Answer:
1008.33 W.
Explanation:
Power: This can be defined as the rate at which energy is consumed or dissipated. The S.I unit of power is Watt (W).
P = E/t.......................... Equation 1
Where P = Energy, E = Energy, t = time.
But from the question,
E = 1/2mΔv²................... Equation 2
where m = mass of the sprinter, Δv = change in velocity of the sprinter = final velocity - initial velocity.
Substitute equation 2 into equation 1
P = 1/2mΔv/t....................... Equation 3
Given: m = 50 kg, Δv = 11-0 = 11 m/s, t = 3.0 s.
Substitute into equation 3
P = 1/2(50)(11)²/3
P = 25(121)/3
P = 1008.33 W.
Thus the power output = 1008.33 W.
Final answer:
The power output of a 50-kg sprinter who accelerates from 0 to 11 m/s in 3.0 s is 1008.33 Watts or approximately 1.35 horsepower.
Explanation:
To calculate the power output of a 50-kg sprinter who accelerates from 0 to 11 m/s in 3.0 s, we use the work-energy principle and the definition of power. The work done on the sprinter is equal to the change in kinetic energy, and power is the work done per unit time.
The kinetic energy (KE) at the start is 0 since the sprinter starts from rest. The kinetic energy at the end is KE = 1/2 x mass x velocity² = 1/2 x 50 kg x (11 m/s)². After calculating the kinetic energy, we get KE = 1/2 x 50 x 121 = 3025 Joules.
Now, we find the power by dividing the work by the time interval: Power = Work / Time = 3025 J / 3.0 s = 1008.33 Watts.
To convert watts to horsepower, use the conversion 1 horsepower = 746 watts. Therefore, Power in horsepower = 1008.33 W / 746 W/hp = approximately 1.35 hp.
A butterfly flies from the top of a tree in the center of a garden to rest on top of a red flower at the garden's edge. The tree is 9.0 m taller than the flower, and the garden is 17 m wide.
Determine the magnitude of the butterfly's displacement.
Express your answer using two significant figures.
Answer:
R=12 m
Explanation:
I have drawn a vector form which I attached here.Please first go through that
Given Data
y (Tree Taller) = 9.0 m
x(Distance from center to edge is)= 8.5 m
To find
R(magnitude of the butterfly's displacement
)
Solution
[tex]R=\sqrt{x^{2} +y^{2} }\\R=\sqrt{9^{2}+8.5^{2} }\\ R=12.379m[/tex]
Convert to two significant figures
R=12 m
The butterfly's displacement, when it moves from the tree to the flower, is approximately 19m. This is determined using the Pythagorean theorem, which is applied to the '9m' height difference between the tree and flower, and '17m' width of the garden.
Explanation:To calculate the magnitude of the butterfly's displacement, we need to use the Pythagorean theorem, which establishes a relationship in any right triangle. The vertical height of the tree is 9m, which forms one side of the right triangle. The horizontal distance across the garden (or width of the garden) is 17m, forming the other side of the right triangle.
Applying the Pythagorean theorem (a^2 + b^2 = c^2), where 'a' and 'b' are the sides and 'c' is the hypotenuse (in this case, the butterfly's displacement), we get:
Displacement = √[(9m)^2 + (17m)^2]
So, the displacement of the butterfly is approximately 19 m, when rounded to two significant figures.
Learn more about Displacement here:
https://brainly.com/question/33459975
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Find the magnitude of the electric force on a 2.0 uC charge in a 100n/C electric field.
Answer:
Electric force will be equal to [tex]2\times 10^{-4}N[/tex]
Explanation:
We have given charge [tex]q=2\mu C=2\times 10^{-6}C[/tex]
Electric field will be [tex]E=100N/C[/tex]
We have to find the magnitude of force
Electric force is the multiplication of electric field and charge
So electric force [tex]F=qE=2\times 10^{-6}\times 100=2\times 10^{-4}N[/tex]
Electric force will be equal to [tex]2\times 10^{-4}N[/tex]
Jessie creates a model of two waves that have the same amplitude as shown.
Which description BEST compares the two waves?
Wave B has less energy because it has a lower frequency and a longer wavelength.
Wave A has less energy because it has a higher frequency and a shorter wavelength.
Wave B has greater energy because it has a higher frequency and a shorter wavelength.
