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Lapatulllka [165]
3 years ago
9

The efficiency of a machine is always less than 100% because part of the energy input is used to

Physics
1 answer:
Travka [436]3 years ago
5 0

Answer:

C. Overcome Friction

Explanation:

When using any machine usually those with moving parts, you may notice heat forming near the areas where most movement occurs. As friction continues, more energy is used up and released as heat. For that reason, the efficiency of a machine will forever be less than 100%

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You have landed on an unknown planet, newtonia, and want to know what objects will weigh there. you find that when a certain too
AleksandrR [38]
Refer to the diagram shown.

Because the surface is frictionless, the resistive for, R, is zero.

Let m = the mass of the object.
Let a =  acceleration due to the applied force.
Therefore
12.7 N = (m kg)*(a m/s²)
a = 12.7/m  m/s²

The object travels  16.1 m in 2.5 s, starting from rest. Therefore
16.1 N = (1/2)*(12.7/m m/s²)*(2.5 s)² = 39.6875/m  N
m = 16.1/39.6875 = 0.4057 kg

For freefall, let g =  acceleration due to gravity.
The time to fall from 10.3 m is 2.88 s, therefore
10.3 m = (1/2)*(g m/s²)*(2.88 s)² = 4.1472g m
g = 10.3/4.1472 = 2.484 m/s²

Answer:
The gravitational acceleration on the planet is 2.5 m/s² (nearest tenth)


3 0
3 years ago
Automobile A and B are initially 30 m apart travelling in adjacent highway lanes at speeds VA = 14.4 km/hr., VB 23.4 km/hr. at t
marshall27 [118]

Answer:

        x = 240 m

Explanation:

This is a kinematics exercise

Let's fix our frame of reference on car A

           x = x₀ₐ+ v₀ₐ t + ½ aₐ t²

         

the initial position of car a is zero

           x = 0 + v₀ₐ t + ½ 0.8 t²

for car B

          x = x_{ob} + v_{ob} t - ½ a_b t²

     

car B's starting position is 30 m

         x = 30 + v_{ob} t - ½ 0.4 t²

at the point where they meet, the position of the two vehicles is the same

         0 + v₀ₐ t + ½ 0.8 t² = 30 + v_{ob} t - ½ 0.4 t²

let's reduce the speeds to the SI system

        v₀ₐ = 14.4 km / h (1000 m / 1 km) (1h / 3600s) = 4 m / s

        v_{ob} = 23.4 km / h = 6.5 m / s

        4 t + 0.4 t² = 30 + 6.5 t - 0.2 t²

        0.2 t² - 2.5 t - 30 = 0

        t² - 12.5 t - 150 = 0

we solve the quadratic equation

       t = \frac{12.5 \pm \sqrt{12.5^2 + 4 \ 150}  }{2}

       t = \frac{12.5 \  \pm 27.5}{2}

       t₁ = 20 s

       t₂ = -7.5 s

time must be a positive quantity so the correct result is t = 20 s

let's look for the distance

        x = 4 t + ½ 0.8 t²

        x = 4 20 + ½ 0.8 20²

        x = 240 m

8 0
3 years ago
What is the minimum frequency with which a 200-turn, flat coil of cross sectional area 300 cm2 can be rotated in a uniform 30-mT
sergiy2304 [10]

Answer:

The minimum frequency of the coil is 7.1 Hz

Explanation:

Given;

number of turns, N = 200 turns

cross sectional area, A = 300 cm² = 300 x 10⁻⁴ m²

magnitude of magnetic field strength, B = 30 x 10⁻³ T

maximum value of the induced emf, E = 8 V

Maximum induced emf is given as;

E = NBAω

where

ω is angular velocity (ω = 2πf)

E = NBA2πf

where;

f is the minimum frequency, measured in hertz (Hz)

f = E / (NBA2π)

f = 8 / (200 x 30 x 10⁻³  x 300 x 10⁻⁴ x 2 x 3.142)

f = 7.073 Hz

f = 7.1 Hz

Therefore, the minimum frequency of the coil is 7.1 Hz

8 0
3 years ago
D=1/2at^2 <br> solve for a
nignag [31]

Answer:

a = 2d / t²

Explanation:

d = ½ at²

Multiply both sides by 2:

2d = at²

Divide both sides by t²:

a = 2d / t²

4 0
3 years ago
What is a lol diagram.
Novosadov [1.4K]

Explanation:

where is your diagram? lol

7 0
2 years ago
Read 2 more answers
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