Answers
Answer:
Explanation:
While each of these can cause erosion and weathering, lightning is probably the least important as it occurs less frequently and affects a much smaller surface area when it strikes.
Wind is not very effective by itself, but it can carry abrasives which work to degrade rock surfaces. It covers a very large area at once so the net effect can be moderate to large especially desert areas where plants are not readily available to disrupt the flow.
Rain covers huge areas and is quite common.
Freezing/Thawing cycles cover large areas and are quite common in the temperate and arctic latitudes and even in tropical altitudes.
Attached is a photo taken atop Half Dome in Yosemite National Park showing two of thousands of divots in the rock there caused by lightning strikes. The current in the lightning heats the stone causing water trapped in it to flash to steam. The increased pressure inside the stone can overwhelm the material strength and blow rock chunks over a fairly good sized area. This is a fairly rapid weathering and erosion when it occurs, but that is typically limited to a few dozen days per year and occurs mostly on high ground where lightning is more likely to strike earth.
Related Questions
A car accelerates from rest at a rate of 8.1 m/s2. How much time does it take the car to travel a distance of 65 meters?
Answers
Answer:
ोोञककतकषगकषषमषघ
Explanation:
zzjjjfkgkzgkggkkkammmmmmatAllzy##pglCsg
An object floats in water with 40% of its volume submerged . a)If the object was placed in methanol with a density of 0.79g/cm^3, what percentage would be submerged? b)If it was placed in liquid carbon tetrachloride with a density of 1.58g/cm^3, what percentage would be submerged?
Answers
Let v be the object's volume. The object displaces 0.4v cm³ of water, which, at a density of about 0.997 g/cm³, has a weight of
b = (0.000997 kg/cm³) (0.4 v cm³) g ≈ 0.000391v N
(and b is also the magnitude of the buoyant force). Then the net force on the object while it's floating in water is
∑ F = b - mg = 0
so that b = mg, where mg is the object's weight. This weight never changes, so the object feels the same buoyant force in each liquid.
(a) In methanol, we have
b = 0.000391v N = (0.00079 kg/cm³) (pv cm³) g
where p is the fraction of the object's volume that is submerged. Solving for p gives
p = (0.000391 N) / ((0.00079 kg/cm³) g) ≈ 0.0505 ≈ 5.05%
(b) In carbon tetrachloride, we have
b = 0.000391v N = (0.00158 kg/cm³) (pv cm³) g
==> p ≈ 0.0253 ≈ 2.53%
For the following vector fields, find its curl and determine if it is a gradient field.
F= (4xz+ y^2) i + 2xyj+ 2x^2k
Answers
Given
[tex]\vec F = (4xz+y^2)\,\vec\imath + 2xy\,\vec\jmath + 2x^2\,\vec k[/tex]
its curl would be
[tex]\nabla\times\vec F = \left(\dfrac{\partial(2xy)}{\partial z} - \dfrac{\partial(2x^2)}{\partial y}\right)\,\vec\imath - \left(\dfrac{\partial(4xz+y^2)}{\partial z} - \dfrac{\partial(2x^2)}{\partial x}\right)\,\vec\jmath + \left(\dfrac{\partial(4xz+y^2)}{\partial y} - \dfrac{\partial(2xy)}{\partial x}\right)\,\vec k[/tex]
which reduces to the zero vector. Since the curl is zero, and [tex]\vec F[/tex] doesn't have any singularities, [tex]\vec F[/tex] is indeed a gradient field.
To determine what it is a gradient of, we look for a scalar function f(x, y, z) such that [tex]\nabla f = \vec F[/tex]. This entails solving for f such that
[tex]\dfrac{\partial f}{\partial x} = 4xz+y^2 \\\\ \dfrac{\partial f}{\partial y} = 2xy \\\\ \dfrac{\partial f}{\partial z} = 2x^2[/tex]
Integrate both sides of the first equation with respect to x, which gives
[tex]f(x,y,z) = 2x^2z+xy^2 + g(y,z)[/tex]
Differentiate both sides of this with respect to y :
[tex]\dfrac{\partial f}{\partial y} = 2xy = 2xy+\dfrac{\partial g}{\partial y} \\\\ \implies \dfrac{\partial g}{\partial y} = 0 \implies g(y,z) = h(z)[/tex]
Differentiate f with respect to z :
[tex]\dfrac{\partial f}{\partial z} = 2x^2 = 2x^2 + \dfrac{\mathrm dh}{\mathrm dz} \\\\ \implies \dfrac{\mathrm dh}{\mathrm dz} = 0\implies h(z) = C[/tex]
So it turns out that [tex]\vec F[/tex] is the gradient of
[tex]f(x,y,z) = 2x^2z+xy^2+C[/tex]
Calculus-based Physics I, can someone explain this to me?
