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3 years ago

What type of machine are wire cutter pliers?

Physics

2 answers:

love history [14]3 years ago

4 0

It a compound machine

ludmilkaskok [199]3 years ago

4 0

Answer: Compound machine

Explanation :

The machines that are made up of simple machines are called as Compound machines.

Lever, pulley, and wedge are some examples of simple machines.

A wire cutter pliers consist of a fulcrum of the first class lever. It is used to cut wires.

So, Wire cutter pliers is a compound machine.

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A skier starts from rest at the top of a hill that is inclined 10.5° with respect to the horizontal. The hillside is 200 m long,

Svetlanka [38]

Answer:

d) 289.31 m

Explanation:

Energy provided by potential energy = mgh = m x 9.8x 200 sin10.5 = 357.18m

Energy used by friction = μmgcos 10.5 x 200 = .075 x m x 9.8 x cos 10.5 x200 = 144.54 m .

Energy used by friction on plain surface = μmg x d.( dis distance covered on plain ) =.075x m x 9.8 xd = .735 m d

To equate

357.18 m -144.54 m = .735 m d

d = 289.31 m .

4 0

2 years ago

Given: F = k· m. g<br> Solve for "k"

gulaghasi [49]

Answer:

$F = kmg \\ k = \frac{F}{mg}$

8 0

3 years ago

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You are riding a bicycle. If you apply a forward force of 150 N, and you and the bicycle have a combined mass of 90 kg, what wil

bekas [8.4K]

<h3>Hello There!!</h3>

<h3><u>Given</u>,</h3>

Force(F) = 150N

Mass(m) = 90kg

Acceleration(a) = ?

F= m×a

$150 = 90 \times \text{a} \\ \\ \text{a} = \frac{150}{90} \\ \fbox{cancelling by 3} \\ \\ \text{ a} = \cancel \frac{150}{90} \\ \\ \text{ a} = \frac{5}{3} = 1.67 \text{m/s} {}^{2}$

$\therefore \text{Option A= 1.67 m/s² is the correct answer}$

<h3>Hope this helps</h3>

6 0

2 years ago

I really need help! Please no one has helped me! Will give BRAINLIEST, THANKS, and RATE!! Please help if you are good at science

adell [148]

I took this test and the correct answer is c
I hope this helps

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2 years ago

Read 2 more answers

A sound source A and a reflecting surface B move directly toward each other. Relative to the air, the speed of source A is 28.7

aleksandrvk [35]

(a) 1440.5 Hz

The general formula for the Doppler effect is

$f'=(\frac{v+v_r}{v+v_s})f$

where

f is the original frequency

f is the apparent frequency

$v$ is the velocity of the wave

$v_r$ is the velocity of the receiver (positive if the receiver is moving towards the source, negative otherwise)

$v_s$ is the velocity of the source (positive if the source is moving away from the receiver, negative otherwise)

Here we have

f = 1110 Hz

v = 334 m/s

In the reflector frame (= on surface B), we have also

$v_s = v_A = -28.7 m/s$ (surface A is the source, which is moving towards the receiver)

$v_r = +62.2 m/s$ (surface B is the receiver, which is moving towards the source)

So, the frequency observed in the reflector frame is

$f'=(\frac{334 m/s+62.2 m/s}{334 m/s-28.7 m/s})1110 Hz=1440.5 Hz$

(b) 0.232 m

The wavelength of a wave is given by

$\lambda=\frac{v}{f}$

where

v is the speed of the wave

f is the frequency

In the reflector frame,

f = 1440.5 Hz

So the wavelength is

$\lambda=\frac{334 m/s}{1440.5 Hz}=0.232 m$

(c) 1481.2 Hz

Again, we can use the same formula

$f'=(\frac{v+v_r}{v+v_s})f$

In the source frame (= on surface A), we have

$v_s = v_B = -62.2 m/s$ (surface B is now the source, since it reflects the wave, and it is moving towards the receiver)

$v_r = +28.7 m/s$ (surface A is now the receiver, which is moving towards the source)

So, the frequency observed in the source frame is

$f'=(\frac{334 m/s+28.7 m/s}{334 m/s-62.2 m/s})1110 Hz=1481.2 Hz$

(d) 0.225 m

The wavelength of the wave is given by

$\lambda=\frac{v}{f}$

where in this case we have

v = 334 m/s

f = 1481.2 Hz is the apparent in the source frame

So the wavelength is

$\lambda=\frac{334 m/s}{1481.2 Hz}=0.225 m$

8 0

3 years ago