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Juli2301 [7.4K]
3 years ago
6

Does mechanical advantage have units? Explain your answer.

Physics
2 answers:
Citrus2011 [14]3 years ago
7 0

Answer:

NO!!!!!

Explanation:

The mechanical advantage of a machine is the ratio of the load (the resistance overcome by a machine) to the effort (the force applied). For an ideal (without friction) mechanism, it is also equal to: There is no unit for mechanical advantages since the unit for both input and output forces cancel out.

motikmotik3 years ago
6 0

Answer:

The mechanical advantage of a machine is the ratio of the load (the resistance overcome by a machine) to the effort (the force applied). For an ideal (without friction) mechanism, it is also equal to: There is no unit for mechanical advantages since the unit for both input and output forces cancel out.

Explanation:

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A car speeds up from rest to +16 m/ s in 4s. calculate the acceleration
Pani-rosa [81]

The magnitude of acceleration is (change in speed) / (time for the change).

Change in speed = (speed at the end) - (speed at the beginning) =

                                   (16 m/s)  -  (0)  =  16 m/s .

Time for the change  =  4 s .

Magnitude of acceleration = (16 m/s) / (4 s) = 4 m/s per sec = 4 m/s² .


6 0
3 years ago
Read 2 more answers
A 20 m high filled water tank develops a 0.50 cm hole in the vertical wall near the base. With what speed does the water shoot o
Rina8888 [55]

Answer:

The speed of the water shoot out of the hole is 20 m/s.

(d) is correct option.

Explanation:

Given that,

Height = 20 m

We need to calculate the velocity

Using formula Bernoulli equation

\dfrac{1}{2}\rho v_{1}^2+\rho gh_{1}=\dfrac{1}{2}\rho v_{2}^2+\rho gh_{2}

Where,

v₁= initial velocity

v₂=final velocity

h₁=total height

h₂=height of the hole from the base

Put the value into the formula

v_{1}^2=2g(h_{2}-h_{1})

v_{1}=\sqrt{2g(h_{2}-h_{1})}

v_{1}=\sqrt{2\times9.8\times(20-0.005)}

v_{1}=19.7\ m/s= approximate\ 20\ m/s

Hence, The speed of the water shoot out of the hole is 20 m/s.

7 0
3 years ago
Which nuclei is not radioactive? <br> A. Am-241<br> B. Mg-24<br> C. Pu-241<br> D. U-238
Helen [10]

Hello!

Which nuclei is NOT radioactive?

A) Am-241 B) Mg-24 C) Pu-241 D) U-238

Solving:

It is noteworthy that chemical elements located on the periodic table in the lanthanide and actinide groups are radioactive.  

Am-241 (americium) belongs to the group of actinides and is a heavy and radioactive metal.  

Mg-24 (magnesium) is an essential element for the body, mainly for the nervous system, in addition to synthesizing proteins and serves for hormonal control, belongs to the group of alkaline earth metals and is a non-radioactive nucleus.

Pu-241 (plutonium) is an element that is isotope of fission by plutonium, belongs to the group of actinides and is a heavy and radioactive metal.  

U-238 (uranium) is an element that is isotope of non-fission uranium, belongs to the group of actinides and is a heavy and radioactive metal.

Answer:

B) Mg-24

_______________________

I Hope this helps, greetings ... Dexteright02! =)

6 0
3 years ago
A meteoroid is in a circular orbit 600 km above the surface of a distant planet. The planet has the same mass as Earth but has a
AVprozaik [17]
<h2>Answer:</h2><h2>The acceleration of the meteoroid due to the gravitational force exerted by the planet = 12.12 m/s^{2}</h2>

Explanation:

A meteoroid is in a circular orbit 600 km above the surface of a distant planet.

Mass of the planet = mass of earth = 5.972 x 10^{24} Kg

Radius of the earth = 90% of earth radius = 90% 6370 = 5733 km

The acceleration of the meteoroid due to the gravitational force exerted by the planet = ?

By formula, g = \frac{GM}{r^{2} }

where g is the acceleration due to the gravity

G is the universal gravitational constant = 6.67 x 10^{-11} m^{3} kg^{-1} s^{-2}

M is the mass of the planet

r is the radius of the planet

Substituting the values, we get

g =  \frac{(6.67 * 10^{-11}) (5.972 * 10^{24})  }{5733^{2} }

g = 12.12 m/s^{2}

The acceleration of the meteoroid due to the gravitational force exerted by the planet = 12.12 m/s^{2}

6 0
3 years ago
The frequency of the second harmonic of a certain musical instrument is 100 Hz. What is the fundamental frequency of the instrum
ruslelena [56]
The harmonic frequency of a musical instrument is the minimum frequency at which a string that is fixed at both ends in the instrument may vibrate. The harmonic frequency is known as the first harmonic. Each subsequent harmonic has a frequency equal to:
n*f, where n is the number of the harmonic and f is the harmonic frequency. Therefore, the harmonic frequency may be calculated using:
f = 100 / 2
f = 50 Hz
4 0
3 years ago
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