Answer:
The giraffe is the tallest of all mammals. It reaches an overall height of 18 ft (5.5 m) or more. The legs and neck are extremely long. The giraffe has a short body, a tufted tail, a short mane, and short skin-covered horns
If you double the mass, what happen to the acceleration?
Answer: The math behind this is quite simple. If you double the force, you double the acceleration, but if you double the mass, you cut the acceleration in half.
Explanation:
A sound wave is produced when a medium begins to
Answer:
when a medium begins to vibrate
What best describes the relationship between sleep and memory? The quantity of sleep impacts memory, while the quality of sleep does not. The quantity of sleep does not impact memory, while the quality of sleep does. The quantity and quality of sleep have no impact on memory. The quantity and quality of sleep both impact memory.
Answer:
The quantity and quality of sleep both impact memory.
Explanation:
An airplane of mass 270,000 kg that is sitting on a runway has its center of gravity located at 10.0 m in front of the main landing wheels as shown. Find the magnitude of the force exerted by the runway on the nose wheel located 25.0 m from the main landing wheels.
For an airplane of mass 270,000 kg , the magnitude of the force exerted is mathematically given as
N2=1584431.13N
N1=1061568.87N
What is the magnitude of the force exerted by the runway on the nose wheel located 25.0 m from the main landing wheels.?Generally, the equation for the Force balance is mathematically given as
N1+N2=270000*g
Therefore
N2*10=N1*15
N1=10N2/15
N1=0.67N2
Hence
0.67N2+N2=270000*g
N2=1584431.13N
N1=1061568.87N
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What best describes the speed of light waves in solids, liquids, and gases?
O The speed is fastest in solids.
O The speed is fastest in liquids.
O The speed is fastest in gases.
O The speed is the same in all matter.
Answer:
C
Explanation:
Generally, the speed of light slows down when passing through a medium that is not a vacuum. This is not always the case, but I will be ignoring the rare/exotic exceptions. Light has a harder time traveling through solids and liquids than it does with gases.
Answer:
its c
Explanation:
The time period of 2 pendulums are 1.22 s and 0.54 s. Calculate the ratio of their lengths.
[tex]~~~~~\dfrac{T_1}{T_2}= \dfrac{1.22}{0.54}\\\\\\\implies \dfrac{2\pi \sqrt{\dfrac{L_1}{g}}}{2\pi \sqrt\dfrac{L_2}g}}= 2.259\\\\\\\implies \dfrac{\sqrt{L_1}}{\sqrt g} \times \dfrac{\sqrt g}{\sqrt{L_2}} = 2.259\\\\\\\implies \sqrt{\dfrac{L_1}{L_2}} = 2.259\\\\\\\implies \dfrac{L_1}{L_2} = 5.104\\\\\\\implies L_1:L_2 = 5.104[/tex]
The position x of a particle moving along x axis varies with time t as x = Asin(wt) , where A and w are constants . The acceleration of the particle varies with its position as
[tex] \small \red{ \rm Nonsense = Report} \checkmark[/tex]
Let's see
[tex]\\ \rm\Rrightarrow x=Asin(\omega t)\dots(1)[/tex]
Now we know the formula of acceleration
[tex]\\ \rm\Rrightarrow \alpha=-A\omega^2sin(\omega t)[/tex]
[tex]\\ \rm\Rrightarrow \alpha=-Asin(\omega t)\times \omega^2[/tex]
From eq(1)[tex]\\ \rm\Rrightarrow \alpha=-x\omega^2[/tex]
Or
[tex]\\ \rm\Rrightarrow \alpha=-\omega^2x[/tex]
Given that the position x of a particle along X-axis varies with time t by the equation:
[tex]{:\implies \quad \sf x=A\sin (\omega t)}[/tex]
As it defines the position, so x is just displacement here, and we need to find the acceleration first for telling with what if varies, so by definition, the second differential coefficient of displacement is acceleration, so differentiating both sides w.r.t.x of the above equation in accordance with chain rule we have:
[tex]{:\implies \quad \sf \dfrac{dx}{dt}=A\cos (\omega t)\cdot \omega \cdot \dfrac{dt}{dt}}[/tex]
[tex]{:\implies \quad \sf \dfrac{dx}{dt}=A\omega \cos (\omega t)}[/tex]
Differentiating both sides w.r.t.x by chain rule again to get the 2nd order derivative
