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NemiM [27]
3 years ago
7

If the IMA of a machine is 2 and the effort force is 50 newtons, then the force applied to the resistance is _____. 48 N 50 N 25

N 100 N
Physics
2 answers:
yawa3891 [41]3 years ago
8 0

Answer:

100N

Explanation:

he is right

zysi [14]3 years ago
7 0

IMA stands for ideal mechanical advantage, which is the theoretical force amplification factor on an ideal mechanical device free of friction, deformations, etc.


If the applied force (effort) is 50N, then the force applied to the resistance is multiplied by the IMA=2 to get 100N.

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A tennis player swings her 1000 g racket with a speed of 12 m/s. she hits a 60 g tennis ball that was approaching her at a speed
tensa zangetsu [6.8K]

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Explanation:

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Which of these galaxies would you most likely find at the center of a large cluster of galaxies?
Arada [10]

Answer:

b. a large elliptical galaxy

Explanation:

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4 0
3 years ago
You collect some more data on that horse at a later time interval, but now you are measuring thehorse’s velocity, not its positi
Monica [59]

Answer:

a)  x(t) = 10t + (2/3)*t^3

b) x*(0.1875) = 10.18 m

Explanation:

Note: The position of the horse is x = 2m. There is a typing error in the question. Otherwise, The solution to cubic equation holds a negative value of time t.

Given:

- v(t) = 10 + 2*t^2 (radar gun)

- x*(t) = 10 + 5t^2 + 3t^3  (our coordinate)

Find:

-The position x of horse as a function of time t in radar system.

-The position of the horse at x = 2m in our coordinate system

Solution:

- The position of horse according to radar gun:

                              v(t) = dx / dt = 10 + 2*t^2

- Separate variables:

                              dx = (10 + 2*t^2).dt

- Integrate over interval x = 0 @ t= 0

                             x(t) = 10t + (2/3)*t^3

- time @ x = 2 :

                              2 = 10t + (2/3)*t^3

                              0 = 10t + (2/3)*t^3 + 2

- solve for t:

                              t = 0.1875 s

- Evaluate x* at t = 0.1875 s

                              x*(0.1875) = 10 + 5(0.1875)^2 + 3(0.1875)^3

                              x*(0.1875) = 10.18 m

3 0
4 years ago
A man 2 meters tall walks at the rate of 2 meters per second toward a streetlight that's 5 meters above the ground. At what rate
GaryK [48]

Answer:

It changes at a rate of 4/3 meter per second

Explanation:

In the given figure below we have

\Delta OBD\simeq \Delta ABC\\\\\therefore \frac{5}{X+Y}=\frac{2}{Y}\\\\

Solving for Y given  X=2m/s we get

\frac{5}{2+Y}=\frac{2}{Y}\\\\5Y=4+2Y\\\\Y=\frac{4}{3}m/s

8 0
3 years ago
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