Answer: Remember speed is distance divided by time, so if he travels 1000 m in 7.045 s, his speed is
(1000 m)/(7.045 s) = 141.9 m/s.
Note there are 1609 metres in a mile, or 1 mi = 1609 m, so m = 1/1609 mi, or
141.9/1609 mi/s = 0.08822 mi/s. Now, note that 1 h = 3600 s, so the speed is
0.08822*3600 mi/h = 317.6 mi/h.
Answer:
Time period of the osculation will be 2.1371 sec
Explanation:
We have given mass m = (B+25)
And the spring is stretched by (8.5 A )
Here A = 13 and B = 427
So mass m = 427+25 = 452 gram = 0.452 kg
Spring stretched x= 8.5×13 = 110.5 cm
As there is additional streching of spring by 3 cm
So new x = 110.5+3 = 113.5 = 1.135 m
Now we know that force is given by F = mg
And we also know that F = Kx
So 

Now we know that 
So 


Answer:
Explanation:
When the box is on the ramp , component of its weight along the ramp
= mg sinθ
Friction force acting on it in upward direction
=μ mg cosθ
For sliding
μ mg cosθ < mg sinθ
μ cosθ < sinθ
.5 x cos35 < sin35
.41 < .57
So the box will slide
When sliding starts , kinetic friction acts
Net force in downward direction
mgsinθ - μ mg cosθ
acceleration
= gsinθ - μ g cosθ
= 5.62 - .3 x 9.8 x cos35
= 5.62 - 2.4
= 3.22 m /s²
Explanation:
The given data is as follows.
= 57 kg,
= 79 kg
= 6.5 m,
= (6.5 - 1.9) m = 4.6 m
(a) The sum of torque ends about far end is as follows.
= 0
= 0
T = 828 N
Therefore, 828 N is the tension in the cable closer to the painter.
(b) Now, we will calculate the sum about close ends as follows.
= 0
T= 506 N
Therefore, 506 N is the tension in the cable further from the painter.