The question involves analyzing the projectile motion of a rock thrown from a building using physics principles like motion decomposition and energy conservation. It requires breaking down the initial velocity into horizontal and vertical components and applying kinematic equations to determine parameters such as max height, range, and flight time.
Explanation:This question involves the principles of projectile motion and energy conservation in physics. When the man throws the rock at an angle of 28 degrees above the horizontal with an initial velocity of 26.0 m/s from a height of 14.0 meters, we need to analyze the horizontal and vertical components of the motion separately to determine various aspects of the rock's trajectory, such as its range, maximum height, and time of flight. However, since the specific request is missing in this context, we'll focus on the general approach to solving such problems.
To solve problems involving objects thrown at an angle, we first decompose the initial velocity into its horizontal (vx = v*cos(θ)) and vertical (vy = v*sin(θ)) components, where v is the magnitude of the initial velocity and θ is the angle of projection. The horizontal motion is uniform, meaning the velocity remains constant, whereas the vertical motion is affected by gravity, leading to acceleration in the opposite direction of the initial vertical velocity component.
Energy conservation or kinematic equations can be used to find specific details like maximum height reached, time of flight, and range. For example, the formula s = ut + 0.5at² can be applied where s is displacement, u is initial velocity, a is acceleration (gravity in the case of vertical motion), and t is time. Remember, acceleration due to gravity (a) is -9.8 m/s², indicating it acts downwards. Ignoring air resistance simplifies calculations by omitting drag force considerations.
The maximum horizontal distance is approximately 61.7 meters. This is determined using the projectile motion equations for horizontal distance with initial velocity, angle, and height given.
To find the maximum horizontal distance the rock travels, we can analyze the projectile motion. The initial velocity of 26.0 m/s is broken down into horizontal and vertical components. The horizontal component is [tex]\( v_x = v \cdot \cos(\theta) \)[/tex], where \( v \) is the magnitude of the velocity (26.0 m/s) and \( \theta \) is the angle (28.0 degrees). The vertical component is [tex]\( v_y = v \cdot \sin(\theta) \).[/tex]
The vertical motion is affected by gravity, and the time it takes for the rock to hit the ground can be calculated using the equation [tex]\( h = v_y \cdot t - \frac{1}{2} g t^2 \),[/tex] where \( h \) is the initial height (14.0 m), \( g \) is the acceleration due to gravity (approximately 9.8 m/s\(^2\)), and \( t \) is the time of flight.
Using the quadratic formula to solve for \( t \), we find two solutions: one when the rock is at the initial height and one when it hits the ground. We use the positive solution for the time of flight to calculate the horizontal distance traveled using the equation [tex]\( d = v_x \cdot t \).[/tex]
Substituting the values, we find the maximum horizontal distance to be approximately 61.7 meters.
The question probably maybe: What is the maximum horizontal distance the rock travels before hitting the ground, given that a man stands on the roof of a building of height 14.0 m and throws a rock with a velocity of magnitude 26.0 m/s at an angle of 28.0 degrees above the horizontal, ignoring air resistance?
A 0.1 m by 0.1 m sheet of cardboard is placed in a uniform electric field of 10 N/C. At first, the plane of the sheet is oriented perpendicular to the electric field vector so that the electric flux through the sheet is 0.01 N-m2/C. By what angle do you need to rotate the sheet to reduce the electric flux by 1/2?
Answer:
The angle is 89°.
Explanation:
Given that,
Electric field = 10 N/C
Electric flux = 0.01 N-m²/C
Area [tex]A=\pi\times(0.1)^2[/tex]
We need to calculate the angle
Using formula of electric flux
[tex]\phi=EA\cos\theta[/tex]
[tex]\cos\theta=\dfrac{\phi}{EA}[/tex]
Where, E = electric field
[tex]\phi[/tex] = electric flux
A = area
Put the value into the formula
[tex] \cos\theta=\dfrac{\dfrac{0.01}{2}}{10\times\pi\times(0.1)^2}[/tex]
[tex]\theta=\cos^{-1}(0.01592)[/tex]
[tex]\theta=89.0^{\circ}[/tex]
Hence, The angle is 89°.
The angle required to rotate the sheet to reduce the electric flux by 1/2 is 89 degrees.
What is electric flux?The number of electric lines that interact the area of a given object or space.
It can be given as,
[tex]\phi=ES\cos \theta[/tex]
Here, [tex]E[/tex] is the magnitude of electric field, [tex]S[/tex] is the area of surface and [tex]\theta[/tex] is the angle between electric field and perpendicular.
Given information-
The dimensions of sheet of cardboard is 0.1 by 0.1.
The sheet is placed between the uniform electricity field of 10 N/C.
Put the values in the above formula as,
[tex]\dfrac{0.01}{2} =10\times\pi\times{0.1^2}\times\cos \theta\\\theta=cos^-(0.01592)\\\theta=89^o[/tex]
Hence the angle required to rotate the sheet to reduce the electric flux by 1/2 is 89 degrees.
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You are lost at night in a large, open field. Your GPS tells you that you are 122.0 m from your truck, in a direction 58.0o east of south. You walk 72.0 m due west along a ditch. How much farther, and in what direction, must you walk to reach your truck?
Answer:
The person is 187[m] farther and 70° south to east.
Explanation:
We can solve this problem by drawing a sketch of the location of the person and the truck, then we will draw the displacement vectors and finally the length of the vector and the direction of the vector will be measured in order to give the correct indication of where the person will have to move.
First we establish an origin of a coordinate system.
We can see in the attached schema that the red vector is the displacement vector from the last point to where the truck is located.
The length of the vector is 187 [m], and the direction is 70 degrees south to East.
This problem involves using vectors and trigonometry to calculate the direct distance and direction to the truck from the new location after walking due west. This is achieved by adding the horizontal and vertical components of the vectors representing the initial location and the walking path.
Explanation:To solve this, one may use vectors and trigonometry. Initially, you are 122.0m from your truck, 58.0 degrees east of south. This can be treated as a vector from your truck to your original location. Then, you walk 72.0m due west, which is another vector in the opposite direction.
To find the resulting vector, i.e., the direct distance and direction to the truck from your new location, we have to add these vectors. While the mathematics is somewhat complex, the concept involves adding the horizontal (east-west) and vertical (north-south) components of each vector. Once the resulting vector is calculated, the remaining distance to the truck can be found from its magnitude, and the direction from its angle relative to south.
This involves math calculations, including trigonometry and Pythagorean theorem.
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Carbon dioxide in a piston-‐‐cylinder is expanded in a polytropic manner. The initialtemperature and pressure are 400 K and 550 kPa. The polytropic exponent, n, is equalto 1.2. The final temperature (at the end of the expansion process) is 350 K. Determinethe heat transfer per kg in the process (i.e., in units of kJ/kg).
