Answer:
[tex]P=2.57\times 10^{-7}\ N/m^2[/tex]
Explanation:
Given that,
A radio wave transmits 38.5 W/m² of power per unit area.
A flat surface of area A is perpendicular to the direction of propagation of the wave.
We need to find the radiation pressure on it. It is given by the formula as follows :
[tex]P=\dfrac{2I}{c}[/tex]
Where
c is speed of light
Putting all the values, we get :
[tex]P=\dfrac{2\times 38.5}{3\times 10^8}\\\\=2.57\times 10^{-7}\ N/m^2[/tex]
So, the radiation pressure is [tex]2.57\times 10^{-7}\ N/m^2[/tex].
Suppose that the hatch on the side of a Mars lander is built and tested on Earth so that the internal pressure just balances the external pressure. The hatch is a disk 50.0 cm in diameter. When the lander goes to Mars, where the external pressure is 650 N/m2, what will be the net force (in newtons and pounds) on the hatch, assuming that the internal pressure is the same in both cases? Will it be an inward or outward force?
Answer:
The value is [tex]F_{net} = 4444 lb[/tex]
The force will be outward
Explanation:
From the question we are told that
The diameter of the disk is [tex]d = 50.0 \ cm = \frac{50}{100} = 0.5 \ m[/tex]
The external pressure on Mars is [tex]P = 650 \ N/m^2[/tex]
From the question we are told that
Internal pressure = External pressure
Generally the external Force on earth is
[tex]F_E = P_{atm} * A[/tex]
Here [tex]P_{atm}[/tex] is the atmospheric pressure with value [tex]P_{atm} = 1.013*10^{5}\ Pa[/tex]
So
[tex]F_E = 1.013 *10^{5} * \pi * \frac{d^2}{4}[/tex]
=> [tex]F_E = 1.013 *10^{5} *3.142 * \frac{0.50 ^2}{4}[/tex]
=> [tex]F_E = 19893 \ N[/tex]
Generally the external Force on Mars is
[tex]F= P * A[/tex]
[tex]F = 650 * \pi * \frac{d^2}{4}[/tex]
=> [tex]F = 650 *3.142 * \frac{0.5^2}{4}[/tex]
=> [tex]F = 127.6 \ N[/tex]
Net force is mathematically represented as
[tex]F_{net} = F_E -F[/tex]
=> [tex]F_{net} = 19893 -127.6[/tex]
=> [tex]F_{net} = 19765.6 \ N[/tex]converting to pounds
[tex]F_{net} = \frac{19765.6}{4.448}[/tex]
=> [tex]F_{net} = 4444 lb[/tex]
Given that that the value is positive then the force will be outward
The ability to do work or cause a change is the definition for
energy
motion
inertia
force
What is the SI unit for energy?
Newton meter
Joule
all of the above
What are the only two things you can do with energy?
use and waste
store and transfer
store and use
transfer and waste
Energy stored as motion is also known as
chemical potential
elastic potential
kinetic
gravitational potential
Kinetic energy depends upon
mass and velocity
mass and gravity
mass and displacement
mass and height
Energy stored in position is also known as
chemical
potential
kinetic
elastic
Potential energy depends upon
mass and velocity
mass and gravity
mass, height and gravity
mass and displacement
Energy stored in chemical bonds is known as
kinetic
chemical potential
gravitational potential
elastic potential
Energy stored in a rubber band is known as
kinetic
gravitational potential
chemical potential
elastic potential
What is the force of gravity for a 12 kg turkey?
Answer:
1. Energy
2. All of the above
3. Store and use
4. Kinetic
5. mass and velocity
6. Potential
7. mass, height, and gravity
8. Chemical Potential
9. Elastic Potential
10. (`⌒*)O-(`⌒´Q)
Explanation:
Consider a block sliding down a ramp whose motion is opposed by frictional forces. The total energy of this system is modeled by the equation:
Etotal = 1/2mv^2 + mgh + Ff(f is underscore)d
Which part of the equation represents the amount of energy converted to thermal energy?
