Answer:
The initial mass of the rocket is 526.2 kg.
Explanation:
Given that,
Final speed = 105 m/s
Time = 18.0 s
Exhaust initial speed = 1200 m/s
Mass of burned fuel = 110 kg
We need to calculate the initial mass
The velocity change of rocket under gravity is defined as,
[tex]v=u\ ln(\dfrac{m_{i}}{m})-gt[/tex]....(I)
We know that,
[tex]m=m_{i}-m_{bf}[/tex]
Put the value of m in equation (I)
[tex]v=u\ ln(\dfrac{m_{i}}{m_{i}-m_{bf}})-gt[/tex]
[tex]m_{i}=\dfrac{m_{bf}}{1-e^-{\dfrac{v+gt}{u}}}[/tex]
[tex]m_{i}=\dfrac{110}{1- e^-{\frac{105+9.8\times18}{1200}}}[/tex]
[tex]m_{i}=526.2 kg[/tex]
Hence, The initial mass of the rocket is 526.2 kg.
The initial mass of the rocket including the initial fuel is 1257.14 kg.
Conservation of linear momentum
The initial mass of the rocket is determined by applying the principle of conservation of linear momentum as follows;
m₁v₁ = m₂v₂
105m = 1200 x 110
105 m = 132,000
m = 132,000/105
m = 1257.14 kg
Thus, the initial mass of the rocket including the initial fuel is 1257.14 kg.
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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 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]
On a cold day, a heat pump absorbs heat from the outside air at 14°F (−10°C) and transfers it into a home at a temperature of 86°F (30°C). Determine the maximum κ of the heat pump.
Final answer:
The maximum coefficient of performance (COP) of the heat pump absorbing heat from the outside air at -10°C and transferring it into a home at 30°C is 0.75.
Explanation:
The coefficient of performance (COP) of a heat pump is a measure of its efficiency and is defined as the ratio of the heat delivered to the heat absorbed. In this case, we have a heat pump that absorbs heat from the outside air at -10°C and transfers it into a home at 30°C. To determine the maximum COP of the heat pump, we can use the formula:
COP = T_h / (T_h - T_c)
where T_h is the temperature of the hot reservoir (in this case, 30°C) and T_c is the temperature of the cold reservoir (in this case, -10°C).
Plugging in the values, we get:
COP = 30 / (30 - (-10)) = 30 / 40 = 0.75
Therefore, the maximum COP of the heat pump is 0.75.
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 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
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.
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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]
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
A 64.4-kg firefighter climbs a flight of stairs 21.3-m high. How much work (in J) does he do?
Answer:
Work done by firefighter = 13456.57 J
Explanation:
Here work done is equal to potential energy gained by him.
Potential energy, PE = mgh, where m is the mass, g is acceleration due to gravity value and h is the height.
Mass, m = 64.4 kg
Acceleration due to gravity, g = 9.8 m/s²
Height, h = 21.3 m
Substituting
Potential energy, PE = mgh = 64.4 x 9.81 x 21.3 = 13456.57 J
Work done by firefighter = 13456.57 J
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.
13) A pendulum of mass M, length L, amplitude A, and on the earth with acceleration due to gravity of g has a period T. How long would the pendulum need to be in order to to have a period of 2T?
Answer:
4L
Explanation:
mass of pendulum = M, length of pendulum, L1 = L,
acceleration due to gravity = g, Amplitude = A
Use the formula for the time period for simple pendulum
[tex]T = 2\pi \sqrt{\frac{L}{g}}[/tex] ..... (11)
Let the new length be L2 for which the time period is 2T.
[tex]2T = 2\pi \sqrt{\frac{L_{2}}{g}}[/tex] ..... (2)
Divide equation (2) by equation (1)
[tex]\frac{2T}{T} = \sqrt{\frac{L_{2}}{L}}[/tex]
So, L2 = 4 L
Thus, the length of the pendulum is 4 L.
The mass of a hot-air balloon and its cargo (not including the air inside) is 170 kg. The air outside is at 10.0°C and 101 kPa. The volume of the balloon is 530 m3. To what temperature must the air in the balloon be warmed before the balloon will lift off? (Air density at 10.0°C is 1.244 kg/m3.)
Answer:
108.37°C
Explanation:
P₁ = Initial pressure = 101 kPa
V₁ = Initial volume = 530 m³
T₁ = Initial temperature = 10°C = 10+273.15 =283.15 K
P₂ = Final pressure = 101 kPa (because it is open to atmosphere)
V₂ = Final volume = 530 m³
P₁V₁ = n₁RT₁
⇒101×530 = n₁RT₁
⇒53530 J = n₁RT₁
P₂V₂ = n₂RT₂
⇒53530 J = n₂RT₂
[tex]\frac{m_1}{m_2}=\frac{\rho V_1}{\rho V_1-170}\\\Rightarrow \frac{m_1}{m_2}=\frac{1.244\times 530}{1.244\times 530-170}=1.347\\\Rightarrow \frac{m_1}{m_2}=1.347\\\Rightarrow \frac{n_1}{n_2}=1.347[/tex]
Dividing the first two equations we get
[tex]1=\frac{n_1}{n_2}\frac{T_1}{T_2}\\\Rightarrow 1=1.347\frac{283.15}{T_2}\\\Rightarrow T_2=1.347\times 283.15= 381.52\ K[/tex]
∴Temperature must the air in the balloon be warmed before the balloon will lift off is 381.25-273.15 = 108.37°C
To lift off, the air in the balloon must be warmed to a temperature of approximately 916.93 K.
Explanation:To calculate the temperature to which the air in the balloon must be warmed before it lifts off, we can use the ideal gas law equation: PV = nRT.
