (a) If a random sample of 25 specimens is selected, what is the probability that the sample average sediment density is at most 3.00? Between 2.61 and 3.00
Answer: The probability that the sample average sediment density is at most 3.00 = 0.9913
The probability that the sample average sediment density is between 2.61 and 3.00 = 0.4913
Explanation:
Given : Mean : [tex]\mu=2.61 [/tex]
Standard deviation : [tex]\sigma =0.82[/tex]
Sample size : [tex]n=25[/tex]
The value of z-score is given by :-
[tex]z=\dfrac{x-\mu}{\dfrac{\sigma}{\sqrt{n}}}[/tex]
a) For x= 3.00
[tex]z=\dfrac{3.00-2.61}{\dfrac{0.82}{\sqrt{25}}}=2.38[/tex]
The p-value : [tex]P(z\leq2.38)=0.9913[/tex]
b) For x= 2.61
[tex]z=\dfrac{2.61-2.61}{\dfrac{0.82}{\sqrt{25}}}=0[/tex]
The p-value : [tex]P(0<z\leq2.38)=P(2.38)-P(0)=0.9913-0.5=0.4913[/tex]