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mamaluj [8]
2 years ago
10

What is unique regarding the abdominal muscle when is compared to other muscle in the body?

Physics
1 answer:
marissa [1.9K]2 years ago
8 0
Answer:a is the answer
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In a football game, the running back takes a handoff and begins running toward midfield at 3.21 yards/s . As he moves through hi
Inessa [10]

Given Information:

Initial speed = u = 3.21 yards/s

Acceleration = α  = 1.71 yards/s²

Final speed = v = 7.54 yards/s

Required Information:

Distance = s = ?

Answer:

Distance = s = 13.61

Explanation:

We are given the speeds and acceleration of the runner and we want to find out how much distance he covered before being tackled.

We know from the equations of motion,

v² = u² + 2αs

Where u is the initial speed of the runner, v is the final speed of the runner, α is the acceleration of the runner and s is the distance traveled by the runner.

Re-arranging the above equation for distance yields,

2αs = v² - u²

s = (v² - u²)/2α

s = (7.54² - 3.21²)/2×1.71

s = 46.55/3.42

s = 13.61 yards

Therefore, the runner traveled a distance of 13.61 yards before being tackled.

8 0
3 years ago
A photon of wavelength 192 nm strikes an aluminum surface along a line perpendicular to the surface and releases a photoelectron
alex41 [277]

Answer:

KE=3.529\times10^{−27}\ J

Explanation:

Given that

Wavelength λ=192 nm

So energy of photon,E

E=\dfrac{hC}{\lambda }

Now by putting the values

h=6.6\times 10^{-34}\ m^2.kg/s

C=3\times 10^{8}\ m/s

E=\dfrac{6.6\times 10^{-34}\times 3\times 10^{8}}{192\times 10^{-9} }

E=1.03\times 10^{-18} J

We know that

Kinetic energy given as

KE=\dfrac{P^2}{2m}

KE=\dfrac{E^2}{2mC^2}

KE=\dfrac{(1.03\times 10^{-18})^2}{2\times 1.67\times 10^{-27}(3\times 10^8)^2}

KE=3.529\times10^{−27}\ J

5 0
3 years ago
When _______ light passes through a prism or diffraction grating
lana66690 [7]
I don’t know ... sorry
5 0
3 years ago
What was the direction of the ball’s velocity
Tju [1.3M]

Part of the question is missing. Here it is:

<em>A 72 g autographed baseball slides off of a 1.3 m high table and strikes the floor a horizontal distance of 0.7m away from the table.     The acceleration of gravity is 9.81 m/s2. What was the direction of the ball’s velocity  just before it hit the floor? </em>

Answer:

\theta=-75.7^{\circ}

Explanation:

The motion of the ball is a projectile motion, which consists of two separate motions:

- A horizontal motion at constant velocity

- A vertical motion at constant acceleration (free fall)

We start by analyzing the vertical motion, to find the time of flight of the ball. This can be done by using the suvat equation

s=ut+\frac{1}{2}at^2

where, choosing downward as positive direction:

s =1.3 m is the vertical displacement of the ball

u = 0 is the initial vertical velocity

a=g=9.8 m/s^2 is the acceleration of gravity

t is the time

Solving for t,

t=\sqrt{\frac{2s}{a}}=\sqrt{\frac{2(1.3)}{9.8}}=0.52 s

Now we can find the final vertical velocity of the ball, using:

v_y=u+at

And susbtituting t = 0.52 s, we find

v_y = 0 +(9.8)(0.52)=5.1 m/s

It is important to keep in mind that the direction of this velocity is downward, since we chose downward as positive direction.

The horizontal velocity of the ball instead is constant; we know that the ball covers a horizontal distance of

d = 0.7 m

In a time of

t = 0.52 s

So, the horizontal velocity is

v_x = \frac{0.7}{0.52}=1.3 m/s

So now we can find the direction of the ball's velocity using:

\theta=tan^{-1}(\frac{v_y}{v_x})=tan^{-1}(\frac{5.1}{1.3})=75.7^{\circ}

And since the vertical direction is downward, this means that this velocity is below the horizontal, so the answer is

\theta=-75.7^{\circ}

8 0
3 years ago
The force of gravity is (4 points) 1 a contact force 2 a distance force 3 an electrical force 4 a magnetic force
d1i1m1o1n [39]
It is a distance force
4 0
3 years ago
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