A - B = (10i - 2j - 4k) - (i + 7j - k)
A - B = 9i - 9j - 3k
|A - B| = √(9² + (-9)² + (-3)²) = √189 = 3√19
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.
Answer:
HOPE THIS ANSWER WILL HELP YOU
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
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.
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]
Two unequal masses m and 2m are attached to a thin bar of negligible mass that rotates about an axis perpendicular to the bar. When m is a distance 2d from the axis and 2m is a distance d from the axis, the moment of inertia of this combination is "I". If the masses are now interchanged, the moment of inertia will be:_______
A) 2/3 I,
B) I ,
C) 3/2 I ,
D) 2 I ,
E) 4 I
Answer:
C) 3/2 I
Explanation:
Initial condition
I = m(2d)² + 2m(d²)
I = 4md² + 2md² = 6md²
swap masses
I' = 2m(2d)² + m(d²)
I' = 8md² + md² = 9md²
I'/I = 9md²/6md² = 3/2
I' = 3/2I
Scientists use models to represent physical situations that are difficult to explore
firsthand.
True
False
Answer:
I believe it is True
Explanation:
should people eat animals?
Answer:
I hope it's helpful for you ☺️
Answer:
Nó
THEY FEEL PAIN. And dies.For the following vector fields, find its curl and determine if it is a gradient field.
F= (4xz+ y^2) i + 2xyj+ 2x^2k
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]
Which of the following is NOT a physical property?
Reactivity
Density
Conductivity
Malleability
Answer:
Reactivity
Explanation:
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?
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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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?
Answer:
ोोञककतकषगकषषमषघ
Explanation:
zzjjjfkgkzgkggkkkammmmmmatAllzy##pglCsg
b. Describe in general how terminator devices capture the power of waves. In particular, explain how the oscillating water column works. (3 points)
Answer:
bnkjlji
Explanation:
What is a Standard Unit? Give 3 examples
Answer:
The standard (metric) units that would be discussed at primary school would include: grams and kilograms, centimetres, metres and kilometres, millilitres and litres (though children also learn about imperial units in Year 5 maths).
Answer:
Standard units are commonly used units of measurement. For an example; grams and kilograms, centimetres.
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
Answer:
Please mark brainliest
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!câu 1 vẽ sơ đồ trình bày nguyên lý làm việc các thông số và đặc điểm của mạch chỉnh lưu cầu 1 pha
câu 2 vẽ sơ đồ trình bày nguyên lý làm việc của bộ nghịch lưu áp 1 pha
Answer:hola puto
Explanation:cabroné
1. A leftward force is used to pull a large wagon. In the wagon there is an unknown mass
that makes it very difficult to move, but a force applied is able to set the wagon into
constant motion. Let the force applied to the wagon and its unknown mass be 77.3 N. The
coefficient of friction between the wagon's wheels and the floor is 0.011. Determine the
mass of the wagon if it was empty.
i. Let's say that the unknown mass in the wagon is 95.5% * mass of the
wagon. What is the weight of the wagon when its empty and what is the
weight of the unknown mass?
Answer:
Wagon weight 366.41314(9.81) = 3,594.5129 = 3590 N
Load weight 349.9245(9.81) = 3,432.75982 = 3430 N
Explanation:
Let's assume the surface and the applied force are horizontal.
At constant motion, the friction force must balance the applied force
(from F = ma)
Ff = 77.3
μΝ = 77.3
μmg = 77.3
m = 77.3 / (0.011(9.81))
m = 716.3376... kg
Let M be the mass of the wagon
M + 0.955M = 716.3376
M(1.955) = 716.3376
M = 366.41314...kg
0.955 M = 349.9245... kg
round to the three significant digits of the question numerals.
A probability of breaking a tree is more than a short one during storm.why?
Answer:
Due to interia of motion
What is motion please explain with diagram
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.
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.
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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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)
Answer:
don't know what class are you which subject is this
You push yourself on a skateboard with a force of 30 N east and accelerate at 0.5 m/s² east. Find the mass of theskateboard if your mass is 58 kg.
Answer:
2Kg
first you find the total mass of both the person und skateboard
f = m*a
30 = m* 0.5
m= 30/ 0.5
m= 60 kg
then you subtract the mass of the person from the total mass
m1 - m2
60 - 58
= 2 kg
Taking into account the Newton's second law, the mass of the skateboard is 2 kg.
Newton's Second Law establishes the relationship between the forces acting on an object and the acceleration it experiences. This law says that the acceleration that an object experiences when subjected to a net force (force applied to the body) is proportional to that force.
