PLEASE ANSWER ASAP BEFORE MY TEACHER AND MY MOM KILLES ME PLEASE ASAP
The first person with the right answer gets to be a brainlest
In the attachment there is a density column where there is colour
Question: tell me why is the red at the bottom of the density column if it is the least dense
Answer:
As you know, the denser objects have more weight per unit of volume, this will mean that the force that pulls down these objects is a bit larger.
This will mean that the denser objects will always go to the bottom.
This clearly implies that the red liquid, the one with one of the smaller densities, can not be at the bottom.
There are some cases where a liquid with a small density may become a lot denser as the temperature or pressure changes, and in a case like that, we could see the red liquid at the bottom, but for this case, there is no mention of changes in the temperature nor in the pressure, so this can be discarded.
The only thing that makes sense is that the red part at the bottom is the base of the tube, and has nothing to do with the red liquid.
A vehicle hits a bridge abutment at a speed estimated by
investigations as 20kmph. Skid marks of 30m on pavement
(f=0.35) followed by skid marks of 60m.on the gravel shoulder
approaching the abutment (f=0.50).What was the initial speed of vechile
Answer:
54.5 kmph
Explanation:
From work-kinetic energy principles, work done by friction on both pavement and gravel shoulder = kinetic energy change of vehicle
ΔK = W = -(f₁d₁ + f₂d₂) where f₁ = frictional force due to pavement = μ₁mg where μ₁ = coefficient of friction of pavement = 0.35, m = mass of vehicle and g = acceleration due to gravity = 9.8 m/s² and d₁ = distance moved by vehicle across pavement = 30 m and
f₂ = frictional force due to gravel shoulder = μ₂mg where μ₂ = coefficient of friction of pavement = 0.50, m = mass of vehicle and g = acceleration due to gravity = 9.8 m/s² and d₂ = distance moved by vehicle across gravel shoulder = 60 m
ΔK = 1/2m(v₁² - v₀²) where v₀ = initial velocity of vehicle, v₁ = final velocity of vehicle = 20 kmph = 20 × 1000/3600 = 5.56 m/s and m = mass of vehicle
So,
ΔK = -(f₁d₁ + f₂d₂)
1/2m(v₁² - v₀²) = -(μ₁mgd₁ + μ₂mgd₂)
1/2(v₁² - v₀²) = -(μ₁gd₁ + μ₂gd₂)
v₁² - v₀² = -2g(μ₁d₁ + μ₂d₂)
v₀² = v₁² + 2g(μ₁d₁ + μ₂d₂)
v₀ = √[v₁² + 2g(μ₁d₁ + μ₂d₂)]
substituting the values of the variables into the equation, we have
v₀ = √[(5.56 m/s)² + 2 × 9.8 m/s²(0.35 × 30 m + 0.5 × 60 m]
v₀ = √[30.91 (m/s)² + 4.9 m/s²(10.5 m + 30 m]
v₀ = √[30.91 (m/s)² + 4.9 m/s²(40.5 m]
v₀ = √[30.91 (m/s)² + 198.45 (m/s)²]
v₀ = √[229.36 (m/s)²
v₀ = 15.14 m/s
v₀ = 15.14 × 3600/1000
v₀ = 54.5 kmph
So, the initial speed of the vehicle is 54.5 kmph
If a battery of 9 volts is connected across a resistor of 1000 ohm, what will be the value of current flowing through it?
A battery of 9 volts is connected to a resistor of 1000 ohms, then the value of current flowing through it will be 0.009 A.
What is current?Any flow of electrical charge carriers, such as ions, holes, or subatomic charged particles, is referred to as electrical current.
When electrons serve as the charge transfer in a wire, the amount of charge moving through any point of wire in a given amount of time is measured as electric current. Electric charges' motion is intermittently reversed in the alternating current, but not in direct current.
In many situations, the direction of movement in electric circuits is assumed to be the direction of positive charge flow, which is the way opposite of the real particle drift. When properly specified, the current is known as conventional current.
From the question,
V = ir
i = v/r
i = 9/1000
i = 0.009 A
Therefore, the value of the current flowing through it is 0.009 A.
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If -->B is added to -->A , under what conditions does the resultant vector have a magnitude equal to A + B? Under what conditions is the resultant vector equal to zero?
