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3 years ago

The fact that muscles will improve in strength and conditioning in response to resistance training is called the

1 answer:

3 0

<span>The fact that muscles will improve in strength and conditioning in response to resistance training is called MUSCLE HYPERTROPHY.

This occurs because during resistance training the muscles develop very small tears, called micro-tears, that eventually heal and become scar tissues. As the scar tissues continue to increase, muscle fibers are said to bulk up creating that big and muscular physique that you see in people who are into resistance training.</span>

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Calculate how long

lora16 [44]

Alpha Centauri is 4.4 lightyears away. Traveling at 1/10 speed of light it would take you 44 years.

3 0

3 years ago

A rock is thrown off a cliff at an angle of 53° with respect to the horizontal. The cliff is 100 m high. The initial speed of th

Nadusha1986 [10]

(a) 129.3 m

The motion of the rock is a projectile motion, consisting of two indipendent motions along the x- direction and the y-direction. In particular, the motion along the x- (horizontal) direction is a uniform motion with constant speed, while the motion along the y- (vertical) direction is an accelerated motion with constant acceleration $g=-9.8 m/s^2$ downward.

The maximum height of the rock is reached when the vertical component of the velocity becomes zero. The vertical velocity at time t is given by

$v(t) = v_0 sin \theta +gt$

where

$v_0 = 30 m/s$ is the initial velocity of the rock

$\theta=53^{\circ}$ is the angle

t is the time

Requiring $v(t)=0$ , we find the time at which the heigth is maximum:

$0=v_0 sin \theta + gt\\t=\frac{-v_0 sin \theta}{g}=-\frac{(30)(sin 53^{\circ})}{(-9.8)}=2.44 s$

The heigth of the rock at time t is given by

$y(t) = h+(v_0 sin\theta) t + \frac{1}{2}gt^2$

Where h=100 m is the initial heigth. Substituting t = 2.44 s, we find the maximum height of the rock:

$y=100+(30)(sin 53^{\circ})(2.44)+\frac{1}{2}(-9.8)(2.44)^2=129.3 m$

(b) 44.1 m

For this part of the problem, we just need to consider the horizontal motion of the rock. The horizontal displacement of the rock at time t is given by

$x(t) = (v_0 cos \theta) t$

where

$v_0 cos \theta$ is the horizontal component of the velocity, which remains constant during the entire motion

t is the time

If we substitute

t = 2.44 s

Which is the time at which the rock reaches the maximum height, we find how far the rock has moved at that time:

$x=(30)(cos 53^{\circ})(2.44)=44.1 m$

(c) 7.58 s

For this part, we need to consider the vertical motion again.

We said that the vertical position of the rock at time t is

$y(t) = h+(v_0 sin\theta) t + \frac{1}{2}gt^2$

By substituting

y(t)=0

We find the time t at which the rock reaches the heigth y=0, so the time at which the rock reaches the ground:

$0=100+(30)(sin 53^{\circ})t+\frac{1}{2}(-9.8)t^2\\0=100+23.96t-4.9t^2$

which gives two solutions:

t = -2.69 s (negative, we discard it)

t = 7.58 s --> this is our solution

(d) 136.8 m

The range of the rock can be simply calculated by calculating the horizontal distance travelled by the rock when it reaches the ground, so when

t = 7.58 s

Since the horizontal position of the rock is given by

$x(t) = (v_0 cos \theta) t$

Substituting

$v_0 = 30 m/s\\\theta=53^{\circ}$

and t = 7.58 s we find:

$x=(30)(cos 53^{\circ})(7.58)=136.8 m$

(e) (36.1 m, 128.3 m), (72.2 m, 117.4 m), (108.3 m, 67.4 m)

Using the equations of motions along the two directions:

$x(t) = (v_0 cos \theta) t$

$y(t) = h+(v_0 sin\theta) t + \frac{1}{2}gt^2$

And substituting the different times, we find:

$x(2.0 s)=(30)(cos 53^{\circ})(2.0)=36.1 m$

$y(2.0 s)= 100+(23.96)(2.0)-4.9(2.0)^2=128.3 m$

$x(4.0 s)=(30)(cos 53^{\circ})(4.0)=72.2 m$

$y(4.0 s)= 100+(23.96)(4.0)-4.9(4.0)^2=117.4 m$

$x(6.0 s)=(30)(cos 53^{\circ})(6.0)=108.3 m$

$y(6.0 s)= 100+(23.96)(6.0)-4.9(6.0)^2=67.4 m$

3 0

3 years ago

If light of wavelength 700 nm strikes such a photocathode, what will be the maximum kinetic energy, in eV , of the emitted elect

Oksana_A [137]

If the light of wavelength 700 nm strikes such a photocathode the maximum kinetic energy, in eV, of the emitted electrons is 0.558 eV.

so - $KE_{max} = hc/lembda} work

threshold when KE = 0

hc/lambda = work = 1240/900=1.38 eV

b) Kemax = hc/lambda - work = 1240/640 -1.38=0.558 eV

What is photocathode?

A photocathode electrolyte interface can be used in a photoelectrolysis cell as the primary light-harvesting junction (in conjunction with an appropriate electrochemical anode) or as an optically complementary photoactive half-cell in a tandem photoelectrode photoelectrolysis cell (Hamnett, 1982; Kocha et al, 1994).
In the case of the former, the electrode should ideally harvest photon energy across the majority of the solar spectrum in order to achieve the highest energy conversion efficiency possible.
In the latter case, however, the photocathode may only be active in a specific band of the solar spectrum in order to generate a cathodic photocurrent sufficient to match the current generated in the photoanodic half-cell.

To learn more about Photocathode from the given link:

brainly.com/question/9861585

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3 0

2 years ago

You drop a rock off of a cliff at exactly the edge after 25 seconds exactly you hear a splash your physics friend tells you the

Reptile [31]

The speed of sound is 340.29 meters per second.

Knowing that, we can calculate how high this cliff is by 340.29 * .4

The cliff is 340.29 * .4 = 136.12 meters

8 0

4 years ago

Assume the first dot for animal A represents t = 0. At which time does animal B overtake animal A?

Maslowich

The time in which the animal B overtake animal A according to the motion map is at 2 seconds.

<h3>What is a Motion map?</h3>

This is referred to as one-dimensional plots built off of a position line and depicts velocity and acceleration.

The time in which Animal B is ahead of animal A is at 2 seconds which therefore makes it the most appropriate choice.

Read more about Motion map here brainly.com/question/5064705

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8 0

2 years ago