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3 years ago

PLZ HELP 150 POINTS AND BRAINLIEST

Physics

1 answer:

kondor19780726 [428]3 years ago

4 0

Answer:

A wheel and axle may either increase or decrease the input force, depending on whether the input force is applied to the axle or the wheel. ... Because the output force is less than the input force, the mechanical advantage is less than 1. However, the wheel turns over a greater distance, so it turns faster than the axle.

If you apply a force to the wheel (the handle), the wheel spins and multiplies the effort to make the output force of the axle (shaft) greater.

The drive shaft, which on most cars runs the length of the vehicle to the rear wheels, turns as the combustion engine burns gasoline. The turning drive shaft sends power to the rear axle and wheels, which cause them to turn as well, moving the car forward

There some moments when the steering wheel becomes stiff and hard to turn. From irregular servicing, lack of fluid oil to low tyre pressure or bad wheel alignment, any of these could be the cause

hope it helps plz Mark me brainliest :)

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1. A car travels at 55 km/h for 6.0 hours. How far does it travel?

Veronika [31]

Answer:

330 km

55*6= 330 km. An easy formula for this is multiplying time with speed.

Explanation:

330 kilometros

55 * 6 = 330 km. Una fórmula fácil para esto es multiplicar el tiempo por la velocidad.

4 0

3 years ago

Read 2 more answers

Two particles oscillate in simple harmonic motion along a common straight-line segment of length 1.0 m. Each particle has a peri

igor_vitrenko [27]

Answer:

a) the particles are 0.217 m apart

b) the particles are moving in the same direction.

Explanation:

a) The amplitude of the oscillations is A/2 and the period of each particle is

T = 1.5 s however, they differ by a phase of π/6 rad. Let the phase of the first particle be zero so that the phase of the second particle is π/6. So we can write the coordinates of each of the particles as,

x₁ = A/2 cos(ωt)

x₂ = A/2 cos(ωt + π/6)

we can write the angular frequency ω, as

ω = 2π / T

so,

x₁ = A/2 cos(2π / T)

x₂ = A/2 cos(2π / T + π/6)

Thus, the coordinates of the particles at t = 0.45 s are,

x₁ = A/2 cos((2π × 0.45) / 1.5)) = -0.155 A

x₂ = A/2 cos((2π × 0.45) / 1.5) + π/6) = -0.372 A

Their separation at that time is, therefore,

Δx = x₁ - x₂

= -0.155 A + 0.372 A

= 0.217 A

since A = 1 m

Thus,

Δx = 0.217 m

b) In order to find their directions, we must take the derivatives at t = 0.45 s.

Therefore,

v₁ = dx₁ / dt

= (-πA / T) sin(2πt / T)

= -(π(1) / 1.5) sin(2π(0.45) / 1.5)

= -1.99

and,

v₂ = dx₂ / dt

= (-πA / T) sin((2πt / T) + π/6)

= -(π(1) / 1.5) sin((2π(0.45) / 1.5) + π/6)

= -1.40

Since both v₁ and v₂ are negative, this shows that the particles are moving in the same direction.

6 0

3 years ago

Which kind of road surface is easier to see when driving at night, a pebbled uneven surface or a mirror smooth surface?

pshichka [43]

A pebbled, uneven road would be easier to see at night because it minimizes the reflection of light from car’s light coming in the opposite direction. It is difficult to see when driving on the rainy day because the roadway reflects light from cars coming in the opposite directions.

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3 years ago

When a falling meteoroid is at a distance above the Earth's surface of 3.40 times the Earth's radius, what is its acceleration d

Alchen [17]

Answer:

g = 0.85 m $s^{-2}$

Explanation:

g = $\frac{GM}{h^{2} }$

were; g is the acceleration due to Earth's gravity, G is Newton's gravitation constant (6.674 x $10^{-11}$ N $m^{2}$ $kg^{-2}$ ), M is the mass of the earth (5.972 x $10^{24}$ kg), and h is the distance of meteoroid to the earth.

h = 3.40 x R

= 3.40 x 6371 km

h = 21661.4 km

= 21661400 m

Thus,

g = $\frac{6.674*10^{-11}*5.972*10^{24} }{(21661400)^{2} }$

= $\frac{3.9857 *10^{14} }{4.6922*10^{14} }$

= 0.84944

g = 0.85 m $s^{-2}$

The acceleration due to the Earth's gravitation is 0.85 m $s^{-2}$ .

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3 years ago

An automobile starter motor has an equivalent resistance of 0.055 Ï and is supplied by a 12.0 v battery which has a 0.0305 Ï int

LiRa [457]

12 V is the f.e.m. $\epsilon$ of the battery. The potential difference that is applied to the motor is actually the fem minus the voltage drop on the internal resistance r:
$\epsilon - Ir$
this is equal to the voltage drop on the resistance of the motor R:
$RI$
so we can write:
$\epsilon - Ir = RI$
and using $r=0.0305~\Omega$ and $R=0.055~\Omega$ we can find the current I:
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