My apologies for the broadness of my question. I especially don't understand the notation being used here, but I know this is about data collection, specifically standard deviation and standard error. I mostly need help with the data collection of multiple variables, the formulae for standard deviation and standard error make no sense to me.
I could also use some examples.
Answers
2: For a sample of data [tex]x_1,x_2,\ldots,x_N[/tex], the mean of this sample denoted by [tex]\overline x[/tex] is the sum of the data divided by the number of data points,
[tex]\overline x = \dfrac{x_1+x_2+\cdots+x_N}N = \displaystyle\frac1N\sum_{i=1}^N x_i[/tex]
As an example, consider [tex]x_1=-1[/tex], [tex]x_2=1[/tex], and [tex]x_3=3[/tex]. Then
[tex]\overline x = \dfrac{-1+1+3}3 = 1[/tex]
3: Standard deviation is a measure of how dispersed a given data sample is relative to the mean. Consult the plot: for a normal distribution, approximately 68% of it lies within 1 standard deviation of the mean, approx. 95% within 2 standard deviations, and approx. 99.7% within 3 standard deviations.
For instance, if the data is pulled from a normally distributed population with mean 0 and standard deviation 1, if you were to randomly select any data from the population, then 68% of the time it will fall in the range (-1, 1); 95% of the time it will fall within (-2, 2); 99.7% of the time it fall within (-3, 3).
To compute the standard deviation for a sample, for each [tex]x_i[/tex] in [tex]x_1,x_2,\ldots,x_N[/tex], you
• take the difference between [tex]x_i[/tex] and the mean [tex]\overline x[/tex]
• square this difference
• sum all the squared differences
• divide the sum by N - 1 (for a sample) or N (for a population)
• take the square root
Here the standard deviation is denoted [tex]\sigma^x_{N-1}[/tex], which I would read as "the sample standard deviation of the data x" - sample because of the N - 1 subscript.
Continuing with the previous example, we'd have
[tex]\sigma^x_{N-1} = \displaystyle \sqrt{\frac{\left(-1-1\right)^2+\left(1-1\right)^2+\left(3-1\right)^2}{3-1}} = \sqrt4 = 2[/tex]
4: Not much more to say here, the standard error is basically a measure of how accurate a given estimate is about the population based on the sample data. It's analogous to uncertainty in measuring length with a ruler, for instance.
In our example,
[tex]\alpha^x = \dfrac2{\sqrt3}[/tex]
5: If x, y, and z are random variables, then I suppose ρ is meant to denote a function of these random variables (so that ρ itself is just another random variable). For instance, you could have ρ = x + 3y - 2z. Then [tex]\overline\rho[/tex] is the sample mean of ρ.
I'm not entirely sure about the notation [tex]x(\overline x,\sigma^x_{N-1},\alpha^x)[/tex], but I suspect it's just referring to sample x with mean [tex]\overline x[/tex] and standard deviation [tex]\sigma^x_{N-1}[/tex] with standard error [tex]\alpha^x[/tex].
∆ρ is just the differential of ρ, essentially capturing how ρ changes with respect to small changes in x, y, and z. The expression you see here follows from the chain rule for differentiation.
The formula you see for [tex]\sigma^\rho_{N-1}[/tex] is the sample standard deviation of ρ. Think of ∆ρ as a vector with 3 components. Then [tex]\sigma^\rho_{N-1}[/tex] is the magnitude of this vector.
Similarly, [tex]\alpha^\rho[/tex] is the standard error for ρ, and corresponds to the magnitude of the vector whose components are the standard errors of x, y, and z.
