Answer:
1 and 3
Explanation:
Scientists can use the following techniques to determine the characteristics of Earth's layers: a) conduct experiments on how minerals change under high pressure; and b) investigate how seismic waves travel through different layers.
Characteristics of Earth's layersMechanically and chemically, the Earth is divided into two categories. The lithosphere, asthenosphere, mesospheric mantle, outer core, and inner core are mechanically separated. However, chemically, which is the more popular of the two, it can be divided into the crust, mantle, and core - which can be further subdivided into outer core and inner core. The outer core is liquid, the inner core is solid, and the mantle is solid/plastic. This is due to the relative melting points of the different layers, as well as the increase in temperature and pressure as depth increases. Because they are cool enough, nickel-iron alloys and silicates are solid at the surface.To learn more about Earth's layers refer :
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The speed of sound in steel is 5000 m/s. What is the wavelength of a sound wave of frequency 660 Hz in steel?
Answer:7.58 m
Explanation:
The required wavelength of a sound wave of frequency 660 Hz in steel is approximately 7.58 meters.
What is wavelength?Wavelength is a fundamental property of waves, including electromagnetic waves like light and radio waves, as well as other types of waves, such as sound waves or water waves.
Here,
The wavelength (λ) of a sound wave is given by the formula,
λ = v/f
where v is the speed of sound in the medium and f is the frequency of the wave.
In this case, the speed of sound in steel is v = 5000 m/s, and the frequency of the sound wave is f = 660 Hz. Substituting these values into the formula, we get:
λ = 5000 m/s / 660 Hz
λ = 7.58 meters (rounded to two decimal places)
Therefore, the wavelength of a sound wave of frequency 660 Hz in steel is approximately 7.58 meters.
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A car has a weight of 10,000 N. What is the mass of the car on Earth (g = 10
N/kg)?
A. 1000 kg
B.1000 N
C. 10,000 kg
D. 10,000N
Given :
Weight of car is, W = 10000 N.
Acceleration due to gravity, g = 10 N/kg.
To Find :
The mass of car on Earth.
Solution :
We know, mass of an object is given by :
[tex]m = \dfrac{W}{g}\\\\m = \dfrac{10000}{10}\ kg\\\\m = 1000\ kg[/tex]
Therefore, mass of the car on Earth is 1000 kg.
Two cars collide and then stick together in an accident.
Car A has a mass of 1000. kg and is moving 5.00 m/s east, and
car B has a mass of 2000. kg and is moving 2.00 m/s west.
What is the speed of the cars after the collision?
b. 1.53m/s
c. 2.43m/s
a. 0.33m/s
d. 3.93m/s
What’s the answer
Answer:
Option A. 0.33 m/s
Explanation:
From the question given above, the following data were obtained:
Mass of A (mₐ) = 1000 kg
Velocity of A (uₐ) = 5 m/s
Mass of B (m₆) = 2000 kg
Velocity of B (u₆) = 2 m/s
Velocity (v) after collision =?
mₐuₐ – m₆u₆ = v (mₐ + m₆)
(1000 × 5) – (2000 × 2) = v (1000 + 2000)
5000 – 4000 = 3000v
1000 = 3000v
Divide both side by 3000
v = 1000 / 3000
v = 0.33 m/s
Thus, the speed of the cars after collision is 0.33 m/s
The speed of the cars after the collision is 0.33 m/s.
The right option is a. 0.33 m/s
To calculate the speed of the cars after the collision, we use the formula below.
Formula:
mu+m'u' = V(m+m')............... Equation 1Where:
m = mass of car Am' = mass of car Bu = initial velocity of car Au' = initial velocity of car BV = Speed of the cars after the collision.make V the subject of the equation
V = (mu+m'u')/(m+m').............. Equation 2From the question,
Given:
m = 1000 kgm' = 2000 kgu = 5 m/s eastu' = -2 m/s westSubstitute these values into equation 2
V = [(1000×5)+(2000×(-2))]/(1000+2000)V = (5000-4000)/(3000)V = 1000/3000V = 0.33 m/sHence, The speed of the cars after the collision is 0.33 m/s.
The right option is a. 0.33 m/s
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what color is my socks? :0
A color.......................................................................yeah the color of your socks is a color :}
An analog-to-digital converter is a device that can be used with a microphone to take in an audio wave (an analog signal). The device then converts the information to a digitized form that can be stored and used by a computer. The image below shows the steps in the digitization process. The question mark in the image represents how the data is stored after it is processed by the analog-to-digital converter.