A. mg
B. Ffd
C. mgh
D. 1/2 mv^2
Answer:
Energy Flows Quick check answers:
1. Ffd.
2. The kinetic energy decreases, and gravitational potential energy increases.
3. The internal energy of the system increases.
4. KEbox= Etotal-mgh
5. Etotal = 1/2m1(v1)^2+1/2m^2(v2)^2+U
The part of the equation that represents the amount of energy converted to thermal energy is [tex]F_f d[/tex].
The given equation for the total energy of a system;
[tex]E_{total} = \frac{1}{2} mv^2 \ +\ mgh\ + \ \ F_fd[/tex]
The definition of the various terms in the energy equation is given as;
[tex]E_{total}[/tex]: this is the total mechanical energy of the system[tex]\frac{1}{2} mv^2[/tex]: this is the kinetic energy of the system[tex]mgh[/tex]: this is the potential energy of the system[tex]F_f d[/tex]: this is the energy lost due to friction.The energy lost due to friction is equal to the energy converted to thermal energy.
Thus, the part of the equation that represents the amount of energy converted to thermal energy is [tex]F_f d[/tex].
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A coconut is thrown straight downward and explodes into two pieces. One piece moves directly to the right. What direction does the other piece go?
A. left
B. right
C. down
D. up
E. it stays at rest
A skier leaves the end of a horizontal ski jump at 23.0 m/s and falls through a vertical distance of 3.45 m before landing.
(a) Neglecting air resistance, how long does it take the skier to reach the ground?
(b) How far horizontally does the skier travel in the air before landing?
m
Explanation:
Given
Velocity v = 23.0m/s
Distance S = 3.45m
Required
Time it will take the skier to reach the ground;
Using the equation of motion;
S = ut + 1/2gt²
3.45 = 23t + 1/2(9.8)t²
3.45 = 23t + 4.9t²
4.9t²+23t-3.45 = 0
Factorize;
t = -23 ±√23²-4(4.9)(-3.45)/2(4.9)
t = -23 ±√529+67.62/9.8
t = -23±√596.62/9.8
t = -23±24.43/9.8
t = 1.43/9.8
t = 0.146 secs
Hence take the skier 0.146 secs to reach the ground.
b) Horizontal distance covered is the range;
Range = U√2H/g
Range = 23√2(3.45)/9.8
Range = 23√6.9/9.8
Range = 23√0.7041
Range = 23(0.8391)
Range = 19.29m
Hence the horizontal distance travelled in air is 19.29m
(a).The time taken by the skier to reach the ground is 0.145 second.
(b).The skier travel in the air before landing is 19.29 meter.
a. Given that A skier leaves the end of a horizontal ski jump at 23.0 m/s and falls through a vertical distance of 3.45 m before landing.
Using equation of motion.
[tex]S=ut+\frac{1}{2}gt^{2}[/tex]
Where S is vertical distance , u is initial velocity and g is gravitational acceleration.
Substitute S = 3.45 m, u = 23m/s and g = 9.8 in above equation.
[tex]3.45=23t+\frac{1}{2} (9.8)t^{2}\\\\4.9x^{2} +23t-3.45=0\\\\t=0.145,-4.83[/tex]
Since, time can not be negative.
So that, [tex]t=0.145s[/tex]
b. The horizontal distance travel before landing is known as Range.
Horizontal distance ,
[tex]=v*\sqrt{\frac{2S}{g} }[/tex]
Substitute v = 23m/s , S = 3.45 and g = 9.8 meter per second square.
[tex]Distance=23*\sqrt{\frac{2*3.45}{9.8} } \\\\Distance=23*\sqrt{0.7041} \\\\Distance=23*0.8391=19.29m[/tex]
Thus, The skier travel in the air before landing is 19.29 meter.
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What is energy ? Use in your own words.
Answer:
the potential for causing changes
Explanation:
Write Radar gun summary
Answer:
A radar gun is a device for measuring the speed of moving objects. ... The radar gun is a Doppler radar unit that can be static, vehicle-mounted or hand-held. It measures the
Explanation:
1.045cm and a height of 30.48cm what is the volume of the cylinder
Answer:
104.57
Explanation:
V = πr^2h
-3.14, 1.045^2, and 30.48=104.567666431
in half wave rectifier circuit the diode and load resistance are connected in ...to ac power source
Explain how the net impact forces can be reduced for basketball players through the
application of scientific and engineering concepts in sneaker design.
