Idk can u say it in a another way
<span>If there isn't any force then the normal contact force will be
N=m*g=7.5*9.81=73.58N
which is 73.58-23=50.58N less
so, there the person must pull at 23 degree upward
break down the tension in two components, vertical and horizontal.
vertical tension= 50.58=T*sin23
T=50.58/sin23=129.45N</span>
Answer: 24.97 kg
Explanation:
The gravitational force between two objects of masses M1, and M2 respectively, and separated by a distance R, is:
F = G*(M1*M2)/R^2
Where G is the gravitational constant:
G = 6.67*10^-11 m^3/(kg*s^2)
In this case, we know that
R = 0.002m
F = 0.0104 N
and that M1 = M2 = M
And we want to find the value of M, then we can replace those values in the equation to get
0.0104 N = (6.67*10^-11 m^3/(kg*s^2))*(M*M)/(0.002m)^2
(0.0104 N)*(0.002m)^2/(6.67*10^-11 m^3/(kg*s^2)) = M^2
623.69 kg^2 = M^2
√(623.69 kg^2) = M = 24.97 kg
This means that the mass of each object is 24.97 kg
Answer:
1. 3 m
2. 27 s
Explanation:
1. "A car traveling at +33 m/s sees a red light and has to stop. If the driver can accelerate at -5.5 m/s², how far does it travel?"
Given:
v₀ = 33 m/s
v = 0 m/s
a = -5.5 m/s²
Unknown: Δx
To determine the equation you need, look for which variable you don't have and aren't solving for. In this case, we aren't given time and aren't solving for time. So look for an equation that doesn't have t in it.
Equation: v² = v₀² + 2aΔx
Substitute and solve:
(0 m/s)² = (33 m/s)² + 2(-5.5 m/s²) Δx
Δx = 3 m
2. "A plane starting from rest at one end of a runway accelerates at 4.8 m/s² for 1800 m. How long did it take to accelerate?"
Given:
v₀ = 0 m/s
a = 4.8 m/s²
Δx = 1800 m
Unknown: t
Equation: Δx = v₀ t + ½ a t²
Substitute and solve:
1800 m = (0 m/s) t + ½ (4.8 m/s²) t²
t ≈ 27 s
Answer: B
Explanation: This can be easily done by inputting 5280 * 2.2.