Answer:
See explanation below
Explanation:
For two vectors A and B to have a positive resultant, they must both move in the positive direction i.e along the positive x axis
For two vectors A and B to have a zero resultant, they must both have the same magnitude but move in the opposite direction i.e one in the positive direction while the other in the negative direction.
Fie example if A = +5N, B = -5B
So that A+B = +5-5 = 0N
For an object like a planet, with a typical temperature of a few hundred kelvin, what kind of blackbody radiation would it principally emit
Answer:
Low-temperature blackbody
Explanation:
There are 3 types of blackbody temperatures.
Low-temperature blackbody
High temperature extended area blackbody
High-temperature cavity blackbody
A Low-temperature blackbody is a type of black body radiation that has the range of -40° C to 175° C, typically between 233 K and 448 K. A perfect fit for the temperature range mentioned in the question, "a few hundred Kelvin". Therefore, it's the kind of blackbody temperature that the object would emit.
An object is placed at O ona number line. It moves 3 units to the right, then 4 units to the left, and then 6 units to
the right. What is the displacement of the object?
Answer:
You have a displacement of 5 units to the right.
Explanation:
First you go three to the right which lands on the 3 mark. Then you move it 4 to the left which substracts 4, landing the object at -1. Finally you move 6 to the right, and you finish at marker 5. Since displacement is not total distance but just final distance from the start point directly to end point, it is only a displacement of 5.
Answer:
I think it would be +5
Explanation:
I looked on a number line and started at 0.
I then moved the object 3 units to the right and got 3.
I then moved the object 4 units the the left and got -1.
I then moved the object 6 units to the right and my final answer was positive 5. (+5 or just 5)
I hope I was right and helped
At a distance of 10 km from a radio transmitter, the amplitude of the E-field is 0.20 volts/meter. What is the total power emitted by the radio transmitter
Answer:
The total power is [tex]P = 6.665 *10^{4} \ W[/tex]
Explanation:
From the question we are told that
The distance is [tex]r = 10 \ km = 1000 \ m[/tex]
The amplitude of the electric field is [tex]E = 0.20 \ volt/meter[/tex]
Generally the average intensity of the electromagnetic field from the radio transmitter is mathematically represented as
[tex]I = \frac{E^2}{ 2 \mu_o * c }[/tex]
Here c is the speed of light with value [tex]c = 3.0 *10^{8} \ m/s[/tex]
[tex]\mu_o[/tex] is the permeability of free space with value [tex]\mu_o = 4\pi *10^{-7} \ N/A^2[/tex]
So
[tex]I = \frac{0.2^2}{ 2 * 4\pi *10^{-7} * 3.0*10^{8} }[/tex]
=> [tex]I = 5.307 *10^{-5} \ W/m^2[/tex]
Generally this intensity can also be mathematically represented as
[tex]I = \frac{P }{ 4 \pi r^2 }[/tex]
=> [tex]P = I ( 4 \pi r^2 )[/tex]
=> [tex]P = 5.307 *10^{-5} ( 4 * 3.142 * 1000^2 )[/tex]
=> [tex]P = 6.665 *10^{4} \ W[/tex]
The total power emitted by the radio transmitter is [tex]6.67\times 10^4 \ W[/tex].
The given parameters;
amplitude of the electric field, E = 0.2 V/mdistance of the transmitter, = 10 km = 10,000 mThe intensity of the radio wave from the transmitter is calculated as follows;
[tex]I = \frac{E^2}{2\mu_0 c} \\\\I = \frac{0.2^2 }{2\times 4\pi \times 10^{-7} \times 3\times 10^8} \\\\I = 5.305 \times 10^{-5} \ W/m^2[/tex]
The total power emitted by the radio transmitter is calculated as follows;
[tex]I = \frac{P}{A} \\\\P = IA\\\\P = I \times 4\pi r^2\\\\P = (5.305\times 10^{-5} )\times 4\pi \times (10,000)^2\\\\P = 6.67\times 10^{4} \ W[/tex]
Thus, the total power emitted by the radio transmitter is [tex]6.67\times 10^4 \ W[/tex].
