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
A car has four tires that are each inflated to an absolute pressure of 2.0 x 10^5 Pa. Each tire has an area of 0.024 m^2 in contact with the ground. How much does the car weigh?
a.
1.9 x 10^3 N
b.
2.9 x 10^4 N
c.
1.2 x 10^3 N
d.
1.9 x 10^4 N
Answer:
If u dig a hole 3 feet how big is that hole
Explanation:
It’s about 20 feet
what is a molecule??
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
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]
Water is flowing in a pipe with a varying cross-sectional area, and at all points the water completely fills the pipe. At point 1 the cross-sectional area of the pipe is 0.070 m^2, and the magnitude of the fluid velocity is 3.50 m/s.
Required:
a. What is the fluid speed at point in the pipe where the cross-sectional area is 0.105m^2?
b. What is the fluid speed at point in the pipe where the cross-sectional area is 0.047m^2?
c. Calculate the volume of water discharged from the open end of the pipe in 1.00hour.
Answer:
b
Explanation:
Hey any physicist or engineer around. am giving brainliest to anyone who will answer this question.
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...
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.
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 money be used for space travel when there are so many serious problems on Earth to be addressed? explain your answer.
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
Please mark me brainliest
If Leslie has a hang time of 0.2 seconds, what distance did she jump?
Answer:
um.......I don't know soory
The speed of light will be minimum while passing through
A) water
B) vaccum
C) air
D) glass
Answer:
The correct choice is option D. glass
Glass's optical density is very great. Which is why speed of light will be minimum when passing glass.
Answer:
glass
Explanation:
i think, sana makatulong
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
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
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
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.
Learn more about magnetic force here:
https://brainly.com/question/25932320
#SPJ2
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
(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]
Learn more here: https://brainly.com/question/12409152
Stella's respiratory system is not working well why is this a problem for her ability to exercise
Explanation:
Her cells will not work well when they have low levels of oxygen.
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:
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
An astronaut floating in space is trying to use her jetpack to get back to her space station, but she is being pulled away by a nearby planet, as shown in the image below. Her jetpack provides a constant thrust of 165 N. If she angles her jetpack in such a way that it cancels out the vertical force due to the planet's gravity, what is her net horizontal force?
A. 132.6 N toward the space station
B. 53.3 N away from the space station
C. 98.5 N toward the space station
D. 112.8 N away from the space station
Answer:
C. 98.5N toward the space station
Explanation:
Fx= Fjet×cos(20) - Fplanet × cos(45) = 98.5
Because the result is positive so she is going toward the space station
If a horse can trot with an average velocity of 13 km/h, how far can it travel in
8 hours?
A. 2 km
B. 104 km
C. 35 km
D. 78 k
Answer:
104 km
Explanation:
Displacement = Velocity x Time
= 13 x 8
= 104 km
should people eat animals?
Answer:
I hope it's helpful for you ☺️
Answer:
Nó
THEY FEEL PAIN. And dies.the precision of interferometer of wavelength of light 800 nm would be: (a) 200 nm (b) 100 nm (c) 400 nm (d) 800 nm
Answer:
Explanation:
Did this come with a Picture?
An eccentric emu runs 20 m/s for 5 minutes for the first part of his trip to Hollywood. Once tired, the emu runs slower speed for the next hour. The average velocity of the emu is 15 m/s. what speed was the emu running when he was tired?
Answer:
14.6 m/s
Explanation:
The total run time was 5 + 60 = 65 minute or 65(60) = 3900 s
At his average velocity, emu ran 15 m/s(3900 s) = 58,500 m
Which is a heck of a running distance for ANY emu.
In the first 5 minutes the emu traveled 20 m/s(5 min)(60 s/min) = 6000 m
So in the last hour (3600 s) the emu traveled 58,500 - 6000 = 52,500 m
at a speed of 52,500 m /3600 s = 14.583333333... m/s
The emu was running at a speed of 14.58 m/s when he was tired.
