Answer:
The strong attractions among ions within a crystal lattice.
Explanation:
The electrical force between the charges would: D) It will decrease to one-ninth the original force.
Answer:
5572.8 N
Explanation:
Applying,
F = ma.............. Equation 1
Where F = Force, m = mass of the car, a = acceleration.
We can find a by applying,
v² = u²+2as............. Equation 2
Where v = final velocity, u = initial velocity, a = acceleration, = distance.
From the question,
Given: v = 0 m/s (come to rest), u = 1.7 m/s, s = 0.210 m
Substitute these value into equation 2
0² = 1.7²+2×0.21×a
a = -1.7²/(2×0.21)
a = -2.89/0.42
a = -6.88 m/s²
Also given: m = 810 kg
Substitute these value into equation 1
F = 810(-6.88)
F = -5572.8 N
Hence the force on the bumber is 5572.8 N
The person's horizontal position is given by
![x=v_0\cos40^\circ t](https://tex.z-dn.net/?f=x%3Dv_0%5Ccos40%5E%5Ccirc%20t)
and the time it takes for him to travel 56.6 m is
![56.6\,\mathrm m=\left(24.0\,\dfrac{\mathrm m}{\mathrm s}\right)\cos40^\circ t\implies t=3.08\,\mathrm s](https://tex.z-dn.net/?f=56.6%5C%2C%5Cmathrm%20m%3D%5Cleft%2824.0%5C%2C%5Cdfrac%7B%5Cmathrm%20m%7D%7B%5Cmathrm%20s%7D%5Cright%29%5Ccos40%5E%5Ccirc%20t%5Cimplies%20t%3D3.08%5C%2C%5Cmathrm%20s)
so your first computed time is the correct one.
The question requires a bit of careful reading, and I think there may be a mistake in the problem. The person's vertical velocity
at time
is
![v_y=v_{0y}-gt](https://tex.z-dn.net/?f=v_y%3Dv_%7B0y%7D-gt)
which tells us that he would reach the ground at about
. In this time, he would have traveled
![x=v_{0x}(3.15\,\mathrm s)=57.9\,\mathrm m](https://tex.z-dn.net/?f=x%3Dv_%7B0x%7D%283.15%5C%2C%5Cmathrm%20s%29%3D57.9%5C%2C%5Cmathrm%20m)
But we're told that he is caught by a net at 56.6 m, which would mean that the net cannot have been placed at the same height from which he was launched. However, it's possible that the moment at which he was launched doesn't refer to the moment the cannon went off, but rather the moment at which the person left the muzzle of the cannon a fraction of a second after the cannon was set off. After this time, the person's initial vertical velocity
would have been a bit smaller than
.