Answer:
q_poly = 14.55 KJ/kg
Explanation:
Given:
Initial State:
P_i = 550 KPa
T_i = 400 K
Final State:
T_f = 350 K
Constants:
R = 0.189 KJ/kgK
k = 1.289 = c_p / c_v
n = 1.2 (poly-tropic index)
Find:
Determine the heat transfer per kg in the process.
Solution:
-The heat transfer per kg of poly-tropic process is given by the expression:
q_poly = w_poly*(k - n)/(k-1)
- Evaluate w_poly:
w_poly = R*(T_f - T_i)/(1-n)
w_poly = 0.189*(350 - 400)/(1-1.2)
w_poly = 47.25 KJ/kg
-Hence,
q_poly = 47.25*(1.289 - 1.2)/(1.289-1)
q_poly = 14.55 KJ/kg
Final answer:
To determine the heat transfer per kg in the polytropic expansion of carbon dioxide gas in a piston-cylinder system, we can use the first law of thermodynamics. By plugging in the given values and solving the equation, we can find the heat transfer per kg in units of kJ/kg.
Explanation:
In this problem, we are given the initial temperature and pressure of a carbon dioxide gas in a piston-cylinder system. The gas undergoes a polytropic expansion process with a given polytropic exponent. The final temperature is also given.
To determine the heat transfer per kg in the process, we can use the first law of thermodynamics which states that the change in internal energy of a system is equal to the heat transfer into the system minus the work done by the system.
Since the process is polytropic, we can use the formula Q = ΔU + W = CvΔT + W, where Q is the heat transfer per kg, ΔU is the change in internal energy, Cv is the specific heat at constant volume, ΔT is the change in temperature, and W is the work done.
Plugging in the given values and solving the equation will give us the heat transfer per kg in units of kJ/kg.
Two stones are thrown vertically upward from the ground, one with three times the initial speed of the other. (a) If the faster stone takes 10 s to return to the ground, how long will it take the slower stone to return? (b) If the slower stone reaches a maximum height of H, how high (in terms of H) will the faster stone go? Assume free fall.
Answer:
t₁ = 3.33s, h = 9 H
Explanation:
let the v₁ = initial velocity of the faster stone and v₂ = initial velocity of the slower stone
using equation of motion and displacement equals to zero since the stone returned to the point of projection
y - y₀ = v₁ t - 1/2gt²
- v₁ t = - 1/2gt²
2v₁ t / t² = g
g = 2v₁ / t
repeat the same produce for the slower stone where the time = t₁
y-y₀ = v₂t₁ - 1/2 gt₁²
- v₂t₁ = - 1/2 gt₁²
t₁ = 2v₂ / g = 2v₂ / (2v₁ / t) = (2v₂ / 2v₁) × t
and
v₁ = 3v₂
t₁ = (2v₂ / 2v₁) × t = (v₂ / 3v₂) × 10 = 3.33 s
b) using the equation of motion
vf₂² = v₂² - 2gH
since the body stop momentarily at maximum height
- v₂² = - 2gH
v₂² / 2H = g
repeating the same procedure for the faster stone
vf₁² = v₁² - 2gh
- v₁² = - 2gh
v₁²/ 2g = h
substitute for g
h = v₁² / 2(v₂² / 2H ) = (v₁² / v₂²) × H = (3v₂)² / (v₂² ) × H = 9H
A yo-yo with a mass of 0.075 kg and a rolling radius of 2.50 cm (the distance from the axis of the pulley to where the string comes off the spool) rolls down a string with a linear acceleration of 6.50 m/s2. Approximate the rotational inertia of the yo-yo with that of disk with mass, m, and radius, r, rotating about its center (mr2/2). Calculate the tension in the string.
Answer:
0.24825 N
0.0000238701923077 kgm²
Explanation:
m = Mass of yo yo = 0.075 kg
a = Acceleration = 6.5 m/s²
g = Acceleration due to gravity = 9.81 m/s²
The net force is given by
[tex]F_n=mg-T[/tex]
[tex]\Rightarrow T=mg-ma[/tex]
[tex]\Rightarrow T=m(g-a)[/tex]
[tex]\Rightarrow T=0.075(9.81-6.5)[/tex]
[tex]\Rightarrow T=0.24825\ N[/tex]
The tension in the string is 0.24825 N
Angular acceleration is given by
[tex]\alpha=\dfrac{a}{r}\\\Rightarrow \alpha=\dfrac{6.5}{2.5\times 10^{-2}}\\\Rightarrow \alpha=260\ rad/s^2[/tex]
Torque is given by
[tex]\tau=I\alpha\\\Rightarrow Tr=I\alpha\\\Rightarrow I=\dfrac{Tr}{\alpha}\\\Rightarrow I=\dfrac{0.24825\times 2.5\times 10^{-2}}{260}\\\Rightarrow I=0.0000238701923077\ kgm^2[/tex]
The moment of inertia is 0.0000238701923077 kgm²
The tension in the string is equal to 0.2475 Newton.
Given the following data:
Mass of yo-yo = 0.075 kgRadius = 2.50 cm to m = [tex]\frac{2.5}{100} = 0.0025 \;m[/tex]Linear acceleration = 6.50 [tex]m/s^2[/tex]To determine the tension in the string:
First of all, we would determine the downward force applied by the yo-yo's weight:
[tex]F_y = mg[/tex]
Where:
[tex]F_y[/tex] is the yo-yo's weight. m is the mass of the yo-yo. g is acceleration due to gravity.
Substituting the given parameters into the formula, we have;
[tex]F_y = 0.075 \times 9.8\\\\F_y = 0.735 \; Newton[/tex]
Next, we would determine the force acting on the string:
[tex]F_s = 0.075 \times 6.5\\\\F_s = 0.4875\;Newton[/tex]
Now, we can find the tension in the spring:
[tex]Tension = F_y - F_s\\\\Tension = 0.735 - 0.4875[/tex]
Tension = 0.2475 Newton.
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A Pitot-static tube is used to measure the velocity of helium in a pipe. The temperature and pressure are 44oF and 24 psia. A water manometer connected to the Pitot-static tube indicates a reading of 3.0 in. (a) Determine the helium velocity. (b) Is it reasonable to consider the flow as incompressible?
Answer:
Part A:
[tex]V_1=\sqrt{\frac{2\gamma_{h20}h}{\rho_H}}\\ V_1=\sqrt{\frac{2*62.43*3}{5.5210*10^{-4}*12}}\\ V_1=237.778 ft/s[/tex]
Part B:
Ma=0.0737
Since Ma<0.3, it means the flow is in compressible.
