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7nadin3 [17]
3 years ago
12

Sam has been working to improve his muscular fitness. He jumps rope and trains with weights. What will most likely be the result

of Sam's efforts?
Physics
2 answers:
irina [24]3 years ago
7 0

Answer:

increase bone density

Explanation:

Sam in an effort to improve his muscular fitness jumps rope and trains weights. The result of the Sam's effort would his increased bone density.

Several research have shown a positive impact of exercise on bone density.A regular Pelvic exercise for 30 days increase the hip bone density significantly.

Similar is case of other exercises as well. Doctors recommend exercise to improve bone mass density

const2013 [10]3 years ago
4 0
Increased Bone Density
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A sinusoidal transverse wave travels along a long, stretched string. The amplitude of this wave is 0.08190.0819 m, its frequency
Anvisha [2.4K]

Answer:

The shortest transverse distance between a maximum and a minimum of the wave is 0.1638 m.

Explanation:

Given that,

Amplitude = 0.08190 m

Frequency = 2.29 Hz

Wavelength = 1.87 m

(a). We need to calculate the shortest transverse distance between a maximum and a minimum of the wave

Using formula of distance

d=2A

Where, d = distance

A = amplitude

Put the value into the formula

d=2\times0.08190

d=0.1638\ m

Hence, The shortest transverse distance between a maximum and a minimum of the wave is 0.1638 m.

5 0
3 years ago
What is the frequency of light with a wavelength of 7.9 x 10^-9 m? ( the speed of light is 3.00 x 10^8)
nlexa [21]
Using the formula v=f times lambada
then v=the speed of light.
and f=what’s we’re looking for
and lambada=the wavelength.

so then you sub what you have (v and lambada) in the formula.
then multiply the frequency(f) by the given wavelength and then solve for f

3 0
2 years ago
Sometimes a person cannot clearly see objects close up or far away. To correct this type of vision, bifocals are often used. The
Rudik [331]

Answer:

1)   P₁ = -2 D,   2) P₂ = 6 D

Explanation:

for this exercise in geometric optics let's use the equation of the constructor

          \frac{1}{f} = \frac{1}{p} + \frac{1}{q}

where f is the focal length, p and q are the distance to the object and the image, respectively

1) to see a distant object it must be at infinity (p = ∞)

          \frac{1}{f_1} = \frac{1}{q}

           q = f₁

2) for an object located at p = 25 cm

            \frac{1}{f_2} = \frac{1}{25} + \frac{1}{q}

We can that in the two expressions we have the distance to the image, this is the distance where it can be seen clearly in general for a normal person is q = 50 cm

we substitute in the equations

1) f₁ = -50 cm

2)  

        \frac{1}{f_2} = \frac{1}{25} + \frac{1}{50}

        \frac{1}{f_2} = 0.06

         f₂ = 16.67 cm

the expression for the power of the lenses is

          P = \frac{1}{f}

where the focal length is in meters

           

1)       P₁ = 1/0.50

        P₁ = -2 D

2)     P₂ = 1 /0.16667

        P₂ = 6 D

4 0
2 years ago
Which warning sign suggests being bullied?
Vikentia [17]

Answer:

when you tell someone over and over to stop bothering you and they dont so i think you should tell a teacher

Explanation:

6 0
3 years ago
QUESTION 1
Sveta_85 [38]

distance from the Sun of 2.77 astronomical units or about 414 million km 257 million miles and orbiting period of 4.62 years

7 0
3 years ago
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