His total displacement from his original position is -1 m
We know that total displacement of an object from a position x to a position x', d = final position - initial position.
d = x' - x
If we assume the lad's initial position in front of her house is x = 0 m. The lad then moves towards the positive x-axis, 5 m. He then ends up at x' = 5 m. He then finally goes back 6 m.
Since displacement = final position - initial position, and his displacement is d' = -6 m (since he moves in the negative x - direction or moves back) from his initial position of x' = 5 m.
His final position, x" after moving back 6 m is gotten from
x" - x' = -6 m
x" = -6 + x'
x" = -6 + 5
x" = -1 m
Thus, his total displacement from his original position is
d = final position - initial position
d = x" - x
d = -1 m - 0 m
d = -1 m
So, his total displacement from his original position is -1 m
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Answer:
Psm = 30.66 [Psig]
Explanation:
To solve this problem we will use the ideal gas equation, recall that the ideal gas state equation is always worked with absolute values.
P * v = R * T
where:
P = pressure [Pa]
v = specific volume [m^3/kg]
R = gas constant for air = 0.287 [kJ/kg*K]
T = temperature [K]
<u>For the initial state</u>
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P1 = 24 [Psi] + 14.7 = 165.47[kPa] + 101.325 = 266.8 [kPa] (absolute pressure)
T1 = -2.6 [°C] = - 2.6 + 273 = 270.4 [K] (absolute Temperature)
Therefore we can calculate the specific volume:
v1 = R*T1 / P1
v1 = (0.287 * 270.4) / 266.8
v1 = 0.29 [m^3/kg]
As there are no leaks, the mass and volume are conserved, so the volume in the initial state is equal to the volume in the final state.
V2 = 0.29 [m^3/kg], with this volume and the new temperature, we can calculate the new pressure.
T2 = 43 + 273 = 316 [K]
P2 = R*T2 / V2
P2 = (0.287 * 316) / 0.29
P2 = 312.73 [kPa]
Now calculating the manometric pressure
Psm = 312.73 -101.325 = 211.4 [kPa]
And converting this value to Psig
Psm = 30.66 [Psig]
Answer:
v = 2.94 m/s
Explanation:
When the spring is compressed, its potential energy is equal to (1/2)kx^2, where k is the spring constant and x is the distance compressed. At this point there is no kinetic energy due to there being no movement, meaning the net energy in the system is (1/2)kx^2.
Once the spring leaves the system, it will be moving at a constant velocity v, if friction is ignored. At this time, its kinetic energy will be (1/2)mv^2. It won't have any spring potential energy, making the net energy (1/2)mv^2.
Because of the conservation of energy, these two values can be set equal to each other, since energy will not be gained or lost while the spring is decompressing. That means
(1/2)kx^2 = (1/2)mv^2
kx^2 = mv^2
v^2 = (kx^2)/m
v = sqrt((kx^2)/m)
v = x * sqrt(k/m)
v = 0.122 * sqrt(125/0.215) <--- units converted to m and kg
v = 2.94 m/s
Answer:
Tension, T = 87.63 N
Explanation:
Given that,
Mass of the object, m = 6.9 kg
The string is acting in the upward direction, a = 2.9 m/s²
Acceleration due to gravity, g = 9.8 m/s²
As the lift is accelerating upwards, it means the net force acting on it is given by :
T = m(a+g)
= 6.9 (2.9+9.8)
= 6.9(2.7)
= 87.63 N
So, the tension in the string is 87.63 N.
Intensity of sunlight at given position is defined as power received per unit area
so here we can say
area on which photons are received is given as
now we can find the power received due to sunlight
now we can say this power is due to photons that strikes on surface of earth
so here we can say
given here that
so it will strike 2.47 * 10^18 photons on given area per second