To solve this question, we'll begin by calculating the distance travelled in the first part of the trip. This can be obtained as follow:
Time (t₁) = 5 min = 5 × 60 = 300 s
Speed 1 (S₁) = 20 m/s
Distance 1 (d₁) =?Speed = distance / time
S₁ = d₁ / t₁
20 = d₁ / 300
Cross multiply
d₁ = 20 × 300
Distance 1 (d₁) = 6000 mNext, we shall determine the total distance travelled by the emu.
Average speed = 15 m/s
Time 1 (t₁) = 300 s
Time 2 (t₂) = 1 h = 60 mins = 60 × 60 = 3600 s
Total time (T) = t₁ + t₂ = 300 + 3600 = 3900 s
Total distance (D) =?Average speed = Total distance / total time
15 = D / 3900
Cross multiply
D = 15 × 3900
Total distance (D) = 58500 mNext, we shall determine the distance travelled in the second part (i.e when he was tired) of the trip.
Total distance (D) = 58500 m
Distance 1 (d₁) = 6000 m
Distance 2 (d₂) =?D = d₁ + d₂
58500 = 6000 + d₂
Collect like terms
58500 – 6000 = d₂
Distance 2 (d₂) = 52500 mFinally, we shall determine the speed of the emu in the second part of the trip.
Distance 2 (d₂) = 52500 m
Time 2 (t₂) = 3600 s
Speed 2 (S₂) =?Speed = distance / time
S₂ = 52500 / 3600
S₂ = 14.58 m/sTherefore, the emu was running at a speed of 14.58 m/s when he was tired.
Learn more: https://brainly.com/question/24502267
Which of the following is NOT a physical property?
Reactivity
Density
Conductivity
Malleability
Answer:
Reactivity
Explanation:
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
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!A pulse has a speed of 5cm.s^-1. How far does it travel in 2,5s?
Answer:
12.5
Explanation:
Its 5m per second times 2.5 seconds so it's 12.5m
Answer:
[tex]\boxed {\boxed {\sf 12.5 \ centimeters}}[/tex]
Explanation:
We are asked to find how far a pulse travels or the distance. Distance is the product of speed and time.
[tex]d= s \times t[/tex]
The speed of the plus is 5 centimeters per second and the time is 2.5 seconds.
s= 5 cm/s t= 2.5 sSubstitute the values into the formula.
[tex]d= 5 \ cm/s \times 2.5 \ s[/tex]
Multiply. Note that the units of seconds will cancel each other out.
[tex]d= 12.5 \ cm[/tex]
The pulse travels 12.5 centimeters.
multiples and submultiples of kg
Answer:
Submultiples of kilogram are decagram ,centigram etc. And multiples of kilogram is megagram(1 tonne) , gigagram etc.
hope this helps you:)
Help meh in this question plzzz
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:
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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differentiate between air pressure and liquid pressure in table .
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.
a ball is thrown straight up into the air with an initial speed of 3.1 m/s. a. After 0.24 seconds what is the ball's velocity and b. what is it's acceleration?
a. The ball's velocity after 0.24 s is 0.75 m/s
b. The acceleration of the ball is given by the acceleration due to gravity
a. The ball's velocity can be calculated with the following equation:
[tex] v_{f} = v_{0} - gt [/tex]
Where:
[tex] v_{f} [/tex]: is the final speed =?
[tex]v_{0}[/tex]: is the initial speed = 3.1 m/s
g: is the acceleration due to gravity = 9.81 m/s²
t: is the time = 0.24 s
The minus sign is because the acceleration is in the opposite direction (downward) of the motion of the ball (upward).
The final speed is:
[tex] v_{f} = v_{0} - gt = 3.1 m/s - 9.81 m/s^{2}*0.24 s = 0.75 m/s [/tex]
Hence, the ball's velocity after 0.24 s is 0.75 m/s.
b. The acceleration of the ball is given by the acceleration due to gravity because the ball is thrown straight up (the motion of the ball is in the y-direction). The velocity of the ball in the x-direction is zero so the acceleration in the same direction is also zero.
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I hope it helps you!