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3 years ago

What are some of the physical benefits to be derived from aerobic

Physics

1 answer:

VladimirAG [237]3 years ago

6 0

Some of the benefits are increased heart muscles, increase in blood flow, and reduced body fat

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Hi :) In a pendulum experience , what affects the period & explanation too? Thank you sm :))

stich3 [128]

Answer:

T=2π $\sqrt{l/g}$ (simple pendulum)

length of the string and gravity mainly effect the pendulum period

Explanation:

4 0

2 years ago

A rubber toy duck is at rest on an inclined plane. When the angle of inclination of the plane is increased to 36.0°, the toy duc

ch4aika [34]

Answer

toy begins to slide when the plane is at an angle of 36.0°

at this point the toy has just overcome the static friction

$f_s = \mu_s N$

$f_s = \mu_s mgcos \theta$

since toy begins to slide

$f_s = mgsin \theta$

now on comparing

$mgsin \theta = \mu_s mgcos \theta$

$\mu_s = tan \theta$

$\mu_s = tan (36^0)$

$\mu_s = 0.726$

as the body is moving at constant velocity acceleration will be zero

so now,

$m g sin \theta - f_k = ma$

$m g sin \theta - \mu_k mg cos\theta = 0$

$\mu_k = tan \theta$

$\mu_k = tan (29^0)$

$\mu_k = 0.554$

3 0

2 years ago

What is the relationship between temperature, dew point temperature, and air pressure

olchik [2.2K]

Dew point is the temperature at which the water vapor in the air condenses , then evaporates. The barometric or air pressure is independent from the dew point.

6 0

2 years ago

Two 1.9 kg masses are 1.1 m apart (center to center) on a frictionless table. Each has + 9.6 μC of charge.

Anni [7]

Answer:

F= 0.6 N

Explanation:

Fe(electrical force)= $k q1q2/r^2$

$k=9*10^9\\q1=-9.6 *10^-6 C\\q2= -9.6*10^-6 C\\r= 1.1m$

So,

$F=\frac{9*10^9*-9.6 *10^-6 C* -9.6*10^-6 C}{1.1^2}$

$F= 0.6 N$

5 0

3 years ago

Can you help me please?

egoroff_w [7]

<h3>Answer</h3>

option B)

19N

<h3>Explanation</h3>

If the object is at equilibrium, then the net force acting upon the object should be 0 N. Thus, if all the forces are added together, horizontal and vertical forces separately, then the resultant force (the vector sum) should be 0 Newton.

As we only need to find the magnitude of x-component of force F

so find all x component/horizontal forces acting on the object.

50cos(40) - 40cos(25) + 30cos(55) + x = 0

38.30 - 36.25 + 17.21 + x + = 0

19.26 + x = 0

x = - 19.26

x ≈ 19 (magnitude only)

7 0

3 years ago