Answer:
the answers the correct one is: the time increases
Explanation:
When an insulating material is placed in the cavity of a wall the thermal conductivity of the system increases, therefore the temperature changes take longer.
P= dQ / dt = K A dT / dx
when placing insulation the value of k is less
When reviewing the answers the correct one is: the time increases
The average force on the squid during the ejection of 0.60 kg of water at a velocity of 15.0 m/s in 0.15 seconds is 60 N.
We can calculate the average force with the average acceleration as follows:
(1)
Where:
- m: is the mass of water = 0.60 kg
: is the average acceleration
The <em>average acceleration</em> is given by the change of velocity in an interval of time
(2)
Where:
: is the initial velocity = 0 (the squid is at rest)
: is the final velocity = 15.0 m/s
: is the initial time = 0
: is the final time = 0.15 s
Now we can find the <em>average force</em> after entering equation (2) into (1)
Therefore, the average force on the squid during the propulsion is 60 N.
Find more about average force here:
I hope it helps you!
Definitely the last one, brain stem
Answer:
P₂ = 2 P₁
we conclude that in the second time the power used is double that in the first rise
Explanation:
In this exercise we are asked the power to climb the stairs, if we assume that we go up with constant speed, we use an energy equal to the potential energy due to the difference in height of the stairs, as this height is constant the potential energy does not change and therefore therefore the energy used by us does not change either.
Now we can analyze the required power,
P = W / t
From the analysis of the previous paragraph the work is equal to the energy used, according to the work energy theorem,
therefore the first time the power is
P₁ = E / 10
P₁ = 0.1 E
for the second time the power is
P₂ = E / 5
P₂ = 0.2 E
we see that the power in the second case is
P₂ = 2 P₁
Therefore, we conclude that in the second time the power used is double that in the first rise.
The force exerted on his torso by his legs during the deceleration is 4365 N.
<u>Explanation:</u>
Mass of the torso m=45kg
Height of the building s=3.5 m
Decelerating distance=0.71 m
when he jumps to the ground, the only acceleration is acceleration due to gravity g
<u>motion1 from top to ground </u>
initial velocity u=0
we have to calculate final velocity v using the following equation of motion.

use height of the building as the distance s as the jump from top to the ground is only described here.
<u>Motion 2 on the ground</u>
v=0
u=8.3(final velocity of motion 1)
The deceleration after striking the ground can be calculated from the equation of motion

The decelerating distance is used in the place of s since since the motion after hitting the ground is described in this case.
The equation of force is
