A characteristic common to waves is that they

For exercise, an athlete lifts a barbell that weighs 400 N from the ground to a height of 2.0 m in a time of 1.6 s. Assume the e

Sati [7]

Answer:

Explanation:

(ΔK + ΔUg + ΔUs + ΔEch + ΔEth = W)

ΔK is increase in kinetic energy . As the athelete is lifting the barbell at constant speed change in kinetic energy is zero .

ΔK = 0

ΔUg is change in potential energy . It will be positive as weight is being lifted so its potential energy is increasing .

ΔUg = positive

ΔUs is change in the potential energy of sportsperson . It is zero since there is no change in the height of athlete .

ΔUs = 0

ΔEth is change in the energy of earth . Here earth is doing negative work . It is so because it is exerting force downwards and displacement is upwards . Hence it is doing negative work . Hence

ΔEth = negative .

b )

work done by athlete

= 400 x 2 = 800 J

energy output = 800 J

c )

It is 25% of metabolic energy output of his body

so metalic energy output of body

= 4x 800 J .

3200 J

power = energy output / time

= 3200 / 1.6

= 2000 W .

d )

1 ) Since he is doing same amount of work , his metabolic energy output is same as that in earlier case .

2 ) Since he is doing the same exercise in less time so his power is increased . Hence in the second day his power is more .

5 0

3 years ago

Riders in an amusement park ride shaped like a Viking ship hung from a large pivot are rotated back and forth like a rigid pendu

erica [24]

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 = $\sqrt{2gh}$

let's calculate

v = $\sqrt{2 \ 9.8 \ 14.0}$

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

6 0

2 years ago

The force between a pair of charges is 900 newtons. The distance between the charges is 0.01 meters. If one of the charges is 2e

Maksim231197 [3]

Answer:

$\fbox{strength \: of \: the \: other \: charge = - 0.0196 Ke \: Coulomb}$

Explanation:

Given:

Force between pair of charges= 900 newtons

The distance between the charges = 0.01 meters

Strength of Charge first q1 = 2e-10 Coulomb

To find:

Strength of Charge second q2 = ____ Coulomb?

Solution:

We know that,

Force between two charges separate by distance r is given by the equation,

$|F| = K_e \frac{q1 \cdot \: q2}{ {r}^{2} } \\ 900 =K_e \frac{(2e - 10)\cdot \: q2}{ {0.01}^{2} } \\ 900 \times {10}^{ - 4} = K_e {(2e - 10)\cdot \: q2} \\ q2 = \frac{9 \times {10}^{ - 2} }{(2e - 10) K_e} \\ \\ \fbox{We \: know \: that \: e = 2.71 } \\ substituting \: the \: value \: \\ q2 = \frac{9 \times {10}^{ - 2} }{(2 \times 2.71 - 10)K_e} \\ q2 = \frac{0.09}{ - 4.58 K_e} \\ q2 = \frac{-0.0196}{K_e}\: coulomb$