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NemiM [27]
2 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]2 years ago
8 0

Answer:

100N

Explanation:

he is right

zysi [14]2 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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What are the concepts of dynamically continuous innovation and discontinuous innovation? Can you share examples to illustrate th
KiRa [710]

Answer:

Explained

Explanation:

Dynamically continuous innovation:

- Falls in between continuous and discontinuous innovation.

-Changes in customer habits are not as large as in discontinuous innovation and not as negligeble as in continuous innovation.

best example can as simple as transformation in  Television. New HD TVs have flat panels, wide screens and improved performance The Added features are considered dynamically improved.

Discontinuous innovation:

- discontinuous innovation comprise of new to world product only so they are discontinuous to every customer segment.

- these product are so fundamentally different from the the product that already exist that they reshape market and competition.

For example- the mobile and the internet technology are reshaping the market through regular innovation and change.

7 0
3 years ago
Determine the energy required to accelerate an electron between each of the following speeds. (a) 0.500c to 0.900c MeV (b) 0.900
Aleonysh [2.5K]

Answer:

The energy required to accelerate an electron is 0.582 Mev and 0.350 Mev.

Explanation:

We know that,

Mass of electron m_{e}=9.11\times10^{-31}\ kg

Rest mass energy for electron = 0.511 Mev

(a). The energy required to accelerate an electron from 0.500c to 0.900c Mev

Using formula of rest,

E=\dfrac{E_{0}}{\sqrt{1-\dfrac{v_{f}^2}{c^2}}}-\dfrac{E_{0}}{\sqrt{1-\dfrac{v_{i}^2}{c^2}}}

E=\dfrac{0.511}{\sqrt{1-\dfrac{(0.900c)^2}{c^2}}}-\dfrac{0.511}{\sqrt{1-\dfrac{(0.500c)^2}{c^2}}}

E=0.582\ Mev

(b). The energy required to accelerate an electron from 0.900c to 0.942c Mev

Using formula of rest,

E=\dfrac{E_{0}}{\sqrt{1-\dfrac{v_{f}^2}{c^2}}}-\dfrac{E_{0}}{\sqrt{1-\dfrac{v_{i}^2}{c^2}}}

E=\dfrac{0.511}{\sqrt{1-\dfrac{(0.942c)^2}{c^2}}}-\dfrac{0.511}{\sqrt{1-\dfrac{(0.900c)^2}{c^2}}}

E=0.350\ Mev

Hence, The energy required to accelerate an electron is 0.582 Mev and 0.350 Mev.

4 0
3 years ago
A circular loop of wire of cross-sectional area 0.12 m2 consists of 200 turns, each carrying 0.50 A. It is placed in a magnetic
defon

Answer:

0.52 Nm

Explanation:

A = 0.12 m^2, N = 200, i = 0.5 A, B = 0.050 T

Angle between the plane of loop and magnetic field = 30 Degree

Angle between the normal of loop and the magnetic field = 90 - 30 = 60 degree

θ = 60°

Torque = N i A B Sinθ

Torque = 200 x 0.5 x 0.12 x 0.050 x Sin 60

Torque = 0.52 Nm

4 0
3 years ago
Explain the origin of the magnitude designation for determining the brightness of stars. Why does it seem to go backward, with s
Mashcka [7]

Answer:

Hipparchus was an ancient Greek who classified stars based on the brightness in 129 B.C. He grouped the brightest stars and ranked them 1 (first magnitude) and dimmest stars as 6 (sixth magnitude). Thus, the smaller numbers indicated brighter stars. Now, the scale extends in negative axis as well. More the negative number, brighter is the star. For example, Sun has magnitude -26.74.

This the apparent magnitude which means the classification is based on the brightness of the star as it appears from the Earth.

5 0
3 years ago
An acorn falls from a tree and hits the ground in 0.8 s. How far did the acorn fall . Use g = 9.8 m/s^2. Round your final result
mylen [45]

The distance covered by the acorn is 3.136 m.

<u>Explanation:</u>

The time taken for the acorn to hit the ground is 0.8 s. As it is a free fall, the acorn will be completely under the influence of gravity. So the acceleration will be acceleration due to gravity.

Then using the second law of equation,

s=ut+\frac{1}{2}gt^{2}

Since the initial velocity and time is zero, then the time taken to reach the ground is stated as 0.8 s, so

   s=0+\left(\frac{1}{2} \times 9.8 \times 0.8 \times 0.8\right)=\frac{6.272}{2}=3.136 \mathrm{m}

So the distance covered by the acorn is 3.136 m.

8 0
3 years ago
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