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kirill115 [55]
3 years ago
13

A 1.10-kg wrench is acting on a nut trying to turn it. The length of the wrench lies directly to the east of the nut. A force 15

0.0 N acts on the wrench at a position 15.0 cm from the center of the nut in a direction 30.0° north of east. What is the magnitude of the torque about the center of the nut?
Physics
1 answer:
Reika [66]3 years ago
3 0
This can be solved with the following procedure:
<span>150N * sin30 *. 15  = 11.25NM.
</span>Remember to u<span>se the triangle method for visualization, draw the right triangle with the properties given. Hope this is useful</span>
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A group of individuals  living in a particular geographic area is termed population.
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4. What is the acceleration of a plane that changes velocity from 75 m/s to
kompoz [17]

Answer:

a =  \frac{v - u}{t}  \\ s =  \frac{ {v}^{2} -  {u}^{2}  }{2a}

Explanation:

Apply SUVAT

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You plan to take a trip to the moon. Since you do not have a traditional spaceship with rockets, you will need to leave the eart
hichkok12 [17]

Answer:

v = 3.5 \times 10^5 m/s

Explanation:

At some distance from the Earth the force of attraction due to moon is balanced by the force due to Moon

so we will have

\frac{GM_em}{r^2} = \frac{GM_m}{(d-r)^2}

now we have

\frac{d - r}{r} = \sqrt{\frac{M_m}{M_e}}

\frac{3.844\times 10^8 - r}{r} = \sqrt{\frac{7.36 \times 10^{22}}{5.9742\times 10^{24}}}

so we will have

r = 3.46 \times 10^8 m

Now by energy conservation

-\frac{GM_e}{R_e} - \frac{GM_m}{d - (R_e + R_m)} + \frac{1}{2}v^2 = -\frac{GM_e}{r} - \frac{GM_m}{d - r}

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7 0
3 years ago
A transverse periodic wave is established on a string. The wave is described by the expression y = 0.005 sin(20.0x + 2ft) where
Crank

Answer:

f =63.64Hz

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The equation of the transverse periodic wave

y = 0.005 sin(20.0x + 2ft)

From the equation we can deduce that

Amplitude A = 0.005

Wave number K = 20

Angular frequency w = 2πf

We are asked to find the frequency (f)

the formular for the wave number is given as K = 2π / lambda

Lambda = 2π/ K

Lamda is the wavelength

= 2 × 3.142 /20

= 0.3142m

The velocity of the wave is expressed as v = frequency × lamba (wavelength)

Frequency = velocity / wavelength

f = 20/ 0.3142

f = 63.65Hz

6 0
2 years ago
if the vessel in the sample problem accelerates fir 1.00 min, what will its speed be after that minute ?
LUCKY_DIMON [66]

Answers:

a) 154.08 m/s=554.68 km/h

b) 108 m/s=388.8 km/h

Explanation:

<u>The complete question is written below: </u>

<u></u>

<em>In 1977 off the coast of Australia, the fastest speed by a vessel on the water was achieved. If this vessel were to undergo an average acceleration of 1.80 m/s^{2}, it would go from rest to its top speed in 85.6 s.  </em>

<em>a) What was the speed of the vessel? </em>

<em> </em>

<em>b) If the vessel in the sample problem accelerates for 1.00 min, what will its speed be after that minute? </em>

<em></em>

<em>Calculate the answers in both meters per second and kilometers per hour</em>

<em></em>

a) The average acceleration a_{av} is expressed as:

a_{av}=\frac{\Delta V}{\Delta t}=\frac{V-V_{o}}{\Delta t} (1)

Where:

a_{av}=1.80 m/s^{2}

\Delta V is the variation of velocity in a given time \Delta t, which is the difference between the final velocity V and the initial velocity V_{o}=0 (because it starts from rest).

\Delta t=85.6 s

Isolating V from (1):

V=a_{av}\Delta t + V_{o} (2)

V=(1.80 m/s^{2})(85.6 s) + 0 m/s (3)

V=154.08 \frac{m}{s} (4)

If 1 km=1000m and 1 h=3600 s then:

V=154.08 \frac{m}{s}=554.68 \frac{km}{h} (4)

b) Now we need to find the final velocity when \Delta t=1 min=60 s:

<em></em>

V=(1.80 m/s^{2})(60 s) + 0 m/s (5)

V=108 \frac{m}{s}=388.8 \frac{km}{h} (6)

5 0
3 years ago
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