An Olympic skier moving at 20.0 m/s down a 30.0o slope encounters a region of wet snow, of
coefficient of friction μk = 0.740. How far down the slope does she go before stopping?
a.119 m
b.145 m
c.170 m
d.199 m
Answer:
Option B
Explanation:
Forces acting on the skier-
F1 [tex]= -mg sin(30)[/tex] down the slope
F2 [tex]= -mg cos(30)[/tex]
F3 = friction force [tex]= 0.74 mg cos(30)[/tex]
Net force, down the slope
[tex]= -mg sin(30)+ 0.74 mg cos(30)\\= mg(.74 cos(30)-sin(30))\\= 0.14mg\\= 1.38m[/tex]
Acceleration [tex]= F/m= 1.38[/tex] m/s2
Acceleration remains constant, initial speed is 20 m/s
Speed at time t is [tex]1.38t- 20[/tex] m/s
Distance down the slope at time t is [tex]0.69t^2- 20t[/tex]
When the skier stops, her speed is 0. Thus,
, [tex]1.38t- 20= 0\\t= 20/1.38\\= 14.5[/tex] seconds
Distance travelled in 14.5 seconds [tex]= (0.69)(14.52- 20(14.5)= -145 m[/tex](negative because it is down the slope).
Option B is correct
13. The percent of Earth's surface covered by high clouds
in January 1987 was closest to which of the following?
A. 13.09
B. 13.5%
C. 14.0%
D. 14.5%
16. Which of the following figures best represents the
monthly average cover of high. middle, and low clouds
in January 19922
E.
H.
cloud cover
okud cover
middle cloud
high clouds
high cloud
low clouds
low clouds
midle clouds
G.
J.
14. Based on Table I. a cosmic ray flux of 440.000 parti-
cles/m he would correspond to a cover of low clouds
that is closest to which of the following?
F. 28.75
G. 29.09
H. 29.35
J. 29.6%
ckud cover
cloud cover
high clouds
low clouds
middle cloud
high clouds
middle clouds
low clouds
PLEASE HELP
A ball, 1.8 kg, is attached to the end of a rope and spun in a horizontal circle above a student's head. As the student rotates the ball in a horizontal clockwise circle their lab partner counts 53 rotations in one minute. What is the ball's angular velocity in radians per second?
Answer:
The angular speed of the ball in radians per second is 5.55 rad/s.
Explanation:
Given;
mass of the ball, m = 1.8 kg
number of the ball's rotation per minute, n = 53 RPM
The angular speed of the ball in radians per second is calculated as follows;
[tex]\omega = 53\frac{rev}{\min} \times \frac{2\pi \ rad}{1 \ rev} \times \frac{1\min}{60 \ s } \\\\\omega = 5.55 \ rad/s[/tex]
Therefore, the angular speed of the ball in radians per second is 5.55 rad/s.
Water is to be heated from 10°C to 80°C as it flows through a 2-cm-internal-diameter, 13-m-long tube. The tube is equipped with an electric resistance heater, which provides uniform heating throughout the surface of the tube. The outer surface of the heater is well insulated, so that in steady operation all the heat generated in the heater is transferred to the water in the tube. If the system is to provide hot water at a rate of 5 L/min, determine the power rating of the resistance heater. Also, estimate the inner surface temperature of the pipe at the exit.
