Answer:
81.6 m
Explanation:
Answer: 81.6 m.
The time it takes gravity to slow 40 m/s to zero when it teaches maximum height is
-v(initial) / -g = t
-40 m/s / -9.8 m/s^2 = 4.08 s
The height reached is the average velocity times this time 4.08 s, with v(avg) = [v(initial) + v(final)] / 2 with v(final) = 0. v(avg) = v(initial) / 2 = 40 m/s / 2 = 20 m/s.
So the distance d of maximum height is
d = v(avg)•t
d = 20 m/s • 4.08 s = 81.6 m.
I can't tell right away. The elevator's limit is given in terms of mass, but the four people are listed in terms of their weights. In order to compare them, we'll have to convert one or the other.
The total of the four people's weights is:
John . . . . . . . 850N
Betty . . . . . . . 530N
Robert . . . . . . 740N
Alice . . . . . . . .610N
Total . . . . . . 2,730N
The elevator can't accept more than 300 kg of mass.
The weight of any mass is (mass) x (gravity).
On Earth, 300kg weighs (300kg) x (9.8 m/s²) = 2,940 N.
Wonderful !
2,730N is less than 2,940N.
<em>Yes !</em> According to Otis' engineers, John, Betty, Robert and Alice <em>can safely ride the elevator together</em>, (although Betty and Alice may have a different opinion).
Answer:
A 26m
Explanation:
firstly the answer will not be part c as distance unit will be meters (m)
next, substitute the values of a and t as shown
![\frac{1}{2} (9.8)(2.3)^2 = 25.921](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%20%20%289.8%29%282.3%29%5E2%20%3D%2025.921)
the answer rounded off to 2 significant figures will be hence 26m
Answer:
The new force is 1/4 of the previous force.
Explanation:
Given
---- ![r_1](https://tex.z-dn.net/?f=r_1)
--- ![r_2](https://tex.z-dn.net/?f=r_2)
Required
Determine the new force
Let the two particles be q1 and q2.
The initial force F1 is:
--- Coulomb's law
Substitute 2 for r1
![F_1 = \frac{kq_1q_2}{2^2}](https://tex.z-dn.net/?f=F_1%20%3D%20%5Cfrac%7Bkq_1q_2%7D%7B2%5E2%7D)
![F_1 = \frac{kq_1q_2}{4}](https://tex.z-dn.net/?f=F_1%20%3D%20%5Cfrac%7Bkq_1q_2%7D%7B4%7D)
The new force (F2) is
![F_2 = \frac{kq_1q_2}{r_2^2}](https://tex.z-dn.net/?f=F_2%20%3D%20%5Cfrac%7Bkq_1q_2%7D%7Br_2%5E2%7D)
Substitute 4 for r2
![F_2 = \frac{kq_1q_2}{4^2}](https://tex.z-dn.net/?f=F_2%20%3D%20%5Cfrac%7Bkq_1q_2%7D%7B4%5E2%7D)
![F_2 = \frac{kq_1q_2}{4*4}](https://tex.z-dn.net/?f=F_2%20%3D%20%5Cfrac%7Bkq_1q_2%7D%7B4%2A4%7D)
![F_2 = \frac{1}{4}*\frac{kq_1q_2}{4}](https://tex.z-dn.net/?f=F_2%20%3D%20%5Cfrac%7B1%7D%7B4%7D%2A%5Cfrac%7Bkq_1q_2%7D%7B4%7D)
Substitute ![F_1 = \frac{kq_1q_2}{4}](https://tex.z-dn.net/?f=F_1%20%3D%20%5Cfrac%7Bkq_1q_2%7D%7B4%7D)
![F_2 = \frac{1}{4}*F_1](https://tex.z-dn.net/?f=F_2%20%3D%20%5Cfrac%7B1%7D%7B4%7D%2AF_1)
![F_2 = \frac{F_1}{4}](https://tex.z-dn.net/?f=F_2%20%3D%20%5Cfrac%7BF_1%7D%7B4%7D)
The new force is 1/4 of the previous force.