How is the information from the audio wave stored after it is processed by the analog-to-digital converter?
A.
It is stored as graphs because it was converted to base ten data.
B.
It is stored as a table of ones and twos because it was converted to a radio wave.
C.
It is stored as a series of ones and zeros because it was converted to binary data.
D.
It is stored as images because it was converted to mathematical equations.
Reset
Answer:
Answer C It is stored as a series of ones and zeros because it was converted to binary data.
Explanation:
After the Analog to Digital Converter (ADC) the data is stored in a series of numbers encoded in binary mode (combinations of Zeros and Ones).
This agrees with answer labeled as C) in the list of options.
Answer: C
Explanation: Study Island correct
Water is entering the prism at a rate of A m^3/hr. The prism is empty at time 0. Express the depth d of the water in meters in terms of A, the length of time t the water has been entering the trough, and the length L of the prism.
This question is incomplete, the complete question is;
The picture shows a triangular prism. The end of prism are equilateral triangles with x meters. the other dimension of the prism is L meters
a) Find the volume V in terms of x and L
b) Water is entering the prism at a rate of A m³/hr. The prism is empty at time 0. Express the depth d of the water in meters in terms of A, the length of time t the water has been entering the trough, and the length L of the prism.
Answer:
a) the volume V in terms of x and L is ((√3/4)x²L) m³
b) required expression is (2/(3)^(1/u))√(At/L)
Explanation:
Given that;
form the question and image below;
triangular prism ends are equilateral triangle
side length = x meter
Dimension of the prism = L meter
Area of the equilateral triangle = √3/4 (side)² = √3/4 (x)² meter
Volume of the triangular prism = Area × height
= √3/4 (x)² × L
V = ((√3/4)x²L) m³
Therefore, the volume V in terms of x and L is ((√3/4)x²L) m³
b)
Rate of water entering = A m³/hr
Depth of water tank = d meter
Time = t
Length of prism = L
now Rate of water entering is A m³/hr
dv/d = A [ V = ((√3/4)x²L) m³ ]
and
dv/dt = √3/4 [2x dx/dt ] L { L is constant }
so
A = √3/4 [2x dx/dt ] L
∫A dt = √3/2 [ Lx dx ] { Integrate both sides}
At = √3/2 × Lx × x²/2
x² = uAt / √3L { we find square root of both sides}
x = √( uAt / √3L )
x = (2/(3)^(1/u))√(At/L)
Therefore; required expression is (2/(3)^(1/u))√(At/L)
Order the following from smallest to largest. Place the smallest value at the top of your list.
= 10kg
= 100g
= 1,000,000mg
Answer:
mg, g, kg,
Explanation:
Tip King Henry Died Unit Drinking Chocolate Milk
Mg is in the back so its smallest then unit which is g and finally King also known as Kilogram which is the biggest
What are two main forces that act in gases in a star?
Answer:
hydrogen and helium
Explanation:
the sun is so hot that the huge amount of hydrogen is undergoing a constant star wide nuclear reaction
Answer:
the correct answer is C and D
I took a test hope it helps
Carlota does 2000 J of work on a machine. The machine does 500 J of work. What is the efficiency of the
machine?
4%
O 25%
O 75%
O 400%
Answer:
25%
Explanation:
500 J out of 2000 J is 25%
A student throws a ball upward with a velocity of 35 m/s. What is the acceleration of the ball as it rises to the top of its arc?
Answer:
The acceleration of the ball as it rises to the top of its arc equals 9.807 meters per square second.
Explanation:
Let suppose that maximum height of the arc is so small in comparison with the radius of the Earth.
Since the ball is launched upwards, then the ball experiments a free-fall motion, that is, an uniform accelerated motion in which the element is accelerated by gravity. Then, the acceleration experimented by the motion remains constant at every instant and position.
Besides, the gravitational acceleration in the Earth and, in consequence, the acceleration of the ball as it rises to the top of its arc equals 9.807 meters per square second.
if the tension of the cable is 25.0 N what is the mass of the ball
Answer:576
Explanation:
Which contact force always acts against the direction of movement?
A. spring
В.
friction
C.
impact
D. normal
Answer:
the answer to your question is b
what types of bodies sink in water
Answer:
Rocks, coins, honey, vegetable oil, diamonds, etc.
Explanation:
Object that are more dense than water.
A ballet dancer spins with 2.4 rev/s with her arms outstretched,when the moment of inertia about axis of rotation is 1 .With her arms folded,the moment of inertia about the same axis becomes 0.61 . calculate the new rate of spin.