[tex]{:\implies \quad \sf \dfrac{d^{2}x}{dt^{2}}=-A\omega \sin (\omega t)\cdot \omega \dfrac{dt}{dt}}[/tex]
[tex]{:\implies \quad \sf \dfrac{d^{2}x}{dt^{2}}=-A\omega^{2}\sin (\omega t)}[/tex]
Re-write as :
[tex]{:\implies \quad \sf \dfrac{d^{2}x}{dt^{2}}=-\omega^{2}\{A\sin (\omega t)\}}[/tex]
Can be further written as
[tex]{:\implies \quad \sf \dfrac{d^{2}x}{dt^{2}}=-x\omega^{2}}[/tex]
This is the Required answer
If they ask you the differential equation for the acceleration of a wave (as the given equation was general equation of a wave), you can simply write:
[tex]{:\implies \quad \sf \dfrac{d^{2}x}{dt^{2}}+x\omega^{2}=0}[/tex]
a sound has a wavelength of 0.52028. If the speed is 343.00m/s, what is the musical note?
A small block sits at one end of a flat board that is 2.50 m long. The coefficients of friction between the block and the board are μs = 0.450 and μk = 0.400. The end of the board where the block sits is slowly raised until the angle the board makes with the horizontal is α0, and then the block starts to slide down the board.
Hi there!
In order for a block to begin sliding, the force due to STATIC friction must be overcome.
In this instance, the following forces are acting on the block ALONG the axis of the incline.
Force due to gravity (Fg)Force due to STATIC friction (Fs)Force due to gravity:
On an incline, the component of the force due to gravity contributing to the object's downward movement is equivalent to the horizontal (sine) component.
[tex]F_g = Mgsin\theta[/tex]
Force due to static friction:
The force due to friction is equivalent to the normal force multiplied by the coefficient of friction.
The normal force is the cosine component (perpendicular to the incline), so:
[tex]N = Mgcos\theta\\\\F_s = \mu_sMgcos\theta[/tex]
To find the minimum angle for the block to begin sliding, we can set the two forces equal to 0. They work in opposite directions (let down the incline be negative and up the incline be positive).
[tex]\Sigma F = F_s - F_g\\\\0 = F_s - F_g\\\\0 = \mu_sMgcos\theta - Mgsin\theta\\\\Mgsin\theta = \mu_sMgcos\theta[/tex]
Cancel out 'Mg' and rearrange to solve for theta.
[tex]sin\theta = \mu_scos\theta\\\\tan\theta = \mu_s\\\\\theta = tan^{-1}(\mu_s) = tan^{-1}(.45) = \boxed{24.228^o}[/tex]
You pull on a crate using a rope as in (Figure 1), except the rope is at an angle of 20.0 ∘ above the horizontal. The weight of the crate is 245 N , and the coefficient of kinetic friction between the crate and the floor is 0.270. What must be the tension in the rope to make the crate move at a constant velocity?
Express your answer with the appropriate units.
What is the normal force that the floor exerts on the crate?
Express your answer with the appropriate units.
Hi there!
For an object on an incline with friction being pulled, the following forces are present.
Force due to Gravity Force due to FrictionForce due to tensionThe force due to friction opposes the force due to gravity which would cause the object to slide down. (The force due to friction acts up the incline). Additionally, the force due to the rope is also upward.
Let up the incline be positive, and down the incline be negative.
Doing a summation of forces:
[tex]\Sigma F = F_T + F_f - F_g[/tex]
For the crate to be moving at a constant velocity, there is NO net force acting on the crate, so:
[tex]0 = F_T + F_f - F_g[/tex]
Now, we can express each force as an equation.
Force due to tension:
Must be solved for.Force due to gravity:
On an incline, this is equivalent to the SINE component of its weight. (Force of gravity is STILL THE WEIGHT, but on an incline, it contains a horizontal component that contributes to the net force)This is expressed as:
[tex]F_g = Mgsin\theta[/tex]
Force due to friction:
Equivalent to the normal force and coefficient of friction. The normal force is the VERTICAL component of the object's weight, so:[tex]N = Mgcos\theta[/tex]
[tex]F_f = \mu Mgcos\theta[/tex]
Now, plug these expressions into the above equation.
