<u>Answer</u>
95
<u>Explanation</u>
One box has 85 ball bearings.
17 boxes = 17 × 85
= 1,445 ball bearing
One contains 90 springs
15 boxes ⇒ 15 × 90
⇒ 1,350 spring
Number of ball bearing more than springs are;
1445 - 1350 =<em> 95</em>
Answer:
25.06s
Explanation:
Remaining part of the question.
(A large stone sphere has a mass of 8200 kg and a radius of 90 cm and floats with nearly zero friction on a thin layer of pressurized water.)
Solution:
F = 60N
r = 90cm = 0.9m
M = 8200kg
Moment of inertia for a sphere (I) = ⅖mr²
I = ⅖ * m * r²
I = ⅖ * 8200 * (0.9)²
I = 0.4 * 8200 * 0.81
I = 2656.8 kgm²
Torque (T) = Iα
but T = Fr
Equating both equations,
Iα = Fr
α = Fr / I
α = (60 * 0.9) / 2656.8
α = 0.020rad/s²
The time it will take her to rotate the sphere,
Θ = w₀t + ½αt²
Angular displacement for one revolution is 2Π rads..
θ = 2π rads
2π = 0 + ½ * 0.02 * t²
(w₀ is equal to zero since sphere is at rest)
2π = ½ * 0.02 * t²
6.284 = 0.01 t²
t² =6.284 / 0.01
t² = 628.4
t = √(628.4)
t = 25.06s
Answer:
The time elapsed at the spacecraft’s frame is less that the time elapsed at earth's frame
Explanation:
From the question we are told that
The distance between earth and Retah is 
Here c is the peed of light with value 
The time taken to reach Retah from earth is 
The velocity of the spacecraft is mathematically evaluated as

substituting values


The time elapsed in the spacecraft’s frame is mathematically evaluated as

substituting value
![T = 90000 * \sqrt{ 1 - \frac{[2.4*10^{8}]^2}{[3.0*10^{8}]^2} }](https://tex.z-dn.net/?f=T%20%20%3D%20%2090000%20%2A%20%20%5Csqrt%7B%201%20-%20%20%5Cfrac%7B%5B2.4%2A10%5E%7B8%7D%5D%5E2%7D%7B%5B3.0%2A10%5E%7B8%7D%5D%5E2%7D%20%7D)

=> 
So The time elapsed at the spacecraft’s frame is less that the time elapsed at earth's frame
Given that,
Puck slide total Deltax = 12m
Puck and board Mk = 0.10
Find the initial speed = ?
We know that,
Vf^2-Vi^2 = 2a Deltax
-Vi^2 = -2MkgDeltax ............(1)
then,
fk = -Mk mg = ma
a = -Mkg .........(2)
From equation (1),
Vi^2 = 2MkgDeltax
Vi = âš2MkgDeltax [g=9.8m/s^2]
Vi = âš2(0.10)(0.98)(12)
Vi = âš0.2(9.8)(12)
Vi = âš1.92(12)
Vi = âš23.52
Vi = 4.8 m/s
Therefore the initial speed of the puck Vi = 4.8 m/s
I think it is B hope this helps
God Bless