Wave A has greater energy because it has a lower frequency and a longer wavelength.
Answer:
b
Explanation:
Answer:
B
Explanation:
egde test 2023
What is the magnitude of the minimum magnetic field that will keep the particle moving in the earth's gravitational field in the same horizontal, northward direction
Answer:
The magnitude of the magnetic field vector is 1.91T and is directed towards the east.
The steps to the solution can be found in the attachment below.
Explanation:
For the charge to remain in the the earth' gravitational field the magnetic force on the charge must be equal to the earth's gravitational force on the charge and must act opposite the direction of the earth's gravitational force.
Fm = Fg
qvBSin(theta) = mg
Where q = magnitude of charge
v = magnitude of the velocity vector = 4 x10^4 m/s
B = magnitude of the magnetic field vector
theta = the angle between the magnetic field and velocity vectors = 90°
m = mass of the charge = 0.195g
g = acceleration due to gravity =9.8m/s²
On substituting the respective values of all variables in the equation (1) above
B = 1.91T
The direction of the magnetic field vector was found by the application of the right hand rule: if the thumb is pointed in the direction of the magnetic force and the index finger is pointed in the direction of the velocity vector, the middle finger points in the direction of the magnetic field.
Below is the step by step procedure to the solution.
Final answer:
The magnitude of the minimum magnetic field that will keep the particle moving in the Earth's gravitational field in the same horizontal, northward direction can be calculated using the equation for magnetic force. We need to find the minimum value of B that produces a force greater than or equal to the gravitational force acting on the particle. By setting up the equation qvBsinθ ≥ mg and solving for B, we can find the minimum magnetic field strength required.
Explanation:
The magnitude of the minimum magnetic field that will keep the particle moving in the Earth's gravitational field in the same horizontal, northward direction can be calculated using the equation for magnetic force. The force on a moving charged particle in a magnetic field is given by the equation F = qvBsinθ, where F is the magnetic force, q is the charge of the particle, v is the velocity of the particle, B is the magnetic field strength, and θ is the angle between the velocity and the magnetic field. In this case, we can assume that the angle between the velocity and the magnetic field is 90 degrees, as the particle is moving horizontally northward. Therefore, to find the minimum magnetic field that keeps the particle moving in this direction, we need to find the minimum value of B that produces a force greater than or equal to the gravitational force acting on the particle.
Let's take the example given in Example 22.1, where a glass rod with a positive charge of 20 nC is thrown with a horizontal velocity of 10 m/s due west in a place where the Earth's magnetic field is due north parallel to the ground. The force on the rod due to the Earth's magnetic field can be calculated using the equation F = qvBsinθ, where q is the charge of the rod (20 nC), v is the velocity of the rod (10 m/s), B is the magnetic field strength (unknown), and θ is the angle between the velocity and the magnetic field (90 degrees). The force on the rod should be equal to or greater than the gravitational force acting on it to keep it moving in the same horizontal, northward direction. Therefore, we can set up the equation F ≥ mg, where F is the force on the rod, m is the mass of the rod, and g is the acceleration due to gravity. Substituting the values, we get qvBsinθ ≥ mg. Solving for B, we get B ≥ mg / qvsinθ. Plugging in the given values, we can find the minimum magnetic field strength required to keep the particle moving in the Earth's gravitational field in the same horizontal, northward direction.
A body weighs 1000 lbf when exposed to a standard earth gravity of g=32.174 ft/s^2. a. What is its mass in kg? b. What is its weight in N if on moon (gmoon = 1.62 m/s^2)? c. How fast will it accelerate if under a force of 400 lbf on moon and on the earth in m/s.
Answer
given,
weight = 1000 lbf
g = 32.174 ft/s²
mass =[tex] \dfrac{1000}{32.174}[/tex]
m = 31 slug
1 slug = 14.593 kg
m = 31 x 14.593
m = 452.383 Kg
b) weight on moon
W = m g
W = 452.383 x 1.62
W= 732.86 N
c) we know,
F = m a
400 lbf = 31 slugs x a
a = 12.90 ft/s²
1 ft = 0.304 m
a = 3.92 m/s²
Final answer:
This answer covers calculating mass in kg, weight on the moon, and acceleration with a force on both Earth and the moon.
