Answer:
Average speed 3m/sAverage velocity is 4m/s
Explanation:
In this problem, we are going to solve for the average speed and average velocity for the entrire trip made by the dolphin.
Given data
speed v= 2 m/s
time t= 24s
the distance travelled by the dolphin is
speed= distance/time
distance= speed*time
distance= 2*24
distance= 48m
now the dolpine made a reverse and covered this 48m in 8s to return to it initial position
the speed it took is
speed= distance/time
speed= 48/8
speed= 6m/s
It average speed is the total distance divided by the total time taken
average speed= (48+48)/(24+8)= 96/32= 3m/s
average speed= 3 m/s
it average velocity is the sum of the initial and final velocity is divided by 2 to find the average
(2+6)/2= 8/2= 4m/s
average velocity is 4m/s
How is a scientific theory different from a scientific law?
Convert the following measurements. Only type numbers and decimals into the answer boxes:
1. 34mm =
cm
2.234cm =
m
3.3km =
m
4.35 m =
mm
Answer:
2m 4m2Explanation:rteji3ujnrej
1. .134
2. 22.34
3. 330
4. 4350
In Einstein's special theory of relativity, mass and energy are equivalent. An expression of this equivalence can be made in terms of electron volts ( units of energy) and kilograms, with one electron volt (eV) being equal to 1.78*10^ -36 kg. Using this ratio, express the mass of the heaviest mammal on earth, the blue whale, which has an average mass of 1.90*10^ 5 kg , in mega electron volts and tera electron volts
Answer:
In MeV: 10.674 × 10^(34) MeV
In TeV: 10.674 × 10^(28) TeV
Explanation:
We are given that;
1.78 × 10^(-36) kg = 1 eV
We are now told that the blue whale has an average mass of 1.90 × 10^(5) kg
Thus, converting this mass of the blue whale to eV, we have;
(1.90 × 10^(5) × 1)/(1.78 × 10^(-36)) = 10.674 × 10^(40) eV
Now, converting to mega electron volts;
From conversions;
1 eV = 10^(-6) MeV
Thus,
10.674 × 10^(40) eV will be;
(10.674 × 10^(40) × 10^(-6))/1 = 10.674 × 10^(34) MeV
Also, converting to Tera electron Volts;
From conversion, we know that;
1 eV = 10^(-12) TeV
Thus;
10.674 × 10^(40) eV will give;
(10.674 × 10^(40) × 10^(-12))/1 = 10.674 × 10^(28) TeV
Is anyone interested in a pile of bricks, it's free on craigslist?
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"A student bikes to school by traveling first dN = 0.900" and "Similarly, let d⃗ W be the displacement vector corresponding to the second leg of the student's trip. Express d⃗ W in component form. Express your answer as two numbers separated by a comma. Be careful with your signs"
Complete Question
A student bikes to school by traveling first dN = 0.900 miles north, then dW = 0.300 miles west, and finally dS = 0.100 miles south.
Similarly, let d⃗ W be the displacement vector corresponding to the second leg of the student's trip. Express d⃗ W in component form.
Express your answer as two numbers separated by a comma. Be careful with your signs.
Answer:
The value is [tex]dT = ( -0.3, 0.8)[/tex]
Explanation:
From the question we are told that
The first displacement is [tex]dN = 0.900 \ due \ North[/tex] i.e positive y-axis
The second displacement is [tex]dW = 0.300 \ miles \ due \ west[/tex] i.e negative x-axis
The final displacement is [tex]dS = 0.100 \ miles \ due \ south[/tex] i.e negative y-axis
Generally dW in component for is
[tex]dW = (-0.3 , 0)[/tex]
Generally the total displacement of the student is mathematically represented as
[tex]dT = ( -0.3, (0.90 - 0.10))[/tex]
[tex]dT = ( -0.3, 0.8)[/tex]
What year will y'all graduate
Answer:
I'm in college and i'm a senior so i will graduate this year.
Explanation: