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valkas [14]
3 years ago
12

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?
Physics
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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Your car's 30.0 W headlight and 2.50 kW starter are ordinarily connected in parallel in a 12.0 V system. What power (in W) would
Tatiana [17]

Answer:

<h3>The power of headlight in series connection is 29.64 W</h3>

Explanation:

Given :

Power of headlight P_{1} = 30 W

Power of starter P_{2} = 2500 W

Voltage of headlight and starter V = 12 V

From equation of power,

 P = \frac{V^{2} }{R}

 R = \frac{V^{2} }{P}

For finding the resistance of headlight and starter,

⇒ For headlight,

 R_{1}  = \frac{144}{30} = 4.8 Ω

⇒ For starter,

R_{2} = \frac{144}{2500} = 0.057 Ω

Since equivalent resistance,

R_{eq} = R_{1} + R_{2} + ........

R_{eq} = 4.8 +0.057 = 4.857 Ω

So power in series is given by,

 P_{s } = \frac{V^{2} }{R_{eq} }  = \frac{144}{4.857}

 P_{s } = 29.64 W

8 0
3 years ago
A child is trying to throw a ball over a fence. She gives the ball an initial speed of 8.0 m/s at an angle of 40° above the hori
EastWind [94]

Answer:

the child is 1.581 m far from the fence

Explanation:

The diagrammatic illustration that give a better view of what the question denote can be seen in the image attached below.

From the image attached below, let assume that the release point is the origin, then equation of the motion (x) is as follows:

x - x_o = u_xt

\mathtt{x = u_xt  \ \  \ since (x_o = 0)}  ---- (1)

the equation of the motion y is :

\mathtt{y - y_o =u_yt - 0.5 gt^2}

\mathtt{y = u_yt-4.9t^2     \ \ \  since (y_o =0)}

\mathtt{ 1= (u \ sin 40^0)t -4.9 \ t^2        }

\mathtt{1 = 8 sin 40^0 t - 4.9 t^2}

\mathtt{1 = 5.14t - 4.9t^2}

\mathtt{4.9t^2 - 5.14t +1 = 0}

By using the quadratic formula, we have;

\mathtt{ \dfrac{ -b \pm \sqrt{b^2 - 4ac}}{2a}}     }

where;

a = 4.9,   b = -5.14     c = 1

= \mathtt{ \dfrac{ -(-5.14) \pm \sqrt{(-5.14)^2 - 4(4.9)(1)}}{2(4.9)}}     }

= \mathtt{ \dfrac{ 5.14 \pm \sqrt{26.4196 -19.6}}{9.8}}     }

= \mathtt{ \dfrac{ 5.14 \pm \sqrt{6.8196}}{9.8}}     }

= \mathtt{ \dfrac{ 5.14+ \sqrt{6.8196}}{9.8}  \  \ OR \  \  \dfrac{ 5.14- \sqrt{6.8196}}{9.8}}    }

= \mathtt{ \dfrac{ 5.14+ 2.6114}{9.8}  \  \ OR \  \  \dfrac{ 5.14- 2.6114}{9.8}}    }

= \mathtt{ \dfrac{ 7.7514}{9.8}  \  \ OR \  \  \dfrac{ 2.5286}{9.8}}    }

= \mathbf{ 0.791 \  \ OR \  \  0.258}    }

In as much as the ball is traveling upward, then we consider t= 0.258sec.

From equation (1)

\mathtt{x = u_x(0.258)}

\mathtt{x = ucos 40^0 (0.258)}

\mathtt{x = 8 \ cos 40^0 (0.258)}

\mathbf{x = 1.581  \ m}

Thus, the child is 1.581 m far from the fence

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3 years ago
After watching a video about submarines, Jamil wants to learn more about the ocean. which question could be answered through sci
SOVA2 [1]
<span>Submarines are equipped with water tanks called ballast tanks that fill up to submerge the vessel. Emptying the tanks and filling them with air causes the submarine to surface.</span>
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3 years ago
A rolling ball has an initial velocity of 1.6 meters per second. if the ball has a constant acceleration of 0.33 meters per seco
lukranit [14]
We have:

Initial velocity (u) = 1.6 m/s
Constant acceleration (a) = 0.33 m/s²
Time (t) = 3.6 sec

There are five constant acceleration equations that would help us to find the velocity:

v=u+at
s=ut+ \frac{1}{2}at^2
s= \frac{1}{2}(u+v)t
v^2=u^2+2as
s=vt- \frac{1}{2}at^2

Since we have u, a, t and we want v
We will use the first formula v=u+at

v=1.6+(0.33)(3.6)
v= 2.788 m/s
7 0
3 years ago
The force of gravity acting on an object is the object's ______. A. acceleration B. mass C. weight D. matter
ryzh [129]

Answer:

d

Explanation:

3 0
2 years ago
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