Answer:
m = 0.51[kg]
Explanation:
Potential energy is defined as the product of mass by gravity by height.
where:
Epot = potential energy = 15 [J]
m = mass [kg]
g = gravity acceleration = 9.8 [m/s²]
h = elevation = 3 [m]
Now replacing:
![E_{pot}=m*g*h\\15=m*9.8*3\\m = 0.51[kg]](https://tex.z-dn.net/?f=E_%7Bpot%7D%3Dm%2Ag%2Ah%5C%5C15%3Dm%2A9.8%2A3%5C%5Cm%20%3D%200.51%5Bkg%5D)
Yes, therefore the object would then slow down
Answer:
a) I = 2279.5 N s
, b) F = 3.80 10⁵ N, c) I = 3125.5 N s and d) F = 5.21 10⁵ N
Explanation:
The impulse is equal to the variation in the amount of movement.
I =∫ F dt = Δp
I = m
- m v₀
Let's calculate the final speed using kinematics, as the cable breaks the initial speed is zero
² = V₀² - 2g y
² = 0 - 2 9.8 30.0
= √588
= 24.25 m/s
a) We calculate the impulse
I = 94 24.25 - 0
I = 2279.5 N s
b) Let's join the other expression of the impulse to calculate the average force
I = F t
F = I / t
F = 2279.5 / 6 10⁻³
F = 3.80 10⁵ N
just before the crash the passenger jumps up with v = 8 m / s, let's take the moments of interest just when the elevator arrives with a speed of 24.25m/s down and as an end point the jump up to vf = 8 m / n
c) I = m - m v₀
I = 94 8 - 94 (-24.25)
I = 3125.5 N s
d) F = I / t
F = 3125.5 / 6 10⁻³
F = 5.21 10⁵ N
Sound waves are a type of classical waves and so they transport only energy without transporting matter through the medium.
The question is incomplete, the complete question is;
A body initially at a all 100 degree centigarde cools to 60 degree centigarde in 5 minutes and to 40 degree centigarde in 10 minutes . What is the temperature of surrounding? What will be the temperature in 15 minutes?
Answer:
See explanation
Explanation:
From Newton's law of cooling;
θ1 - θ2/t = K(θ1 + θ2/2 - θo]
Where;
θ1 and θ2 are initial and final temperatures
θo is the temperature of the surroundings
K is the constant
t is the time taken
Hence;
100 - 60/5 = K(100 + 60/2 - θo)
100 - 40/10 = K(100 + 40/2 - θo)
8= (80 - θo)K -----(1)
6= (70 - θo)K -----(2)
Diving (1) by (2)
8/6 = (80 - θo)/(70 - θo)
8(70 - θo) = 6(80 - θo)
560 - 8θo = 480 - θo
560 - 480 = -θo + 8θo
80 = 7θo
θo = 11.4°
Again from Newton's law of cooling;
θ = θo + Ce^-kt
Where;
t= 0, θ = 60° and θo = 11.4°
60 = 11.4 + C e^-K(0)
60 - 11.4 = C
C= 48.6°
To obtain K
40 = 11.4 + 48.6e^-10k
40 -11.4 = 48.6e^-10k
28.6/48.6 = e^-10k
0.5585 = e^-10k
-10k = ln0.5585
k= ln0.5585/-10
K= 0.0583
Hence, the temperature in 15 minutes;
θ= 11.4 + 48.6e^(-0.0583 × 15)
θ= 31.7°