[tex]0 = F_T + \mu Mgcos\theta - Mgsin\theta\\\\Mgsin\theta - \mu Mgcos\theta = F_T[/tex]
Mg = 245 N (weight). Plug in all values:
[tex]245sin(20) - 0.270(245)cos(20) = F_T\\\\F_T = \boxed{21.634 N}[/tex]
The normal force is equivalent to the vertical component (PERPENDICULAR TO THE INCLINE) of the weight (cosine), so:
[tex]N = Mgcos\theta\\\\N = 245cos(20) = \boxed{230.225 N}[/tex]
To find the tension in the rope needed to make the crate move at a constant velocity, we need to consider the forces acting on the crate.
1. Tension in the rope (T): This force is applied upwards at an angle of 20 degrees above the horizontal.
2. Weight of the crate (W): This force acts downward vertically and is equal to 245 N.
3. Normal force (N): This force is exerted by the floor on the crate and acts perpendicular to the surface of the floor.
4. Kinetic friction force (f_k): This force opposes the motion of the crate and acts parallel to the surface of the floor. The magnitude of kinetic friction is given by: f_k = μ_k * N, where μ_k is the coefficient of kinetic friction (0.270 in this case).
Since the crate is moving at a constant velocity, the net force on the crate must be zero. In other words, the forces pulling the crate forward (T and the horizontal component of the weight) must balance the forces opposing its motion (kinetic friction). The vertical forces (the vertical component of the weight and the normal force) must also balance.
Let's proceed with the calculations:
Horizontal Forces:
T * cos(20°) - f_k = 0
T * sin(20°) + N - W = 0
We can now solve for T and N:
From the horizontal forces equation:
T * cos(20°) = f_k
vertical force:
T = f_k / cos(20°)
T = (0.270 * N) / cos(20°)
From the vertical forces equation:
T * sin(20°) + N = W
T * sin(20°) + N = 245 N
Now, substitute the expression for T from the horizontal forces equation:
(0.270 * N) / cos(20°) * sin(20°) + N = 245 N
Now, solve for N:
N * (0.270 * tan(20°) + 1) = 245 N
N = 245 N / (0.270 * tan(20°) + 1)
Now, plug in the values and calculate N:
N = 245 N / (0.270 * tan(20°) + 1)
N ≈ 208.75 N
Now, we can calculate the tension in the rope (T):
T = (0.270 * N) / cos(20°)
T ≈ (0.270 * 208.75 N) / cos(20°)
T ≈ 67.24 N
So, the tension in the rope needed to make the crate move at a constant velocity is approximately 67.24 N, and the normal force exerted by the floor on the crate is approximately 208.75 N.
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A sample of ore containing radioactive strontium 38Sr90 has an activity of 8.2 × 105 Bq. The atomic mass of strontium is 89.908 u, and its half-life is 28.8 yr (1 yr = 3.156 × 107 s). How many grams of strontium are in the sample?
From the activity values and the decay constant, the mass of of Strontium in the sample is:
[tex]1.62 × 10^{-7}g[/tex]
What is the decay constant of an element?The decay constant of an element is the probability of decay of a nucleus per unit time.
{λ = ln 2 / t1/2
where;
t1/2 is the half-life of the isotope.
The half-life is converted to seconds since the decay constant is asked in per seconds.
[tex]28.8 years = 28.8 × 3.156 × 10^{7} = 908928000 s \\ [/tex]
Hence;
[tex]λ = \frac{ln2}{90892800s} = 7.626 s^{-1}[/tex]
The activity of the element, A, the decay constant, λ and the number of nuclei, N are related as follows:
A = (–) λN[tex]N = \frac{8.25 ×10^{5}}{7.626×10^{-10}} = 1.082 × 10^{15} [/tex]
Molar mass of Strontium-90 is 90 g.
1 mole of Strontium-90 contains 6.02×10^23 nuclei.
