Answer:
The velocity of a ball will be "-70.13 m/s".
Explanation:
The given values are:
u = 70 m
t = 0.0 s
g = a = -9.8 m/s²
s = -1 m
v = ?
As we know,
The equation of motion will be:
⇒ 
On substituting the estimated values, we get
⇒ 
⇒ 
⇒ 
⇒ 
⇒ 
⇒ 
In downward direction, it will be:
⇒ 
I dont know from option
Because SI Unit of acceleration is m/s^2
Answer:
(A) 10132.5Pa
(B)531kJ of energy
Explanation:
This is an isothermal process. Assuming ideal gas behaviour then the relation P1V1 = P2V2 holds.
Given
m = 10kg = 10000g, V1 = 0.1m³, V2 = 1.0m³
P1 = 101325Pa. M = 102.03g/mol
P2 = P1 × V1 /V2 = 101325 × 0.1 / 1 = 10132.5Pa
(B) Energy is transfered by the r134a in the form of thw work done in in expansion
W = nRTIn(V2/V1)
n = m / M = 10000/102.03 = 98.01mols
W = 98.01 × 8.314 × 283 ×ln(1.0/0.1)
= 531kJ.
Hey.. the virtual image formed by a plane mirror would be the same size. If formed by a convex mirror it would be diminished and if formed by a concave mirror it would be magnified.
the negative sign in the value of magnification indicates the type of mirror, that the mirror being used is convex, and if it is positive then the mirror being used is concave mirror. The negative sign denotes that the image projected is in the inverted direction.
Answer: Your question is missing below is the question
Question : What is the no-friction needed speed (in m/s ) for these turns?
answer:
20.1 m/s
Explanation:
2.5 mile track
number of turns = 4
length of each turn = 0.25 mile
banked at 9 12'
<u>Determine the no-friction needed speed </u>
First step : calculate the value of R
2πR / 4 = πR / 2
note : πR / 2 = 0.25 mile
∴ R = ( 0.25 * 2 ) / π
= 0.159 mile ≈ 256 m
Finally no-friction needed speed
tan θ = v^2 / gR
∴ v^2 = gR * tan θ
v = √9.81 * 256 * tan(9.2°) = 20.1 m/s