In order for these statistics to make sense, each of x, y, and z must be samples of the same number of data. Say we take x as before [tex](x_1=-1,x_2=1,x_3=3)[/tex], along with [tex]y_1=0,y_2=4,y_3=-2[/tex] and [tex]z_1=-3,z_2=\frac12,z_3=10[/tex]. Suppose ρ = x + 3y - 2z. Then
• the sample means of y and z :
[tex]\overline y = \dfrac{0+4-2}3 = \dfrac23 \\\\ \overline z = \dfrac{-3+\frac12+10}3 = \dfrac52[/tex]
• the standard deviations of y and z :
[tex]\sigma^y_{N-1} = \sqrt{\dfrac{\left(0-\frac23\right)^2+\left(4-\frac23\right)^2+\left(2-\frac23\right)^2}{3-1}} = 2\sqrt{\dfrac73} \approx 3.06\\\\ \sigma^z_{N-1} = \sqrt{\dfrac{\left(-3-\frac52\right)^2+\left(\frac12-\frac52\right)^2+\left(10-\frac52\right)^2}{3-1}} = \dfrac{\sqrt{181}}2 \approx 6.73[/tex]
• the values of ρ :
[tex]\rho_1 = x_1+3y_1-2z_1 = -1+2\times0-2\times(-3) = 5 \\\\ \rho_2 = x_2+3y_2-2z_2 = 1+3\times4-2\times\dfrac12=12 \\\\ \rho_3 = x_3+3y_3-2z_3 = 3+3\times(-2)-2\times10 = -23[/tex]
• the sample mean of ρ :
[tex]\overline\rho = \dfrac{5+12-23}3 = -2[/tex]
• by the chain rule,
[tex]\Delta\rho = \Delta x+3\Delta y-2\Delta z[/tex]
so the standard deviation of ρ :
[tex]\sigma^\rho_{N-1} = \sqrt{\left(\sigma^x_{N-1}\right)^2 + \left(3\sigma^y_{N-1}\right)^2 + \left(-2\sigma^z_{N-1}\right)^2} \\\\\sigma^\rho_{N-1}= \sqrt{2^2 + 9\left(2\sqrt{\dfrac73}\right)^2 + 4\left(\dfrac{\sqrt{181}}2\right)^2} = \dfrac12\sqrt{\dfrac{703}3} \approx 7.65[/tex]
• the standard errors of y and z :
[tex]\alpha^y = \dfrac{2\sqrt{\frac73}}{\sqrt3} = \dfrac23\sqrt7 \approx 1.76 \\\\ \alpha^z = \dfrac{\frac{\sqrt{181}}2}{\sqrt3} = \dfrac12\sqrt{\dfrac{181}3} \approx 3.88[/tex]
• the standard error of ρ :
[tex]\alpha^\rho=\sqrt{\left(\alpha^x\right)^2+\left(3\alpha^y\right)^2+\left(-2\alpha^z\right)^2}\\\\\alpha^\rho=\sqrt{\left(\dfrac2{\sqrt3}\right)^2+9\left(\dfrac23\sqrt7\right)^2+4\left(\dfrac12\sqrt{\dfrac{703}3}\right)^2}=\sqrt{269}\approx16.40[/tex]
1. Desde un piso horizontal, un proyectil es lanzado con una velocidad inicial de 10 m/s formando 30o con la horizontal. Si consideramos que la aceleración de la gravedad es 10 m/s2 . Calcular: a) El tiempo que tarda en llegar al piso. b) La máxima altura que alcanza. c) ¿A qué distancia del punto de lanzamiento choca con el piso?
Answers
Un proyectil se define como cualquier objeto que traza una trayectoria parabólica. Los parámetros importantes en el movimiento de proyectiles son; Tiempo de vuelo, rango y altura máxima.
a) 1 segundo
b) 1,25 m
c) 10 m
Deje que el tiempo de vuelo sea T
T = 2usinθ / g
u = velocidad inicial
g = aceleración debida a la gravedad
θ = ángulo
T = 2 * 10 m / s * sin (30) / 10
T = 1 segundo
Sea la altura máxima H
H = u ^ 2 sin ^ 2 θ / 2g
H = (10) ^ 2 (sin30) ^ 2/2 * 10
Alto = 1,25 m
Sea el rango R
R = u ^ 2sin 2θ / g
R = (10) ^ 2 sin 2 (30) / 10
R = 8.66 m
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a) El tiempo que tarda el proyectil en llegar al piso es:
[tex]t_{vuelo}=1\: s[/tex]
b) La máxima altura alcanzada por el proyectil es:
[tex]y_{max}=1.25\: m[/tex]
c) La máxima distancia alcanzada por el proyectil es:
[tex]x_{max}=8.66\: m[/tex]
a)
Para calcular el tiempo que le toma al proyectil llegar al piso, se puede usar la siguiente ecuación de tiro parabólico.
[tex]y=y_{i}+v_{iy}t-0.5gt^{2}[/tex] (1)
Donde:
y es la altura finaly(i) es la altura inicialv(iy) es la velocidad inicial en el eje y g es la gravedad (10 m/s²)t el tiempoRecordemos que la componente de la velocidad incial (v(i)=10 m/s) en el eje y esta dada por:
[tex]v_{iy}=v_{i}sin(30)[/tex]
[tex]v_{iy}=10sin(30)[/tex]
Sabemos que y(i) = 0 y ademas haciendo que y = 0, sabremos el tiempo total de vuelo. De la ecuación (1):
[tex]0=0+10sin(30)t-0.5(10)t^{2}[/tex]
Despejando t.
