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2 years ago

Which method of measurement would be accurate but lack precision?

Physics

1 answer:

kodGreya [7K]2 years ago

7 0

Answer:

Explanation:

reading the volume of water in a graduated cylinder which can be read to the nearest mL is accurate, it lacks precision due to the bottom meniscus formed.

the bottom meniscus may cause a wrong reading due to refraction of light

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Two gliders on an air track collide in a perfectly elastic collision. Glider A has a mass of 1.1 kg and is initially travelling

Eva8 [605]

m1= mass 1 = 1.1 kg

Vi1 = initial velocity 1 = 2.7 m/s

m2= 2.4 kg

V2i = -1.9 m/s

We assume east as positive and west as negative.

Apply the formulas:

Vf1 = ?

$vf1=(\frac{m1-m2}{m1+m2})Vi1+(\frac{2m2}{m1+m2})Vi2$

Replacing:

$Vf1=\frac{(1.1-2.4)}{(1.1+2.4)}2.7+\frac{(2\times2.4)}{(1.1+2.4)}-1.9$ $Vf1=(\frac{-1.3}{3.5})2.7+(\frac{4.8}{3.5})-1.9$ $Vf1=-1-2.6=-3.6\text{ m/s}$

Answer: 3.6 m/s west

6 0

1 year ago

When you stretch a spring 20 cm past its natural length, it exerts a force of 8

Nastasia [14]

Answer:

40 N/m

Explanation:

F = -kx (This is the Hooke's Law equation)

F is the force the spring exerts = 8 N

-k = spring constant

x = displacement (The distance stretched past it's natural length) = 20cm

x needs to be in meters, and 20 cm is = to 0.2 meters

Finally:

8N = -k (0.2m)

-k = 8N / 0.2 m

k = -40 N/m

6 0

3 years ago

Read 2 more answers

A 5-g lead bullet traveling in 20°C air at 300 m/s strikes a flat steel plate and stops.

densk [106]

To solve this problem it is necessary to apply the concepts related to the Kinetic Energy and the Energy Produced by the heat loss. In mathematical terms kinetic energy can be described as:

$KE = \frac{1}{2} mv^2$

Where,

m = Mass

v = Velocity

Replacing we have that the Total Kinetic Energy is

$KE = \frac{1}{2} mv^2$

$KE = \frac{1}{2} (5*10^{-3})(300)^2$

$KE = 225J$

On the other hand the required Energy to heat up t melting point is

$Q_1 = mC_p \Delta T$

$Q_2 = L_f m$

Where,

m = Mass

$C_p =$ Specific Heat

$\Delta T =$ Change at temperature

$L_f =$ Latent heat of fussion

Heat required to heat up to melting point,

$Q = Q_1+Q_2$

$Q = mC_p \Delta T+L_f m$

$Q = 5*0.128*(327-20) + 5*24.7$

$Q = 310J$

The energy required to melt is larger than the kinetic energy. Therefore the heat of fusion of lead would be 327 ° C: The melting point of lead.

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3 years ago

Based on what you have learned from this unit, construct a tri-fold brochure instructing a new freshman as to the best way to le

Rzqust [24]

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3 years ago

In a low-density cloud, the internal pressure is much too low to support the weight, so the cloud must __________, eventually pr

andrey2020 [161]

The answer to this question is flatten

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3 years ago