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igomit [66]
2 years ago
7

A machine raised a load of 360N through a distance of 0.2m. The effort, a force of 50N moved 1.8m during the process. Calculate

the efficiency of the machine.
Select one:
a. 40%
b. 100%
c. 80%
d. 60%
Physics
1 answer:
frosja888 [35]2 years ago
6 0

Answer:

c. 80%

Explanation:

Given;

load raised by the machine, L = 360 N

distance through which the load was raised, d = 0.2 m

effort applied, E = 50 N

distance moved by the effort, e = 1.8 m

The efficiency of the machine is calculated as follows;

Efficiency = \frac{ 0utput \ work }{1nput \ work} \times 100\% \\\\Efficiency = \frac{Load \ \times \ distance \ moved \ by \ load }{Efort \ \times \ distance \ moved \ by \ effort} \times 100\%\\\\Efficiency = \frac{360 \times 0.2}{50 \times 1.8} \times 100\%\\\\Efficiency =80\%

Therefore, the efficiency of the machine is 80%

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What is the magnitude of fs on an object lying on a flat surface without moving, on
Degger [83]

The magnitude of the force acting on the object lying on a flat surface without moving is 10 N.

The given parameters;

  • magnitude of force on the object, F = 10 N
  • angle between the object and the horizontal flat surface = 0⁰

Apply Newton's second law of motion to determine the magnitude of the force on the object.

Due to the position of the object, the magnitude of the force acting on it is calculated as;

\Sigma F_{net} = F\sin(\theta ) + F cos(\theta)\\\\\Sigma F_{net} = 10 sin(0) + 10cos(0)\\\\\Sigma F_{net} = 10 \ N

Therefore, the magnitude of the force acting on the object is 10 N.

Learn more here: brainly.com/question/19887955

7 0
2 years ago
Arrange the distances between Earth and various celestial objects in order from least to greatest. Use the conversion table to h
Kaylis [27]

distance to the star Betelgeuse: 640 ly

As we know that

1 ly = 63000 AU

also we know that

1AU = 1.5 \times 10^8 km

1 ly = 63000 (1.5 \times 10^8) = 9.45 \times 10^{12} km

So the distance of Betelgeuse = 640 ly

d_1 = 640 \times 9.45 \times 10^{12} = 6.05 \times 10^{15} m

distance to the star VY Canis Majoris: 3.09 × 10^8 AU

d_2 = 3.09\times 10^8 \times 1.5 \times 10^8 km

d_2 = 4.64 \times 10^{16} km

distance to the galaxy Large Magellanic Cloud: 49976 pc

1 pc = 3.262 ly = 3.262 \times 9.45 \times 10^{12} km

1pc = 3.08 \times 10^{13} km

now we have

d_3 = 49976 \times 3.08 \times 10^{13}

d_3 = 1.54 \times 10^{18} km

distance to Neptune at the farthest: 4.7 billion km

d_4 = 4.7 \times 10^9 km

now the order of distance from least to greatest is as following

1. distance to Neptune at the farthest

2. distance of Betelgeuse

3. distance to the star VY Canis Majoris

4. distance to the galaxy Large Magellanic Cloud

6 0
3 years ago
What is the name of Newton's first law of motion?
Ne4ueva [31]
A) law of inertia. an object in motion stays in motion
3 0
3 years ago
1. An excited lithium atom emits a red light with wavelength a = 671nm. What is the corresponding photon energy? hc (6.63 x 10-3
Rudiy27

Answer:

 E = 2,964 10⁻¹⁹ J

Explanation:

The energy of the photons is given by the Planck relation

          E = h f

the speed of light is related to wavelength and frequency

          c = λ f

we substitute

          E = h c /λ

let's reduce the magnitude to the SI system

          λ = 671 nm = 671 10⁻⁹ m

let's calculate

          E = 6.63 10⁻³⁴ 3 10⁸ /671 10⁻⁹

          E = 2,964 10⁻¹⁹ J

6 0
3 years ago
A car is moving at 19 m/s along a curve on a horizontal plane with radius of curvature 49m.
JulsSmile [24]

Answer:

\mu =0.75

Explanation:

<u>Frictional Force </u>

When the car is moving along the curve, it receives a force that tries to take it from the road. It's called centripetal force and the formula to compute it is:

F_c=m.a_c

The centripetal acceleration a_c is computed as

\displaystyle a_c=\frac{v^2}{r}

Where v is the tangent speed of the car and r is the radius of curvature. Replacing the formula into the first one

F_c=m.\frac{v^2}{r}

For the car to keep on the track, the friction must have the exact same value of the centripetal force and balance the forces. The friction force is computed as

F_r=\mu N

The normal force N is equal to the weight of the car, thus

F_r=\mu .m.g

Equating both forces

\displaystyle \mu .m.g=m.\frac{v^2}{r}

Simplifying

\displaystyle \mu =\frac{v^2}{rg}

Substituting the values

\displaystyle \mu =\frac{19^2}{(49)(9.8)}

\boxed{\mu =0.75}

7 0
3 years ago
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