1.- Al aplicar una fuerza de 60 Newtons sobre un cuerpo, se produce una aceleración de 4 m/s2 calcula el valor de su masa.
1 answer:
You might be interested in
Answer:
The position of the car at t = 1.5 s is at -8.1625 meters
Explanation:
The initial position of the car is 3.2 meters
The initial velocity is -8.4 m/s
The constant acceleration is 1.1 m/s²
We need to find the final position of the car at the time t = 1.5 seconds
The displacement <em>s</em> = final position - initial position
, where <em>u</em> is the initial velocity, <em>a</em> is the
constant acceleration and <em>t</em> is the time
So we can find the final velocity by using the rule:
final position - initial position =
initial position = 3.2 meters , u = -8.4 m/s , a = 1.1 ²m/s , t = 1.5 s
Substitute these values in the rule
final position - 3.2 =
final position - 3.2 = -12.6 + 1.2375
final position - 3.2 = -11.3625
add 3.2 for both sides
final position = -8.1625
<em>That means the car is at 8.1625 meters in opposite direction</em>
<em>The position of the car at t = 1.5 s is at -8.1625 meters </em>
Answer:
a) -0.0934 kJ/kg. K
b) The direction of flow is from right to left.
Explanation:
A free flow diagram of the horizontal insulated duct is as shown below.
NOW,
Let assume that the direction of flow is from left to right and consider the following relation for the entropy rate balance equation for a control volume as:
Now; if the value for this relation is greater than zero; then we conclude that our assumption is correct.
If the value is less than zero; then we conclude that the assumption is wrong.
Then, the flow is said to be in the opposite direction
Formula for the change in specific entropy can be calculated as:
where;
are specific entropies
R = universal gas constant
= pressure at location 1
= pressure at location 2
We obtain the specific properties of air at temperature at = (67°C + 273)K = 340 K from the table A-22 ( Ideal gas properties of air)
= 1.8279 kJ/kg.K
We also obtain the specific properties of air at temperature = 22°C + 273) K = 295 K
From the table A- 22
= 1.68515 kJ/kg . K
R =
0.95 bar
0.8 bar
Now replacing our values into equation (2) from above; we have;
Equating our result to equation (1)
Therefore , our assumption is wrong and the direction of flow is said to be from right to left.
We therefore conclude that the direction of flow is from right to left.
