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Archy [21]
3 years ago
8

Ari uses a tool called calipers to measure the diameter of different cylinders he is using in an experiment. His old calipers co

uld measure down to 0.01 mm, but his new calipers only measure down to 0.1 mm. Are his new calipers less precise, less accurate, both less accurate and less precise, or neither less precise or less accurate?
Physics
1 answer:
Vladimir [108]3 years ago
7 0

Answer:

Explanation:

Different measuring instruments have different level of precision . For example , our usual scale can measure upto 1 mm. Callipers can measure upto .1 mm. screw gauge can measure upto .01 mm.etc The deeper it can measure , the more precise it is. Hence screw gauge is most precise among the three.

On this basis we can say that , new caliper is less precise than old one.

Accuracy tells us about our result , how close our result is to the true measurement. It is dependent on  the manner in which  we take reading , So measuring methods , rather than measuring instruments , decide the accuracy of our result.

Hence we can say that the new caliper is less precise only.

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A large, 34.0 kg bell is hung from a wooden beam so it can swing back and forth with negligible friction. The bell's center of m
Lorico [155]

Answer:

length L of the clapper rod for the bell to ring silently = 0.756m

Explanation:

We are given;

Mass of Bell;m_b = 34 kg

Distance of centre of mass from pivot;d = 0.7m

The bells moment of inertia about an axis at the pivot;I = 18 kg.m²

Mass of clapper;m_c = 1.8 kg

Length of slender rod is L

Now, the formula for period of physical pendulum having small amplitude is given as;

T_b = 2π√(I/mgd)

Where;

I is moment of inertia

m is mass

g is acceleration due to gravity = 9.8 m/s²

d is distance from rotation axis to centre of gravity

Plugging in the relevant values and using mass of bell, we have;

T_b = 2π√(18/(34*9.81*0.7)

T_b = 2π√(18/(34*9.81*0.7)

T_b = 1.745 s

Now, the formula for period for a simple pendulum which is essentially what the clapper rod is would be;

T_c = 2π√(L/g)

Now, we want to find length of clapper L.

Thus, let's make it the subject;

L = g(T_c/2π)²

Now, we are told that for the bell to ring silently, T_b = T_c.

Thus, T_c = 1.745 s.

So,

L = 9.8(1.745/2π)²

L = 0.756m

7 0
3 years ago
A reducing elbow in a horizontal pipe is used to deflect water flow by an angle θ = 45° from the flow direction while accelerati
sleet_krkn [62]

Answer:

915 N is the anchoring force needed to hold the elbow in place.

Explanation:

Momentum equation:

∑

7 0
4 years ago
Infrared radiation from young stars can pass through the heavy dust clouds surrounding them, allowing astronomers here on Earth
Semmy [17]

Answer:

\lambda=6.83\times 10^{-5}\ m

Explanation:

Given that,

An infrared telescope is tuned to detect infrared radiation with a frequency of 4.39 THz.

We know that,

1 THz = 10¹² Hz

So,

f = 4.39 × 10¹² Hz

We need to find the wavelength of the infrared radiation.

We know that,

\lambda=\dfrac{c}{f}\\\\\lambda=\dfrac{3\times 10^8}{4.39\times 10^{12}}\\\\=6.83\times 10^{-5}\ m

So, the wavelength of the infrared radiation is 6.83\times 10^{-5}\ m.

5 0
3 years ago
A golf club rotates 215 degrees and has a length (radius) equal to 29 inches. The time it took to swing the club was 0.8 seconds
vichka [17]

Answer:

The average linear velocity (inches/second) of the golf club is 136.01 inches/second

Explanation:

Given;

length of the club, L = 29 inches

rotation angle, θ = 215⁰

time of motion, t = 0.8 s

The angular speed of the club is calculated as follows;

\omega = (\frac{\theta}{360} \times 2\pi, \ rad) \times \frac{1}{t} \\\\\omega =  (\frac{215}{360} \times 2\pi, \ rad) \times \frac{1}{0.8 \ s} \\\\\omega = 4.69 \ rad/s

The average linear velocity (inches/second) of the golf club is calculated as;

v = ωr

v = 4.69 rad/s  x  29 inches

v = 136.01 inches/second

Therefore, the average linear velocity (inches/second) of the golf club is 136.01 inches/second

8 0
3 years ago
I really don’t know the answer for this
EleoNora [17]
The correct answer is the reverse wave I took the test
6 0
3 years ago
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