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3 years ago

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When the palmaris longus muscle in the forearm is flexed, the wrist moves back and forth. If the muscle generates a force of 45.

5 N and it is acting with an effective lever arm of 2.45 cm, what is the torque that the muscle produces on the wrist?

1 answer:

SpyIntel [72]3 years ago

3 0

Answer:

Torque on the rocket will be 1.11475 N -m

Explanation:

We have given that muscles generate a force of 45.5 N

So force F = 45.5 N

This force acts on the is acting on the effective lever arm of 2.45 cm

So length of the lever arm d = 2.45 cm = 0.0245 m

We have to find torque

We know that torque is given by $\tau =F\times d=45.5\times 0.0245=1.11475N-m$

So torque on the rocket will be 1.11475 N -m

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A projectile is launched from ground level at angle u and speed v0 into a headwind that causes a constant horizontal acceleratio

ale4655 [162]

Answer:

Explanation:

Given

Launch angle =u

Initial Speed is $v_0$

Horizontal acceleration is $a_x=a$

At maximum height velocity is zero therefore

$v_f=v_i-gt$

$0=v_0\sin u-gt$

$t=\frac{v_0\sin u}{g}$

Total time of flight $T=2t=\frac{2v_0\sin u}{g}$

During this time horizontal range is

$R=v_o\cos u\cdot 2t-\frac{a(2t)^2}{2}$

$R=\frac{2v_0^2\sin u\cos u}{g}-\frac{2av_0^2\sin ^u}{g^2}$

For maximum range $\frac{\mathrm{d} R}{\mathrm{d} u}=0$

$\frac{\mathrm{d} R}{\mathrm{d} u}=\frac{2v_0^2\cos 2u}{g}-\frac{4av_0^2\sin u\cos u}{g^2}$

$\frac{\mathrm{d} R}{\mathrm{d} u}=\frac{2v_0^2}{g}\left [ \cos 2u-\frac{a}{g}\sin 2u\right ]=0$

$\tan 2u=\frac{g}{a}$

$u=\frac{1}{2}tan ^{-1}\frac{g}{a}$

(b)If a =10% g

$a=0.1g$

thus $u=\frac{1}{2}tan^{-1}\frac{g}{0.1g}$

$u=42.14^{\circ}$

7 0

2 years ago

A crude approximation for the x component of velocity in an incompressible laminar boundary layer is a linear variation from u =

slega [8]

Answer:

2.5 * 10^-3

Explanation:

<u>solution:</u>

The simplest solution is obtained if we assume that this is a two-dimensional steady flow, since in that case there are no dependencies upon the z coordinate or time t. Also, we will assume that there are no additional arbitrary purely x dependent functions f (x) in the velocity component v. The continuity equation for a two-dimensional in compressible flow states:

<em>δu/δx+δv/δy=0</em>

so that:

<em>δv/δy= -δu/δx</em>

Now, since u = Uy/δ, where δ = cx^1/2, we have that:

<em>u=U*y/cx^1/2</em>

and we obtain:

<em>δv/δy=U*y/2cx^3/2</em>

The last equation can be integrated to obtain (while also using the condition of simplest solution - no z or t dependence, and no additional arbitrary functions of x):

v=∫δv/δy(dy)=U*y/4cx^1/2

=y/x*(U*y/4cx^1/2)

=u*y/4x

which is exactly what we needed to demonstrate.

Also, using u = U*y/δ in the last equation we can obtain:

v/U=u*y/4*U*x

=y^2/4*δ*x

which obviously attains its maximum value for the which is y = δ (boundary-layer edge). So, finally:

(v/U)_max=δ^2/4δx

=δ/4x

=2.5 * 10^-3

7 0

2 years ago

Jerry the mouse is running along a straight desert road at a constant velocity of 18 m/s. If a certain Tom cat wants to capture

Kruka [31]

Answer:

a) t = 1.75 s

b) x = 31.5 m

Explanation:

a) The time at which Tom should drop the net can be found using the following equation:

$y_{f} = y_{0} + v_{oy}t - \frac{1}{2}gt^{2}$

Where:

$y_{f}$ : is the final height = 0

y₀: is the initial height = 15 m

g: is the gravity = 9.81 m/s²

$v_{0y}$ : is the initial vertical velocity of the net = 0 (it is dropped from rest)

$0 = 15m - \frac{1}{2}9.81 m/s^{2}*t^{2}$

$t = \sqrt{\frac{2*15 m}{9.81 m/s^{2}}} = 1.75 s$

Hence, Tom should drop the net at 1.75 s before Jerry is under the bridge.

b) We can find the distance at which is Jerry when Tom drops the net as follows:

$v = \frac{x}{t}$

$x = v*t = 18 m/s*1.75 m = 31.5 m$

Then, Jerry is at 31.5 meters from the bridge when Jerry drops the net.

I hope it helps you!

3 0

2 years ago

Watt (w) is a drived unit why

ValentinkaMS [17]

Answer:

because it is from a mathematical combination of SI base units

Explanation:

4 0

2 years ago

Master of physics needed

Delicious77 [7]

Hey JayDilla, I get 1/3. Here's how:
Kinetic energy due to linear motion is:
$E_{linear}= \frac{1}{2}mv^2$
where
$v=r \omega$
giving
$E_{linear}= \frac{1}{2}mr^2 \omega ^2$

The rotational part requires the moment of inertia of a solid cylinder
$I_{cyl} = \frac{1}{2}mr^2$
Then the rotational kinetic energy is
$E_{rot}= \frac{1}{2}I \omega ^2= \frac{1}{4}mr^2 \omega ^2$
Adding the two types of energy and factoring out common terms gives
$\frac{1}{2}mr^2 \omega ^2(1+ \frac{1}{2})$
Here the "1" in the parenthesis is due to linear motion and the "1/2" is due to the rotational part. Since this gives a total of 3/2 altogether, and the rotational part is due to a third of this (1/2), I say it's 1/3.

8 0

3 years ago