[tex]10sin(30)=0.5(10)t[/tex]
[tex]t=\frac{10sin(30)}{5}[/tex]
El tiempo de vuelo sera:
[tex]t_{vuelo}=1\: s[/tex]
b)
Para calcular la máxima altura, usamos la siguiente ecuación.
[tex]v_{fy}^{2}=v_{iy}^{2}-2gy[/tex]
Si hacemos que veocidad final en y v(fy) sea 0, encontraremos la máxima altura.
[tex]0=v_{iy}^{2}-2gy_{max}[/tex]
[tex]0=(v_{i}sin(30))^{2}-2gy_{max}[/tex]
Despejamos y(max):
[tex]y_{max}=\frac{(v_{i}sin(30))^{2}}{2g}[/tex]
[tex]y_{max}=\frac{(10sin(30))^{2}}{2(10)}[/tex]
La máxima altura alcanzada por el proyectil será:
[tex]y_{max}=1.25\: m[/tex]
c)
Sabemo que la velocidad del proyectil en la dirección x es constante, por lo tanto, la ecuacion cinemática será:
[tex]x=vt[/tex]
La máxima distancia se determina con el tiempo de vuelo:
[tex]x_{max}=v_{ix}t_{vuelo}[/tex]
La componente de la velocidad en la direccion x es:
[tex]v_{ix}=v_{i}cos(30)[/tex]
Por lo tanto, el máximo desplazamiento será:
[tex]x_{max}=10cos(30)*1[/tex]
[tex]x_{max}=8.66\: m[/tex]
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A 67 kg man stands at the front end of a uniform boat of mass 179 kg and of length, L = 2.5 m. Assume there is no friction or drag between the boat and water.
(a) What is the location of the center of mass of the system when the origin of our coordinate system (i) on the man's original location (ii) on the back end of the boat? [4 marks]
(b) If the man walks from the front end to the back end of the boat, by how much is the boat displaced? [3 marks]
(c) Now consider the man and his friend with identical mass of 67 kg are rowing the boat on a hot summer afternoon when they decide to go for a swim. The man jumps off the front of the boat at speed 3 m/s and his friend jumps off the back at speed 4 m/s. If the boat was moving forward at 1.5 m/s when they jumped, what is the speed of the boat after their jump?
Answers
Answer:
Explanation:
ai. com=(67*0+179*1.25)/(179+67)=0.91m
ii. com=(67*2.5+179*1.25)/(179+67)=1.59m
b. 0.91=(67(2.5-d)+179(1.25-d))/(179+67)
d=0.68m
c. 1.5=(67*-3+67*-4+179*v)/(67+67+179)
v=5.24m/s
The answers to your question are ;
A) Location of the center of the mass of system when origin of coordinate system is
i) On the original location of man = 0.91 m
ii) On back end of boat = 1.59 m
B) The boat is displaced by = 0.68 m
C) speed of boat after their jump ( V ) = 5.24 m/s
Given data :
mass of man = 67 kg
mass of boat = 179 kg
length of boat = 2.5 m
assumptions : No friction/drag force
Solutions
A) determine location of center of mass
i) On the man's original location = 0
= ( mass of man * 0 + mass * 1.25 ) / ( mass of boat + mass of man)
= ( 0 + 179 * 1.25 ) / ( 179 + 67 )
= ( 0 + 223.75 ) / ( 246 ) = 0.9096 ≈ 0.91 m
ii) On the back end of the boat
= ( mass of man * length of boat + mass of boat * 1.25 ) / ( mass of boat + mass of man )
= ( 67 * 2.5 + 179 * 1.25 ) / ( 179 + 67 )
= ( 167.5 + 223.75 ) / ( 246 ) = 1.590 m
B) By How much is the boat displaced ( d )
0.91 ( location of center of mass from man's original location )
0.91 = ( 67 * (2.5 - d) + 179 ( 1.25 - d ) ) / ( 179 + 67 )
= ( 67 * ( 2.5 - d ) + 223.75 - 179d )) / ( 246 )
∴ d ≈ 0.68 m
C) Determine speed of boat after the jump
Initial speed of boat = 1.5 m/s
hence speed after the jump ( v )
speed of first man = 3 m/s
speed of second man = 4 m/s
1.5 m/s = ( 67 * -3 + 67 * -4 + 179 * v ) / ( 67 + 67 + 179 )
= ( - 201 + -268 + 179v ) / ( 313 )
∴ v ≈ 5.24 m/s
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Which of the following phrases best describes the term scientific model?
