Answer:
2.87m
Explanation:
Using the law of gravitation to solve this question
F = GMm/r²
G is the gravitational constant
M and m are the masses
r is the distance between the masses
Substitute the given values
G = 6.67×10^-11 m³/kgs²
M =8.8 x 10^6 kg
m = 5.6 x 10^5 kg
F =440N
400 = 6.67×10^-11×8.8 x 10^6 ×5.6 x 10^5/r²
400r² = 328.698×10
400r² = 3286.98
r² = 3286.98/400
r² = 8.21745
r = √8.21745
r = 2.87m
Hence the distance of separation is 2.87m
Volume= Length X width X height.
Plug in the values for each and solve for the volume.
V= (L)(W)(H)
V=(4cm)(5cm)(10cm).
Explanation:
V=u+at
where,
v=final speed
u=initial speed,(starting speed)
a=acceleration
t=time
- v=u+at = 6=2+a*2
6=2+2a
2a=6-2
2a=4
a=4/2 = 2
a =2
2. to find time taken
v=u+at
25=5*2t
2t=25-5
2t=20
t=20/2
t=10sec
3. finding final speed
v=u+at
v=4+10*2
=4+20
v=24m/sec
5.v=u+at
=5+8*10
=5+80
V=85m/sev
6. v=u+at
8=u+4*2
8=u+8
U=8/8
u=1
these are your missing values
Here, you can calculate it's potential energy with respect to ground.
We know, U = mgh
Here, m = 75 Kg
g = 9.8 m/s² [ constant value for earth system ]
h = 300 m
Substitute their values into the expression:
U = 75 × 9.8 × 300
U = 220500 J
In short, Your Final Answer would be 220,500 J
Hope this helps!
Answer:
W = 55.12 J
Explanation:
Given,
Natural length = 6 in
Force = 4 lb, stretched length = 8.4 in
We know,
F = k x
k is spring constant
4 = k (8.4-6)
k = 1.67 lb/in
Work done to stretch the spring to 10.1 in.
![W = \dfrac{k}{2}[x^2]_6^{10.1}](https://tex.z-dn.net/?f=W%20%3D%20%5Cdfrac%7Bk%7D%7B2%7D%5Bx%5E2%5D_6%5E%7B10.1%7D)
W = 55.12 J
Work done in stretching spring from 6 in to 10.1 in is equal to 55.12 J.