Answer:
- the power rating of the resistance heater is 24139.5 W
- the inner surface temperature of the pipe at the exit is 96.34°C
Explanation:
Given the data in the question;
Flow rate of water in the tube V" = 5L/min = 8.333 × 10⁻⁵ m³/s
The water is to be heated from 10°C to 80°C;
so Average or mean temperature [tex]T_{avg[/tex] will be;
[tex]T_{avg[/tex] = (T₁ + T₂) / 2 = (10 + 80) / 2 = 90/2 = 45°C
Now, from the Table " Properties of Water " at average temperature;
at [tex]T_{avg[/tex] = 45°C
density p = 990.1 kg/m³
specific heat [tex]C_p[/tex] = 4180 J/kg-k
thermal conductivity k = 0.637 W/m-°C
Now, we determine the mass flow;
m" = pV"
we substitute
m" = 990.1 × 8.333 × 10⁻⁵
m" = 0.08250 kg/s
we know that the power rating of the resistance heater is equal to the heat transfer rate to the water;
Q' = m"[tex]C_p[/tex]( T₂ - T₁ )
we substitute
Q' = (0.08250 × 4180 ) ( 80 - 10 )
Q' = 344.85 × 70
Q' = 24139.5 W
Hence, the power rating of the resistance heater is 24139.5 W
Next, we determine the average velocity of water in the tube;
[tex]V_{avg[/tex] = V" / [tex]A_c[/tex]
[tex]V_{avg[/tex] = V" / ( [tex]\frac{1}{4}[/tex]πD² )
given that; flows through a 2-cm-internal-diameter; D = 0.02 m
we substitute
[tex]V_{avg[/tex] = (8.333 × 10⁻⁵) / ( [tex]\frac{1}{4}[/tex]π × (0.02)² )
[tex]V_{avg[/tex] = (8.333 × 10⁻⁵) / ( 3.14159 × 10⁻⁴ )
[tex]V_{avg[/tex] = 0.265 m/s
Also, from table " saturated water property table "
At 45°C
viscosity μ = 0.596 × 10⁻³ kg/m-s
Prandtl number Pr = 3.91
Now, we determine the kinematic viscosity
v = μ / p
we substitute
v = ( 0.596 × 10⁻³ ) / 990.1
v = 6.01959 × 10⁻⁷ m²/s
so, Reynolds number in the flow region will be;
Re = ([tex]V_{avg[/tex] × D) / v
we substitute
Re = ( 0.265 × 0.02) / (6.01959 × 10⁻⁷)
Re = 8804.586
we can see that our Reynolds number ( 8804.586 ) more than 2300 and less than 10,000.
Hydraulic and thermal entry length are equal in this flow region,
such that;
[tex]L_h[/tex] = [tex]L_t[/tex]
⇒ 10 × D = 10 × 0.02 = 0.2 m
we can see that the entry length ( 0.2 m ) is smaller than the given length ( 13 m ) in the question; the flow is a turbulent flow.
So we the Nuddelt number
Nu = [tex]0.023Re^{0.8} Pr^{0.4[/tex]
Nu = 0.023 × [tex]8804.586^{0.8[/tex] × [tex]3.91^{0.4[/tex]
Nu = 56.8
Hence, the heat transfer coefficient h will be;
h = [tex]\frac{k}{D}[/tex] × Nu
we substitute
h = [tex]\frac{0.637}{0.02}[/tex] × 56.8
h = 31.85 × 56.8
h = 1809.1 W/m²-°C
Now, area of the heat transfer will be
A[tex]_s[/tex] = πDL
we substitute
A[tex]_s[/tex] = π × 0.02 × 13
A[tex]_s[/tex] = 0.8168 m²
Finally we determine the inner temperature of the pipe at exit. using the relation;
Q' = hA[tex]_s[/tex]( T₃ - T₂ )
we substitute
24139.5 = 1809.1 × 0.8168( T₃ - 10 )
24139.5 = 1477.67288( T₃ - 80 )
24139.5 = 1477.67288T₃ - 118213.8304
24139.5 + 118213.8304 = 1477.67288T₃
1477.67288T₃ = 142353.3304
T₃ = 142353.3304 / 1477.67288T
T₃ = 96.34°C
Therefore, the inner surface temperature of the pipe at the exit is 96.34°C
A rigid body consists of four bodies joined together, as drawn below to scale. The point is at one corner of the rectangle and the component bodies are: A uniform disk of radius and mass . A uniform rod of length and mass . A uniform rectangle with side lengths and , and mass . A point mass at with mass . What is the moment of inertia about the axis through the point
Answer:
I_total = ½ m R² + 1/12 m L² + ½ m (a² + b²) + m (x² + y²) ^ {3/2}
Explanation:
The moment of inertia is a scalar quantity, therefore additive, therefore we can find the moments of inertia of each body with respect to the point and add them.
Let's use the parallel axis theorem for the moment of inertia.