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What magnitude point charge creates a 9,000 N/C electric field at a distance of 1.00 m? a. 1.00 C b. 1.00 mC c. 1.00 ?C d. 1.00 nC
Answer:
1 µC
Explanation:
From the question given above, the following data were obtained:
Electric field intensity (E) = 9000 N/C
Distance (r) = 1 m
Charge (Q) =?
NOTE: Electric force constant (K) = 9×10⁹ Nm²C¯²
The magnitude of the point charge can be obtained as follow:
E = KQ/r²
9000 = 9×10⁹ × Q / 1²
9000 = 9×10⁹ × Q
Divide both side by 9×10⁹
Q = 9000 / 9×10⁹
Q = 1×10¯⁶ C
Recall:
1 micro charge (µC) = 1×10¯⁶ C
Hence, the magnitude of the point charge is 1 µC
Answer:
1ηCExplanation:
The magnitude of the point charge is expressed as;
V = kQ/r
k is the coulombs constant = 9.0×10^9 N⋅m²/C²
V is the electric field = 9000N/C
Q is the charge
r is the distance = 1.00m
Get the required charge
Substitute the given parameters into the formula;
9000 = 9.0×10^9Q/1.0
9000 = 9.0×10^9Q
Q = 9000/9.0×10^9
Q = 9.0×10^3/9.0×10^9
Q = 9.0/9.0×10^(3-9)
Q = 1.0×10^6C
Q = 1ηC
hence the magnitude of the point charge is 1ηC
Determine the displacement and distance covered by a man if he walks 10 m north, turns east and walks 20 m, and then turns right and walks 10 m.
Answer:
20m
Explanation:
The two tens cancel each other out, as they are in opposite directions. Now we only care about the 20m, which if we have no 10's, will end up 20m away.
The distance covered by man is 40 m and the displacement is 20 m towards the east.
Distance and DisplacementThe distance can be defined as the measurement of the total ground covered by an object during its motion. It is a scalar quantity.
The displacement can be defined as the measurement of the shortest path over the ground is covered by an object during its motion. It is a vector quantity.
Given that a man starts walking towards the north and covered 10 m. he turns east and walks for 20 m then turns right and walks for 10 m. The attachment shows the total ground area covered by the man.
The total distance covered by the man is given below.
[tex]D = 10 + 20 + 10[/tex]
[tex]D = 40 \;\rm m[/tex]
The displacement of the man is the difference between its final position and initial position.
[tex]d = 20 - 0[/tex]
[tex]d = 20 \;\rm m[/tex]
Hence we can conclude that the distance covered by man is 40 m and the displacement is 20 m towards the east.
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Which statements explain the special theory of relativity? Check all that apply.
- Time and space are relative.
- The speed of light in a vacuum is constant for all observers.
- Physical laws change based on an observer's motion.
- Physical laws remain constant regardless of an observer's motion.
- The special theory of relativity applies to objects with constant velocity.
- The special theory of relativity applies to accelerating objects.
Answer:
A. Time and space are relative.
B. The speed of light in a vacuum is constant for all observers.
D. Physical laws change based on an observer's motion.
E. Physical laws remain constant regardless of an observer's motion.
Explanation:
The statements that explain the special theory of relativity are a),b) and e) respectively.
The special theory of relativity, developed by Albert Einstein, is based on the idea that the laws of physics are the same for all observers who are moving uniformly relative to each other. This means that there is no "absolute" or "preferred" frame of reference in the universe. Instead, all frames of reference are equally valid, and any observer can use their own frame of reference to describe physical phenomena.
One consequence of this idea is that time and space are relative. This means that the measurements of time and distance depend on the observer's frame of reference. For example, if two observers are moving relative to each other, they will measure different lengths and times for the same event. This effect is known as time dilation and length contraction.
Finally, the special theory of relativity applies only to objects with constant velocity. It does not apply to accelerating objects, which require the more general theory of relativity. The special theory of relativity is a fundamental theory in modern physics, and it has important implications for our understanding of space, time, and the nature of reality.
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The correct question is :
Which statements explain the special theory of relativity? Check all that apply.
a) Time and space are relative.
b) The speed of light in a vacuum is constant for all observers.
c) Physical laws change based on an observer's motion.
d) Physical laws remain constant regardless of an observer's motion.
e) The special theory of relativity applies to objects with constant velocity.
f) The special theory of relativity applies to accelerating objects.
