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Below is the answers:
Fullback running
<span>Mo = mass * velocity </span>
<span>Mo = 95kg * 8.2 m/s =779 kg*m/s (a </span>
<span>He got stopped Change in Mo = 779 kg*m/s (b </span>
<span>Both stopped ===> Tackle's mo = - Halfback's Mo = - 779 kg*m/s (c & d </span>
<span>- 779 = 128 * v </span>
<span>v= - 6.09 m/s (e</span>
Answer:
805.48N/m
Explanation:
According to Hookes law
F = Ke
F is the force = mg
F = 2.4×9.8 = 23.52N
e is the extension = 2.92cm = 0.0292m
Force constant K = F/e
K = 23.52/0.0292
K = 805.48N/m
Hence the force constant of the spring is 805.48N/m
I'd say the answer to this on is d.Facts as they are using the temperatures in the graph and the temperatures are not just estimates
Answer:
2.40 x 10⁻¹³ C
Explanation:
= number of electrons = 6.25 x 10⁶
= charge on electron = - 1.6 x 10⁻¹⁹ C
= number of protons = 7.75 x 10⁶
= charge on proton = 1.6 x 10⁻¹⁹ C
Net charge is given as
Q =
+
![q_{p}](https://tex.z-dn.net/?f=q_%7Bp%7D)
Q = (- 1.6 x 10⁻¹⁹) (6.25 x 10⁶) + (1.6 x 10⁻¹⁹) (7.75 x 10⁶)
Q = 2.40 x 10⁻¹³ C
Answer:
The combined velocity is 8.61 m/s.
Explanation:
Given that,
The mass of a truck, m = 2800 kg
Initial speed of truck, u = 12 m/s
The mass of a car, m' = 1100 kg
Initial speed of the car, u' = 0
We need to find the combined velocity the moment they stick together. Let it is V. Using the conservation of momentum.
![m_1v_1+m_2v_2=(m_1+m_2)V\\\\V=\dfrac{m_1v_1+m_2v_2}{(m_1+m_2)}\\\\V=\dfrac{2800\times 12+0}{2800+1100}\\\\V=8.61\ m/s](https://tex.z-dn.net/?f=m_1v_1%2Bm_2v_2%3D%28m_1%2Bm_2%29V%5C%5C%5C%5CV%3D%5Cdfrac%7Bm_1v_1%2Bm_2v_2%7D%7B%28m_1%2Bm_2%29%7D%5C%5C%5C%5CV%3D%5Cdfrac%7B2800%5Ctimes%2012%2B0%7D%7B2800%2B1100%7D%5C%5C%5C%5CV%3D8.61%5C%20m%2Fs)
So, the combined velocity is 8.61 m/s.