The volume of the balloon is given as 530 m3, the pressure is atmospheric pressure (101 kPa), and the mass is 170 kg (which can be converted to moles using the molar mass of air). We can rearrange the equation to solve for temperature:
T = PV/(nR)
Substituting the values, we have:
T = (101,000 Pa) × (530 m3) / (n × 8.31 J/mol·K)
Where n is the number of moles of air. To find the number of moles, we can use the density of air at 10.0°C (1.244 kg/m3):
n = mass / molar mass = 170 kg / (1.244 kg/m3 × 530 m3)
Substituting this value for n and solving for T:
T = (101,000 Pa) × (530 m3) / ((170 kg / (1.244 kg/m3 × 530 m3)) × 8.31 J/mol·K)
Calculating, we find T ≈ 916.93 K.
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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.
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 5.40 uF parallel-plate, air capacitor has a plate separation of 3.50 mm and is charged to a potential difference of 480 V. Calculate the energy density the region between the plates.
Answer:
Energy density = 0.0831 J/m³
Explanation:
Energy density of a capacitor is given by the expression
[tex]u=\frac{1}{2}\epsilon E^2[/tex]
We have electric field ,
[tex]E=\frac{V}{d}=\frac{480}{3.5\times 10^{-3}}=1.37\times 10^5V/m[/tex]
Here there is no dielectric
So energy density,
[tex]u=\frac{1}{2}\epsilon_0 E^2=\frac{1}{2}\times 8.85\times 10^{-12} \times (1.37\times 10^5)^2=8.31\times 10^{-2}=0.0831 J/m^3[/tex]
Energy density = 0.0831 J/m³
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.
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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.
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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.
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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
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.
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.
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]
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
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At what frequency will a 3.0 μF capacitor have a reactance of 7.0 kΩ?
Answer:
Frequency, f = 7.57 Hz
Explanation:
It is given that,
Capacitance, [tex]C=3\ \mu F=3\times 10^{-6}\ F[/tex]
Capacitive reactance, [tex]X_C=7\ k\Omega=7\times 10^3\ \Omega[/tex]
We need to find the frequency. The capacitive reactance of the capacitor is given by :
[tex]X_C=\dfrac{1}{2\pi fC}[/tex]
f is the frequency
[tex]f=\dfrac{1}{2\pi CX_C}[/tex]
[tex]f=\dfrac{1}{2\pi \times 3\times 10^{-6}\ F\times 7\times 10^3\ \Omega}[/tex]
f = 7.57 Hz
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.
An Earth satellite moves in a circular orbit 561 km above Earth's surface with a period of 95.68 min. What are (a) the speed and (b) the magnitude of the centripetal acceleration of the satellite?
Explanation:
It is given that,
Radius of earth, r = 6371 km
An earth satellite moves in a circular orbit above the Earth's surface, d = 561 km
So, radius of satellite, R = 6371 km + 561 km = 6932 × 10³ m
Time taken, t = 95.68 min = 5740.8 sec
(a) Speed of the satellite is given by :
[tex]v=\dfrac{d}{t}[/tex]
d = distance covered
For circular path, d = 2πR
[tex]v=\dfrac{2\pi \times 6932\times 10^3\ m}{5740.8\ sec}[/tex]
v = 7586.92 m/s
(b) Centripetal acceleration is given by :
[tex]a=\dfrac{v^2}{R}[/tex]
[tex]a=\dfrac{(7586.92\ m/s)^2}{6932\times 10^3\ m}[/tex]
[tex]a=8.3\ m/s^2[/tex]
Hence, this is the required solution.
a) The speed of the satellite is approximately 7584.38 m/s
b) The magnitude of the centripetal acceleration is approximately 8.29 m/s².
To solve this problem, we will calculate both the speed and the magnitude of the centripetal acceleration of the satellite.
(a) Speed of the Satellite
First, we need to determine the radius of the satellite's orbit. The radius of the Earth is R = 6378 km. Since the satellite orbits 561 km above the surface, the radius of the orbit (r) is:
[tex]r = 6378 km + 561 km = 6939 km = 6939000 m[/tex]
The period (T) of the satellite is 95.68 minutes, which we convert to seconds:
[tex]T = 95.68 min \times 60 s/min = 5740.8 s[/tex]
Using the formula for the speed (v) of an object in circular motion:
[tex]v = 2\pi r/T[/tex]
We can plug in the values:
[tex]v = 2\pi \times 6939000 m / 5740.8 s = 7584.38 m/s[/tex]
(b) Magnitude of the Centripetal Acceleration
The centripetal acceleration (ac) is given by:
[tex]a_c = v^2/r[/tex]
Substituting the values we have:
[tex]a_c = (7584.38 m/s)^2 / 6939000 m = 8.29 m/s^2[/tex]
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
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..
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]
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.
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 %
A series circuit that is connected to a 50 V, 60 Hz source is made up of 25 ohm resistor, capacite wieh X= 18 ohms, and inductor with X 24 ohms, what is the circuit impedance? a. 25.70 b. 29.2Q 32.4Q c. d. 35.90 5
Answer:
the impedance of the circuit is 25.7 ohms.
Explanation:
It is given that,
Voltage, V = 50 volts
Frequency, f = 60 Hz
Resistance, R = 25 ohms
Capacitive resistance, [tex]X_C=18\ ohms[/tex]
Inductive resistance, [tex]X_L=24\ ohms[/tex]
We need to find the impedance of the circuit. It is given by :
[tex]Z=\sqrt{R^2+(X_L-X_C)^2}[/tex]
[tex]Z=\sqrt{25^2+(24-18)^2}[/tex]
Z = 25.7 ohms
So, the impedance of the circuit is 25.7 ohms. Hence, this is the required solution.