What was your train of thought as you navigated the picture of the candle?
Answer:
Where is the picture
Explanation:
WHERE IS THE PICTURE
The small size of cells _____.
allows for efficient transport of materials
causes food to be broken down slowly
slows down the removal of wastes
allows for fewer cell organelles
Answer:
allows for efficient transport of materials
Explanation:
Living cells relatively have a small size, so small that a microscope is needed to see their structure. However, this small size is for their benefitting as it increases their surface area to volume ratio (SA:V).
A large surface area to volume ratio (SA:V) enables easy movement of molecules to and fro the cell membrane via the process of DIFFUSION. Therefore, having a small size allows for efficient transport of materials needed in the cell.
You are driving at 25 m/s when an ambulance passes you and pulls into your lane, just in front of you, and speeds ahead at 35 m/s. The ambulance driver hears a siren sound of 850 Hz. What frequency do you hear
Answer:
The value is [tex]f_1 = 828 \ Hz[/tex]
Explanation:
From the question we are told that
The speed is [tex]v = 25 \ m/s[/tex]
The speed of the ambulance is [tex]u = 35 \ m/ s[/tex]
The frequency of the siren is [tex]f = 850 \ Hz[/tex]
Generally from Doppler effect equation we have that
[tex]f_1 = \frac{v_s - v }{ v_s + u } * f[/tex]
Here [tex]v_s[/tex] is the velocity of sound with the value [tex]v_s = 343 \ m/s[/tex]
=> [tex]f_1 = \frac{343 - 25 }{ 343 + 35 } * 850[/tex]
=> [tex]f_1 = 828 \ Hz[/tex]
A potter's wheel moves uniformly from rest to an angular speed of 0.17 rev/s in 32.0 s.
a. Find its angular acceleration in radians per second per second.
b. Would doubling the angular acceleration during the given period have doubled final angular speed?
Answer:
a) α = 0.0334 rad / s² , b) w = 2.14 rad/s see that the angular velocity doubles.
Explanation:
This is a magular kinematics exercise
Let's reduce the magnitudes to the SI system
w = 0.17 rev /s (2π rad / 1rev) = 1.07 rad / s
a) as part of rest its initial velocity is zero w or = 0
w = w₀ + α t
α = [tex]\frac{\omega -\omega_{o} }{t}[/tex]
α = [tex]\frac{1.07-0}{32}[/tex]
α = 0.0334 rad / s²
b) If we double the angular relation what will be the final velocity
w = w₀ + (2α) t
w = 0 + 2 0.0334 32
w = 2.14 rad/s
We see that the angular velocity doubles.
In the attachment there is a density column where there is colour
Question: tell me why is the red at the bottom of the density column if it is the least dense
Answer: The red at the bottom represents the base of the tube, not the red liquid.
Explanation:
The densest materials have more weight per unit of volume.
This means that those elements will always flow to the bottom of the containers, like the one in the image.
The red liquid being the least dense one, can not go to the bottom by its own means.
There could be some cases, like:
The red liquid when solid, is way denser than in its liquid phase, and then the red at the bottom could be solid phase of the red liquid, but there is no mention of this in the question, then we can discard this idea.
Another trivial idea is that the red liquid at the bottom could be trapped by some kind of wall, but again, there is no mention of this, so again we can discard this idea.
The thing that makes sense is that the red at the bottom represents the base of the tube and not the red liquid.
How many words can be made with five letters
Answer:
agree eager eagre ragee
Explanation:
hope this help
An ideal spring with spring constant k is hung from the ceiling. The initial length of the spring, with nothing attached to the free end, is 15.0 cm. When an object of mass m is attached to the free end of the spring, the spring stretches to a final length of 30.0 cm. If k = 3.5 N/cm, what is the mass m of the object?
