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3 years ago

13

Jazz is a 172 lb athlete who exercised at 7.6 METs. At this workload, what is his energy expenditure in kcals/min.? Round to the

nearest whole number.\

1 answer:

dezoksy [38]3 years ago

8 0

Answer:

Energy expenditure in K cals/min = 10 K cals /min (approximately)

Explanation:

As we know

Energy expenditure in Kcal/min= METs x 3.5 x Body weight (kg) / 200

Given is METs=7.6

Weight of Jazz= 172lb=78.02kg

putting the values in formula,

Energy expenditure in K cals/min= 7.6 x 3.5 x 78.02 / 200

=10.38 K cals /min

=10 K cals /min (approximately)

Therefore, Energy expenditure in K cals/min by Jazz will be approximately 10 K cals /min

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In Example 2.12, two circus performers rehearse a trick in which a ball and a dart collide. Horatio stands on a platform 6.4 m a

pickupchik [31]

Answer:

time of collision is

t = 0.395 s

$h = 5.63 m$

so they will collide at height of 5.63 m from ground

Explanation:

initial speed of the ball when it is dropped down is

$v_1 = 0$

similarly initial speed of the object which is projected by spring is given as

$v_2 = 16.2 m/s$

now relative velocity of object with respect to ball

$v_r = 16.2 m/s$

now since we know that both are moving under gravity so their relative acceleration is ZERO and the relative distance between them is 6.4 m

$d = v_r t$

$6.4 = 16.2 t$

$t = 0.395 m$

Now the height attained by the object in the same time is given as

$h = v_2 t - \frac{1}{2}gt^2$

$h = 16.2(0.395) - \frac{1}{2}(9.81).395^2$

$h = 5.63 m$

so they will collide at height of 5.63 m from ground

4 0

3 years ago

Water flows through a garden hose which is attached to a nozzle. The water flows through hose with a speed of 1.81 m/s and throu

Serggg [28]

Answer:

a) 17.086m

b) 0.1671 m

Explanation:

Given data: speed of water through the hose = 1.81 m/s

through the nozzle = 18.3 m/s

We know that maximum height of an object with upward velocity v is given by,

a) H = v^2/2g

where H is the maximum height water emerges

= 18.3^2/(2×9.8) = 17.086 m answer

b) Again,

H = v^2/2g

= 1.81^2/(2×9.8) = 0.1671 m

6 0

2 years ago

Which colors allowed the most electricity to be generated

Naddik [55]

Yellow and red hope that helped

3 0

3 years ago

A beam of light strikes a sheet of glass at an angle of 56.6° with the normal in air. You observe that red light makes an angle

Yuri [45]

Answer:

(a). Index of refraction are $n_{red}$ = 1.344 & $n_{violet}$ = 1.406

(b). The velocity of red light in the glass $v_{red} =$ 2.23 × $10^{8} \ \frac{m}{s}$

The velocity of violet light in the glass $v_{violet} =$ 2.13 × $10^{8} \ \frac{m}{s}$

Explanation:

We know that

Law of reflection is

$n_1 \sin\theta_{1} = n_2 \sin\theta_{2}$

Here

$\theta_1$ = angle of incidence

$\theta_2$ = angle of refraction

(a). For red light

1 × $\sin 56.6$ = $n_{red}$ × $\sin 38.4$

$n_{red}$ = 1.344

For violet light

1 × $\sin 56.6$ = $n_{violet}$ × $\sin 36.4$

$n_{violet}$ = 1.406

(b). Index of refraction is given by

$n = \frac{c}{v}$

$n_{red}$ = 1.344

$v_{red} = \frac{c}{n_{red} }$

$v_{red} = \frac{3(10^{8} )}{1.344}$

$v_{red} =$ 2.23 × $10^{8} \ \frac{m}{s}$

This is the velocity of red light in the glass.

The velocity of violet light in the glass is given by

$v_{violet} = \frac{3(10^{8} )}{1.406}$

$v_{violet} =$ 2.13 × $10^{8} \ \frac{m}{s}$

This is the velocity of violet light in the glass.

8 0

2 years ago

Why might surface mining be less risky for miners than underground mining?

aev [14]

Surface miners work aboveground. (Apex) ^-^

6 0

2 years ago

Read 2 more answers