Answer:
C = (7 + 8) i + (5 - 1) j adding vectors
C = 15 i + 4 j
theta = arctan (4 / 15) = 14.9 deg
Note that this is the same as adding x and y components
Answer:
The acceleration is
Explanation:
Given that,
Speed
Time t = 60.0 sec
We need to calculate the acceleration
Using formula off acceleration
We know that,
Missile goes from rest
So, Initial velocity =0
Put the value into the formula
On right hand side multiplying and dividing by g = 9.8m/s²
Put the value of g
Hence, The acceleration is
From the definition of entropy, the entropy change of an object is
where
Q is the heat absorbed
T is the absolute temperature
in our problem, we have Q=25.0 J, while the absolute temperature is (converting in Kelvin)
and so the entropy change is
Answer:
We know that ΔK = Kf - Ki = 1/2 m Vf^2 - 1/2 m Vi^2 = 1/2m(Vf^2-Vi^2) = 1/2 m ΔV^2.
The mass remains the same, just calculate the difference of squared velocities and multiply it by half of the mass.
Answer:
its terminal velocity is 19.70 m/s
the velocity of a 58.0-kg person hitting the ground, assuming no drag contribution in such a short distance is 8.85 m/s
Explanation:
Firstly,
given that
m = 580g = 0.58kg
Area A = 0.11 * 0.22 = 0.0242m
g = 9.8
idensity constant p = 1.21 kg/m^3
the terminal velocity of the sphere Vt is ;
Vt = √ ( 2mg / pCA)
we substitute
Vt = √ ( (2*0.58*9.8) / (1.21*1*0.0242)
Vt = √ (11.368 / 0.029282)
Vt = √ ( 388.22)
Vt = 19.70 m/s
its terminal velocity is 19.70 m/s
What will be the velocity of a 58.0-kg person hitting the ground, assuming no drag contribution in such a short distance?
The Velocity of the person is;
V2 = √ 2ax
V2 = √ ( 2 * 9.8 * 4 )
V2 = √ (78.4)
V2 = 8.85 m/s
the velocity of a 58.0-kg person hitting the ground, assuming no drag contribution in such a short distance is 8.85 m/s