Answer:
* If we increase the distance traveled before stopping, the forces decrease significantly,
* If we decrease the height of the fall by grabbing the net to slow down.
* Creating a process of absorption of force in the shoes, through the deformation of the elastic soles.
Explanation:
When basketball players jump to place the ball in the net they see from a good height, when they hit the ground, strong tensions are produced in the bones and bonds.
Applying science to this case of fall, we can suggest several processes to decrease the forces on impact.
* If we increase the distance traveled before stopping, the forces decrease significantly,
* If we decrease the height of the fall by grabbing the net to slow down.
* Creating a process of absorption of force in the shoes, through the deformation of the elastic soles.
By designing shoes that have some degree of force absorption, it is possible to significantly reduce the stress on the player.
Solar radiation makes several other energy sources possible, including ________
a. wind energy
b. tidal energy
c. geothermal energy
d. hydrothermal energy
e. nuclear energy
Answer:
a
Explanation:
wind energy results from solar radiation
Answer:
♡ madeline here ♡
goodmorning!
it's a, have a great day! ☆
- madeline/madi
✧･ﾟ: *✧･ﾟ:･ﾟ✧*:･ﾟ✧･ﾟ
Explanation:
Two identical wind-up cars A and B are released. Car B has a 2 kilogram weight strapped to the back of the car. Which will have the greatest average speed towards the end of the motion?
A)Car A
B)They will both have an average speed of zero.
C)They will have the same average speed.
D)Car B
E)There is not enough information to answer.
Answer:
Car A would have a better average speed
Explanation:
added weight to a object that is self propelled will be slower than a identical object with no added weight
Write an algorithm and draw the flowchart of finding the area and perimeter of a circle. The
radius (float) need to be entered by the user and if the user enters negative radius or zero radius,
the algorithm will ask the user to be entered again. (The algorithm will ask the user to enter the
radius until it is entered as positive.) At the end, the algorithm will find and print area and
perimeter of the circle as output.
Answer:
The algorithm
Step 1: Start
Step 2: Input Radius
Step 3: While Radius < = 0
Input Radius
Step 4: Pi = 3.14
Step 5: Perimeter = 2 * Pi * Radius
Step 6: Area = Pi * r^2
Step 7: Print Area, Perimeter
Step 8: Stop
Explanation:
This indicates the start of the algorithm
Step 1: Start
This inputs the radius
Step 2: Input Radius
The following loop is repeated until user input for radius is greater than 0
Step 3: While Radius < = 0
Input Radius
This initializes Pi to 3.14
Step 4: Pi = 3.14
This calculates the Perimeter or Circumference
Step 5: Perimeter = 2 * Pi * Radius
This calculates the Area
Step 6: Area = Pi * r^2
This prints the calculated Perimeter ad Area
Step 7: Print Area, Perimeter
This indicates the end of the algorithm
Step 8: Stop
I've added the flowchart as an attachment
The mass of the Earth is about 6 x 1024 kg
and the mass of the moon is about 7 x 1022 kg.
The distance between the Earth and the Moon
is about 3.8 x 108
The magnitude of the gravitational force
between is closest to which of these?
m.
Well it seems that you did not give answer choices, but that its fine since we can use newtons law of universal gravitational, Fg = GM1M2/r^2. So G is the gravitational constant, which is 6.67*10^-11, we can plug in 6*1024 for M1, and 7*1022 for M2, and 3.8*108 for r. Which then we get 1.74 * 10^8 N as the force of attraction between the Earth and the moon.
The magnitude of the gravitational force between the earth and the moon is equal to 20 ×10¹⁹ N.
What is gravitational force?Gravitational force can be described as a force that attracts a body toward the center of the earth or any physical system that has mass. Every particle with mass exerts a gravitational pull on every other object with mass.
Mathematically gravitational force can be written as:
[tex]{\displaystyle F = G\frac{mM}{r^2}[/tex]
Where F is the force between objects, m and M are their masses, r is the distance between them, and G is the universal gravitational constant.