Explanation:
Mass Calculation:
Given weight = 1000 lbf and gravitational acceleration on Earth, g = [tex]32.174 ft/s^2.[/tex]Convert weight to Newtons: 1000 lbf × 4.448 N/lbf = 4448 N.Calculate mass using weight formula: Weight = mass × acceleration due to gravity.Mass = Weight / acceleration = [tex]4448 N / 9.8 m/s^2[/tex] = 454.7 kg.Weight on Moon:
Weight on the moon = mass × moon gravity acceleration = 454.7 kg × [tex]1.62 m/s^2[/tex] = 736.63 N.Acceleration with Force:
On the Moon: acceleration = force / mass = 400 lbf × 4.448 N/lbf / 454.7 kg = 3.91 m/s.On Earth: acceleration = 400 lbf × 4.448 N/lbf / 1000 kg = 1.79 m/s.Pedas hot water spring is located 15 km from Seremban town. The government has decided to build geothermal power plant using hot water from Pedas hot water spring. A geothermal power plant uses geothermal water at 150°C at a rate 210 kg/s as the heat source and produces 8000kW of net power. The geothermal water leaves the plant at 90°C. If the environment temperature is around 25°C. Determine:
a. The actual rate of heat input to this power plant.
b. The actual thermal efficiency and the maximum possible thermal efficiency of this power plant.
c. The actual rate of heat rejection from this power plant.
d The power output if the geothermal water leaves the plant at 40°C, and maintain it thermal efficiency.
Final answer:
To determine the actual rate of heat input to the geothermal power plant, calculate the heat input using the provided formula. The actual thermal efficiency can be found by dividing the net power output by the heat input. The actual rate of heat rejection can then be calculated as the difference between the heat input and the net power output. To determine the power output if the geothermal water leaves the plant at a different temperature while maintaining its thermal efficiency, use the provided formula with the appropriate values.
Explanation:
The actual rate of heat input to the geothermal power plant can be calculated using the formula:
Heat input = mass flow rate * specific heat capacity * (hot water temperature - environment temperature)
Given that the mass flow rate is 210 kg/s, the specific heat capacity is 4.186 J/g°C, the hot water temperature is 150°C, and the environment temperature is 25°C, we can substitute these values into the formula to find the heat input.
The actual thermal efficiency of the power plant can be calculated using the formula:
Thermal efficiency = (Net power output / Heat input) * 100%
Given that the net power output is 8000 kW and we have already calculated the heat input, we can substitute these values into the formula to find the actual thermal efficiency. The maximum possible thermal efficiency can be calculated as the Carnot efficiency, which is the maximum possible efficiency for a heat engine operating between two temperatures.
The actual rate of heat rejection from the power plant can be calculated as the difference between the heat input and the net power output.
To find the power output if the geothermal water leaves the plant at 40°C and maintain its thermal efficiency, we can use the formula:
Power output = (Heat input - Heat rejection) * (Net power output / Heat input)
Substituting the appropriate values into the formula will give us the desired power output.
Calculate the average density of the earth in g/cm3, assuming our planet to be a perfect sphere.
To develop this problem we will apply the relationship of density, such as the unit of mass per unit volume of a body. For this relationship we will use the known constants of the value of the land's mass and its respective radius. These values will be converted from kilograms to grams and meters to centimeters respectively. We will find the volume through the geometric relationship of the sphere using the radius of the earth.
The mass of the Earth is given as
[tex]m_E = 5.9722*10^{24}kg (\frac{1000g}{1kg}) = 5.9722*10^{27}g[/tex]
The radius of the Earth is
[tex]r_E = 6.3781*10^6m (\frac{100cm}{1m}) = 6.3781*10^8cm[/tex]
Using the geometric value of volume for a sphere, and using the radius of the earth, as the radius of that sphere we have to
[tex]V_e = \frac{4}{3} \pi r^3[/tex]
Replacing,
[tex]V_E = \frac{4}{3} \pi (6.3781*10^8)^3[/tex]
[tex]V_E = 1.09*10^{27}cm^3[/tex]
The expression for the density of the Earth is
[tex]\rho = \frac{m_E}{V_E}[/tex]
Replacing,
[tex]\rho = \frac{5.9722*10^{27}}{1.09*10^{27}}[/tex]
[tex]\rho = 5.48g/cm^3[/tex]
Therefore the average density of the earth is [tex] 5.48g/cm^3[/tex]