From our physics class, we know that Work is a product of
Force and Displacement. The force must be the force in the same direction with
the displacement. So force and displacement must be parallel.
A. How much work in J does the string do on the boy if the
boy stands still?
Since we know that: Work = Force * Displacement
and in this case displacement is equal to zero, therefore
Work must also be equal to zero.
Work = 0
B. How much work does the string do on the boy if the boy
walks a horizontal distance of 11m away from the kite?
Now are given a displacement of 11 m which is a horizontal
distance. Therefore we must first calculate the horizontal component of force:
Fx = 4.5 N * cos 30
Fx = 3.897 N
So work is:
Work = 3.897 N * 11 m
<span>Work = 42.87 J ~ 43 J</span>
True. If the amount displaced is more than the mass, it floats. If the amount is less than the mass, it will sink.
Answer:
<em>time = 6.9 s</em>
<em></em>
Explanation:
velocity of the plane v = 22.7 m/s
acceleration of the train a = -3.3 m/s^2
it finally comes to res, final velocity u = 0 m/s
time t = ?
using the equation
v = u + at
substituting, we have
0 = 22.7 + (-3.3)t
0 = 22.7 - 3.3t
3.3t = 22.7
t = 22.7/3.3 = <em>6.9 s</em>
The sl unit of work is the joule (j)
Given: Heat given to mercury (Q) = 100 J
Mass of mercury (m) = 80 g = 80 × 10⁻³ kg
Specific Heat capacity of mercury (c) = 138 J/kg°C
Let the rise of temperature be Δt
Formula Used: Q = mcΔt
Here, all alphabets are in their usual meanings.
Calculation:
Q = mcΔt
or, Δt = Q / mc
or, Δt = 100 J /[(80 × 10⁻³ kg)×(138 J/kg°C)]
or, Δt = 9.058 °C
or, Δt = 9.1 °C
Hence, the required increase in temperature will be 9.1 °C.