Answer:
350 ft/s²
Explanation:
First, convert mph to ft/s.
58 mi/hr × (5280 ft/mi) × (1 hr / 3600 s) = 85.1 ft/s
Given:
v₀ = 85.1 ft/s
v = 0 ft/s
t = 0.24 s
Find: a
v = at + v₀
a = (v − v₀) / t
a = (0 ft/s − 85.1 ft/s) / 0.24 s
a = -354 ft/s²
Rounded to two significant figures, the magnitude of the acceleration is 350 ft/s².
To solve this problem we will apply the concepts related to load balancing. We will begin by defining what charges are acting inside and which charges are placed outside.
PART A)
The charge of the conducting shell is distributed only on its external surface. The point charge induces a negative charge on the inner surface of the conducting shell:
. This is the total charge on the inner surface of the conducting shell.
PART B)
The positive charge (of the same value) on the external surface of the conducting shell is:
The driver's net load is distributed through its outer surface. When inducing the new load, the total external load will be given by,
Answer:
2.71 m
Explanation:
Force is the product of mass and acceleration
F=m*a
Work done is the product of force and distance
Work done=F*d
In this case;
F= 35 N
Work done = 95 J
95 =35 * d
95 /35 = d
2.71 m= d
The sum of potential energy<span> and kinetic </span><span>energy.
Hope I helped!</span>
<span>AS T1,T2,T3 are the tensions in the ropes,assuming that there are Three blocks of mass 3m, 2m, and m.T3 is the string between 3m and 2m,T2 is the string between 2m and m ,T1 is the string attached to m thus T1 pulls the whole set of blocks along, so it must be the largest. T2 pulls the last
two masses, but T3 only pulls the last mass, so T3 < T2 < T1.</span>
