Answer:
Answer:
the amount of energy flowing is 1.008x10⁹J
Explanation:
To calculate how much heat flows, the expression is the following:
Where
K=thermal conductivity=0.81W/m°C
A=area=6.2*12=74.4m²
ΔT=30-8=22°C
L=thickness=8cm=0.08m
t=time=16.9h=60840s
Replacing:
Explanation:
The student who did the most work is student 2 with 2500 Joules.
<u>Given the following data:</u>
To determine which of the students did the most work:
Mathematically, the work done by an object is given by the formula;
<u>For </u><u>student 1</u><u>:</u>
Work done = 600 Joules
<u>For </u><u>student 2</u><u>:</u>
Work done = 2500 Joules.
Therefore, the student who did the most work is student 2 with 2500 Joules.
Read more: Read more: brainly.com/question/13818347
Answer:
-5 m/s
Explanation:
The linear velocity of B is equal and opposite the linear velocity of E.
vB = -vE
vB = -ωE rE
10 m/s = -ωE (12 m)
ωE = -0.833 rad/s
The angular velocity of E is the same as the angular velocity of D.
ωE = ωD
ωD = -0.833 rad/s
The linear velocity of Q is the same as the linear velocity of D.
vQ = vD
vQ = ωD rD
vQ = (-0.833 rad/s) (6 m)
vQ = -5 m/s
Answer:
35 mph
Explanation:
The key of this problem lies in understanding the way that projectile motion works as we are told to neglect the height of the javelin thrower and wind resistance.
When the javelin is thown, its velocity will have two components: a x component and a y component. The only acceleration that will interact with the javelin after it was thown will be the gravety, which has a -y direction. This means that the x component of the velocity will remain constant, and only the y component will be affected, and can be described with the constant acceleration motion properties.
When an object that moves in constant acceleration motion, the time neccesary for it to desaccelerate from a velocity v to 0, will be the same to accelerate the object from 0 to v. And the distance that the object will travel in both desaceleration and acceleration will be exactly the same.
So, when the javelin its thrown, it willgo up until its velocity in the y component reaches 0. Then it will go down, and it will reach reach the ground in the same amount of time it took to go up and, therefore, with the same velocity.