A. The application of scientific knowledge to make predictions about
an object, system, or process
B. An experiment in which variables are controlled
C. A physical copy of a scientific object, system, or process
D. A simplified representation used to explain or make predictions
about something
SUBMIT
Answers
Stella's respiratory system is not working well why is this a problem for her ability to exercise
Answers
Explanation:
Her cells will not work well when they have low levels of oxygen.
Hey any physicist or engineer around. am giving brainliest to anyone who will answer this question.
Answers
Answer:
N = 167 Newtons
R = 727 Newtons
Explanation:
i) For static equilibrium, moments about any convenient point must sum to zero.
A moment is the product of a force and a moment arm length. Only the force acting perpendicular to a moment arm passing through the pivot point makes a moment.
ii) I will ASSUME the two moment arms are 0.05m and 0.15 m
CCW moments about the fulcrum are
190 N(0.2 m) + 280 N(0.05 m) = 52 N•m
CW moments are (N)N(0.15 m + 90 N(0.3 m) = 27 + 0.15N N•m
For static equilibrium, these must be equal
27 + 0.15N = 52
0.15N = 25
N = 166.6666666...
Sum moments about N to zero
(Same as saying CW and CCW moments must balance)
190(0.2 + 0.15) + 280(0.05 + 0.15) - R(0.15) - 90(0.3 - 0.15) = 0
R = 726.6666666...
We could verify this by summing vertical forces to zero.
R - 190 - 280 - 166.666666 - 90 = 0
R = 726.6666666...
Two 22.7 kg ice sleds initially at rest, are placed a short distance apart, one directly behind the other, as shown in Fig. 1. A 3.63 kg cat, standing on the left sled, jumps across to the right one and immediately comes back to the first. Both jumps are made horizontally at a speed of 3.05 m/s relative to the ice. Ignore the friction between the sled and ice.
(a) Find the final speeds of the two sleds. [6 marks]
(b) Calculate the impulse on the cat as it lands on the right sled. [2 marks]
(c) Find the average force on the right sled applied by the cat while landing. Consider that the cat takes 12 ms to finish the landing.
Answers
Newton's third law of motion sates that force is directly proportional to the rate of change of momentum produced
(a) The final speeds of the ice sleds is approximately 0.49 m/s each
(b) The impulse on the cat is 11.0715 kg·m/s
(c) The average force on the right sled is 922.625 N
The reason for arriving at the above values is as follows:
The given parameters are;
The masses of the two ice sleds, m₁ = m₂ = 22.7 kg
The initial speed of the ice, v₁ = v₂ = 0
The mass of the cat, m₃ = 3.63 kg
The initial speed of the cat, v₃ = 0
The horizontal speed of the cat, v₃ = 3.05 m/s
(a) The required parameter:
The final speed of the two sleds
For the first jump to the right, we have;
By the law of conservation of momentum
Initial momentum = Final momentum
∴ m₁ × v₁ + m₃ × v₃ = m₁ × v₁' + m₃ × v₃'
Where;
v₁' = The final velocity of the ice sled on the left
v₃' = The final velocity of the cat
Plugging in the values gives;
22.7 kg × 0 + 3.63 × 0 = 22.7 × v₁' + 3.63 × 3.05
∴ 22.7 × v₁' = -3.63 × 3.05
v₁' = -3.63 × 3.05/22.7 ≈ -0.49
The final velocity of the ice sled on the left, v₁' ≈ -0.49 m/s (opposite to the direction to the motion of the cat)
The final speed ≈ 0.49 m/s
For the second jump to the left, we have;
By conservation of momentum law, m₂ × v₂ + m₃ × v₃ = m₂ × v₂' + m₃ × v₃'
Where;
v₂' = The final velocity of the ice sled on the right
v₃' = The final velocity of the cat
Plugging in the values gives;
22.7 kg × 0 + 3.63 × 0 = 22.7 × v₂' + 3.63 × 3.05
∴ 22.7 × v₂' = -3.63 × 3.05
v₂' = -3.63 × 3.05/22.7 ≈ -0.49
The final velocity of the ice sled on the right = -0.49 m/s (opposite to the direction to the motion of the cat)
The final speed ≈ 0.49 m/s
(b) The required parameter;
The impulse of the force
The impulse on the cat = Mass of the cat × Change in velocity
The change in velocity, Δv = Initial velocity - Final velocity
Where;
The initial velocity = The velocity of the cat before it lands = 3.05 m/s
The final velocity = The velocity of the cat after coming to rest =
∴ Δv = 3.05 m/s - 0 = 3.05 m/s
The impulse on the cat = 3.63 kg × 3.05 m/s = 11.0715 kg·m/s
(c) The required information
The average velocity
Impulse = [tex]F_{average}[/tex] × Δt
Where;
Δt = The time of collision = The time it takes the cat to finish landing = 12 ms
12 ms = 12/1000 s = 0.012 s
We get;
[tex]F_{average} = \mathbf{\dfrac{Impulse}{\Delta \ t}}[/tex]
∴ [tex]F_{average} = \dfrac{11.0715 \ kg \cdot m/s}{0.012 \ s} = 922.625 \ kg\cdot m/s^2 = 922.625 \ N[/tex]
The average force on the right sled applied by the cat while landing, [tex]\mathbf{F_{average}}[/tex] = 922.625 N
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Scientists use models to represent physical situations that are difficult to explore
firsthand.