Explanation:
Part A:
According to Bernoulli equation:
[tex]P_1+\frac{\rho_H}{2}V^2_1 =P_{2}+\frac{\rho_H}{2}V^2_2\\ V_2=0,\\P_1+\frac{\rho_H}{2}V^2_1 =P_{2}[/tex]
Velocity will become:
[tex]V_1=\sqrt{\frac{2(P_2-P_1)}{\rho_H}}[/tex].........Eq (1)
Now,[tex]P_2-P_1[/tex] can be calculated from the specific weight of water and helium[tex]P_2-P_1[/tex][tex]=(\gamma_{h2}o-\gamma_H)h[/tex]
Since the specific weight of helium is much smaller than specific weight of water we can neglect the specific weight of helium.
[tex]P_2-P_1[/tex]=[tex]=(\gamma_{h2o})h[/tex]
For water,[tex]\gamma_{h2o}=62.43 lb/ft^3[/tex]
h=3.0 in
Density of helium:
[tex]\rho_H=\frac{P}{RT}[/tex]
T=460+44=504 degree R
R=[tex]1.242*10^4 ft.lb/R.slug[/tex]
[tex]\rho_H=\frac{24*12^2}{1.242*10^4*504}\\ \rho_H=5.5210*10^{-4} lb/ft^3[/tex]
From Eq (1):
[tex]V_1=\sqrt{\frac{2\gamma_{h20}h}{\rho_H}}\\ V_1=\sqrt{\frac{2*62.43*3}{5.5210*10^{-4}*12}}\\ V_1=237.778 ft/s[/tex]
Part B:
Checking Ma:
[tex]Ma=\frac{V}{c}[/tex]
c is speed of sound:
k=1.66 for helium, In ideal gases:
[tex]c=\sqrt{kRT}\\ c=\sqrt{1.66*1.242*10^4*504}\\ c=3223.51 ft/s\\Ma=\frac{237.778}{3223.51}\\ Ma=0.0737[/tex]
Since Ma<0.3, it means the flow is in compressible.
(a) The helium velocity is approximately 2811.62 ft/s. (b) No, it's not reasonable to consider the flow as incompressible due to the high Mach number[tex](Ma ≈ 2.58)[/tex], indicating compressibility effects.
To determine the helium velocity using a Pitot-static tube, you can use the Bernoulli's equation, assuming steady, incompressible flow:
\[P + 0.5 * ρ * V^2 + ρ * g * h = constant\]
Where:
- [tex]\(P\)[/tex] is the pressure in the pipe (psia)
- [tex]\(ρ\)[/tex]is the density of the fluid (helium in this case, lb/ft^3)
- [tex]\(V\)[/tex] is the velocity of the fluid (ft/s)
- [tex]\(g\)[/tex] is the acceleration due to gravity (32.2 ft/s^2)
- [tex]\(h\)[/tex] is the height difference in the manometer (inches)
Given:
- Temperature[tex](\(T\))[/tex] = 44°F = 44 + 460 = 504 Rankine (R)
- Pressure[tex](\(P\))[/tex] = 24 psia
- Reading in manometer [tex](\(h\))[/tex] = 3.0 inches
First, we need to find the density [tex](\(ρ\))[/tex] of helium at the given conditions. You can use the ideal gas law:
[tex]\[PV = nRT\][/tex]
Where:
-[tex]\(P\)[/tex] is pressure (psia)
- [tex]\(V\)[/tex]is volume ([tex]ft^3[/tex])
- [tex]\(n\)[/tex]is the number of moles
- [tex]\(R\)[/tex]is the specific gas constant[tex](for helium, \(R = 53.34\)[/tex][tex]ft·lb/(lbmol·R))[/tex]
- [tex]\(T\)[/tex] is temperature (Rankine)
Rearrange the equation to find [tex]\(ρ\):[/tex]
[tex]\[ρ = \frac{n}{V} = \frac{P}{RT}\][/tex]
Substitute the values:
[tex]\[ρ = \frac{24}{53.34 * 504} = 0.0008933 lb/ft^3\][/tex]
Now, we can calculate the velocity[tex](\(V\))[/tex]using Bernoulli's equation:
[tex]\[P + 0.5 * ρ * V^2 + ρ * g * h = constant\][/tex]
[tex]\[V = \sqrt{\frac{2 * (P - \text{manometer correction})}{ρ}}\][/tex]
The manometer correction accounts for the density of the manometer fluid, which is typically water in this case. Since we're given that the manometer reading is in inches, we need to convert it to feet:
Manometer Correction = 3.0 inches / 12 = 0.25 ft
Now, calculate the velocity:
[tex]\[V = \sqrt{\frac{2 * (24 - 0.25 * 62.4)}{0.0008933}} = 2811.62 ft/s\][/tex]
(a) The helium velocity is approximately 2811.62 ft/s.
(b) No, it's not reasonable to consider the flow as incompressible because helium is a compressible gas, and at high velocities and pressure differentials, compressibility effects become significant. To consider the flow as incompressible, the Mach number (Ma) should be much less than 0.3. To calculate Ma:
[tex]\[Ma = \frac{V}{a}\][/tex]
Where[tex]\(a\)[/tex] is the speed of sound in helium, which can be calculated using:
[tex]\[a = \sqrt{\gamma * R * T}\][/tex]
Where \(\gamma\) is the specific heat ratio for helium (approximately 1.66).
Calculate \(a\) and then Ma:
[tex]\[a = \sqrt{1.66 * 53.34 * 504} = 1087.92 ft/s\][/tex]
[tex]\[Ma = \frac{2811.62}{1087.92} \approx 2.58\][/tex]
Since the Mach number is significantly greater than 0.3, the flow of helium in this case cannot be considered incompressible.
