Since Saturn's volume is equivalent to 763.59 earths and Mercury is only 0.055 of one earth, then Saturn's volume is over 10,000 times larger than Mercury's.
Answer: TRUE
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
what color is my socks? :0
A color.......................................................................yeah the color of your socks is a color :}
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
By experiment, determine what makes a force attractive or repulsive. Describe your experiments and observations with some examples. 4. What evidence do you see that Newton's third law applies to electrostatic forces
Answer:
he combs are brought closer, each one has to move, therefore a force must be applied to each one, and the only force is electrostatic. force must have the same magnitude in each one and in the opposite direction; this description is equivalent to Newton's third law.
Explanation:
A very simple experiment that we can do is to rub a brush or plastic comb with a piece of paper, this creates some electrical charges in the comb, we hang the comb by a thread.
We take another comb and we form it with the same paper, when we bring the combs closer, we see that they repel each other, this is proof that the electrical charges created are of the same type.
If now we rub the comb with another material, for example plastic, and bring it closer to the comb that is hanging, we see that they are approaching, therefore, it is proven that the load created from another type
With these two simple experiments it can be proved that electric charges are of two types, which are called positive and negative.
4. In the previous experiment we see that when the combs are brought closer, each one has to move, therefore a force must be applied to each one, and the only force is electrostatic. If we observe carefully we see that the breaking of the combs is the same, therefore the force must have the same magnitude in each one and in the opposite direction; this description is equivalent to Newton's third law.
An 11.0kg object is thrown vertically into the air with an applied force of 145N. What is the
initial acceleration of the object?
Answer:
13.18m/s²
Explanation:
According to Newton's second law of motion
Force = Mass * acceleration
Given
Mass = 11kg
Force = 145N
Required
acceleration
From the formula
Acceleration = Force/Mass
Acceleration = 145/11
Acceleration = 13.18m/s²
Hence the initial acceleration of the object is 13.18m/s²
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.
Learn more about wavelength here:
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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
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]
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
Help me and get points..
When a basketball player jumps to make a shot, once his or her feet are off the ground, the jumper's acceleration
A) Depends on the time in the air.
B) Depends on launch speed.
C) Is usually greater for taller players (but not always).
D) All the above are true
E) Is 10 m/s/s
Answer:
B) Depends on launch speed.
Explanation:
This is true when considering the basketball player in a given basketball game. The speed at which the player jumps up is a strong factor which determines the acceleration of his acceleration. The direct co-relation show that, speed and acceleration of the basketball player are interrelated.
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 .
What force is required to give an object with mass 300 kg an acceleration of 2 m/s^2
A football is thrown from the edge of a cliff from a height of 22 m at a velocity of 18 m/s [39degrees above the horizontal]. A player at the bottom of the cliff is 12 m away from the base of the cliff and runs at a maximum speed of 6.0 m/s to catch the ball.
Is it possible for the player to catch the ball? Support your answer with calculations.
Please and thankyou!
Answer:
The ball will get to the bottom of the cliff in 1.26 s, while the player will reach the bottom of the cliff 2 s later. Thus, it is not possible for the player to catch the ball.
Explanation:
Given;
vertical height of the cliff, h = 22 m
velocity of the ball, u = 18 m/s at an angle 39⁰
vertical component of the velocity, [tex]u_y = u \ sin \theta[/tex]
The time for the ball to get to the bottom of the cliff is calculated as;
h = ut + ¹/₂gt²
h = (u sinθ)t + ¹/₂ x 9.8 x t²
22 = (18 sin 39)t + 4.9t²
22 = 11.328t + 4.9t²
4.9t² + 11.328t - 22 = 0
Solve the above equation with formula method;
a = 4.9, b = 11.328, c = -22
[tex]t = \frac{-b \ \ +/- \ \ \sqrt{b^2 - 4ac} }{2a} \\\\t = \frac{-11.328\ \ +/- \ \ \sqrt{(-11.328)^2 - 4(4.9\times -22)} }{2(4.9)}\\\\t = 1.26 \ s[/tex]
The time for the player to get to the bottom of the cliff;
Given maximum speed, Vx = 6.0 m/s and horizontal distance, X = 12 m;
[tex]t = \frac{X}{V_x} \\\\t = \frac{12}{6} \\\\t = 2 \ s[/tex]
The ball will get to the bottom of the cliff in 1.26 s, while the player will reach the bottom of the cliff 2 s later. Thus, it is not possible for the player to catch the ball.
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
a substance that turns litmus paper blue and contains hydroxide ions is most likely
Answer:
If it turns the paper blue it would have to be a base.
Explanation:
I hope that answers your question!
Answer:
If it turns the paper blue it would have to be a base.
Explanation:
I hope that answers your question!
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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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
At t = 0, a car registers at 30 miles/hr. Forty seconds later, the car’s velocity is now at 50 miles/hr. Assuming constant acceleration, what is the distance covered after 40s?
Select one:
a. 0.442 mi
b. .33 mi
c. 0.11 mi
d. 10.33
e. None of the choices given.
Answer: Choice A) 0.44 miles
=================================================
Work Shown:
1 hour = 60 minutes
1 minute = 60 seconds
1 hour = 60*60 = 3600 seconds
40 seconds = 40*(1/3600) = 40/3600 = 1/90 hours
The car travels 1/90 of an hour. Let t = 1/90.
The initial velocity is vi = 30 mph.
The final velocity is vf = 50 mph.
Apply one of the kinematics equations as given below.
x = distance traveled
x = 0.5*(vi+vf)*t
x = 0.5*(30+50)*(1/90)
x = 0.44444444444444 which is approximate
x = 0.44
The car traveled roughly 0.44 miles. This matches with choice A.
I'm assuming choice A is supposed to say 0.44 instead of 0.442; otherwise, I think your teacher made a typo by putting that 2 in there. Each of the other answer choices are accurate to 2 decimal places, so it would make sense that choice A is also accurate to 2 decimal places as well.
If 0.442 was intended by your teacher, and it's not a typo, then the answer would be E) None of the choices given.
what types of bodies sink in water
Answer:
Rocks, coins, honey, vegetable oil, diamonds, etc.
Explanation:
Object that are more dense than water.
if the tension of the cable is 25.0 N what is the mass of the ball
Answer:576
Explanation:
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
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.
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]
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.
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 sculpture is suspended in equilibrium by two cables, one from a wall and the other
from the ceiling of a museum gallery. Cable 1 applies a horizontal force to the right of
the sculpture and has a tension, Fn. Catble 2 applies a force upward and to the left at an
angle of 37.0° to the negative x-axis and has a tension, Fn. The gravitational force on
the sculpture is 5.00 '10`N. What is Fn?
Answer:
[tex]T_1=6655.295917 \approx 6655.3N[/tex]
Explanation:
From the question we are told that
Angle of cable 2 [tex]\theta=37.0\textdegree[/tex]
Weight of sculpture [tex]W=5000 N[/tex]
Generally the Tension from cable 2 [tex]T_2[/tex] is mathematically given by
[tex]T_2sin37\textdegree=5000N[/tex]
[tex]T_2=5000N/sin37\textdegree[/tex]
[tex]T_2=8308.2N[/tex]
Generally the Tension from Cable 1 [tex]T_1[/tex] is mathematically given by
[tex]T_1=T_2 cos37\textdegree[/tex]
[tex]T_1=8308.2* cos 37\textdegree[/tex]
[tex]T_1=6655.295917 \approx 6655.3N[/tex]
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)