Given, the mass of the moon, M = 7 ×10²² Kg
The mass of earth, M = 6 ×10²⁴ Kg
The distance between the earth and the moon, r = 3.8 ×10⁸ m
The value of G = 6.7 ×10⁻¹¹ Nm²/Kg
The magnitude of the gravitational force can be calculated as:
[tex]{\displaystyle F = 6.7\times 10^{-11}\frac{6\times 10^{24}\times 7\times 10^{22}}{(3.8 \times 10^8)^2}[/tex]
F = 20 ×10¹⁹ N
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A man walks south at a speed of 2.00 m/s for 60.0 minutes. He then turns around and walks north a distance 3000 m in 25.0 minutes. What is the average velocity of the man during his entire motion?
Answer:
v = 0.823 m/s
Explanation:
A man walks south at a speed of 2.00 m/s for 60.0 minutes.
The distance covered in South = 60 × 60 × 2 = 7200 m
He then turns around and walks north a distance 3000 m in 25.0 minutes.
As they moved in opposite direction, net displacement will be : 7200 - 3000 = 4200 m
Average velocity of the man = net displacement/time
[tex]v=\dfrac{4200\ m}{(60+25)\times 60}\\\\=0.823\ m/s[/tex]
So, the average velocity of the man is 0.823 m/s.
A 227 g ball is tied to a string. It is pulled to an angle of 9.62° and released to swing as a pendulum. A student with a stopwatch finds that 5 oscillations take 31 s. How long is the string?
Answer:
9.55m
Explanation:
From T = 2π√L/g ........ 1
Where T is the period
L is the length of the string and g is the acceleration due to gravity = 9.8m/s^2
Given that,
Mass of ball = 227g
The ball makes 5 oscillations in 31 s.
T = 31/5
T = 6.2s
Solving for L from equation 1
T = 2π√L/g
Square both sides
T^2 = 4π^2×L/g
T^2×g = 4π^2×L
Make L the subject
L = T^2*g/4π^2 ......... 2
Now, substitute the values into equation 2
L = 6.2^2 × 9.8/4×3.14^2
L = 38.44*9.8/39.4384
L = 376.712/39.4384
L = 9.55m
Hence, the length of the string is 9.55m
Position c
Position B
Position D
Position A
Position E
Position
Kinetic Energy
Potential Energy
A
B
С
D
Answer:
B porque tiene energía n..
An airtight box, having a lid of area 80.0 cm^2, is partially evacuated. Atmospheric pressure is 1.01 Times 10^5 Pa. A force of 108 lb is required to pull the lid off the box. The pressure in the box was:_________.
Answer:
5×10^4Pa
Explanation:
Given force of 108 lb is required to pull the lid off the box,
To convert "Ib"to Newton ,we use conversation rate below
1 pounds = 4.4482216282509 newtons
Then 108 lb=x Newton
Cross multiply we have
X= 480.41Newton
The force that is needed to open the lid is F and pressure P.
We know that Pressure= Force/Area
Area is given as 80.0 cm^2, we can convert to m^2 for unit consistency since 1cm^2= 0.001m^2 then
80.0 cm^2 = 80×10^-4m^2
Substitute to the equation of the pressure we have
P= 480.41Newton/(80×10^4m^2)
P=6×10^4 Pa
The pressure in the box will be difference between the initial pressure and final pressure
=( 1.01 ×10^5 Pa)-(6×10^4 Pa)
= 50100Pa
= 5×10^4Pa
Therefore, The pressure in the box was
5×10^4Pa
An object with a mass of 3.0 kg has a
force of 9.0 newtons applied to it. What
is the resulting acceleration of the
object?
[tex] \LARGE{ \underline{ \tt{Required \: answer:}}}[/tex]
We have:
Mass of the object = 3 kgForce on the object = 9 NWe need to find:
Resulting accleration of the object?Solution:
According to Newton's 2nd law of motion, or quantitative measure of Force:
Force = Mass × AcclerationUsing this,
➝ F = ma
➝ 9N = 3 kg × a
➝ a = 9/3 m/s²
➝ a = 3 m/s²
Hence,
The resulting accleration of the object is 3 m/s². And we are done! :D⛱️ [tex] \large{ \blue{ \bf{FadedElla}}}[/tex]
An ant crawled from a hole to the food. For the first 6 minutes,it travelled at an average speed of 15m/min. For The next 9 minutes it travelles at an average speed of 10m/min what was the average speed of the whole journey?