True
False
Answers
Answer:
I believe it is True
Explanation:
Help meh in this question plzzz
Answers
The Moment of Inertia of the Disc is represented by [tex]I = \frac{15}{32}\cdot M\cdot R^{2}[/tex]. (Correct answer: A)
Let suppose that the Disk is a Rigid Body whose mass is uniformly distributed. The Moment of Inertia of the element is equal to the Moment of Inertia of the entire Disk minus the Moment of Inertia of the Hole, that is to say:
[tex]I = I_{D} - I_{H}[/tex] (1)
Where:
[tex]I_{D}[/tex] - Moment of inertia of the Disk.[tex]I_{H}[/tex] - Moment of inertia of the Hole.Then, this formula is expanded as follows:
[tex]I = \frac{1}{2}\cdot M\cdot R^{2} - \frac{1}{2}\cdot m\cdot \left(\frac{1}{2}\cdot R^{2} \right)[/tex] (1b)
Dimensionally speaking, Mass is directly proportional to the square of the Radius, then we derive the following expression for the Mass removed by the Hole ([tex]m[/tex]):
[tex]\frac{m}{M} = \frac{R^{2}}{4\cdot R^{2}}[/tex]
[tex]m = \frac{1}{2}\cdot M[/tex]
And the resulting equation is:
[tex]I = \frac{1}{2}\cdot M\cdot R^{2} -\frac{1}{2}\cdot \left(\frac{1}{4}\cdot M \right) \cdot \left(\frac{1}{4}\cdot R^{2} \right)[/tex]
[tex]I = \frac{1}{2} \cdot M\cdot R^{2} - \frac{1}{32}\cdot M\cdot R^{2}[/tex]
[tex]I = \frac{15}{32}\cdot M\cdot R^{2}[/tex]
The moment of inertia of the Disc is represented by [tex]I = \frac{15}{32}\cdot M\cdot R^{2}[/tex]. (Correct answer: A)
Please see this question related to Moments of Inertia: https://brainly.com/question/15246709
Answer:
Help meh in this question
Explanation:
A plane leaves Houston and flies 400 km north to Dallas. The pilot realizes he has forgotten his
golf clubs and returns to Houston to pick them up on the way back to Houston, the plane runs
out of gas and is forced to land in Huntsville, 100 km North of Houston. The trip from Houston to
Dallas took 120 min and the trip from Dallas to Huntsville took 70 min.
A. What is the total distance (in meters) covered by the airplane?
B. What is the plane's total displacement (in meters)?
C What is the average velocity for: 190 MM
i Houston to Dallas
ii. Dallas to Huntsville
Answers
(A) The total distance covered by the plane is 500,000 m
(B) The plane's total displacement is 300,000 m
(C) The average velocity of the plane is 100 m/s
The given parameters:
initial displacement of the plane = 400 km Dallas
final displacement of the train = 100 km Huntsville
the time taken for the initial displacement = 120 min
final time for the second trip = 70 min
A sketch of the plane's trip is as follows:
Dallas
400 km ↑ ↓
↑ ↓ Huntsville 100 km
↑
↑
Houston
(A) The total distance covered by the plane:
The total distance is the total path travelled by the plane.
Total distance = 400 km + 100 km = 500 km, = 500,000 m
(B) The plane's total displacement:
The total displacement is the change in the position of the plane.