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A container in the shape of a cube 10.0 cm on each edge contains air (with equivalent molar mass 28.9 g/mol) at atmospheric pressure and temperature 300 K. Find (a) the mass of the gas, (b) the gravitational force exerted on it, and (c) the force it exerts on each face of the cube. (d) Why does such a small sample exert such a great force? (6%)
Answer:
a) m = 1.174 grams
b) F_g = 0.01151 N
c) F_c = 1013 N
Explanation:
Given:
- The length of a cube L = 10.0 cm
- The molar mass of air M = 28.9 g/mol
- Pressure of air P = 101.3 KPa
- Temperature of air T = 300 K
- Universal Gas constant R = 8.314 J/kgK
Find:
(a) the mass of the gas
(b) the gravitational force exerted on it
(c) the force it exerts on each face of the cube
(d) Why does such a small sample exert such a great force? (6%)
Solution:
- Compute the volume of the cube:
V = L^3 = 0.1^3 = 0.001 m^3
- Use Ideal gas law equation and compute number of moles of air n:
P*V = n*R*T
n = P*V / R*T
n = 101.3*10^3 * 0.001 / 8.314*300
n = 0.04061 moles
- Compute the mass of the gas:
m = n*M
m = 0.04061*28.9
m = 1.174 grams
- The gravitational force exerted on the mass of gas is due to its weight:
F_g = m*g
F_g = 1.174*9.81*10^-3
F_g = 0.01151 N
- The force exerted on each face of cube is due its surface area:
F_c = P*A
F_c = (101.3*10^3)*(0.1)^2
F_c = 1013 N
- The molecules of a gas have high kinetic energy; hence, high momentum. When they collide with the walls they transfer momentum per unit time as force. Higher the velocity of the particles higher the momentum higher the force exerted.
(a) The mass of the gas is 1.174g
(b) Gravitational force on it is 0.0115N
(c) Force exerted on the container walls is 1.01×10³N
(d) High kinetic energy of the molecules
Ideal gas in a container:The sides of the cubical container are, L = 10cm = 0.1m
Pressure of the gas, P = 1 atm = 1.01×10⁵ Pa
Temperature of the gas, T = 300K
Molas mass, M = 28.9 g/mol
(a) The volume of the container:
V = L³ = 0.1³
V = 0.001m³
From the idea gas equation:
PV = nRT
n = PV/RT
n = (1.01×10⁵)(0.001)/8.314×300
n = 0.04061 moles
So, the mass of the gas is:
m = nM = 0.04061×28.9
m = 1.174g
(b) the gravitaitonal force on the gas is given by:
f = mg
f = 1.174×10⁻³×9.8
f = 0.0115N
(c) the force exerted on each wall of the cube is :
F = PA
where A is the area of each wall
F = 1.01×10⁵×0.1×0.1
F = 1.01×10³ N
(d) The atoms/molecules of the gas have high kinetic energies, they constantly collide with the walls of the container and transfer large momentum which results in such large force.
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A 120 g, 8.0-cm-diameter gyroscope is spun at 1000 rpm and allowed to precess. What is the precession period?
The precession period of the gyroscope is approximately [tex]\(_0._6_2_8 seconds\).[/tex]
Explanation:The precession period [tex](\(T_p\))[/tex] of a gyroscope can be determined using the formula:
[tex]\[ T_p = \frac{2\pi I}{mgh} \][/tex]
Where:
[tex]\( I \)[/tex] is the moment of inertia,
[tex]\( m \)[/tex] is the mass of the gyroscope,
[tex]\( g \)[/tex] is the acceleration due to gravity, and
[tex]\( h \)[/tex] is the height of the center of mass.
Firstly, calculate the moment of inertia [tex](\( I \))[/tex] using the formula:
[tex]\[ I = \frac{1}{2} m r^2 \][/tex]
Given that the mass [tex](\( m \))[/tex] of the gyroscope is [tex]\( 0.120 \, kg \)[/tex] and the diameter [tex](\( d \)) is \( 0.08 \, m \), the radius (\( r \)) is \( 0.04 \, m \).[/tex]Substitute these values into the formula to find [tex]\( I \).[/tex]
Next, plug the values of [tex]\( I \), \( m \), \( g \), and \( h \[/tex]) into the precession period formula. The acceleration due to gravity [tex](\( g \))[/tex] is approximately [tex]\( 9.8 \, m/s^2 \), and \( h \)[/tex] is the height of the center of mass, which is not provided but typically considered as the radius of the gyroscope [tex](\( r \)). Finally, solve for \( T_p \).[/tex]
After the calculations, the precession period [tex](\( T_p \))[/tex] is found to be approximately [tex]\(_0._6_2_8 seconds\).[/tex]This represents the time it takes for the gyroscope to complete one precession cycle. The precession period is a crucial parameter in understanding the behavior of gyroscopes and their applications in various fields.
An elevator in a tall building is allowed to reach a maximum speed of 3.3 m/s going down. What must the tension be in the cable to stop this elevator over a distance of 3.4 m if the elevator has a mass of 1320 kg including occupants?
To calculate the tension in the cable required to stop the elevator, we multiply the mass of the elevator by its acceleration. The tension in the cable is -4356 N, indicating it acts in the opposite direction of the weight of the elevator.
Explanation:To calculate the tension in the cable required to stop the elevator over a distance of 3.4 m, we need to consider the force required to decelerate the elevator from its maximum speed of 3.3 m/s to a stop. The tension in the cable must equal the force needed to stop the elevator, which is equal to the mass of the elevator multiplied by its acceleration. The mass of the elevator, including occupants, is given as 1320 kg. Since the elevator is going down, its acceleration will be negative. Therefore, the tension in the cable can be calculated using the formula:
Tension = mass * acceleration = 1320 kg * (-3.3 m/s^2) = -4356 N
Therefore, the tension in the cable to stop the elevator over a distance of 3.4 m is -4356 N. The negative sign indicates that the tension is acting in the opposite direction of the weight of the elevator.
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How much heat is absorbed by a 28g iron skillet when its temperature rises from 10oC to 27oC?
_____ ___
Answer units
Amount of heat absorbed: 214 J
Explanation:
When an object absorbs heat, its temperature increases according to the equation
[tex]Q=mC\Delta T[/tex]
where
Q is the heat absorbed
m is the mass of the object
C is the specific heat capacity of the material
[tex]\Delta T[/tex] is the change in temperature
For the iron skillet in this problem:
m = 28 g = 0.028 kg is the mass
[tex]C=450 J/kg^{\circ}C[/tex] is the iron specific heat capacity
[tex]\Delta T = 27-10=17^{\circ}C[/tex] is the increase in temperature
Solving for Q, we find the amount of heat absorbed:
[tex]Q=(0.028)(450)(17)=214 J[/tex]
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Radio wave radiation falls in the wavelength region of 10.0 to 1000 meters. What is the energy of radio wave radiation that has a wavelength of 784 m?
Radio waves have lower energy compared to other types of waves in the electromagnetic spectrum. The energy of a radio wave with a wavelength of 784 m can be calculated using the equation E = hf, where E is the energy, h is Planck's constant, and f is the frequency.
Explanation:Radio waves fall within the electromagnetic spectrum which consists of various types of waves ranging from gamma rays to radio waves. The energy of a wave is related to its wavelength and frequency through the equation E = hf, where E is the energy, h is Planck's constant, and f is the frequency. Since radio waves have a longer wavelength, they have a lower frequency and therefore lower energy compared to other types of waves in the spectrum.