I = [tex]I_{cm}[/tex] + m d²
the moments and inertia of the bodies are
disk Icm = ½ m R²
rod Icm = 1/12 m L²
rectangle Icm = 1/12 m (a² + b²)
where a and b are the sides of the rectangle
Let's fix a reference frame on the point body, the length of the rectangle is x and its height y, the total moment of inertia is
I_total = I_point + I_disco + I_rod + I_rectangular
the moment of inertia of the point is
I = m r² = m 0
I_point = 0
disk moment of inertia
suppose it is on the y-axis with x = 0
I_disco = ½ m R² + m y²
moment inertia rod
located in the opposite corner
The distance from the point to the center of the mass of the rod is
R = [tex]\sqrt{x^2 +y^2 }[/tex]
I_varrilla = 1/12 m L² + m ([tex]\sqrt{ x^2 + y^2 }[/tex])
rectangle moment inertia
located on the x axis
I_rectangle = ½ m (a² + b²) + m x²
we substitute
I_total = 0 + ½ m R² + m y² + 1/12 m L² + m (Ra x ^ 2 + y²) + ½ m (a² + b²) + m x²
I_total = ½ m R² + 1/12 m L² + ½ m (a² + b²) + m (x² + y²) + m √(x^2 + y²)
I_total = ½ m R² + 1/12 m L² + ½ m (a² + b²) + m (x² + y²) ^ {3/2}
Explain how to make aluminum magnetic
Answer:
Use some Glue.. . . Aluminium is not magnetic, so your magnet can't sticl to aluminium without using Glue. You cannot get a magnet to stick to aluminum, no matter how hard you try. Aluminum is diamagnetic, which means it repels a magnetic field.
Explanation:
Hope It Help
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Question 3
By what volume would 25 L of alcohol increase if its temperature was
increased from 20°C to 30°C? (3 marks)
Answer:
V2 = 37.5 L
Explanation:
Given the following data;
Initial volume, V1 = 25 L
Initial temperature, T1 = 20°C
Final temperature,T2 = 30°C
To find the final volume V2, we would use Charles' law;
Charles states that when the pressure of an ideal gas is kept constant, the volume of the gas is directly proportional to the absolute temperature of the gas.
Mathematically, Charles is given by;
[tex] \frac {V}{T} = K[/tex]
[tex] \frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}} [/tex]
Making V2 as the subject formula, we have;
[tex] V_{2}= \frac{V_{1}}{T_{1}} * T_{2}[/tex]
[tex] V_{2}= \frac{25}{20} * 30 [/tex]
[tex] V_{2}= 1.25 * 30 [/tex]
V2 = 37.5 L
Exposure to a sufficient quantity of ultraviolet will redden the skin, producing erythema - a sunburn. The amount of exposure necessary to produce this reddening depends on the wavelength. For a 1.0 cm2 patch of skin, 3.7 mJ of ultraviolet light at a wavelength of 254 nm will produce reddening; at 300 nm wavelength, 13 mJ are required. Part A What is the photon energy corresponding to each of these wavelengths
Answer:
Energy = 7.83 x 10⁻¹⁹ J
Energy = 6.63 x 10⁻¹⁹ J
Explanation:
The energy of a photon in terms of wavelength can be calculated by the following formula:
[tex]Energy = \frac{hc}{\lambda}\\[/tex]
where,
h = Plank's Constant = 6.63 x 10⁻³⁴ Js
c = speed of light = 3 x 10⁸ m/s
λ = wavelength of light
Now, for λ = 254 nm = 2.54 x 10⁻⁷ m:
[tex]Energy = \frac{(6.63\ x\ 10^{-34}\ Js)(3\ x\ 10^8\ m/s)}{2.54\ x\ 10^{-7}\ m}\\[/tex]
Energy = 7.83 x 10⁻¹⁹ J
Now, for λ = 300 nm = 3 x 10⁻⁷ m:
[tex]Energy = \frac{(6.63\ x\ 10^{-34}\ Js)(3\ x\ 10^8\ m/s)}{3\ x\ 10^{-7}\ m}\\[/tex]
Energy = 6.63 x 10⁻¹⁹ J
1. How fast is a radio wave traveling if it is 4.0 meters long and its frequency is
82 Hz?
Answer:
Speed = 328 m/s
Explanation:
Given the following data:
Wavelength = 4 m
Frequency = 82 Hz
To find the speed of the radio wave;
Speed = wavelength * frequency
Speed = 4 * 82
Speed = 328 m/s
Therefore, the radio wave is travelling at 328 meters per seconds.
