Answer:
Option D - 0.2 s
Explanation:
We are given;
Initial velocity; u = 7 m/s
Height of table; h = 1.8m
Now,since we want to find the time the car spent in the air, we will simply use one of Newton's equation of motion.
Thus;
h = ut + ½gt²
Plugging in the relevant values, we have;
1.8 = 7t + ½(9.8)t²
4.9t² + 7t - 1.8 = 0
Using quadratic formula to find the roots of the equation gives us;
t = -1.65 or 0.22
We can't have negative t value, thus we will pick the positive one.
So, t = 0.22 s
This is approximately 0.2 s
Answer:
v_f = 0.87 m/s
Explanation:
We are given;
F_avg = -17700 N (negative because it's backward)
m = 117 kg
Δt = 5.50 × 10^(−2) s
v_i = 7.45 m/s
Now, formula for impulse is given by;
I = F•Δt = - 17700 x 5.50 × 10^(−2) = - 973.5 kg.m/s
From impulse momentum theory, we know that;
Change in momentum of particle is equal to impulse.
Thus,
Δp = I = m•v_f - m•v_i
Thus,
-973.5= 117(v_f - 7.45)
Thus,
-973.5/117 = (v_f - 7.45)
-8.3205 + 7.45 = v_f
v_f = - 0.87 m/s
We'll take absolute value as;
v_f = 0.87 m/s
Answer:
The answer is 10Nm
Explanation: I ended up just messing around with the numbers, I multiplied 5 and 2 got 10 as my answer and it was right.
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.
