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3 years ago

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3. You are measuring a distance using a caliper to measure something and you measure it to the nearest 0.01 mm. However, you hav

e misread the scale and listed the measurement as 8.62 mm when it really should have been 9.62 mm. Is this a mistake in accuracy or precision, or both?

1 answer:

Anna007 [38]3 years ago

7 0

Answer

Vernier caliper measure nearest to 0.01 mm.

If the correct reading of the experiment is 9.62 mm.

But you misread it and listed it as 8.62 mm.

This is Clearly the case of mistake in Accuracy.

Because the reading is wrong and answer is not accurate hence it is mistake in accuracy.

To Judge weather the reading is precise or not we should have a range where the result is acceptable but in the question no range is given. So, we cannot comment on precision.

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3 years ago

Q1: A runner is jogging in a straight line at a steady vr= 6.8 km/hr. When the runner is L= 2.4 km from the finish line, a bird

serious [3.7K]

Answer:

Q1: 3.2km

Q2: 4.8K

Explanation:

Q1:

So db is the distance of bird, and dr is the distance of runner

db = 2vr and the distance of bird is going to be 2 times greater than the runner.

formulas: db = 2vr & db = 2dr

db = 2dr
L + (L - x) = 2x
2L - x = 2x
2L = 3x
x = $\frac{2}{3}$ L

Insert it in x = $\frac{2}{3}$ L

$\frac{2}{3}$ (2.4km) = 1.6km

Now we use formula db = 2dr

db = 2L - x
db = 2(2.4km) - 1.6km
<u>db = 3.2km</u>

Q2:

Formulas: Vr = L /Δt & Vb = db/Δt

Vr = L/ Δt ⇒ Δt = $\frac{L}{Vr}$
$\frac{2.4km}{6.8km/hr}$
$\frac{6}{17}hr$

(Km cancel each other)

Vb = db/Δt ⇒ db = VbΔt
13.6km/hr $(\frac{6}{17}hr )$
<u>4.8km</u>

(hr cancel each other)

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3 years ago

An unstable particle at rest breaks up into two fragments of unequal mass. The mass of the lighter fragment is equal to 2.90 ✕ 1

motikmotik

Answer:

The speed of the heavier fragment is 0.335c.

Explanation:

Given that,

Mass of the lighter fragment $M_{l}=2.90\times10^{-28}\ kg$

Mass of the heavier fragment $M_{h}=1.62\times10^{-27}\ Kg$

Speed of lighter fragment = 0.893c

We need to calculate the speed of the heavier fragment

Let v is the speed of the second fragment after decay

Using conservation of relativistic momentum

$0=\drac{m_{1}v_{1}}{\sqrt{1-\dfrac{v_{1}^2}{c^2}}}-\drac{m_{2}v_{2}}{\sqrt{1-\dfrac{v_{1}^2}{c^2}}}$

$\drac{m_{1}v_{1}}{\sqrt{1-\dfrac{v_{1}^2}{c^2}}}=\drac{m_{2}v_{2}}{\sqrt{1-\dfrac{v_{1}^2}{c^2}}}$

$\dfrac{2.90\times10^{-28}\times0.893c}{\sqrt{1-(0.893)^2}}=\dfrac{1.62\times10^{-27}v_{2}}{\sqrt{1-\dfrac{v_{2}^2}{c^2}}}$

$\dfrac{v_{2}}{\sqrt{1-\dfrac{v_{2}^2}{c^2}}}=\dfrac{2.90\times10^{-28}\times0.893c}{1.62\times10^{-27}\times0.45}$

$\dfrac{v_{2}}{\sqrt{1-\dfrac{v_{2}^2}{c^2}}}=0.355c$

$\dfrac{v_{2}}{1-\dfrac{v_{2}^{2}}{c^2}}=(0.355c)^2$

$\dfrac{1-\dfrac{v_{2}^2}{c^2}}{v_{2}^2}=\dfrac{1}{(0.355c)}$

$\dfrac{1}{v_{2}^2}-\dfrac{1}{c^2}=\dfrac{1}{(0.355c)^2}$

$\dfrac{1}{v_{2}^2}=\dfrac{1}{c^2}+\dfrac{1}{0.126c^2}$

$\dfrac{1}{v_{2}^2}=\dfrac{1}{c^2}(1+\dfrac{1}{0.126})$

$\dfrac{1}{v_{2}^2}=\dfrac{8.93}{c^2}$

$v_{2}^2=\dfrac{c^2}{8.93}$

$v_{2}=0.335c$

Hence, The speed of the heavier fragment is 0.335c.

7 0

3 years ago