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ira [324]
3 years ago
13

The sensitivity of a measuring instrument is the value of the smallest quantity that can be read or estimated with it. What is t

he smallest part of a centimeter that can be read with the following?
(a) a meterstick with markings at each millimeter.

(b) a digital caliper with a four-digit display

I've got an answer for (a) which was 0.05cm, and tried the following for (b) but none was correct:

0.0001cm, 0.01mm, 0.002mm
Physics
1 answer:
nirvana33 [79]3 years ago
6 0

Answer:

The smallest part of a millimeter that can be read with a digital caliper with a four digit display is 0.02mm. Thus, it has to be converted to centimetre. So, divide by 10, we then have 0.02/10= *0.002cm* not mm.

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You shoot an arrow into the air. Two seconds later (2.00 s) the arrow has gone straight upward to a height of 35.0 m above its l
sdas [7]

This question can be solved by using the equations of motion.

a) The initial speed of the arrow is was "9.81 m/s".

b) It took the arrow "1.13 s" to reach a height of 17.5 m.

a)

We will use the second equation of motion to find out the initial speed of the arrow.

h= v_it + \frac{1}{2}gt^2\\

where,

vi = initial speed = ?

h = height = 35 m

t = time interval = 2 s

g = acceleration due to gravity = 9.81 m/s²

Therefore,

35\ m = (v_i)(2\ s)+\frac{1}{2}(9.81\ m/s^2)(2\ s)^2\\\\v_i(2\ s)=19.62\ m\\\\v_i = \frac{19.62\ m}{2\ s}

<u>vi =  9.81 m/s</u>

b)

To find the time taken by the arrow to reach 17.5 m, we will use the second equation of motion again.

h= v_it + \frac{1}{2}gt^2\\

where,

g = acceleration due to gravity = 9.81 m/s²

h = height = 17.5 m

vi = initial speed = 9.81 m/s

t = time = ?

Therefore,

17.5 = (9.81)t+\frac{1}{2}(9.81)t^2\\4.905t^2+9.81t-17.5=0

solving this quadratic equation using the quadratic formula, we get:

t = -3.13 s (OR) t = 1.13 s

Since time can not have a negative value.

Therefore,

<u>t = 1.13 s</u>

Learn more about equations of motion here:

brainly.com/question/20594939?referrer=searchResults

The attached picture shows the equations of motion in the horizontal and vertical directions.

4 0
1 year ago
Consider an aircraft powered by a turbojet engine that has a pressure ratio of 12. The aircraft is stationary on the ground, hel
agasfer [191]

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3 0
3 years ago
A circular coil of radius 5.0 cm and resistance 0.20 Ω is placed in a uniform magnetic field perpendicular to the plane of the c
Romashka [77]

Answer:

Explanation:

Given that,

Assume number of turn is

N= 1

Radius of coil is.

r = 5cm = 0.05m

Then, Area of the surface is given as

A = πr² = π × 0.05²

A = 7.85 × 10^-3 m²

Resistance of

R = 0.20 Ω

The magnetic field is a function of time

B = 0.50exp(-20t) T

Magnitude of induce current at

t = 2s

We need to find the induced emf

This induced voltage, ε can be quantified by:

ε = −NdΦ/dt

Φ = BAcosθ, but θ = 90°, they are perpendicular

So, Φ = BA

ε = −NdΦ/dt = −N d(BA) / dt

A is a constant

ε = −NA dB/dt

Then, B = 0.50exp(-20t)

So, dB/dt = 0.5 × -20 exp(-20t)

dB/dt = -10exp(-20t)

So,

ε = −NA dB/dt

ε = −NA × -10exp(-20t)

ε = 10 × NA exp(-20t)

Now from ohms law, ε = iR

So, I = ε / R

I = 10 × NA exp(-20t) / R

Substituting the values given

I = 10×1× 7.85 ×10^-3×exp(-20×2)/0.2

I = 1.67 × 10^-18 A

6 0
3 years ago
Suppose that a nascar race car is moving to the right with a constant velocity of +86 m/s. What is the average acceleration of t
Stels [109]

(a) As the car is moving with constant velocity, it means the rate change of velocity does not change, therefore the average acceleration of the car is zero.

Thus, there is no acceleration, when velocity is constant.

(b) Average acceleration,

a =\frac{ v-u}{\Delta t}

Here, v is final velocity and u is the initial velocity and \Delta t is the time interval.

As twelve seconds later, the car is halfway around the track and traveling in the opposite direction with the same speed, therefore

a =\frac{ -86 \ m/s-86 \ m/s }{12 \ s} = - 14 .3 \ m/s^2

Thus, the average acceleration of the car is 14 .3 \ m/s^2 in the direction to the left.

6 0
3 years ago
Joe and Max shake hands and say goodbye. Joe walks east 0.50 km to a coffee shop, and Max flags a cab and rides north 3.45 km to
timama [110]

Answer:

3.486 km

Explanation:

Suppose Joe and Max's directions are perfectly perpendicular (east vs north). We can calculate their distance at the destinations using Pythagorean theorem:

s = \sqrt{J^2 + M^2}

where J = 0.5 km and M= 3.45 km are the distances between Joe and Max to their original parting point, respectively. s is the distance between them.

s = \sqrt{0.5^2 + 3.45^2} = \sqrt{12.1525} = 3.486 km

8 0
3 years ago
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