The displacement = 400 km - 100 km = 300 km = 300,000 m
(C) The average velocity of the plane
[tex]Average \ velocity = \frac{change \ in \ displacement }{change \ in \ time} \\\\Average \ velocity = \frac{400 \ km - 100 \ km}{120 \ \min - \ 70 \ \min} \\\\Average \ velocity =\frac{300 \ km}{2\ hr - 1.167 \ hr} \\\\Average \ velocity = \frac{300 \ km}{0.833 \ hr} \\\\Average \ velocity = 360.14 \ km/hr\\\\Average \ velocity = \frac{360.14 \ km/h}{3.6 \times \frac{km/hr}{m/s} } = 100 \ m/s[/tex]
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what is a molecule??
Answers
Answer:
molecule, a group of two or more atoms that form the smallest identifiable unit into which a pure substance can be divided and still retain the composition and chemical properties of that substance
Explanation:
Here are examples of common molecules:
H2O (water)
N2 (nitrogen)
O3 (ozone)
CaO (calcium oxide)
C6H12O6 (glucose, a type of sugar)
NaCl (table salt)
Explanation:
molecule, a group of two or more atoms that form the smallest identifiable unit into which a pure substance can be divided and still retain the composition and chemical properties of that substance
A Sling Shot accelerates a 12 g Stone to a velocity of
35m/s within a distance of 5.0cm to what (constant force
is the stone subjected during the acceleration
Answers
Answer:
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Answer:
[tex]\huge\boxed{\sf F = 147 N}[/tex]
Explanation:
Given:
Mass = m = 12 g = 0.012 kg
Final Velocity = Vf = 35 m/s
Initial Velocity = Vi = 0 m/s
Distance = S = 5 cm = 0.05 m
Required:
Force = F = ?
Formula:
3rd equation of motion
2aS = Vf²-Vi²
Solution:
2aS = Vf²-Vi²
a = (35)²- (0)² / 2S
a = 1225 / 2(0.05)
a = 1225 / 0.1
a = 12250 m/s²
Force = Mass x Acceleration
F = 0.012 x 12250
F = 147 N
[tex]\rule[225]{225}{2}[/tex]
Hope this helped!
~AH1807Peace!Use the following information to calculate the speed of sound. A 2m long tube of air, closed at one end and open at the other, is excited using a speaker. Successive resonances/harmonics are observed to occur at 129, 215 and 301 Hz (Note: 129 Hz is not necessarily the fundamental frequency). Make sure you clearly show how you get your answer
Answers
Successive resonances/harmonics are observed to occur at 129, 215 and 301 Hz (Note: 129 Hz is not necessarily the fundamental frequency). the speed of sound is 76 added to 129 is 215
what is sound ?The pitch of the sound is a major property where the frequency of sound occur by human ear within the range of human hearing.
It is higher than the frequency of the sound where the higher is its pitch and a lower frequency means a lower pitch.
The loudness of sound is another property which can be determined by amplitude of the sound is a measure of the magnitude of the maximum disturbance of sound.
The speed is another property can be detrained as the sound waves which can travel through the medium is called the speed of sound and it is different for different mediums, travels fastest in solids.
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the precision of interferometer of wavelength of light 800 nm would be: (a) 200 nm (b) 100 nm (c) 400 nm (d) 800 nm
Answers
Answer:
Explanation:
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A footballer kicks a ball at an angle of 45° with the horizontal. If the ball was in the air
for 10 s and lands 4000 m away determine its initial speed.
Answers
Answer:
HOPE THIS ANSWER WILL HELP YOU
1. A car moving with a velocity of 10 m s-1 accelerates uniformly at 1 m s-2 until it reaches a velocity of 15 m s-1. Calculate (i) the time taken (ii) the distance traveled during the acceleration (ui) the velocity reached 100 m from the place where the acceleration began. m/sec
Answers
v (final velocity) = 15ms-1
a = 1ms-2
(i)
a = v-u/t
Rearranging the above equation
t = v - u / a
t = 15 - 10 / 1
t = 5 / 1 = 5 s
t = 5s
(ii)
Distance travelled = s = ut + 1/2 at^2
s = 10 * 5 + 1/2 * 1 * 5 * 5
s = 50 + 12.5
s = 62.5m
Umm I don’t really get the last question. Soo can you retype the question in the comment section so that I can answer it.
Anyways, I you have any doubts plz let me know.
differentiate between air pressure and liquid pressure in table .
Answers
Air pressure or atmospheric pressure is the pressure as the force exerted by the collisions of particles in the air.
The key difference between air pressure and liquid pressure is that air pressure allows the gaseous state of matter to be compressible, whereas liquid pressure makes a liquid incompressible
Answer:
Air pressureAtmospheric pressure, also known as barometric pressure, is the pressure within the atmosphere of Earth. The standard atmosphere is a unit of pressure defined as 101,325 Pa, which is equivalent to 760 mm Hg, 29.9212 inches Hg, or 14.696 psi.
