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1 year ago

15

Suppose a length has been reported as 3.4 cm. What is the minimum uncertainty implied by this measurement in cm?

1 answer:

Aliun [14]1 year ago

6 0

From the measure of 3.4 cm we can see that the minimum uncertainty implied is in the milimeters, that is in the range of 0.1 cm.

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A block of mass m moving due east at speed v collides with and sticks to a block of mass 2m that is moving at the same speed vv

lapo4ka [179]

Answer:

30.36°

Explanation:

By using linear momentum; linear momentum can be expressed by the relation:

$mv_xi + mv_yj$

where ;

m= mass

$v_x$ = velocity of components in the x direction

$v_y$ = velocity of components in the y direction

If we consider the east as the positive x and north as positive y which is synonymous to what we usually have on a graph.

Then;

Initial momentum = $mvi + 2mvcos 45i + 2mvsin45 j$

= $(mv+2mvcos45)i + (2mvsin45)j$

However, the masses stick together after collision and move with a common velocity: $V_xi +V_yj$

∴ Final momentum = $3mv (V_xi +V_yj)$

= $3mV_xi + 3mV_yj$

From the foregoing ;

initial momentum = final momentum

$3mV_xi + 3mV_yj = (mv+2mvcos45)i+(2mvsin45j)$

So;

$3mV_x = mv + 2mv cos 45 \\\\3mV_y = 2mV sin 45$

$V_x = \frac{mv+2mvcos45 }{3m}\\\\V_x = \frac{v+2vcos45}{3}$

$V_y = \frac{2mvsin45}{3m} \\\\V_y = \frac{2vsin45}{3}$

Finally;

The required angle θ = $tan^{-1} = \frac{V_y}{V_x}$

θ = $tan^{-1} = \frac{\frac{2vsin45}{3}}{\frac{v+2v.cos45}{3}}$

θ = $tan^{-1} = \frac{2sin 45}{1+2cos45}\\\\$

θ = 30.36°

7 0

3 years ago

A 3.00 kg object is fastened to a light spring, with the intervening cord passing over a pulley. The pulley is frictionless, and

finlep [7]

Answer:

Part a)

k = 588.6 N/m

Part b)

v = 0.7 m/s

Explanation:

As we know that initially block is at rest

now if block is released from rest then it will go down by 10 cm and again comes to rest

so here we have

Part a)

Work done by gravity + work done by spring force = change in kinetic energy

$W_g + W_{spring} = 0 - 0$

$mg(0.10) + \frac{1}{2}k(0^2 - 0.10^2) = 0$

$3(9.81)(0.10) - \frac{1}{2}k(0.10)^2 = 0$

$k = 588.6 N/m$

Part b)

Now when spring is stretch by x = 5 cm then the speed of the block is given as

$mgx' + \frac{1}{2}k(0^2 - x'^2) = \frac{1}{2}mv^2 - 0$

here we have

$x' = 0.05 m$

$3(9.81)(0.05) - \frac{1}{2}(588.6)(0 - 0.05^2) = \frac{1}{2}(3) v^2$

$1.4715 - 0.736 = 1.5 v^2$

$v = 0.7 m/s$

3 0

3 years ago

What is it called when the right side of a design is reflected across a central axis and mirrored on the left side of the design

guapka [62]

Answer:

What is it called when the right side of a design is reflected across a central axis and mirrored on the left side of the design?

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3 years ago

A solenoid of length 0.250 m and radius 0.0250 m is comprised of 440 turns of wire. Determine the magnitude of the magnetic fiel

Bad White [126]

Answer: 0.02654

Explanation:

in the attachment

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4 years ago

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Calculate the pressure of an enclosed fluid on which a force of 150 N is exerted over an area of 10 cm2. give the answer in pasc

jek_recluse [69]

The pressure of the enclosed fluid is 150000Pa

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4 years ago