Radio waves can be defined as an electromagnetic wave that has its frequency ranging from 30 GHz to 300 GHz and its wavelength between 1mm and 3000m. Therefore, radio waves are a series of repetitive valleys and peaks that are typically characterized of having the longest wavelength in the electromagnetic spectrum.
An 800 kg car finishes a 5000-meter road in 300 seconds. Considering the car was moving at constant speed, what was its momentum?
Answer:
Momentum(P) = 1336kgm/s
Explanation:
Momentum is the product of the mass and velocity, therefore Momentum(P) is = mass*velocity. since the mass is given 800kg, all we are left with is the displacement 500m and the time 300s which can be used to find the velocity. so velocity is the displacement divided by the time, ie 500/300 which will give us the velocity 1.67m\s. hence we multiply the mass and the velocity and we'll get 1336kgm/s.
How does a wind turbine make electricity?
1) The wind turns a generator inside.
2) Magic
3) Waves turn the generator.
4) Hamster turns the generator.
Answer:
1
Explanation:
A long, uninsulated steam line with a diameter of 89 mm and a surface emissivity of 0.8 transports steam at 200C and is exposed to atmospheric air and large surroundings at an equivalent temperature of 20C. (a) Calculate the heat loss per unit length for a calm day. (b) Calculate the heat loss on a breezy day when the wind speed is 8 m/s. (c) For the conditions of part (a), calculate the heat loss with a 20-mm-thick layer of insulation (k 0.08 W/m K). Would the heat loss change significantly with an appreciable wind speed
Answer:
[tex]Q_net=534.67\frac{w}{m}[/tex]
Explanation:
From the question we are told that:
Steam line diameter [tex]D=89mm \approx 0.089m[/tex]
Surface emissivity [tex]\mu=0.8[/tex]
Steam temp [tex]T_s=200\textdegree C \approx 200+273=473k[/tex]
Surrounding temp [tex]T_a=20 \textdegree C \aprrox 20+273= 293k[/tex]
Generally the equation for heat loss per unit length due to radiation [tex]Q_{net}[/tex] is mathematically given by
[tex]Q_net=\sigma*\mu>(\pi *d)*(T_s^4-T_a^4)[/tex]
[tex]Q_net=5.6*10^8*0.8*(\pi *0.089)*(473^4-293^4)[/tex]
[tex]Q_net=534.67\frac{w}{m}[/tex]
Does an infrared wave or an x-ray travel faster in the vacuum of space?
Answer:
All electromagnetic radiation, of which radio waves and X-rays are examples, travels at the speed c in a vacuum. The only difference between the two is that the frequency of X-rays is very much higher than radio waves
describe the concept of force represent it quantiatively and derive unit of force
important please answer it as the question says and only if you know the answer
[tex]\bcancel{\huge\red{\fbox{\tt{࿐αɴѕωєя࿐}}}}[/tex]
Force is defined as the rate of change of momentum. For an unchanging mass, this is equivalent to mass x acceleration. So, 1 N = 1 kg m s-2, or 1 kg m/s2.
A car of mass 800kg is travelling at 10m/s. How much work must be done to stop it?
Answer:
80
Explanation:
800/10
=80
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45
А ______ is a solid, liquid or gas that a wave travels through.
Answer:
medium
Explanation:
I am not really sure
Which wave has the smallest wave
period? What is its period?
Answer:
C
Explanation:
The said wave takes the shortest time to move/get transmitted
The Earth Science students are making a human scale model of the solar system out on the school playground. The school itself represents the Sun. In the model 1 AU = 100 meters (the length of a football field). How far from the school will the student representing Mars stand? A) 50 meters B) 105 meters 150 meters D) 1500 meters NEED HELP ASAP
In the model 1 AU = 100 meters (the length of a football field) , then mars would be 150 meters far from the school, therefore the correct answer is option C.
What is a unit of measurement?A unit of measurement is a specified magnitude of a quantity that is established and used as a standard for measuring other quantities of the same kind. It is determined by convention or regulation.
