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3 years ago

What do fast-twitch and slow-twitch muscle fibers share in common

Physics

1 answer:

leva [86]3 years ago

7 0

Answer:

Explanation:Slow-twitch muscle fibers support long distance endurance activities like marathon running, while fast-twitch muscle fibers support quick, powerful movements such as sprinting or weightlifting.

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An 82.0 kg spacewalking astronaut pushes off a 655 kg satellite, exerting a 95.0 N force for the 0.530 s it takes him to straigh

Tcecarenko [31]

Answer:

41.41 m

Explanation:

When force F is applied on an object of mass m for time t and velocity v₁ is created

F X t = mv₁

F = 95 N , t = .53 s, m = 655 kg

95 x .53 = 655 x v₁

v₁ = .0768 m/s

Applying conservation of momentum on man and satellite

m₁ v₁ = m₂v₂

655 x .0768 = 82 xv₂

v₂ = .6134 m/s

their relative velocity

= .6134 + .0768

= .6902 ( they are in opposite direction )

After 60 second distance between them

= 60 x .6902 m

= 41.41 m

6 0

3 years ago

Jail

kykrilka [37]

Answer:

c.o Kg • ms 08

Explanation:

put all together

7 0

3 years ago

A 0.150 kg stone rests on a frictionless, horizontal surface. A bullet of mass 9.50 g, traveling horizontally at 380 m/s, strike

Anvisha [2.4K]

Answer:

(a)Magnitude=28.81 m/s

Direction=33.3 degree below the horizontal

(b) No, it is not perfectly elastic collision

Explanation:

We are given that

Mass of stone, M=0.150 kg

Mass of bullet, m=9.50 g= $9.50\times 10^{3} kg$

Initial speed of bullet, u=380 m/s

Initial speed of stone, U=0

Final speed of bullet, v=250m/s

a. We have to find the magnitude and direction of the velocity of the stone after it is struck.

Using conservation of momentum

$mu+ MU=mv+ MV$

Substitute the values

$9.5\times 10^{-3}\times 380 i+0.150(0)=9.5\times 10^{-3} (250)j+0.150V$

$3.61i=2.375j+0.150V$

$3.61 i-2.375j=0.150V$

$V=\frac{1}{0.150}(3.61 i-2.375j)$

$V=24.07i-15.83j$

Magnitude of velocity of stone

= $\sqrt{(24.07)^2+(-15.83)^2}$

|V|=28.81 m/s

Hence, the magnitude and direction of the velocity of the stone after it is struck, |V|=28.81 m/s

Direction

$\theta=tan^{-1}(\frac{y}{x})$

= $tan^{-1}(\frac{-15.83}{24.07})$

$\theta=tan^{-1}(-0.657)$

=33.3 degree below the horizontal

(b)

Initial kinetic energy

$K_i=\frac{1}{2}mu^2+0=\frac{1}{2}(9.5\times 10^{-3})(380)^2$

$K_i=685.9 J$

Final kinetic energy

$K_f=\frac{1}{2}mv^2+\frac{1}{2}MV^2$

= $\frac{1}{2}(9.5\times 10^{-3})(250)^2+\frac{1}{2}(0.150)(28.81)^2$

$K_f=359.12 J$

Initial kinetic energy is not equal to final kinetic energy. Hence, the collision is not perfectly elastic collision.

5 0

2 years ago

PLS HELP WILL MARK BRAINLIEST

MariettaO [177]

Answer:

2.835 Watts

Explanation:

P = I²R

P = 1.5² × 1.26

P = 2.835 Watts

3 0

3 years ago

Read 2 more answers

A 42.0-kg parachutist is moving straight downward with a speed of 3.85 m/s. (a) If the parachutist comes to rest with constant a

RideAnS [48]

Answer:

-414.96 N

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

$v^2-u^2=2as\\\Rightarrow a=\frac{v^2-u^2}{2s}\\\Rightarrow a=\frac{0^2-3.85^2}{2\times 0.75}\\\Rightarrow a=-9.88\ m/s^2$

$F=ma\\\Rightarrow F=42\times -9.88\\\Rightarrow F=-414.96\ N$

The force the ground exerts on the parachutist is -414.96 N

If the distance is shorter than 0.75 m then the acceleration will increase causing the force to increase

5 0

3 years ago