Liquid pressureLiquid pressure is the increase in pressure at increasing depths in a liquid. This pressure increases because the liquid at lower depths has to support all of the water above it. We calculate liquid pressure using the equation liquid pressure = mass x acceleration due to g density x depth in fluid.
Two long straight parallel wires seperated by 20cm apart carry current of 20A and 1OA in opposite directions. What is the magnitude of the force on 1m length of the 20A wire
Answers
Answer:
Force is 2 × 10^-4 N
Explanation:
[tex]{ \bf{F = \frac{ \gamma_{o} I _{1} I _{2} l}{2\pi r} }}[/tex]
F is force
I is current
r is separation distance
gamma is a constant
l is wire length
[tex]F = \frac{(4\pi \times {10}^{ - 7}) \times (20) \times (10) \times 1 }{2\pi \times 0.2} \\ \\ F = 2 \times {10}^{ - 4} \: \: newtons[/tex]
The magnitude of the force on 1m length of the 20A wire is 2×10⁻⁴N.
To determine the answer, we need to know about magnetic field due to a current carrying wire and magnetic force.
What is the magnetic field produced due to a straight current carrying wire of current I₁?The magnetic field due to a straight wire having current I₁ at a perpendicular distance (say) 'd' is μ₀I₁ / 2[tex]\pi[/tex]d.
What is the magnetic force experienced by a straight wire of current I₂?Magnetic force on a current carrying wire of length 'L' and having current I₂ is I₂(L×B). Where B is the magnetic field at that wire.
What is the magnetic field on 20A wire due to 10A wire?d= 0.2m and I₁= 10AB= μ₀I₁ / 2[tex]\pi[/tex]d = 4[tex]\pi[/tex] ×10⁻⁷× 10/ 2[tex]\pi \\[/tex]×0.2 = 10⁻⁵ TWhat is the force on 20A wire?Current (I₂) = 20A, B = 10⁻⁵T and L= 1mMagnitude of force on the 20 A wire of length 1m = I₂×L×B( Here L and B are perpenicular to each other so L×B= L·B)
= 20×1×10⁻⁵
=2×10⁻⁴N
Thus, we can conclude that the magnitude of the force on 1m length of the 20A wire is 2×10⁻⁴N.
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of a ball traveling 24 m in 0.6 seconds.
A baseball pitcher throws a fastball across home plate. Calculate the speed of a
baseball that takes 0.01 seconds to travel 0.30 meters across the entire length of home
plate.
Answers
Answer:
Below
Explanation:
To find the speed of an object you can use this formula
speed = displacement / elapsed time
For the first one
24 m / 0.6 seconds = 40 m/s
For the second one
0.3 m / 0.01 seconds = 30 m/s
Hope that helped
should people eat animals?
Answers
Answer:
I hope it's helpful for you ☺️
Answer:
Nó
THEY FEEL PAIN. And dies.what is population compostion
Answers
Answer :
Population composition is the description of a,population according to characteristics such as age and sex . These data are often compared over time using population pyramids.
A mass of 10kg is suspended from the end of a steel rod of length 2m and radius 1mm. What is the elongation of the rod beyond it's original length (Take E=200*10⁹Nm²)
Answers
Answer:
don't know what class are you which subject is this
b. Describe in general how terminator devices capture the power of waves. In particular, explain how the oscillating water column works. (3 points)
Answers
Answer:
bnkjlji
Explanation:
Which of the following is NOT a physical property?
Reactivity
Density
Conductivity
Malleability
Answers
Answer:
Reactivity
Explanation:
should money be used for space travel when there are so many serious problems on Earth to be addressed? explain your answer.
Answers
Answer:
Anyway, space exploration absolutely does give us a direct benefit. When space technology has advanced far enough, we will be able to leave this planet in large numbers and live among the stars. This will solve our population/environment/resource/energy problems for a long, long time. Even a fraction of the money spent annually on space exploration could save millions of people in poverty-stricken countries, and improve living conditions for future generations. The foundations of the world we live in are largely based on science and it is indeed vital to extend our knowledge of the universe.
Thank You
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multiples and submultiples of kg
Answers
Answer:
Submultiples of kilogram are decagram ,centigram etc. And multiples of kilogram is megagram(1 tonne) , gigagram etc.
hope this helps you:)
What is motion please explain with diagram
Answers
Answer:
An object in a motion when it is continuously changing its position based on a reference point and observed by a person or a device.