To calculate the energy of a radio wave with a wavelength of 784 m, we can use the equation c = λf, where c is the speed of light. Rearranging the equation to solve for f, we get f = c/λ. Plugging in the given wavelength of 784 m and the speed of light which is approximately 3 x 10^8 m/s, we can calculate the frequency as 3 x 10^8 m/s / 784 m = 3.83 x 10^5 Hz. Substitute this frequency value into the equation E = hf to calculate the energy of the radio wave.
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The energy of radio wave radiation for a 784-meter wavelength is approximately 2.53 x 10^-34 joules, calculated using Planck's equation for energy.
The energy E of radio wave radiation with a particular wavelength λ can be calculated using the equation:
E = hc/λ
Where:
E is the energy in Joules (J)
h is Planck's constant (6.626 x 10-34 J·s)
c is the speed of light in vacuum (about 3 x 108 m/s)
λ is the wavelength in meters (m)
For a radio wave with a wavelength of 784 meters, we use these values to compute the energy:
E = (6.626 x 10-34 J·s) * (3 x 108 m/s) / 784 m
Calculating using the numerical values:
E = (6.626 x 10-34) * (3 x 108) / 784
E ≈ 2.53 x 10-34 J
Therefore, the energy of radio wave radiation with a wavelength of 784 meters is approximately 2.53 x 10-34 joules.
What is the final temperature when 71.8 g of water at 78.8°C is mixed with 33.6 g of water at 29.0°C? (The specific heat of water is 4.184 J/g·°C.)
Answer:
62.92°
Explanation:
given,
initial mass of water, m₁ = 71.8 g
initial temperature, T₁ = 78.8°C
another mass of water, m₂ = 33.6 g
another temperature of water, T₂ = 29° C
Final temperature of mix = ?
energy going out of the hot water equal to the energy amount going into the cool water.
[tex]q_{lost} =q_{gain}[/tex]
[tex]m_1 c \Delta T = m_1 c \Delta T[/tex]
[tex]71.8 (78.8-x) = 33.6\times (x - 29)[/tex]
[tex] 105.4 x = 6632.24[/tex]
x = 62.92°
Hence, the final temperature of the mix is equal to 62.92°
Can a goalkeeper at her/ his goal kick a soccer ball into the opponent’s goal without the ball touching the ground? The distance will be about 95 m. A goalkeeper can give the ball a speed of 30 m/s.
Answer:
No she cannot.
Explanation:
Let [tex]v_h[/tex] be the horizontal component of the ball velocity when it's kicked, assume no air resistance, this is a constant. Also let [tex]v_v[/tex] be the vertical component of the ball velocity, which is affected by gravity after it's kicked.
The time it takes to travel 95m accross the field is
[tex]t = 95 / v_h[/tex] or [tex]v_h = 95/t[/tex]
t is also the time it takes to travel up, and the fall down to the ground, which ultimately stops the motion. So the vertical displacement after time t is 0
[tex]s = v_vt + gt^2/2= 0[/tex]
where g = -9.8m/s2 in the opposite direction with [tex]v_v[/tex]
[tex]v_vt - 4.9t^2 = 0[/tex]
[tex]v_vt = 4.9t^2[/tex]
[tex]v_v = 4.9t[/tex]
Since the total velocity that the goal keeper can give the ball is 30m/s
[tex]v = v_v^2 + v_h^2 = 30^2 = 900[/tex]
[tex](4.9t)^2 + \left(\frac{95}{t})^2 = 900[/tex]
[tex]24.01t^2 + \frac{9025}{t^2} = 900[/tex]
Let substitute x = [tex]t^2[/tex] > 0
[tex]24.01 x + \frac{9025}{x} = 900[/tex]
We can multiply both sides by x
[tex]24.01 x^2 + 9025 = 900x[/tex]
[tex]24.01x^2 - 900x + 9025 = 0[/tex]
[tex]t= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}[/tex]
[tex]t= \frac{900\pm \sqrt{(-900)^2 - 4*(24.01)*(9025)}}{2*(24.01)}[/tex]
As [tex](-900)^2 - 4*24.01*9025 = -56761 < 0[/tex]
The solution for this quadratic equation is indefinite
So it's not possible for the goal keeper to do this.
An elevator is moving upward at a constant speed of 2.50 m/s. A bolt in the elevator ceiling 3.00 m above the elevator floor works loose and falls. (a) How long does it take for the bolt to fall to the elevator floor? What is the speed of the bolt just as it hits the elevator floor (b) according to an observer in the elevator? (c) According to an observer standing on one of the floor landings of the building? (d) According to the observer in part (c), what distance did the bolt travel between the ceiling and the floor of the elevator?
To answer the question, we use principles from mechanics, involving the concepts of free fall and relative motion. The time taken by the bolt to hit the floor is around 0.78s and has different velocities to an observer inside and outside the elevator. According to a ground observer, the bolt travels a total distance of 4.95m.
Explanation:To answer these questions, we need to involve the principles of physics, specifically mechanics dealing with motion. Firstly, since the bolt was initially at rest, and it falls under the influence of gravity, we use the formula for time in free fall t = √(2h/g), where h is the initial height (3.00 m) and g is the acceleration due to gravity (9.81 m/s²).
For (b), for an observer in the elevator, the bolt appears to fall straight down, so its velocity just as it hits the floor will be v = gt = 9.81*0.78 = 7.65 m/s.
For (c), for an observer on the floor landing, the elevator is moving upwards, thus the velocity of the bolt relative to the observer on the ground would be the sum of the falling velocity and the velocity of the elevator, so v = gt + elevator speed = 7.65 + 2.50 = 10.15 m/s.
For (d), the bolt would have traveled the height of the fall plus the distance the elevator traveled during the fall according to the ground observer. The distance the elevator moves is d = elevator speed * t = 2.50*0.78 = 1.95 m, therefore, the total distance the bolt travels is 3.00 m + 1.95 m = 4.95 m.
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As you drive away from a radio transmitter, the radio signal you receive from the station is shifted to longer wavelengths. (T/F)
The statement is true due to the Doppler Effect. As you move away from the radio transmitter, the radio signal you receive appears to have been shifted to longer wavelengths due to the changes in the observer-source distance.
Explanation:The statement, 'As you drive away from a radio transmitter, the radio signal you receive from the station is shifted to longer wavelengths,' is True. This is due to a phenomenon known as the Doppler Effect.
The Doppler Effect explains that the observed wavelength of electromagnetic radiation is longer (red shift) when the source moves away from the observer. This means the wavelength of the radiation from the radio station would appear to increase (shift to a longer wavelength) as you drive away from it.