Correct question is;
A ballet dancer spins with 2.4 rev/s with her arms outstretched,when the moment of inertia about axis of rotation is I. With her arms folded,the moment of inertia about the same axis becomes 0.6I about the same axis. Calculate the new rate of spin.
Answer:
4 rev/s
Explanation:
We are given;
Initial Angular velocity; ω_i = 2.4 rev/s
Initial moment of inertia; I_i = I
Final moment of inertia; I_f = 0.6I
From conservation of angular momentum, we have;
I_i × ω_i = I_f × ω_f
Where ω_f is the new rate of spin.
Thus, let's make it the subject to get;
ω_f = (I_i × ω_i/I_f)
Plugging in relevant values, we have;
ω_f = (I × 2.4/0.6I)
I will cancel out to give;
ω_f = 2.4/0.6
ω_f = 4 rev/s
How to detect beta rays
Answer:
For the detection of beta particles, organic scintillators can be used. Pure organic crystals include crystals of anthracene, stilbene, and naphthalene. The decay time of this type of phosphor is approximately 10 nanoseconds. This type of crystal is frequently used in the detection of beta particles.
- Hope this helps!
A 45-year-old man is meeting his doctor for his annual physical. He suffers from Type 2 diabetes and has a family history of coronary disease. Recently his job has become more stressful and he has been eating cookies and drinking soda to cope. What state of change is this man in?
Answer:
The man is on the verge of having a heart attack or a stroke.
Explanation:
If he has a family history of coronary (heart) disease, it means it could normally affect. Normally here means without anything aggravating it. It's already in his lineage so he could have it.
Now, he's past middle age - he's 45. He's past the growing stages of life. His organs are fully developed herefore.
Now also, he suffers from Type 2 diabetes. Although this is sometimes milder than Type 1 diabetes, it increases the risk of having a heart disease or a stroke!
Soda, especially sweetened one, is not to be taken too often because it can cause Diabetes Mellitus. For a diabetes patient, this should be a "no-go-area". Taking this constantly (everyday at work) will now put this 45-year-old man in harm's way.
He is no more at risk of having complications but already on the path to a heart disease or a stroke.
The charge per unit length on a long, straight filament is 290.0 mC/m. Find the electric field a) 10.0 cm andb) 100 cm from the filament, where distances are measured perpendicular to the length of the filament.
Answer:
Explanation:
given linear charge density λ = 290 x 10⁻³ C / m
Expression for electric field at distance d is given as follow .
E = λ / 2πε₀r
1 / 4πε₀ = 9 x 10⁹
1 / 2πε₀ = 18 x 10⁹
E = λ / 2πε₀r = 290 x 10⁻³ x 18 x 10⁹ / r
= 5220 x 10⁶ / r
For r = 10 x 10⁻² m = .1 m
E = 5220 x 10⁶ / .1
= 5.22 x 10¹⁰ N/m
For r = 100 x 10⁻² m = 1 m
E = 5220 x 10⁶ / 1
= 5.22 x 10⁹ N/m .
Lou’s latest invention, aimed at urban dog owners, is the X-R-Leash. It is made of a rubber-like material that exerts a force Fx = (−5.7 N/m) x − (78 N/m2 ) x 2 when it is stretched a distance x. The ad claims, "You’ll never go back to your old dog leash after you’ve had the thrill of an X-R-Leash experience. And you’ll see a new look of respect in the eyes of your proud pooch." Find the work done on a dog by the leash if the person remains stationary, and the dog bounds off, stretching the X-R-Leash from x = 0 m to x = 22 m. Answer in units of kJ.
Answer:
W = 29.06 KJ
Explanation:
The work done while stretching the leash can be calculated by the following formula:
[tex]W = \int\limits^b_a {F_{x}} \, dx \\[/tex]
whee,
W = Work Done = ?