The mass, m of Strontium in the sample is calculated:
[tex]m = 1.082 × 10^{15} × \frac{90 g}{6.02 × 10^{23}} = 1.62 × 10^{-7}g \\ [/tex]
Therefore, the mass of of Strontium in the sample is:
[tex]1.62 × 10^{-7} \: g[/tex]
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When ocean temperature rises,
a. excess gas is released into the atmosphere.
b. the ocean absorbs and holds carbon dioxide.
c. carbon dioxide is equalized between the ocean and atmosphere. d. no change takes place.
Answer:
see below
Explanation:
Kind of like when yo open a warm soda .....excess gas is released
Then, complete the riddle below by finding the matching number and writing the letter.
Newton's Second Law
44k el effe &
forces cause an object to accelerate.
3
10
Acceleration depends on I & I and M a ½ ½.
11 12
13
14 15
16
17
18
19
As the force increases, the
accelera
i on
20
21
29
30 31
increases, too.
Why did the artist paint on cement blocks instead of paper?
"
11
He wanted to create
Answer:
I have no clue I'm just trying to get points
Explanation:
:) sorry
Edwin Hubble discovered that:
Galaxies were moving away from us
Some stars got brighter and dimmer in patterns.
Triangulation was not a reliable method for measuring distances
All of the above.
Answer:
galaxies were moving away from us
Explanation:
i got 100% on my quiz
A car moving south speeds up from 10 m/s to 40 m/s in 15 seconds. What is the car’s acceleration?
Answer:
According to the question,
Initial velocity of car (at t=0) = 10 m/s
Final velocity of car (after 15 sec) = 40m/s
Time taken = 15 sec.
Now use, first equation of motion viz.
Explanation:
v = u + at.
40 = 10 + a *15
30 = a*15
a = 30/15
a = 2 m/sec square.
The cart is then pushed up the ramp, compressing the spring a distance A from equilibrium. When
released from rest, the cart oscillates up and down the ramp with a period of T₁. Assume that any friction
is negligible.
The spring is then replaced with a second spring that has a constant k2 = 2k₁, and the new period, T2, is
measured.
Determine the period of T2 in terms of T₁.
When the spring is replaced with a spring with double of initial spring constant k2 = 2k₁, the new period, T2, is [tex]\sqrt{\frac{1}{2} } \ T_1[/tex].
Period of the mass oscillation
The period of the oscillation of the mass or cart on the spring is given by the following formula;
[tex]T = 2\pi \sqrt{\frac{m}{k} } \\\\[/tex]
at a costant mass;
[tex]T_1\sqrt{k_1} = T_2\sqrt{k_2} \\\\T_2 = \frac{T_1\sqrt{k_1} }{\sqrt{k_2} }[/tex]
when spring constant is doubled, k2 = 2k1. the new period, T2 is determined as follows;
[tex]T_2 = \frac{T_1\sqrt{k_1} }{\sqrt{2k_1} } \\\\T_2 = \frac{T_1\sqrt{k_1} }{\sqrt{2} \times \sqrt{k_1} } \\\\T_2 = \frac{T_1}{\sqrt{2} }\\\\T_2 = \sqrt{\frac{1}{2} } \ T_1[/tex]
Thus, when the spring is replaced with a spring with double of initial spring constant k2 = 2k₁, the new period, T2, is [tex]\sqrt{\frac{1}{2} } \ T_1[/tex].
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A new band sensation is playing a concert and recording it for a live album to be released this summer. The band asks the sound mixer if he can autotune
the singer who has a habit of singing slightly lower than the note.
What would the mixer need to do to the sound wave? How would it affect the characteristics of the wave?
The sound mixer will need to increase the amplitude of the sound wave produced by the singer which will increase the loudness of the sound.
Amplitude of sound waveThe amplitude of a sound wave is the maximum vertical displacement of the sound wave.
The sound mixer will need to increase the amplitude of the sound wave produced by the singer.
The increase in the amplitude of the sound wave produced by the lower tune singer will result in increased loudness of the sound.
Thus, the sound mixer will need to increase the amplitude of the sound wave produced by the singer which will increase the loudness of the sound.
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What are the similarities of life science and applied science?