To maintain that same energy level required for transmission, the radio station emits waves at a higher frequency (shorter wavelengths). But, as you move away, the radio waves appear to have a lower frequency (longer wavelengths). This happens because the waves get 'stretched out' as the distance between you (the observer) and the radio station (the source) increases.
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A 4.89 μC test charge is placed 4.10 cm away from a large, flat, uniformly charged nonconducting surface. The force on the charge is 321 N. The charge is then moved 2.00 cm farther away from the surface. What is the force on the test charge now?
Final answer:
The force on the test charge remains the same at 321 N after it is moved farther away because the electric field produced by a large, flat, uniformly charged surface is constant close to the surface.
Explanation:
The question involves calculating the new force on a test charge after it is moved farther away from a charged surface. The force between the test charge and a charged surface is given by Coulomb's law, but since the surface is large and flat, we assume the field is uniform. Hence, the force experienced by the test charge is directly proportional to the electric field strength. The electric field produced by a charged surface is constant for regions close to the surface and does not depend on the distance from it. Therefore, when the test charge is moved farther away within this region, the force it experiences remains the same because the electric field strength is unchanged. In this scenario, after moving the charge from 4.10 cm to a new distance of 6.10 cm (an additional 2.00 cm), the force on the test charge will remain at 321 N.
The owner of a van installs a rear-window lens that has a focal length of -0.300 m. When the owner looks out through the lens at an object located directly behind the van, the object appears to be 0.250 m from the back of the van, and appears to be 0.350 m tall. (a) How far from the van is the object actually located, and (b) how tall is the object?
Final answer:
The object is actually located 1.5 m from the van, and it is 2.1 m tall when considering the given focal length and image properties.
Explanation:
- Focal length of the lens (f) = -0.300 m (negative sign indicates a diverging lens)
- Image distance = 0.250 m (positive because the image appears behind the lens)
- Height of the image = 0.350 m
We need to find:
(a) Actual distance (d) of the object from the van.
(b) Actual height (h) of the object.
(a) To find the actual distance of the object from the van (d), we use the thin lens equation:
1/f = 1/d + 1/image distance
Plugging in the given values:
1/-0.300 = 1/d + 1/0.250
Solving for d:
-3.33 = 1/d + 4
-3.33 - 4 = 1/d
-7.33 = 1/d
d = 1/-7.33
d ≈ -0.136 m
The negative sign indicates that the object is located behind the lens.
(b) To find the actual height of the object (h), we use the magnification formula:
m = height of the image / height of the object = -(image distance / object distance)
Given m = -(image distance / object distance), and we have m = height of the image / height of the object, we can write:
-(image distance / object distance) = height of the image / height of the object
Plugging in the given values:
-(0.250 / -0.136) = 0.350 / h
Solving for h:
h = (0.350 * -0.136) / 0.250
h ≈ -0.190 m
The negative sign indicates that the height of the object is inverted.
So, the answers are:
(a) The object is actually located approximately 0.136 m behind the van.
(b) The actual height of the object is approximately 0.190 m.
Think about the pencil-dropping activity that you did in the introduction. What did the target finally look like?
Answer:
By dropping a pencil from a certain fixed height again and again it will make the target super messay with marks of dot everywhere on the target and some even out side the target.
Explanation:
In uniform circular motion, how does the acceleration change when the speed is increased by a factor of 3? When the radius is decreased by a factor of 2?
Answer:
The acceleration will become 9/2 times.
a' =9/2 a
Explanation:
We know that acceleration of a particle when it is moving in the circular path is given as
[tex]a=\omega^2\ r[/tex]
r=radius
ω= angular speed
If the speed ω '= 3 ω
If the radius ,[tex]r'=\dfrac{r}{2}[/tex]
The final acceleration =a'
[tex]a'=\omega^2'\ r'[/tex]
[tex]a'=(3\omega)^2\times \dfrac{r}{2}[/tex]
[tex]a'=9\omega^2\times \dfrac{r}{2}[/tex]
[tex]a'=\omega^2\times \dfrac{9r}{2}[/tex]
[tex]a'= \dfrac{9r}{2}\times \omega^2\times r[/tex]
[tex]a'=\dfrac{9}{2}a[/tex]
Therefore the acceleration will become 9/2 times.
A wheel rotates clockwise 6 times per second. What will be its angular displacement after 7 seconds? Answer should be rounded to 2 decimal places
Answer:
The frequency of the wheel is the number of revolutions per second:
f= \frac{N_{rev}}{t}= \frac{10}{1 s}=10 Hz
And now we can calculate the angular speed, which is given by:
\omega = 2 \pi f=2 \pi (10 Hz)=62.8 rad/s in the clockwise direction.
Explanation:
(6 rotations/sec) x (7 sec) = 42 rots
Each rotation is 360 degrees or 2π radians.
42 rotations = 15,120 degrees
or
84π radians .
The output voltage of a power supply is normally distributed with mean 5 V and standard deviation 0.02 V. If the lower and upper specifications for voltage are 4.95 V and 5.05 V, respectively, what is the probability that a power supply selected at random will conform to the specifications on voltage? 34
The probability that a randomly selected power supply will conform to the specified voltage range of 4.95 V to 5.05 V, given a normal distribution with mean 5 V and standard deviation 0.02 V, is roughly 95%.
Explanation:To find the probability that a power supply selected at random will conform to the specifications on voltage, we need to calculate the area under the normal distribution curve between the lower specification (4.95 V) and the upper specification (5.05 V). Given that the mean (μ) is 5 V and the standard deviation (σ) is 0.02 V, we first convert these specifications into their corresponding Z-scores.
The Z-score is calculated by the formula Z = (X - μ) / σ, where X is the value we're converting. For the lower specification, 4.95 V, the Z-score is Z = (4.95 - 5) / 0.02 = -2.5. For the upper specification, 5.05 V, the Z-score is Z = (5.05 - 5) / 0.02 = 2.5.
According to the properties of the normal distribution, approximately 95% of observations fall within two standard deviations of the mean. This means that the probability of a power supply being within the range of 4.95 V to 5.05 V is roughly 95%, as both of these Z-scores fall within the ±2 standard deviations of the mean.
A 30.0-kg girl in a swing is pushed to one side and held at rest by a horizontal force \vec{F} F ⃗ so that the swing ropes are 30.0^\circ30.0 ∘ with respect to the vertical. Calculate the tension in each of the two ropes supporting the swing under these conditions.
Answer:
169.74 N
Explanation:
Given,
Mass of the girl = 30 Kg
angle of the rope with vertical, θ = 30°
equating the vertical component of the tension
vertical component of the tension is equal to the weight of the girl.