Answer:
v = 12 m/min
Explanation:
By definition, the average velocity is the rate of change of the position with respect to time, as follows:[tex]v_{avg} =\frac{x_{f} -x_{o} }{t_{f}-t_{o} } (1)[/tex]
Choosing x₀ = 0 and t₀ =0, (1) reduces to :[tex]v_{avg} = \frac{x_{f} }{t_{f} } (2)[/tex]
From the givens, we have:tif = 6 min + 9 min = 15 min
In order to get xf, we know that during the first part, vavg = 15 m/min, so solving for xf:[tex]x_{f1} = v_{avg1}* t_{1} = 15 m/min * 6 min = 90 m (3)[/tex]
For the following 9 min, we know that the average speed was 10m/min, so the distance traveled during the second part of the trip was simply:[tex]x_{f2} = v_{avg2} *t_{2} = 10m/min * 9 min = 90 m (4)[/tex]
Adding (3) and (4):[tex]x_{f} = 90 m + 90 m = 180 m (5)[/tex]
Replacing xf and tif in (2), we finally get:[tex]v_{avg} =\frac{x_{f} }{t_{f}} =\frac{180m}{15 min} = 12 m/min (6)[/tex]
A baseball is thrown across the field. The ____________is measured from where the ball is thrown to where landed was 75 feet.
motion
direction
distance
reference point
Answer:
distance i think
Explanation:
A Long Jumber leaves the ground at on
a bouche horizontal al speed ilms.
How far does he jume in the horizontal direction ?
Answer:
horizontal velocity vh = 6*cos(30°) = 6*(√3)/2 = 3√3 m/s
initial vertical velocity vv = 6*sin(30°) = 6/2 = 3m/s
Using s = ut + at2/2 for change in vertical distance in time t, with acceleration a (-9.8m/s2) and initial velocity u (vv = 3m/s) we have
0 = 3*t - 9.8*t2/2 or t = 6/9.8 s (ignoring the t = 0 solution, which just represents staying still!).
The horizontal distance in time t is vh*t or 3√3*6/9.8 m
Explanation:
A spring with spring constant 11.5 N/m hangs from the ceiling. A 490 g ball is attached to the spring and allowed to come to rest. It is then pulled down 6.70 cm and released.
What is the time constant if the ball's amplitude has decreased to 2.20 cm after 30.0 oscillations?
Answer:
The time constant is [tex]\tau = 17.5 \ s[/tex]
Explanation:
From the question we are told that
The spring constant is [tex]k = 11.5 \ N/m[/tex]
The mass of the ball is [tex]m_b = 490 \ g = 0.49 \ kg[/tex]
The amplitude of the oscillation t the beginning is [tex]x = 6.70 cm = 0.067 \ m[/tex]
The amplitude after time t is [tex]x_t = 2.20 cm = 0.022 \ m[/tex]
The number of oscillation is [tex]N = 30[/tex]
Generally the time taken to attain the second amplitude is mathematically represented as
[tex]t = N * T[/tex] Here T is the period of oscillation
[tex]t = N * 2\pi \sqrt{\frac{m}{k} }[/tex]
=> [tex]t = 30 * 2 * 3.142 * \sqrt{\frac{ 0.490}{11.5} }[/tex]
=> [tex]t = 38.88 \ s[/tex]
Generally the amplitude at time t is mathematically represented as
[tex]x(t) = x e^{-\frac{at}{2m} }[/tex]
Here a is the damping constant so
at [tex]t = 38.88 \ s[/tex] , [tex]x_t = 2.20 cm = 0.022 \ m[/tex]
So
[tex]0.022 = 0.067 e^{-\frac{a * 38.88}{2 * 0.490} }[/tex]
=> [tex]0.3284 = e^{-\frac{a * 38.88}{2 * 0.490} }[/tex]
taking natural log of both sides
=> [tex]ln(0.3284 ) = -\frac{a * 38.88}{2 * 0.490} }[/tex]
=> [tex]a = 0.028[/tex]
Generally the time constant is mathematically represented as
[tex]\tau = \frac{m}{a}[/tex]
=> [tex]\tau = \frac{0.490}{ 0.028}[/tex]
=> [tex]\tau = 17.5 \ s[/tex]
A pulley is in the form of a uniform solid cylinder of radius 7cm and mass 2kg. One end of a very light rope is fixed to wind the pulley and the other end is
free. When we pull the free end of the rope the pulley starts rotating from rest and accelerates uniformly. If the angular acceleration is 100rad/s2 so the
constant force that we exert on the pulley through the rope is:
Chọn một:
a.
