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3 years ago

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A 70 kg human sprinter can accelerate from rest to 10 m/s in 3.0 s. During the same time interval, a 30 kg greyhound can go from

rest to 20 m/s. What is the average power output of each? Average power over a time interval Δt is ΔE/Δt.

1 answer:

ladessa [460]3 years ago

3 0

Answer:

$P_1 = 1166.7 Watt$

$P_2 = 2000 Watt$

Explanation:

Average power for the human sprinter is given as

$Power = \frac{\Delta E}{\Delta t}$

so we have

$P = \frac{\frac{1}{2}mv^2 - 0}{\Delta t}$

$P = \frac{\frac{1}{2}(70)(10^2) - 0}{3}$

$P_1 = 1166.7 Watt$

Average power for greyhound is given as

$P = \frac{\frac{1}{2}mv^2 - 0}{\Delta t}$

$P = \frac{\frac{1}{2}(30)(20^2) - 0}{3}$

$P_2 = 2000 Watt$

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dolphi86 [110]

Answer:

a) 1504.8 J

b) 991.76 J

c) 0J

d) 0J

Explanation:

(a) The work done by the force P on the box is given by the following formula:

$W_P=Px$

P: applied force = 171N

x: distance in which the for P is applied = 8.80m

you replace the values of P and x and obtain:

$W_P=(171N)(8.80m)=1504.8J$

(b) The work don by the friction force is:

$W_f=F_fx=\mu N x=\mu Mg x$

μ = coefficient of kinetic friction = 0.250

M: mass of the box = 46.0kg

g: gravitational constant = 9.8 m/s^2

$W_f=(0.250)(46.0kg)(9.8m/s^2)(8.80m)=991.76J$

(c) The Normal force is

$N=Mg=(46.0kg)(9,8m/s^2)=450.8N$

but this force does not do work on the box because the direction is perpendicular to the direction of the force P.

$W_N=0J$

(d) the same as before:

$W_g=0J$

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2 years ago

Give four importantnce of recycling

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Answer:

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A soccer player extends her lower leg in a kicking motion by exerting a force with the muscle above the knee in the front of her

lisabon 2012 [21]

Answer:

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Explanation:

Analogously to Newton's second law, torque can be defined as:

$\tau=I\alpha$

Here, I is the moment of inertia and $\alpha$ is the angular acceleration. We have:

$\tau=(0.65kg*m^2)(29.5\frac{rad}{s^2})\\\tau=19.18N*m$

Torque is the vector product of the position vector of the point at which the force is applied by the force vector:

$\vec{\tau}=\vec{r}\times \vec{F}$

Since the effective lever arm is perpendicular to the force, the angle between them is $90^\circ$ . The magnitud of this vector product is defined as:

$\tau=rFsen\theta$ .

Solving for F and replacing the known values:

$F=\frac{\tau}{rsen\theta}\\F=\frac{19.18N*m}{1.9m(sen90^\circ)}\\F=10.09N$

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3 years ago

Jumping up before the elevator hits. After the cable snaps and the safety system fails, an elevator cab free-falls from a height

noname [10]

Answer:

a) I = 2279.5 N s , b) F = 3.80 10⁵ N, c) I = 3125.5 N s and d) F = 5.21 10⁵ N

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I =∫ F dt = Δp

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Let's calculate the final speed using kinematics, as the cable breaks the initial speed is zero

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I = 94 24.25 - 0

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just before the crash the passenger jumps up with v = 8 m / s, let's take the moments of interest just when the elevator arrives with a speed of 24.25m/s down and as an end point the jump up to vf = 8 m / n

c) I = m $v_{f}$ - m v₀

I = 94 8 - 94 (-24.25)

I = 3125.5 N s

d) F = I / t

F = 3125.5 / 6 10⁻³

F = 5.21 10⁵ N

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