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mojhsa [17]
2 years ago
11

What is the advantage of constantly performing endurance training such as Brisk walking or jogging, Yard work, Dancing, Swimming

, Biking, Climbing stairs, Playing tennis or basketball.
Physics
1 answer:
Eduardwww [97]2 years ago
5 0
Exercise or Getting grader Stomata
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A clam of mass 0.12 kg dropped by a seagull takes 3.0 s to hit the ground. [Neglect friction.]
jarptica [38.1K]

-- The acceleration of gravity is 9.8 m/s².
So if there's no air resistance, the speed of a falling object
always increases by 9.8 m/s for every second it falls.

             Speed  =  (original speed) + (gravity  x  falling time)

-- If it has no vertical speed when it started, then at the end
of 3 seconds, its speed is

                         =       (0)            + (9.8 m/s²  x  3 sec)

                       Velocity  =  29.4 m/s downward . 
6 0
3 years ago
Estimate how far apart the rays of deepest red and deepest violet light are as they exit the bottom surface. assume nred = 1.57
Harlamova29_29 [7]
We begin by noting that the angle of incidence is the one that's taken with respect to the normal to the surface in question. In this case the angle of incidence is 30. The material is Flint Glass according to the original question. The refractive indez of air n1=1, the refractive index of red in flint glass is nred=1.57, finally for violet in the glass medium is nviolet=1.60. Snell's Law dictates:
n_1sin(\theta_1)=n_2sin(\theta_2)
Where \theta_2 differs for each wavelenght, that means violet and red will have different refractive indices in the glass.
In the second figure provided details are given on which are the angles in question, \Delta x is the distance between both rays.
\theta_{2red}=Asin(\frac{sin(30)}{1.57})\approx 18.5705
\theta_{2violet}=Asin(\frac{sin(30)}{1.60})\approx 18.21
At what distance d from the incidence normal will the beams land at the bottom?
For violet we have:
d_{violet}=h.tan(\theta_{2violet})\approx 0.0132m
For red we have:
d_{red}=h.tan(\theta_{2red})\approx 0.0134m
We finally have:
\Delta x=d_{red}-d_{violet}\approx2.8\times10^{-4}m


6 0
3 years ago
A seamount is an isolated land mass rising from the ocean floor.<br> a. True<br> b. False
iris [78.8K]
The answer is A, True.
7 0
3 years ago
The vector position of an object is given by r what is the torque acting on the object about the origin when a force f = (−12.5i
attashe74 [19]
Let the vector position of the object in the (x-y) plane be 
\vec{r} = x \hat{i} + y \hat{j}

The applied force is
\vec{f} = -12.5 \hat{i}&#10;

By definition, the applied torque is
\vec{T} = \vec{r} \times \vec{f} = (x\hat{i} + y\hat{j}) \times (-12.5y \hat{i}) = 12.5\hat{k}

Answer: 12.5y \, \hat{k}

7 0
3 years ago
Two identical cars A and B are at rest on a loading dock with brakes released. Car C, of a slightly different style but of the s
Nadusha1986 [10]

Answer:

Explanation:

Let the velocity after first collision be v₁ and v₂ of car A and B . car A will bounce back .

velocity of approach = 1.5 - 0 = 1.5

velocity of separation = v₁ + v₂

coefficient of restitution = velocity of separation / velocity of approach

.8 = v₁ + v₂ / 1.5

v₁ + v₂ = 1.2

applying law of conservation of momentum

m x 1.5 + 0 = mv₂ - mv₁

1.5 = v₂ - v₁

adding two equation

2 v ₂= 2.7

v₂ = 1.35 m /s

v₁ = - .15 m / s

During second collision , B will collide with stationary A . Same process will apply in this case also. Let velocity of B and A after collision be v₃ and v₄.

For second collision ,

coefficient of restitution = velocity of separation / velocity of approach

.5 = v₃ + v₄ / 1.35

v₃ + v₄ = .675

applying law of conservation of momentum

m x 1.35 + 0 = mv₄ - mv₃

1.35 = v₄ - v₃

adding two equation

2 v ₄= 2.025

v₄ = 1.0125 m /s

v₃ = - 0 .3375  m / s

3 0
3 years ago
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