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3 years ago

13

Despite the use of diffrent types of measurement in our country

1 answer:

stealth61 [152]3 years ago

6 0

Answer: For standardization purposes

Explanation:

As the question mentions, there were different types of measurements in the country for the same things so measuring a thing in one type of measurement and then comparing it to another was quite cumbersome.

It also made comparing units with the rest of the world difficult. To combat this the SI system was introduced. It gave an international standard for measuring different things based on the metric system. This allowed for worldwide standardization and comparison as things could now be measured in the same units.

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An ideal spring with spring constant k is hung from the ceiling. The initial length of the spring, with nothing attached to the

hram777 [196]

The mass m of the object = 5.25 kg

<h3>Further explanation</h3>

Given

k = spring constant = 3.5 N/cm

Δx= 30 cm - 15 cm = 15 cm

Required

the mass m

Solution

F=m.g

Hooke's Law

F = k.Δx

$\tt m.g=k.\Delta x\\\\m.10=3.5\times 15\\\\m=5.25~kg$

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3 years ago

A constant net force acts on an object. which of the following best describes the object's motion?

Anton [14]

The object<span> is moving with a decreasing acceleration. The </span>object<span> is moving with </span>a constant<span> velocity.</span>

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3 years ago

What is the momentum of a 45-kg quarterback moving eastward at 15<br> m/s?

puteri [66]

Answer:

Given

mass (m) =45kg

velocity (v) =15m/s

momentum (p) =?

Form

p=mv

=45x 15

p=675kg.m/s

the momentum =675kg.m/s

4 0

3 years ago

A hoop (I = MR2) of mass 3 kg and radius 1.1 m is rolling at a center-of-mass speed of 11 m/s. An external force does 842 J of w

Scilla [17]

Answer:

$v_f = 20 m/s$

Explanation:

Since the hoop is rolling on the floor so its total kinetic energy is given as

$KE = \frac{1}{2}mv^2 + \frac{1}{2} I\omega^2$

now for pure rolling condition we will have

$v = R\omega$

also we have

$I = mR^2$

now we will have

$KE = \frac{1}{2}mv^2 + \frac{1}{2}(mR^2)\frac{v^2}{R^2}$

$KE = mv^2$

now by work energy theorem we can say

$W = KE_f - KE_i$

$842 J = mv_f^2 - mv_i^2$

$842 = 3(v_f^2) - 3\times 11^2$

now solve for final speed

$v_f = 20 m/s$

3 0

3 years ago

An 3.7 lb hammer head, traveling at 5.8 ft/s strikes a nail and is brought to a stop in 0.00068 s. The acceleration of gravity i

CaHeK987 [17]

Answer:

31677.2 lb

Explanation:

mass of hammer (m) = 3.7 lb

initial velocity (u) = 5.8 ft/s

final velocity (v) = 0

time (t) = 0.00068 s

acceleration due to gravity (g) 32 ft/s^{2}

force = m x ( a + g )

where

m is the mass = 3.7 lb
g is the acceleration due to gravity = 32 ft/s^{2}
a is the acceleration of the hammer

from v = u + at

a = (v-u)/ t

a = (0-5.8)/0.00068 = -8529.4 ( the negative sign showa the its decelerating)

we can substitute all required values into force= m x (a+g)

force = 3.7 x (8529.4 + 32) = 31677.2 lb

4 0

3 years ago