Mathematically, Newton's second law is expressed as:
F= m×a
where:
F = Force [N] m = Mass [kg] a = Acceleration [m/s²]In this case, you know:
F= 30 N m= ? a= 0.5 m/s²Replacing:
30 N= m× 0.5 m/s²
Solving:
m= 30 N÷0.5 m/s²
m= 60 kg
The mass of the sistem is the sum of the mass of the skateboard and the mass of the mass of your body:
Mass= mass of the skateboard + mass of your body
In this case:
60 kg= mass of the skateboard + 58 kg
Solving:
mass of the skateboard= 60 kg - 58 kg
mass of the skateboard= 2 kg
Finally, the mass of the skateboard is 2 kg.
Learn more about the Newton's second law:
brainly.com/question/23845187?referrer=searchResults brainly.com/question/13959891?referrer=searchResultsFind voltage in the coil.
(B) 1.58×10³ V
Explanation:
If you divide the input voltage by the number of turns in the primary coil, the result will be equal to the output voltage divided by the number of turns in the secondary coil:
Ui/Np = Uo/Ns
83/250 = Uo/4750 multiply both sides by 4750
394250/250 = Uo
Uo = 1577 = 1.58×10³ V
Answer:
if the input effective voltage is 1,58vHow much time does a train take to travel1800 km if it's velocity is 90km/hr
Explanation:
Time= Distance÷Velocity
Time= 1800÷90
Time= 20 hr
20×60= 1200 min
what are the different systemsof units. plz help me don't scam
fundamental UNIT AND DERIVED UNIT
Explanation:
THERE ARE TWO TYPE OF UNIT
1 FUNDAMENTAL UNIT = THE UNIT OF PHYSICAL QUANTITY LIKE METRE SECOND ETC IS FUNDAMENTAL UNIT.
2 DERIVED UNIT=THE UNIT OF DERIVED PHYSICAL QUANTITY LIKE PASCAL ,NEWTON ETC IS DERIVED UNIT
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
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.
For more details sound, visit
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#SPJ5
what is population compostion
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.
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?
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%
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?
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]
Puedes aprender mas sobre tiro parabólico en:
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This is a model that describes how much energy is transferred from one trophic
level to the next.
energy pyramid
food web
food chain
trophic level
Se lanza un dado 19 veces con las siguientes lecturas 5,1,3,3,6,2,6,4,5,2,1,2,5,3,2,6,1,4,4
Cuál es el promedio de las lecturas obtenidas?
Cuál es la tirada que más se repite (moda)?
Cuál es el mayor valor intermedio de todas las lecturas, ordenas estás de menor a mayor (mediana)?
Teniendo en cuenta la definición de media, mediana y moda:
el promedio de las lecturas obtenidas es 3,42la tirada que más se repite es 2.el mayor valor intermedio de todas las lecturas es 3.La media (también llamada promedio o media aritmética) de un conjunto de datos es una medida de posición central cuyo valor se obtiene al dividir la suma de todos los números entre la cantidad de ellos. Entonces, en este caso, el promedio de las lecturas obtenidas se calcula como:
[tex]Promedio=\frac{5+1+3+3+6+2+6+4+5+2+1+2+5+3+2+6+1+4+4}{19}[/tex]
Resolviendo:
Promedio= 3,42
El promedio de las lecturas obtenidas es 3,42.
Por otro lado, la moda de un conjunto de números es el número que aparece más a menudo dentro del grupo de datos.
En este caso, podes observar que la cantidad de veces que aparece cada número es:
1: 3 veces2: 4 veces3: 3 veces4: 3 veces5: 3 veces6: 3 vecesDebido a que el 2 es el número con mayor cantidad de repeticiones, la tirada que más se repite es 2.
Por último, la mediana es valor que se encuentra en la mitad justa entre los valores máximo y mínimo de los datos con los que se está trabajando; es decir, que al ordenar los número de menor a mayor, éste se encuentra justamente en medio de los valores.
En este caso, ordenando de menor a mayor las lecturas del dado:
1,1,1,2,2,2,2,3,3,3,4,4,4,5,5,5,6,6,6
se observa que el valor el mayor valor intermedio de todas las lecturas es 3.
Aprende más con estos ejemplos:
https://brainly.com/question/24086990?referrer=searchResultshttps://brainly.com/question/15969676?referrer=searchResultshttps://brainly.com/question/17010976?referrer=searchResultsI need some help, this is going to be late.
Answer:
gravitational force
Explanation:
since they seem like they are falling, gravity usually pulls them down
i think its gravitational force