Fₓ = Forcing Function = (-5.7 N/m)x - (78 N/m²)x²
a = starting point of x = 0 m
b = end point of x = 22 m
Therefore,
[tex]W = \int\limits^{22\ m}_{0\ m} {(-(5.7\ N/m)x - (7.8\ N/m^{2})x^{2}}) \, dx \\Integrating\ we\ get:\\W = -\frac{(5.7\ N/m)x^{2}}{2} - \frac{(7.8\ N/m^{2})x^{3}}{3}\\Applying\ limits:\\W = -\frac{(5.7\ N/m)(22\ m)^{2}}{2} - \frac{(7.8\ N/m^{2})(22\ m)^{3}}{3} - 0\\W = - 1379.4\ J - 27684.8\ J\\W = 29064.2\ J[/tex]
W = 29.06 KJ
One of the strongest emission lines observed from distant galaxies comes from hydrogen and has a wavelength of 122 nm (in the ultraviolet region). (a) How fast must a galaxy be moving away from us in order for that line to be observed in the visible region at 366 nm
Answer:
[tex]v=2.4*10^8m/s[/tex]
Explanation:
From the question we are told that
Wavelength of emission [tex]\lambda=122nm[/tex]
Observation distance [tex]d=366nm[/tex]
Generally the s equation is given as
[tex]f'=f\sqrt{\frac{(1-\frac{v}{c} )}{1+\frac{v}{c} }[/tex]
where
F is inversely proportional to T
[tex]d=\lambda\sqrt{\frac{(1-\frac{v}{c} )}{1+\frac{v}{c} }[/tex]
[tex]\frac{v}{c} =\frac{(1-\frac{\lambda}{d})}{(1+\frac{\lambda}{d}}[/tex]
[tex]\frac{v}{c}=\frac{1-(\frac{122}{366} )^2}{1+(\frac{122}{366})^2}[/tex]
[tex]\frac{v}{c}=\frac{0.8888888889}{1.11111111}[/tex]
[tex]\frac{v}{c}=0.80[/tex]
[tex]v=0.80*3*10^6[/tex]
[tex]v=2.4*10^8m/s[/tex]
What force is required to give an object with mass 300 kg an acceleration of 2 m/s^2
A baseball is hit high into the upper bleachers of left field. Over its entire flight the work done by gravity and the work done by air resistance respectively are:
A. positive; positive
B. positive; negative
C. negative; positive
D. negative; negative
E. unknown since vital information is lacking
Answer:
B. positive; negative.
Explanation:
From the viewpoint of Principle of Energy Conservation and Work-Energy Theorem, we notice that gravity represents a conservative force, associated with gravitational potential energy, whereas air resistance is a non-conservative force, associated with dissipated work. Therefore, the work done by gravity is positive and work done by air resistance is negative. Therefore, the correct answer is B.
A 12 N force and a 21 N force are acting from a single point in opposite directions. What additional force must be
added to produce equilibrium?
Answer:
We need to add 9 N in the direction of F(1)
Explanation:
We know that in equilibrium the total force is equal to zero, so we have:
[tex]F_{total}=F_{1}+F_{2}+F_{3}=0[/tex]
F(1) is 12 N
F(2) is 21 N in the opposite direction
F(3) is the additional force.
Then, using these values into the total force equation we have:
[tex]12-21+F_{3}=0[/tex]
[tex]-9+F_{3}=0[/tex]
[tex]F_{3}=9\: N[/tex]
Therefore, we need to add 9 N in the direction of F(1)
I hope it helps you!
The space station in the drawing is rotating to create artificial gravity. The speed of the inner ring is one half that of the outer ring. As an astronaut walks from the inner to the outer ring, what happens to her apparent weight
Answer:
It will double.
Explanation:
Apparent weight is the weight of a physical body affected by a force of gravity different of that of the athmosphere. Thus, the lower the gravity, that is, the lower the attraction of one body to another, the lower the apparent weight of the object on which the gravitational force is exerted; while the opposite occurs if the gravitational force is greater. In other words, the apparent weight of an object is determined by the gravitational force exerted on it: the greater the force, the greater the weight. Therefore, as the outer ring has twice the force of gravity than the inner ring, the apparent weight of the astronaut will double as well.
A 1210 kg car is driving NE
(at 45.0°) at 15.2 m/s when it is
struck by a moving 1540 kg car.
Afterward, they stick together and
move directly east (at 0°) at 23.3
m/s. What was the x-component
of the second car's initial velocity?