The mass m of the object = 5.25 kg
Further explanationGiven
k = spring constant = 3.5 N/cm
Δx= 30 cm - 15 cm = 15 cm
Required
the mass m
Solution
F=m.g
Hooke's Law
F = k.Δx
[tex]\tt m.g=k.\Delta x\\\\m.10=3.5\times 15\\\\m=5.25~kg[/tex]
An incandescent lightbulb is rated at 120 Watts when plugged into a 200 V-rms household outlet. Calculate the resistance of the filament and the rms current. g
Answer:
The resistance of the filament is 333.33 ohmsThe rms current is 0.6 AExplanation:
Given;
output power of the incandescent lightbulb , P = 120 W
input voltage, V = 200 V
The resistance of the filament is calculated as;
[tex]P = \frac{V^2}{R}[/tex]
Where;
R is the resistance of the filament
[tex]R = \frac{V^2}{P} \\\\R = \frac{200^2}{120} \\\\R = 333.33 \ ohms[/tex]
The rms current is given as;
[tex]P = I_{rms} V_{rms}\\\\I_{rms} = \frac{P}{V_{rms}} \\\\I_{rms} =\frac{120}{200}\\\\ I_{rms} = 0.6 \ A[/tex]
A satellite travels with a constant speed |v| as it moves around a circle centered on the earth. How much work is done by the gravitational force F on the satellite after it travels half way around the earth in time t?
Answer:
W = 0
Explanation:
As the satellite moves in a circle the force is perpendicular to the path, therefore the work that is defined by
W = F. r = f r cos θ
Since the force and the radius are perpendicular, the angle θ = 90º and the cosine 90 = 0, therefore there is no work for the circular motion.
W = 0
HELPPPPP fasttttt
Do you think an object can travel faster than the speed of sound? Explain your thinking.
Answer: Depends on the context.
Explanation: While simply throwing a football won’t breach the sound barrier, a specialized jet can breach it easily. It all just depends on your definition of an object. Hope this helps!
A basketball is rolling rightward onto the court with a speed of 4.0 m/s and slows down with a consta leftward acceleration of magnitude 0.50 m/s^2 over 14 m.
What is the velocity of the basketball after rolling for 14 m?
a. 2.0 m/s
b. 5.5 m/s
c. 7.5 m/s
d. 1.4 m/s
Answer:
d. 1.4 m/s
Explanation:
Given;
initial velocity of the `basketball, u = 4.0 m/s
acceleration of the basketball, a = -0.5 m/s² (leftward)
distance the basketball traveled, d = 14 m
the final velocity of the basketball, v = ?
(this final velocity should be less than the initial velocity since the ball is slowing down at a constant rate)
Apply the following the kinematic equation;
v² = u² + 2ad
v² = 4² + 2(-0.5)14
v² = 16 - 14
v² = 2
v = √2
v = 1.41 m/s
Therefore, the final velocity of the basketball is 1.4 m/s.
A spacecraft on its way to Mars has small rocket engines mounted on its hull; one on its left surface and one on its back surface. At a certain time, both engines turn on. The one on the left gives the spacecraft an acceleration component in the x direction of ax = 5.10 m/s2, while the one on the back gives an acceleration component in the ydirection of ay = 7.30 m/s2. The engines turn off after firing for 675 s, at which point the spacecraft has velocity components of vx = 3630 m/s and vy = 4276 m/s.What was the magnitude and direction of the spacecraft's initial velocity, before the engines were turned on?Express the direction as an angle measured counterclockwise from the x -axis.
Answer:
v₀ = 677.94 m / s , θ = 286º
Explanation:
We can solve this exercise using the kinematic expressions, let's work on each axis separately.
X axis
has a relation of aₓ = 5.10 m / s², the motor is on for a time of t = 675 s, reaching the speed vₓ = 3630 m / s, let's use the relation
vₓ = v₀ₓ + aₓ t
v₀ₓ = vₓ - aₓ t
let's calculate
v₀ₓ = 3630 - 5.10 675
v₀ₓ = 187.5 m / s
Y Axis
[tex]v_{y}[/tex] = v_{oy} - a_{y} t
v_{oy} = v_{y} - a_{y} t
let's calculate
v_{oy} = 4276 - 7.30 675
v_{oy} = -651.5 m / s
we can give the speed starts in two ways
a) v₀ = (187.5 i ^ - 651.5 j ^) m / s
b) in the form of module and angle
Let's use the Pythagorean theorem
v₀ = [tex]\sqrt{v_{ox}^{2} + v_{oy}^{2} }[/tex]
v₀ = [tex]\sqrt{187.5^{2} +651.5^{2} }[/tex]
v₀ = 677.94 m / s
we use trigonometry
tan θ = [tex]\frac{v_{oy} }{v_{ox} }[/tex]
θ = tan⁻¹ \frac{v_{oy} }{v_{ox} }
θ = tan⁻¹ ([tex]\frac{-651.5}{187.5}[/tex])
θ = -73.94º
This angle measured from the positive side of the x-axis is
θ‘ = 360 - 73.94
θ = 286º
This table shows statistics about the US population in 2010. Which demographic trend does this table best support? People are marrying later than they used to. There are more single-parent families. The American population is more diverse. There are fewer younger adults than older adults.