As given in the problem the Earth Science students are making a human-scale model of the solar system out on the school playground. The school itself represents the Sun. In the model 1 AU = 100 meters,
As given in the table marks is 1.5 AU far from Mars,
1 AU = 100 meters
1.5 AU = 1.5 × 100
= 150 meters
Thus, mars would be 150 meters far from the school, therefore the correct answer is option C.
Learn more about the unit of measurement here, refer to the link ;
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A compressed spring launches a block up an incline. Which objects should be included within the system in order to make an energy analysis as easy as possible
Answer:
Block, incline, spring and gravity.
Explanation:
For us to have an energy analysis involving gravitational and spring potential energy, we will need to have a block with specific mass that is held at rest on a frictionless incline plane. Now, the block will compress the spring by a specific length from its equilibrium position and then the block is released to travel a distance right up the slope.
So basically, we will need Block, incline, spring and then gravity for it to move.
the titan mission scheduled for 2026 is a reconnaissance to explore the origin of life true or false
Answer:
its true
Explanation:
The titan mission scheduled for 2026 is a reconnaissance to explore the origin of life. Because, some scientific studies based on titan has showed that there is the presence of some biomolecules. Hence, the statement is true.
What is titan mission ?The distinctive, abundantly organic planet of Titan is where we will travel next in the solar system, according to NASA. The Dragonfly expedition will fly numerous sorties to sample and inspect locations around Saturn's icy moon, advancing our hunt for the components of life.
In 2026, Dragonfly will take off, and it will land in 2034. The rotorcraft will travel to a great number of Titan's promising sites in search of prebiotic chemical processes that are similar to those on Titan and Earth.
Dragonfly, which has eight rotors and flies like a big drone, is the first multi-rotor vehicle NASA will operate for science on another world. To become the first spacecraft to ever fly its full science payload, it will take advantage of Titan's dense atmosphere, which is four times denser than Earth's. Hence, the statement is true.
Find more on Titan mission :
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An electron of kinetic energy 1.59 keV circles in a plane perpendicular to a uniform magnetic field. The orbit radius is 35.4 cm. Find (a) the electron's speed, (b) the magnetic field magnitude, (c) the circling frequency, and (d) the period of the motion.
Answer:
a) [tex] v = 2.36 \cdot 10^{7} m/s [/tex]
b) [tex] B = 3.80 \cdot 10^{-4} T [/tex]
c) [tex] f = 1.06 \cdot 10^{7} Hz [/tex]
d) [tex] T = 9.43 \cdot 10^{-8} s [/tex]
Explanation:
a) We can find the electron's speed by knowing the kinetic energy:
[tex] K = \frac{1}{2}mv^{2} [/tex]
Where:
K: is the kinetic energy = 1.59 keV
m: is the electron's mass = 9.11x10⁻³¹ kg
v: is the speed =?
[tex] v = \sqrt{\frac{2K}{m}} = \sqrt{\frac{2*1.59 \cdot 10^{3} eV*\frac{1.602 \cdot 10^{-19} J}{1 eV}}{9.11 \cdot 10^{-31} kg}} = 2.36 \cdot 10^{7} m/s [/tex]
b) The electron's speed can be found by using Lorentz's equation:
[tex] F = q(v\times B) = qvBsin(\theta) [/tex] (1)
Where:
F: is the magnetic force
q: is the electron's charge = 1.6x10⁻¹⁹ C
θ: is the angle between the speed of the electron and the magnetic field = 90°
The magnetic force is also equal to:
[tex] F = ma_{c} = m\frac{v^{2}}{r} [/tex] (2)
By equating equation (2) with (1) and by solving for B, we have:
[tex] B = \frac{mv}{rq} = \frac{9.11 \cdot 10^{-31} kg*2.36 \cdot 10^{7} m/s}{0.354 m*1.6 \cdot 10^{-19} C} = 3.80 \cdot 10^{-4} T [/tex]
c) The circling frequency is:
[tex] f = \frac{1}{T} = \frac{\omega}{2\pi} = \frac{v}{2\pi r} [/tex]
Where:
T: is the period = 2π/ω
ω: is the angular speed = v/r
[tex] f = \frac{v}{2\pi r} = \frac{2.36 \cdot 10^{7} m/s}{2\pi*0.354 m} = 1.06 \cdot 10^{7} Hz [/tex]
d) The period of the motion is:
[tex] T = \frac{1}{f} = \frac{1}{1.06 \cdot 10^{7} Hz} = 9.43 \cdot 10^{-8} s [/tex]
I hope it helps you!