Please help
Answer:
[tex]V_2_X=33.16m/s[/tex]
Explanation:
From the question we are told that
Mass of car [tex]M_1=1210kg \\Angle1=\theta _1 45\textdegree NE[/tex]
Velocity of car [tex]v_1= 15m/s[/tex]
Mass of Truck [tex]M_2= 1540kg \\Angle 2=\theta_2 0\textdegree E[/tex]
Final velocity [tex]v_2= 23.3m/s[/tex]
Generally the the equation of the law of conservation of momentum is mathematically given by
Given the x direction
[tex]m_1v_1cos\theta+m_2v_2=(m_1+m_2)v[/tex]
[tex]1210*15.2cos45+1540*V_2=(1210+1540)*23.3[/tex]
[tex]V_2=\frac{(1210+1540)*23.3}{210*15.2cos45+1540}[/tex]
[tex]V_2=33.16m/s[/tex]
The x-component of the second car's initial velocity is
[tex]V_2_X=33.16m/s[/tex]
There is a current I flowing in a clockwise direction in a square loop of wire that is in the plane of the paper. If the magnetic field is toward the right, and if each side of the loop has length L, the net magnetic torque acting on the loop is:
Answer:
IBL²
Explanation:
Given that
Current flowing in the wire, is I
Magnetic field towards the right, is B
Length of each side of the loop, is L
Number of windings on the loop, is N
Torque exerted on the wire, is T
The torque, T is usually given by the formula
Torque, T = NBIAsinθ,
and if N = 1 and sinθ = 1 also, then we have
Torque, T = BIA
And we know that Area A from this particular question will be given as = L². If we then substitute A for L², we then have
Torque, T = IBL²
PLEASE ANSWER QUESTION 12!!! THANKSSS
Jamie dropped a water bottle from the top of an office building. The water bottle impacted the ground after falling for 5 seconds. Calculate the velocity of the water bottle at impact with the street below.
Answer:
The speed of the water bottle at impact with the street is 49 m/s
Explanation:
Free Fall Motion
A free-falling object falls under the sole influence of gravity. Any object that is being acted upon only by the force of gravity is said to be in a state of free fall. Free-falling objects do not experience air resistance.
If an object is dropped from rest in a free-falling motion, it falls with a constant acceleration called the acceleration of gravity, which value is [tex]g = 9.8 m/s^2[/tex].
The final velocity of a free-falling object after a time t is given by:
vf=g.t
The water bottle dropped by Jamie takes t=5 seconds to impact the ground, thus its speed at that moment is:
vf= 9.8*5
vf = 49 m/s
The speed of the water bottle at impact with the street is 49 m/s
ANSWERRR PLEASEEEE!!!
Answer:
50N to the left
Explanation:
I'll explain if you ask
Answer:
50 newtons to the left
Explanation:
A satellite is in orbit 3.117106 m from the center of Earth. The mass of Earth is 5.9821024 kg. Calculate the orbital
period of the satellite.
Answer:
T = 1733.16 s = 28.88 min
Explanation:
The orbital velocity of a satellite about Earth is given as follows:
[tex]v = \sqrt{\frac{GM}{R}}[/tex]
where,
v = orbital speed = ?
G = Gravitational Constant = 6.67 x 10⁻¹¹ Nm²/kg²
M = Mass of Earth = 5.982 x 10²⁴ kg
R = Orbit Radius = 3.117 x 10⁶ m
Therefore,
[tex]v = \sqrt{\frac{(6.67\ x\ 10^{-11}\ Nm^{2}/kg^{2})(5.982\ x\ 10^{24}\ kg)}{(3.117\ x\ 10^{6}\ m)}}\\\\v = 11.3\ x\ 10^{3}\ m/s[/tex]
but the velocity is given as:
[tex]v = \frac{distance}{time}[/tex]
for distance = circumference = 2πR
time = time period = T = ?
Therefore,
[tex]11.3\ x\ 10^{3}\ m/s = \frac{2\pi(3.117\ x\ 10^{6}\ m)}{T}\\\\T = \frac{2\pi(3.117\ x\ 10^{6}\ m)}{11.3\ x\ 10^{3}\ m/s}\\\\[/tex]
T = 1733.16 s = 28.88 min
If an object falls and ends with a velocity of 58.8 m/s, how long was the
object falling?
Answer:
the time taken for the object to fall is 6 s.
Explanation:
Given;
final velocity of the object, v = 58.8 m/s
initial velocity of the object, u = 0
The height of fall of the object is calculated as;
v² = u² + 2gh
v² = 2gh
[tex]h = \frac{v^2}{2g} \\\\h = \frac{(58.8)^2}{2(9.8)} \\\\h = 176.4 \ m[/tex]
The time to fall through the height is calculated as;
[tex]h =ut+ \frac{1}{2} gt^2\\\\h = 0 + \frac{1}{2} gt^2\\\\h = \frac{1}{2} gt^2\\\\t= \sqrt{\frac{2h}{g} } \\\\t = \sqrt{\frac{2\times 176.4}{9.8} } \\\\t = 6 \ s[/tex]
Therefore, the time taken for the object to fall is 6 s.
