Answer:
v₁ = 3.5 m/s
v₂ = 6.4 m/s
Explanation:
We have the following data:
m₁ = mass of trailing car = 400 kg
m₂ = mass of leading car = 400 kg
u₁ = initial speed of trailing car = 6.4 m/s
u₂ = initial speed of leading car = 3.5 m/s
v₁ = final speed of trailing car = ?
v₂ = final speed of leading car = ?
The final speed of the leading car is given by the following formula:
<u>v₂ = 6.4 m/s</u>
The final speed of the leading car is given by the following formula:
<u>v₁ = 3.5 m/s</u>
Answer:
Orbital speed=8102.39m/s
Time period=2935.98seconds
Explanation:
For the satellite to be in a stable orbit at a height, h, its centripetal acceleration V2R+h must equal the acceleration due to gravity at that distance from the center of the earth g(R2(R+h)2)
V2R+h=g(R2(R+h)2)
V=√g(R2R+h)
V= sqrt(9.8 × (6371000)^2/(6371000+360000)
V= sqrt(9.8× (4.059×10^13/6731000)
V=sqrt(65648789.18)
V= 8102.39m/s
Time period ,T= sqrt(4× pi×R^3)/(G× Mcentral)
T= sqrt(4×3.142×(6.47×10^6)^3/(6.673×10^-11)×(5.98×10^24)
T=sqrt(3.40×10^21)/ (3.99×10^14)
T= sqrt(0.862×10^7)
T= 2935.98seconds
Answer: 2 cm
Explanation:
Given , for a converging lens
Focal length : 
Height of object : 
Object distabce from lens : 
Using lens formula: 
, we get
, where v = image distance from the lens.
On solving aboive equation , we get
Formula of Magnification : 
, where h' is the height of image.
Put value of u, v and h in it , we get
Hence, the height of the image is 2 cm.
I believe that the answer is C. Hope this Helps:)))
Answer:
F = 2(50 N) - (50 N) = 50 N
Explanation:
The direction of F is the direction in which the two students are pushing.
