Answer:
Explanation:
mass of box, m = 60 kg
distance, d = 6.2 m
force, P = 150 N
coefficient of kinetic friction, μk = 0.21
work done by the gravitational force, Wg = Force x distance x cos 90
Wg = 0 J
Work done by the normal force, Wn = force x distance x Cos 90
Wn = 0 J
Work done by the friction force, Wf = - μk mg d
Wf = - 0.21 x 60 x 9.8 x 6.2 = - 765.58 J
Work done by the applied force, Wp = P x d
Wp = 150 x 6.2 = 930 J
Well at the fourth half-life it would be 6.25% so if you continue calculating the answer would be B.
Answer:
V = 0.248 L
Explanation:
To do this, use the following equation:
P1*V1/T1 = P2*V2/T2
This equation is used to find a relation between two differents conditions of a same gas, which is this case. From this equation we can solve for V2.
Solving for V2:
V2 = P1*V1*T2/T1*P2
Temperature must be at Kelvin, so, we have to sum the temperature 273 to convert it in K.
Replacing the data we have:
V2 = 1 * 4.91 * (-196+273) / 5.2 * (20+273)
V2 = 378.07 / 1523.6
V2 = 0.248 L
In collision that are categorized as elastic, the total kinetic energy of the system is preserved such that,
KE1 = KE2
The kinetic energy of the system before the collision is solved below.
KE1 = (0.5)(25)(20)² + (0.5)(10g)(15)²
KE1 = 6125 g cm²/s²
This value should also be equal to KE2, which can be calculated using the conditions after the collision.
KE2 = 6125 g cm²/s² = (0.5)(10)(22.1)² + (0.5)(25)(x²)
The value of x from the equation is 17.16 cm/s.
Hence, the answer is 17.16 cm/s.
