Answer : The correct option is, (D) 278 K
Explanation :
We are given temperature 
.
Now the conversion factor used for the temperature is,
where, K is kelvin and 
is Celsius.
Now put the value of temperature, we get
Therefore, the temperature 278 K is equal to the
The problem seems to be incomplete because there is no question. However, from the problem description, the logical question is to find he acceleration needed by the jet to land on the airplane carrier. The working equation would be:
2ad = v₂² - v₁²
Since the jet stops, v₂ = 0. Substituting the values:
2(a)(95 m) = 0² - [(240 km/h)(1000 m/1 km)(1h/3600 s)]²
Solving for a,
<em>a = -23.39 m/s² (the negative sign indicates that the jet is decelerating)</em>
145,600 it equals 145,600 so you put down the zeros
Answer:
k = 1 700.7 N/m
v0 = 9.8 m/s^2
Explanation:
Hello!
We can answer this question using conservation of energy.
The potential energy of the spring (PS) will transform to kinetic energy (KE) of the ball, and eventually, when the velocity of the ball is zero, all that energy will be potential gravitational (PG) energy.
When the kinetic energy of the ball is zero, that is, when it has reached its maximum heigh, all the potential energy of the spring will be equal to the potential energy of the gravitational field.
PS = (1/2) k x^2 <em>where x is the compresion or elongation of the spring</em>
PG = mgh
a)
Since energy must be conserved and we are neglecting any energy loss:
PS = PG
Solving for k
k = (2mgh)/(x^2) = ( 2 * 1.7 * 9.81 * 4.9 Nm)/(0.31^2 m^2)
k = 1 700.7 N/m
b)
Since the potential energy of the spring transfors to kinetic energy of the ball we have that:
PS = KE
that is:
(1/2) k x^2 = (1/2) m v0^2
Solving for v0
v0 = x √(k/m) = (0.31 m ) √( 1 700.7 N/m / 1.7kg)
v0 = 9.8 m/s^2