Answer:
D. There are fewer younger adults than older adults.
Explanation:
Failed the assignment but got this answer correct.
Answer:
D is correct
Explanation:
In performing rotational motion in an ice capade, which skaters must skate faster, those further in or those further out? Explain the reasoning for your conclusion using concept of rotational inertia.
Answer:
if the angular momentum is constant, the skater is closer, going faster.
w₂ = r₁² / r₂² w₁
Explanation:
For the skaters to perform in the turn in l, the angular number of the same must be equal, so we stammer the concept of conservation of angular momentum
L₁ = L₂
the angular momentum is
L = I w
we substitute
I₁ w₁ = I₂ w₂
w₂ = [tex]\frac{I_{1} }{I_{2} }[/tex] w₁
if we consider the skater as a particle, his moment of inertia is
I = m r²
we substitute
w₂ = r₁² / r₂² w₁
therefore, if the angular momentum is constant, the skater is closer, going faster.
In introductory physics laboratories, a typical Cavendish balance for measuring the gravitational constant G uses lead spheres with masses of 1.60 kg and 16.0 g whose centers are separated by about 3.30 cm. Calculate the gravitational force between these spheres, treating each as a particle located at the center of the sphere.
Answer:
The value is [tex]F = 1.568 *10^{-9} \ N[/tex]
Explanation:
From the question we are told that
The mass of the first lead sphere is [tex]m = 1.60 \ kg[/tex]
The mass of the second lead sphere is [tex]M = 16 \ g = 0.016 \ kg[/tex]
The separation between masses is [tex]r = 3.30 \ cm = 0.033 \ m[/tex]
Generally the gravitational force between each sphere is mathematically represented as
[tex]F = \frac{G * m * M }{r^2 }[/tex]
Here G is the gravitational constant with value [tex]G = 6.67 *10^{-11 } \ m^3 \cdot kg^{-1} \cdot s^{-2}[/tex]
[tex]F = \frac{6.67 *10^{-11 } * 1.60 * 0.016 }{0.033^2 }[/tex]
=> [tex]F = 1.568 *10^{-9} \ N[/tex]
As a projectile moves in its path, is there any point along the path where the velocity and acceleration vectors are:(a) perpendicular to each other (b) Parallel to each other
Answer:
at the highest point the velocity and acceleration are perpendicular but when projectile starts to move are said to be parallel to one another
Explanation:
the clear explanation is shown above.
Suppose you are on a cart that is moving at a constant speed v toward the left on a frictionless track. If you throw a massive ball straight up (from your perspective), how will the speed of the cart change?
a. The speed of the cart will increase
b. The speed of the cart will decrease
c. The speed of the cart will not change
d. You need to know how fast the ball was thrown
Which statement describes a climate condition?
Answer:
While there is no text here are some climate conditions you could go off of, tropical, dry, temperate, cold, and polar, hope this helps you with whatever your trying to do
Explanation:
Choose the best answer. A car traveling at constant speed has a net work of zero done on it.
A. True
B. False
Answer:
Explanation:
we know that,
force = mass × acceleration
∴ since speed/velocity is constant, acceleration should be zero.
∴ f = m × 0
f = 0 N
∴ If we apply this to ,
work = force × displacement
we get ,
w = 0 × s
∴ we can say that the net work is zero.
and hence the answer is true!!!
A plane is flying 700 km/hr to the east into a head wind that is moving at 20 km/hr west.
Calculate the planes velocity.
O 720 km/hr
680 km/hr
0-680 km/hr
0 -720 km/hr
Answer:
getc+d
Explanation:
gger