The solution for this is:
Work done = force * distance = m*a*d and power = energy/time
The vo=0 and vf = 25 m/s and t=7 sec. This gives...
3.6 m/s^2 as acceleration and d=87.5 meters and thus F=ma= 5400 N.
Energy = 5400*87.5 = 4.7E5 Joules (2 sig. figs) and Power = 67,500 Watts or 90 HP (2 sig. figs again).
Answer:
Radius, 
Explanation:
It is given that,
Magnetic field, B = 0.275 T
Kinetic energy of the electron, 
Kinetic energy is given by :
v = 1037749.04 m/s
The centripetal force is balanced by the magnetic force as :
So, the radius of the circular path is 
. Hence, this is the required solution.
Answer:
Option 10. 169.118 J/KgºC
Explanation:
From the question given above, the following data were obtained:
Change in temperature (ΔT) = 20 °C
Heat (Q) absorbed = 1.61 KJ
Mass of metal bar = 476 g
Specific heat capacity (C) of metal bar =?
Next, we shall convert 1.61 KJ to joule (J). This can be obtained as follow:
1 kJ = 1000 J
Therefore,
1.61 KJ = 1.61 KJ × 1000 J / 1 kJ
1.61 KJ = 1610 J
Next, we shall convert 476 g to Kg. This can be obtained as follow:
1000 g = 1 Kg
Therefore,
476 g = 476 g × 1 Kg / 1000 g
476 g = 0.476 Kg
Finally, we shall determine the specific heat capacity of the metal bar. This can be obtained as follow:
Change in temperature (ΔT) = 20 °C
Heat (Q) absorbed = 1610 J
Mass of metal bar = 0.476 Kg
Specific heat capacity (C) of metal bar =?
Q = MCΔT
1610 = 0.476 × C × 20
1610 = 9.52 × C
Divide both side by 9.52
C = 1610 / 9.52
C = 169.118 J/KgºC
Thus, the specific heat capacity of the metal bar is 169.118 J/KgºC
Answer:
Explanation:
An object is at rest along a slope if the net force acting on it is zero. The equation of the forces along the direction parallel to the slope is:
(1)
where
is the component of the weight parallel to the slope, with m being the mass of the object, g the acceleration of gravity, 
the angle of the slope
is the frictional force, with 
being the coefficient of friction and R the normal reaction of the incline
The equation of the forces along the direction perpendicular to the slope is
where
R is the normal reaction
is the component of the weight perpendicular to the slope
Solving for R,
And substituting into (1)
Re-arranging the equation,
This the condition at which the equilibrium holds: when the tangent of the angle becomes larger than the value of , the force of friction is no longer able to balance the component of the weight parallel to the slope, and so the object starts sliding down.
