Answer: It is the product of the (force)multiplied by the (perpendicular) distance from the line of action of the force to the pivot
Explanation:
Answer:
The answer to your question is
Explanation:
Data
mass = 0.5kg
T1 = 35
T2 = ?
Q = - 6.3 x 10⁴ J = - 63000 J
Cp = 4184 J / kg°C
Formula
Q = mCp(T2 - T1)
T2 = T1 + Q/mCp
Substitution
T2 = 35 - 63000/(0.5 x 4184)
T2 = 35 - 63000/2092
T2 = 35 - 30.1
T2 = 4.9 °C
Answer:
1.3 m/s
Explanation:
average speed = total distance/ total time
Answer:
h = 2.49 [m]
Explanation:
In order to solve this problem we must use the definition of potential energy, which tells us that energy is equal to the product of mass by gravity by height.
The potential energy can be calculated by means of this equation:
Ep = m*g*h
where:
Ep = potential energy = 980 [J]
m = mass = 40 [kg]
g = gravity acceleration = 9.81 [m/s^2]
h = elevation [m]
Now replacing:
980 = 40*9.81*h
h = 2.49 [m]
Answer:
The deceleration of the dragster upon releasing the parachute such that the wheels at B are on the verge of leaving the ground is 16.33 m/s²
Explanation:
The additional information to the question is embedded in the diagram attached below:
The height between the dragster and ground is considered to be 0.35 m since is not given ; thus in addition win 0.75 m between the dragster and the parachute; we have: (0.75 + 0.35) m = 1.1 m
Balancing the equilibrium about point A;
F(1.1) - mg (1.25) = 
- 1200(9.8)(1.25) = 1200a(0.35)
- 14700 = 420 a ------- equation (1)
--------- equation (2)
Replacing equation 2 into equation 1 ; we have :

1320 a - 14700 = 420 a
1320 a - 420 a =14700
900 a = 14700
a = 14700/900
a = 16.33 m/s²
The deceleration of the dragster upon releasing the parachute such that the wheels at B are on the verge of leaving the ground is 16.33 m/s²