Wind power produces a lot of air pollution.
O True
O False
Answer:
Wind is a renewable energy source. Overall, using wind to produce energy has fewer effects on the environment than many other energy sources. Wind turbines do not release emissions that can pollute the air or water (with rare exceptions), and they do not require water for cooling.
Explanation:
Part C
Now vary the mass of the skateboarder. Change the mass slider to the lowest mass. Repeat the investigation using the
steepest ramp. How does changing the mass of the skateboarder affect his kinetic energy on a given ramp? How does it
affect his speed? Remember to use the pause button to clearly see his energy and speed. Repeat your trials as many
times as needed.
Answer:
When mass increases, kinetic energy also increases. Changing mass doesn’t affect his speed on a given ramp.
Explanation:
Answer:
When mass increases, kinetic energy also increases. Changing mass doesn’t affect his speed on a given ramp.
Explanation:
pluto
Give me four reasons pluto is a cool planet / dwarf planet
What is a centripetal acceleration of a greyhound running on a circular track with a radius of 50 m at 12.5 m/s
Answer:
Centripetal acceleration = 3.125 m/s²
Explanation:
Given the following data;
Radius, r = 50 m
Velocity, V = 12.5 m/s
To find the centripetal acceleration;
Centripetal acceleration = Velocity²/radius
Centripetal acceleration = 12.5²/50
Centripetal acceleration = 156.25/50
Centripetal acceleration = 3.125 m/s²
Therefore, the centripetal acceleration of the greyhound running on a circular track is 3.125 meters per seconds square.
The photo shows a pair of figure skaters performing a spin maneuver. The
axis of rotation goes through the left foot of the skater on the left. What
action could increase the pair's angular velocity?
An action which could increase the pair's angular velocity is the figure skater on the left pointing his right arm down instead of up.
What is angular velocity?Angular velocity simply refers to the rate of change of angular displacement of a physical object (body) with respect to time.
This ultimately implies that, angular velocity is a measure of how fast and quickly a physical object (body) rotates with respect to another point or how its angular displacement (position) changes with respect to time.
In this scenario, an action which could increase the pair's angular velocity is when the figure skater on the left points his right arm down instead of up.
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the skater on the left pulls the skater on the right closer to him
Explanation:
apex bliches
how much work is done in accelerating a 2000 kg car from rest to a speed of 30 m/s?
After the water has boiled, the temperature of the water decreases by 22 °C.
The mass of water in the kettle is 0.50 kg.
The specific heat capacity of water is 4200 J/kg °C.
Calculate the energy transferred to the surroundings from the
water.
Energy=
Explanation:
Use the formula:
[tex] e = mc \delta \theta[/tex]
e is the energy released,
m is the mass of water,
c is the specific heat capacity,
δθ is the change in temperature ( 100 - 22)
In a Young's double-slit experiment, two parallel slits with a slit separation of 0.165 mm are illuminated by light of wavelength 560 nm, and the interference pattern is observed on a screen located 4.05 m from the slits. (a) What is the difference in path lengths from each of the slits to the location of the center of a fifth-order bright fringe on the screen
Answer:
the difference in path lengths from each of the slits to the location of the center of a fifth-order bright fringe on the screen is 28 × 10⁻⁷ m
Explanation:
Given the data in the question;
slit separation d = 0.165 mm = 0.165 × 10⁻³ m
wavelength λ = 560 nm = 560 × 10⁻⁹ m
distance between the screen and slits D = 4.05 m
now,
for fifth-order bright fringe path difference = mλ
where m is 5
so, the difference in path lengths from each of the slits will be;
Δr = mλ
we substitute
Δr = 5( 560 × 10⁻⁹ m )
Δr = 28 × 10⁻⁷ m
Therefore, the difference in path lengths from each of the slits to the location of the center of a fifth-order bright fringe on the screen is 28 × 10⁻⁷ m
name the three major types of clouds
Answer:
Cumulus, Stratus, and Cirrus. There are three main cloud types.
Explanation:
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