200 N
b.
7N
C
0.49 N
d.
49 N
Answer:
correct is b 7N
Explanation:
The torque is
Στ = I α
torque
τ = F x R
bold, indicate vectors. The magnitude of torque is
τ = F R sin θ
in this case the angle is 90º so sin 90 = 1
τ = F R
The moment of inertia of a cylinder
I = ½ M R²
substitute
F R = ½ M R² α
F = ½ M R α
reduce to the SI system
R = 7 cm (1m / 100cm) = 0.07 m
calculate
F = ½ 2 0.07 100
F = 7 N
"2.40 A pressure of 4 × 106N/m2 is applied to a body of water that initially filled a 4300 cm3 volume. Estimate its volume after the pressure is applied."
Answer:Final volume after pressure is applied=4,292cm3
Explanation:
Using the bulk modulus formulae
We have that The bulk modulus of waTer is given as
K =-V dP/dV
Where K, the bulk modulus of water = 2.15 x 10^9N/m^2
2.15 x 10^9N/m^2= - 4,300 x 4 × 106N/m2 / dV
dV = - 4,300 x 4 × 10^6N/m^2/ 2.15 x 10^9N/m^2
dV (change in volume)= -8.000cm^3
Final volume after pressure is applied,
V= V+ dV
V= 4300cm3 + (-8.000cm3)
=4300cm3 - 8.000cm3
Final Volume, V =4,292cm3
Dolphins rely on echolocation to be able to survive in the ocean. In a 20 °C ocean, a dolphin produces an ultrasonic sound with a frequency of 125 kHz. What is the wavelength of this sound, in meters?While remaining stationary, the dolphin emits a sound pulse and receives an echo after 0.220 s. How far away, in meters, is the reflecting object from the dolphin?
Answer:
wavelength = 0.01 m
distance = 162.8 m
Explanation:
Given that;
Speed of sound in water = 1,480 meters per second
Frequency of ultrasound = 125KHZ
From=
v=λf
v= speed of sound
λ= wavelength of sound
f= frequency of sound
λ= 1,480 ms-1/125 * 10^3 Hz
λ= 0.01 m
From
v = 2x/t
where;
v= velocity of sound in water
x= distance traveled
t = time taken
x = vt/2
x = 1,480 ms-1 * 0.220 s/2
x= 162.8 m
what are the two outcomes for a proto star?
Brown dwarf or main sequence star
Brown dwarf or red dwarf
Red giant or red dwarf
Black hole or main sequence star
Answer:
Brown dwarf or main sequence star
What type of meter would be connected on both sides of a resistor in a circuit?
Type of meter would be connected on both sides of a resistor in a circuit is a a voltmeter , which measures potential difference .
What is a voltmeter ?A voltmeter is an instrument used for measuring the potential difference , or voltage between two points in an electrical circuit .
A voltmeter is always attached in a series combination and an ammeter (which measures current in a circuit ) always attached in parallel combination with the circuit.
Since , in question it is given that a meter would be connected on both sides of a resistor in a circuit that means it must be a series combination
hence , correct answer is B) a voltmeter , which measures potential difference .
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An explanation of the relationships among particular phenomena.
Answer:
Theory
Explanation:
Theory is a term that is used often in academic work or scientific research to explain certain things or conditions established on universal principles or laws.
It is used to describe the "why and how" or the reason behind the occurrence of a situation.
Hence, it is correct to conclude that THEORY is "an explanation of the relationships among particular phenomena."
Answer:
E) Theory
Explanation:
Edge 2020
Brainliest?
When is a magnetic force produced in a copper wire?
O A. Whenever the wire is connected to a magnet
B. Whenever the wire is wrapped around an iron bar
C. Whenever there is current in the wire
D. Whenever the wire is coiled
Answer:
ur answer is d
Explanation:
it has to be coiled to give off magnetic energy