8. What is the weight in newtons of a 10 kg mass on the earth's surface?
1 answer:
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Answer:
a) v = 16.57 m / s, b) a = 19.6 m / s², d) N = 1.76 10³ N, N / W = 3
Explanation:
This exercise looks interesting, but I think you have some problem with the writing, the questions seem a bit disconnected from the initial text.
Let's answer the questions.
a) For this part we can use energy considerations.
Starting point. The upper part of the trajectory indicates that the arm is horizontally
Em₀ = U = m g h
in this case h = r
Final point. For lower of the trajectory
Em_f = K = ½ m v²
as they indicate that there is no friction
Em₀ = em_f
mgh = ½ m v²
v =
let's calculate
v =
v = 16.57 m / s
b) the centripetal acceleration has the formula
a = v² / r
a = 16.57² / 14.0
a = 19.6 m / s²
c) see attached where the diagram is
where N is the normal and w the weight
d) let's use Newton's second law
N-W = m a
N - mg = m ar
N = m (g + a)
let's calculate
N = 60.0 (9.8 + 19.6)
N = 1.76 10³ N
the relationship with weight is
N / W = 1.76 10³/( 60 9.8)
N / W = 3
normal is three times greater than body weight
e) the answer is reasonable since by Newton's first law the body must continue in a straight line, therefore to change its trajectory a force must be applied to deflect it
Answer:
1. 18.25 m/s
2. 0 m/s
Explanation:
1.So the centripetal acceleration of the ball at this lowest point must be, taking gravity into account
The speed at this point would then be
2. Similarly, if T = mg, then the centripetal acceleration must be
As the ball has no centripetal acceleration, its speed must also be 0 as well.
Answer:
v_avg = 37 km/h
Explanation:
To find the average velocity in the complete trajectory you use the following formula:
( 1 )
v1: velocity in the first part of the trajectory = 70 km/h
v2: velocity in the second part of the trajectory = ?
You calculate v2 by using the following equation for a motion with constant velocity:
you replace the values of v1 and v2 in (1) and you obtain:
hence, the average velocity is 37 km/h