T cos θ = m g
T cos 30° = 30 x 9.8
T = 339.48 N
Tension on the two ropes is equal to 339.48 N
Tension in each of the rope = T/2
= 339.48/2 = 169.74 N
Hence, the tension in each of the rope is equal to 169.74 N
Final answer:
To calculate the tension in the ropes supporting the swing, one must account for the weight of the girl and set it equal to the combined vertical components of the tension in the ropes. The tension in each rope supporting the swing is found to be approximately 339.4 N.
Explanation:
The subject of this question is Physics, specifically related to the application of Newton's Laws of Motion to calculate tension in ropes. The question includes a scenario where a 30.0-kg girl in a swing is held at rest in a position where the ropes form a 30.0-degree angle with the vertical.
To find the tension in each rope, we first need to consider the forces acting on the girl and the swing: the force of gravity (weight) pulling her down, and the tension T in the ropes that supports her.
Since the swing is at rest, the net force in each direction must be zero (static equilibrium). Thus, the upward tension components in the ropes must equal the downward weight force. For each rope, the vertical component of the tension (Ty) will be Ty = T*cos (30.0°).
The weight, which is equally distributed across both ropes, is the force of gravity acting on the girl, calculated as mg, where m = 30.0 kg and g = 9.8 [tex]m/s^2[/tex]. Setting the vertical components of tension equal to the weight of the girl, we can solve for T:
2 * T * cos(30.0°) = 30.0 kg * 9.8 [tex]m/s^2[/tex]
This gives us:
T = (30.0 kg * 9.8 [tex]m/s^2[/tex]) / (2 * cos(30.0°))
Performing the calculation:
T = 339.4 N (approximately)
Therefore, the tension in each of the two ropes is approximately 339.4 N.
In a laboratory, the Balmer-beta spectral line of hydrogen has a wavelength of 486.1 nm. If the line appears in a star's spectrum at 485.5 nm, what is the star's radial velocity (in km/s)
Use Doppler's formula to find the radial velocity of star.
[tex]\frac{V_r}{c} = \frac{\Delta \lambda}{\lambda_0}[/tex]
Here,
[tex]V_r[/tex] = Radial Velocity
c = Speed of light
[tex]\Delta \lambda[/tex] = Shift in wavelength
[tex]\lambda_0[/tex] = Laboratory wavelength of spectral line
Rearrange for [tex]V_r[/tex],
[tex]V_r = \frac{\Delta \lambda}{\lambda_0} c[/tex]
Find shift in wavelength, [tex]\Delta \lambda[/tex]
[tex]\Delta \lambda = |485.5nm - 486.1nm|[/tex]
[tex]\Delta \lambda = 0.6nm[/tex]
Replacing our values we have then,
[tex]V_r = \frac{0.6nm}{486.1nm}(3*10^8m/s)[/tex]
[tex]V_r = 370000m/s[/tex]
Therefore the radial velocity of star is [tex]3.7*10^5[/tex]m/s
In this case the symbol of [tex]\Delta \lambda[/tex] implies that the star is receding the observer and the wavelength turns to red, then is red shifted.
Final answer:
To calculate the star's radial velocity based on the observed Doppler shift in the hydrogen Balmer-beta line, we use the shift in wavelength from 486.1 to 485.5 nm and apply the formula for Doppler shift.
Explanation:
The student's question involves calculating the radial velocity of a star based on the Doppler shift observed in the hydrogen Balmer-beta spectral line. The observed shift from 486.1 nm in the laboratory to 485.5 nm in the star's spectrum indicates a movement towards us. We can calculate the radial velocity using the formula v = c × (Δλ / λ), where Δλ is the change in wavelength (486.1 - 485.5 = 0.6 nm), λ is the original wavelength (486.1 nm), and c is the speed of light (3 × 108 m/s). Converting 0.6 nm to meters (0.6 × 10-9 m) and plugging in the values gives us the star's radial velocity.
Three charged particles are placed at each of three corners of an equilateral triangle whose sides are of length 4.0 cm . Two of the particles have a negative charge: q1
Answer:
The question continues ; Two of the particles have a negative charge: q1 = -6.3nC and q2 = -12.6nC . The remaining particle has a positive charge, q3 = 8.0nC . What is the net electric force acting on particle 3 due to particle 1 and particle 2?
Explanation:
The step by step and mathematical interpretation is as shown in the attached file.
This question is about charged particles placed in an equilateral triangle.
The subject of this question is Physics and it is at a High School level.
In this question, three charged particles are placed at each of the three corners of an equilateral triangle. The sides of the triangle are given to be 4.0 cm long. Two of the particles have a negative charge, which is denoted as q1.
This question involves understanding the concept of charged particles and their placement in a geometry, as well as the effects of charges on each other. It requires knowledge in basic physics principles and calculations related to charges and forces between them.
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What is the wavelength of the photons emitted by hydrogen atoms when they undergo n =5 to n =3 transitions?
Answer:
[tex]\lambda=1282nm[/tex]
Explanation:
The wavelength of the photons emitted due to an atomic electron transition in a hydrogen atom, is given by the Rydberg formula:
[tex]\frac{1}{\lambda}=R_H(\frac{1}{n_1^2}-\frac{1}{n_2^2}})[/tex]
Here [tex]R_H[/tex] is the Rydberg constant for hydrogen and [tex]n_1,n_2[/tex] are the lower and higher quantum number for the energy levels of the atomic electron transition, respectively. Replacing the given values and solving for [tex]\lambda[/tex]
[tex]\frac{1}{\lambda}=1.097*10^7m^{-1}(\frac{1}{3^2}-\frac{1}{5^2}})\\\frac{1}{\lambda}=7.81*10^5m^{-1}\\\lambda=\frac{1}{7.81*10^5m^{-1}}\\\lambda=1.282*10^{-6}m\\\lambda=1.282*10^{-6}m*\frac{1nm}{10^{-9}m}\\\lambda=1282nm[/tex]
A small rock is thrown straight up with initial speed v0 from the edge of the roof of a building with height H. The rock travels upward and then downward to the ground at the base of the building. Let +y be upward, and neglect air resistance.For the rock's motion from the roof to the ground, what is the vertical component vav−yvav−y of its average velocity?
Express your answer in terms of acceleration due to gravity ggg, and the variables v0v0v_{0} and HHH.
The average vertical velocity of the rock thrown upwards from the height H to the ground is determined by the total vertical displacement divided by the total time, considering the downward acceleration due to gravity is negative.
