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3 years ago

Name some devices we use to make measurements

Physics

1 answer:

NikAS [45]3 years ago

4 0

Answer:

Explanation:

Ruler. A steel ruler aids the measurement and layout of straight lines. The ruler, also called "straightedge" or "straight-edged ruler," is a long, thin strip of wood, metal or plastic marked with increments of measurement.

Measuring Tape. The modern measuring tape's roughly palm-sized casing contains a coiled strip of metal marked with increments of measurement. The metal strip, called "tape," attaches to a spring which automatically retracts the tape into the casing following use.

Walking Tape Measure. The walking tape measure, also called "surveyor's measure," records the distance traveled by a wheel. An operator pushes the measure's wheel, similar to a bicycle wheel, by a handle as an attached ticker box displays feet or meters in the same format as a car's odometer.

Laser Measure. The laser measure offers point and shoot distance measurement. In its most basic form, a laser measure is a hand-held electronic device with a digital display.

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two identical springs of spring constant 7580 N/m are attached to a block of mass 0.245 kg. What is the frequency of oscillation

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The frequency of oscillation on the frictionless floor is 28 Hz.

<h3>Frequency of the simple harmonic motion</h3>

The frequency of the oscillation is calculated as follows;

f = (1/2π)(√k/m)

where;

k is the spring constant
m is mass of the block

f = (1/2π)(√7580/0.245)

f = 28 Hz

Thus, the frequency of oscillation on the frictionless floor is 28 Hz.

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2 years ago

A car with a mass of 3 Kg and velocity of 40 m/s collided with a truck of a velocity of 60 m/s, if the momentum is conserved wha

Luden [163]

Answer:

the mass of the truck is 2 kg.

Explanation:

Given;

mass of the car, m₁ = 3 kg

initial velocity of the car, u₁ = 40 m/s

initial velocity of the truck, u₂ = 60 m/s

let the mass of the truck = m₂

Apply the principle of conservation of linear momemtum;

m₁u₁ = m₂u₂

m₂ = (m₁u₁) / u₂

m₂ = (3 x 40) / (60)

m₂ = 2 kg

Therefore, the mass of the truck is 2 kg.

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3 years ago

An advertisement for an all-terrain vehicle (ATV) claims that the ATV can climb inclined slopes of 35°. What is the minimum coef

navik [9.2K]

An advertisement for an all-terrain vehicle (ATV) claims that the ATV can climb inclined slopes of 35°. The minimum coefficient of static friction needed for this claim to be possible is 0.7

In an inclined plane, the coefficient of static friction is the angle at which an object slide over another.

As the angle rises, the gravitational force component surpasses the static friction force, as such, the object begins to slide.

Using the Newton second law;

$\sum F_x = \sum F_y = 0$

$\mathbf{mg sin \theta -f_s= N-mgcos \theta = 0 }$

So; On the L.H.S

$\mathbf{mg sin \theta =f_s}$

$\mathbf{mg sin \theta =\mu_s N}$

On the R.H.S

N = mg cos θ

Equating both force component together, we have:

$\mathbf{mg sin \theta =\mu_s \ mg \ cos \theta}$

$\mathbf{sin \theta =\mu_s \ \ cos \theta}$

$\mathbf{\mu_s = \dfrac{sin \theta }{ cos \theta}}$

From trigonometry rule:

$\mathbf{tan \theta= \dfrac{sin \theta }{ cos \theta}}$

∴

$\mathbf{\mu_s =\tan \theta}}$

$\mathbf{\mu_s =\tan 35^0}}$

$\mathbf{\mu_s = 0.700}}$

Therefore, we can conclude that the minimum coefficient of static friction needed for this claim to be possible is 0.7

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3 years ago

Bonus: (It's not that hard, you just have to pay attention to units.) The Saturn V rocket first stage

agasfer [191]

$v = 2.45×10^3\:\text{m/s}$

Explanation:

Newton's 2nd Law can be expressed in terms of the object's momentum, in this case the expelled exhaust gases, as

$F = \dfrac{d{p}}{d{t}}$ (1)

Assuming that the velocity remains constant then

$F = \dfrac{d}{dt}(mv) = v\dfrac{dm}{dt}$

Solving for $v,$ we get

$v = \dfrac{F}{\left(\frac{dm}{dt}\right)}\;\;\;\;\;\;\;(2)$

Before we plug in the given values, we need to convert them first to their appropriate units:

The thrust <em>F</em><em> </em> is

$F = 7.5×10^6\:\text{lbs}×\dfrac{4.45\:\text{N}}{1\:\text{lb}} = 3.34×10^7\:\text{N}$

The exhaust rate dm/dt is

$\dfrac{dm}{dt} = 15\dfrac{T}{s}×\dfrac{2000\:\text{lbs}}{1\:\text{T}}×\dfrac{1\:\text{kg}}{2.2\:\text{lbs}}$

$\;\;\;\;\;= 1.36×10^4\:\text{kg/s}$

Therefore, the velocity at which the exhaust gases exit the engines is

$v = \dfrac{F}{\left(\frac{dm}{dt}\right)} = \dfrac{3.34×10^7\:\text{N}}{1.36×10^4\:\text{kg/s}}$

$\;\;\;= 2.45×10^3\:\text{m/s}$

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2 years ago

How much time would it take for an object to fall 4.7 meters

levacccp [35]

Answer:

4.7 is 10 as much as the number 0.47.

If you multiply 0.47 x 10 it will equal 4.7

Explanation:

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3 years ago