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4 years ago

Is walking just a constant state of falling?

Physics

1 answer:

atroni [7]4 years ago

8 0

Answer:

Yes. Walking is controlled falling because you need to let go in order to move forward. If you never let your foot fall, your movements would be stilted and robotic. And according to medical engineers," When we walk normally we are constantly correcting tiny falls to keep ourselves stable."

Explanation:

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What is the speed over the ground of a mosquito flying 2 m/s relative to the air against a 2 m/s headwind?

salantis [7]

We have that the speed over the ground of a mosquito flying 2 m/s relative to the air against a 2 m/s headwind is

X=0m/s

From the question we are told that

mosquito flying 2 m/s

against a 2 m/s headwind

Generally

The speed over the ground is the Flight Speed minus resistance speed

Generally the equation for the speed over the ground is mathematically given as

X=Flight Speed-resistance speed

Therefore

X=2-2

X=0m/s

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3 years ago

Read 2 more answers

If a capacitor has opposite 4.9 μc charges on the plates, and an electric field of 1.6 kv/mm is desired between the plates, what

Temka [501]

Q = 4.6E-6 C
E = 2.6 kV/mm = 2600 kV/m = 2600000 V/m 
ε = 8.85E-12 F/m 
E = Q/Aε 
A = Q/Eε = 4.6E-6/(2600000*8.85E-12) = 0.2 m^2

3 0

4 years ago

It has been claimed that an insect called the froghopper (Philaenus spumarius) is the best jumper in the animal kingdom. This in

Vladimir [108]

Answer:

v=4m/s

Explanation:

The formulas for accelerated motion are:

$v=v_0+at\\x=x_0+v_0t+\frac{at^2}{2}$

We can derive the formula $v^2=v_0^2+2ad$ from them.

We have:

$v-v_0=at\\t=\frac{v-v_0}{a}$

And substitute:

$x=x_0+v_0(\frac{v-v_0}{a})+\frac{a}{2}(\frac{v-v_0}{a})^2\\x-x_0=\frac{v_0(v-v_0)}{a}+\frac{(v-v_0)^2}{2a}\\2a(x-x_0)=2v_0(v-v_0)+(v-v_0)^2=2v_0v-2v_0^2+v^2+v_0^2-2vv_0=v^2-v_0^2$

Where in the first step of the last row we just multiplied everything by 2a. Since $x-x_0$ is the displacement d, we have proved that $v^2=v_0^2+2ad$

We use then our values to calculate the final velocity when starting from rest, traveling a distance 0.002m with acceleration $4000 m/s^2$ :

$v=\sqrt{v_0^2+2ad}=\sqrt{2(4000m/s^2)(0.002m)}=4m/s$

7 0

3 years ago

A 29.0-kg block is initially at rest on a horizontal surface. A horizontal force of 71.0 N is required to set the block in motio

Julli [10]

Answer:

(a) $\mu_s=0.25$

(b) $\mu_k=0.20$

Explanation:

According to Newton's second law:

$\sum F_y:N=mg\\\sum F_x:F_a=F_f$

Recall that the frictional force is related jointly with the coefficient of friction and normal force $F_f=\mu N$ . Replacing in the above equation, we get the coefficient of friction: