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lakkis [162]
3 years ago
13

A caterpillar tries to climb straight up a wall two meters high, but for every 2 cm up it climbs, it slides down 1 cm. Eventuall

y, it reaches the top. When it reaches the top, it does not pull itself over so it will slide down 1 cm.
In the first box, select the total displacement. In the second box, indicate the direction.
Physics
1 answer:
MaRussiya [10]3 years ago
3 0

Answer:

this question does not make sense could you please make it clearer

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The efficiency of a squeaky pulley system is 73 percent. The pulleys are usedto raise a mass to a certain height. What force is
larisa86 [58]

The efficiency of the machine is defined as

\eta = \frac{W_{out}}{W_{in}}

Here

Work out is the work output and Work in is the work input

To find the Work in we have then

W_{in} = \frac{W_{out}}{\eta}

W_{in} = \frac{mgh}{\eta}

Replacing with our values

W_{in} = \frac{(58)(9.8)(3)}{73\%}

W_{in} = 2335.89J

The work done by the applied force is

W = Fd

Here,

F = Force

d = Distnace

Rearranging to find F,

F = \frac{W}{d}

F = \frac{2335.89J }{18}

F = 129.77N

Therefore the force exerted on the machine after rounding off to two significant figures is 130N

8 0
3 years ago
Convert horsepower to Kilo Newton.
SVEN [57.7K]

1 horsepower is equivalent to 0.7457 Kilo Newton. I am hoping that this answer has satisfied your query and it will be able to help you in your endeavor, and if you would like, feel free to ask another question.

8 0
3 years ago
How are streams and rivers agents of erosion?
seropon [69]
Water wears down the earth and loosening the dirt or rock causing it to erode

have a great day

8 0
3 years ago
Water exits a garden hose at a speed of 1.2 m/s. If the end of the garden hose is 1.5 cm in diameter and you want to make the wa
Katarina [22]

Answer:

A 93%

Explanation:

P_1=P_2 = Pressure will be equal at inlet and outlet

\rho = Density of water = 1000 kg/m³

g = Acceleration due to gravity = 9.81 m/s²

v_1 = Velocity at inlet = 1.2 m/s

v_2 = Velocity at outlet

r_1 = Radius of inlet = \dfrac{1.5}{2}=0.75\ cm

r_2 = Radius of outlet

From Bernoulli's relation

P_1+\dfrac{1}{2}\rho v_1^2+\rho gh_1=P_2+\dfrac{1}{2}\rho v_2^2+\rho gh_2\\\Rightarrow \dfrac{1}{2}\rho v_1^2+\rho gh_1=\dfrac{1}{2}\rho v_2^2+\rho gh_2\\\Rightarrow v_2=\sqrt{2(\dfrac{1}{2}v_1^2+gh_1-gh_2)}\\\Rightarrow v_2=\sqrt{2(\dfrac{1}{2}1.2^2+9.81\times 15)}\\\Rightarrow v_2=17.19709\ m/s

From continuity equation

A_1v_1=A_2v_2\\\Rightarrow \pi r_1^2v_1=\pi r_2^2v_2\\\Rightarrow r_2=\sqrt{\dfrac{r_1^2v_1}{v_2}}\\\Rightarrow r_2=\sqrt{\dfrac{0.0075^2\times 1.2}{17.19709}}\\\Rightarrow r_2=0.00198\ m

The fraction would be

\dfrac{A_1-A_2}{A_1}\times 100=\dfrac{r_1^2-r_2^2}{r_1^2}\times 100\\ =\dfrac{0.0075^2-0.00198^2}{0.0075^2}\times 100\\ =93.0304\ \%

The fraction is 93.0304%

8 0
3 years ago
A train is traveling at 30.0 m/sm/s relative to the ground in still air. The frequency of the note emitted by the train whistle
Zigmanuir [339]

Answer

given,

speed of sound = 344 m/s

speed of train = 30 m/s

frequency emitted by the train = 262 Hz

   Doppler's effect

    f_L = \dfrac{v + v_L}{v + v_s}\ f_S

f_L is the frequency of listener

f_S is the frequency of the source of the sound

v is the speed of the sound

v_L is the speed of listener.

v_S is the speed of the source

a) Speed of the passenger in another train , v = 18 m/s

   another train is moving in opposite direction and approaching

   v_L is positive as the listener is moving forward.

    v_S is negative at the source is moving toward the listener.

      f_L = \dfrac{344 + 18}{344 - 30}\times 262

     f_L = 302\ Hz

b) Speed of the passenger in another train , v = 18 m/s

   another train is moving in opposite direction and receding

    v_L is negative as the listener is moving away from source.

    v_S is positive at the source is moving away the listener.

      f_L = \dfrac{344 - 18}{344 + 30}\times 262

     f_L = 228.37\ Hz

8 0
3 years ago
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