Explanation:The vertical component v_{av-y} of the average velocity of a small rock thrown straight up from the edge of a roof of height H to the ground can be found by using the kinematic equations of motion under constant acceleration. Since the acceleration due to gravity (g) is acting downward, it is represented as negative in the equations. Considering the upwards direction as positive and the downward acceleration as negative ensures that the final answer for v_{av-y} takes into account the total displacement, which includes both the upward and the downward path of the rock. The average velocity is thus the total displacement divided by the total time taken. The rock initially travels upwards to a maximum height before falling to the ground, completing its motion.
Vector A points in the negative y-direction and has a magnitude of 14 units. Vector B has twice the magnitude and points in the positive x-direction.
(a) Find the direction and magnitude of A + B. (degrees counterclockwise from the +x axis)
(b) |A + B| = ?
(c) Find the direction and magnitude of A - B. (degrees counterclockwise from the +x axis)
(d) |A - B| =?
(e) Find the direction and magnitude of B - A. (degrees counterclockwise from the +x axis)
To find the magnitude and direction of vectors A+B, A-B, and B-A, we perform vector addition and subtraction and calculate the magnitudes using the Pythagorean theorem. The direction of the vectors depend on the specific directions of vectors A and B.
Explanation:The subject of this question is in the field of Physics, specifically under vectors. The grade level of this question falls under High School.
(a) The direction of vector A+B can be found by adding the two vectors together. The resulting vector will point diagonally, in a direction between that of vector A (which is in the negative y-direction) and vector B (which is in the positive x-direction). The magnitude of the resulting vector is √((14)^2 + (2*14)^2)= 24.5
(b) The magnitude of vector A+B is also known as the modulus of the vector, and it is the same as the magnitude calculated in part (a), which is 24.5 units.
(c) The direction of vector A-B can be found by subtracting vector B from vector A. The direction will be in the negative x-direction. The magnitude of the resulting vector is √((14)^2 + (2*14)^2) = 24.5
(d) The magnitude of the vector A-B, also known as the modulus of the vector, is given by the same calculation as in part (c), which is 24.5 units.
(e) The direction of vector B-A is the opposite of the direction of vector A-B. Therefore, the direction is in the positive x-direction and the magnitude of the resulting vector is the same as before, i.e., 24.5 units.
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Many satellites orbit Earth at maximum altitudes above Earth's surface of 1000 km or less. Geosynchronous satellites, however, orbit at an altitude of 35790 km above Earth's surface. How much more energy is required to launch a 410 kg satellite into a geosynchronous orbit than into an orbit 1000 km above the surface of Earth?
Answer:
6.26 times more
Explanation:
Given:
- most satellite orbit at height r_1 = 1000 km
- Geosynchronous satellites orbit at height r_2 = 35,790 km
- mass of Geosynchronous satellite m = 410 kg
- The radius of the earth r_e = 6371 km
Find:
- Compare the Energy required to send the satellite to Geosynchronous orbit @ r_2 vs Energy required to send the satellite to normal orbit @ r_1. How much more. ( U_1 / U_2 ).
Solution:
- The gravitational potential energy of any mass m in an orbit around another mass M is given by the following relation:
U_g = - G*m*M / r
Where,
G : Gravitational constant
- We compute the gravitational potential energy U_g of the satellite at both orbits as follows:
-Normal orbit U_1 = - G*m*M / r_e + G*m*M / (r_e+r_1)
U_2 = - G*m*M / r_e + G*m*M / (r_2+r_e)
Now: Take a ratio of the two energies U_1 and U_2 as follows:
U_2 / U_1 = (- G*m*M / r_e + G*m*M / r_2+r_e) / (- G*m*M / r_e + G*m*M / r_1+r_e)
U_2 / U_1 = (1 / (r_2+r_e) - 1 / r_e ) / (1 / (r_1 + r_e) - 1 / r_e )
- Plug values:
U_2 / U_1 = (1 / (35790+6371) - 1 / 6371 ) / (1 / (1000+6371) - 1 / 6371 )
- Evaluate:
U_2 / U_1 = (-1.33242681 * 10^-4) / (2.12944*10^-5)
U_2 / U_1 = 6.26
- Hence The energy required to send the satellite to Geosynchronous orbit is 6.26 times more than that required for normal orbit.
Michael Phelps needs to swim at an average speed of 2.00 m/s in order to set a new world record in the 200 m freestyle. If he swims the first 100 meters at an average speed of 1.80 m/s how fast must he swim the second 100 meters in order to break the record?
Answer:
Explanation:
Given
average speed of Phelps [tex]v_{avg}=2\ m/s[/tex]
total distance [tex]d=200\ m[/tex]
he swims first 100 m at an average speed if [tex]1.8 m/s[/tex]
so time taken is [tex]t_1=\frac{100}{1.8}=55.55\ s[/tex]
suppose [tex]t_2[/tex] is the time taken to swim remaining half
average velocity is [tex]v_{avg}=\frac{displacement}{total\ time}[/tex]
[tex]v_{avg}=\frac{100+100}{55.55+t_2}[/tex]
[tex]t_2+55.55=\frac{200}{2}=100[/tex]
[tex]t_2=44.45\ s[/tex]
so velocity in the second half is
[tex]v_2=\frac{100}{45.45}[/tex]
[tex]v_2=2.19\approx 2.2\ m/s[/tex]
Final answer:
Michael Phelps must swim the second 100 meters at an average speed of 2.25 m/s to break the world record in the 200 m freestyle.
Explanation:
To calculate how fast Michael Phelps must swim the second 100 meters to break the world record, let's use the formula for average speed, which is total distance divided by total time. First, we find the time it takes for him to swim the first 100 meters at 1.80 m/s, which is 100 m / 1.80 m/s = 55.56 seconds (rounded to two decimal places). To set a new world record, Phelps must complete the 200 m in a total time less than or equal to 200 m / 2.00 m/s = 100 seconds. Thus, for the second 100 meters, he has 100 seconds - 55.56 seconds = 44.44 seconds. The speed required for the second 100 meters is then 100 m / 44.44 s, which equals 2.25 m/s.
One possible remnant of a supernova, called a neutron star, can have the density of a nucleus, while being the size of a small city. What would be the radius, in kilometers, of a neutron star with a mass 10 times that of the Sun? The radius of the Sun is 7 × 108 m and its mass is 1.99 × 1030 kg.
The radius of a neutron star with a mass 10 times that of the Sun is approximately 29.62 km.
Explanation:To calculate the radius of a neutron star with a mass 10 times that of the Sun, we can use the equation for the Schwarzschild radius:
R = 2GM / c^2
Where R is the radius, G is the gravitational constant, M is the mass, and c is the speed of light. Plugging in the values, we get:
R = 2 * 6.67x10^-11 (m^3/kg/s^2) * (10 * 1.99x10^30 kg) / (3x10^8 m/s)^2
R ≈ 29.62 km
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