Answer~
Both life science and physical science helps to understand the natural habitat of the earth. Life science deals with the living thing; thus, it is a study of the organic world. On the other hand, physical science deals with nonliving things; thus, it is a study of the inorganic world.
Explanation:
Please help quickly!!!
State the basic rules to draw a vector quantity
Explanation:
i dont to i want to copy and paste on my printer
Holding a pen crosswise in your mouth is likely to improve your mood, a result predicted by
Answer:
the facial-feedback hypothesis
Explanation:
what is the mathematical equation for resistivity
Answer:
See below
Explanation:
rho = R A/l R = resistance A = cross sectional area l = length
North of Carthage is this, it is a shallow evaporate lake:
A. Rub al-Khali
B. Sebkhet Arina
C. Sebkhet Qaddafi
Sebkhet Arina
Explanation:
sorry if it's wrong need more information
What is the angular acceleration experienced by a uniform solid disc of mass 5-kg and radius 20 cm when a net
torque of 50 N.m is applied? Assume the disc spins about its center.
Answer:
7.07 radians per second
A piece of fine fiber with a diameter of =6.5 m is used to prop apart the edges of two perfectly flat 3.3-cm-long pieces of glass (see diagram). When the setup is illuminated from above with light of wavelength =590 nm , an interference pattern of alternating bright and dark bands will be seen in the reflected light. If the setup is viewed from high above, how many dark bands will be seen?
We need frequency:-
[tex]\\ \rm\Rrightarrow \nu=\dfrac{c}{\lambda}[/tex]
[tex]\\ \rm\Rrightarrow \nu=\dfrac{3\times 10^8m/s}{590\times 10^{-9}m}[/tex]
[tex]\\ \rm\Rrightarrow \nu=0.0051\times 10^{17}Hz[/tex]
[tex]\\ \rm\Rrightarrow \nu=5.1\times 10^{14}Hz[/tex]
What emits infrared radiation?
Question 5 options:
Humans
Non-living objects
Anything that contains atoms
Answer:
anything that contains atoms is the correct answer of given statement
Can anyone help me on this? It’s a water cycle.
Answer:
The water cycle shows the continuous movement of water within the Earth and atmosphere. It is a complex system that includes many different processes. Liquid water evaporates into water vapor, condenses to form clouds, and precipitates back to earth in the form of rain and snow.
Explanation:
The cone and the cylinder below have equal surface area.
2r
OA. True
B. False
B. False, the surface area of the cone and the cylinder is not the same.
What is the Surface area?The area or region that an object’s surface occupies is known as its surface area. Volume, on the other hand, refers to how much room an object has. Geometry has numerous shapes and dimensions, including spheres, cubes, cuboids, cones, cylinders, etc. Each form has its own volume and surface area.
If the slant height of the cone is 2r, then we can use the Pythagorean theorem to find the radius of the cone:
l² = (2r)² - r²
l² = 3r²
l = r√3
Using this value for the radius of the cone, the formula for its surface area becomes:
A(cone) = πrl + πr²
A(cone) = π(r)(r√3) + πr²
A(cone) = πr²(1 + √3)
For the cylinder, we can use the given height and radius to find its surface area:
A(cylinder) = 2πrl + 2πr²
A(cylinder) = 2π(r)(r) + 2π(r²)
A(cylinder) = 2πr²(1 + r)
Comparing the two surface area formulas, we can see that they are not equal in general, since each figure's coefficient in front of the second term (πr²) is different.
Therefore, we cannot conclude that the two figures have the same surface area based on the information.
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HELP!! ASAP I DONT KNOW HOW TO THIS
The fundamental frequency of a musical note is 330Hz. What is the frequency of the 3rd overtone?
The 3rd overtone or the third harmonic given that the fundamental frequency is 330Hz is 990Hz
What is an Harmonic?Hamonics defined as the number gotten by multiplying the nth overtone by the fundamental frequecency.
In summary, the expression for the nth Hamonic or overtone is given as
Fn = Fundamental Frequency*N
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what is electronic logic
Answer:
Digital logic is the basis of computing and many other electronic devices as well as control systems found in this continually advancing digital world.
(too short, sorry)