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3 years ago

HEY CAN ANYONE PLS HELP ME OUT IN DIS PLS!

Physics

2 answers:

Leona [35]3 years ago

8 0

The significant figure rules for addition and subtraction basically has you add or subtract as normal, but the rounding is where the rule kicks in. Specifically you round based on the least accurate decimal portion (the term with the least number of digits to the right of the decimal point).

We do this because we want to make sure we don't involve digits we aren't 100% sure of their accuracy.

==========================================================

Part A

Add up the values like normal

32.567+135.0+1.4567 = 169.0237

Now we look at the decimal portions of each number. We look at the number of digits to the right of each decimal point.

32.567 has three such digits (5,6 and 7)

135.0 only has one digit (the 0)

1.4567 has four digits (4,5,6,7)

The least accurate is 135.0 since we only know this to the tenths place. So we'll be rounding the sum we got above (169.0237) to the nearest tenth

169.0237 rounds to 169.0

<h3>Answer before rounding = 169.0237</h3><h3>Answer after rounding = 169.0</h3>

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Part B

Add like normal

246.24+238.278+98.3 = 582.818

Then round to one decimal place since this is the fewest number of digits after the decimal point (in the number 98.3)

582.818 rounds to 582.8

<h3>Answer before rounding = 582.818</h3><h3>Answer after rounding = 582.8</h3>

==========================================================

Part C

Subtraction is handled in a similar fashion

4.829-0.26 = 4.569

Which rounds to 4.57 since two digits is the smallest number of digits after the decimal point. Put another way, 0.26 is the least of the accurate measurements comparing 4.829 and 0.26; we always go with the least accurate or weakest link when it comes to rounding sig figs. If we went the other way around, then we could be introducing data that isn't really there.

<h3>Answer before rounding = 4.569</h3><h3>Answer after rounding = 4.57</h3>

ryzh [129]3 years ago

3 0

Answer:

A) 169.0167

Explanation:

ive done this before

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3 years ago

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3 years ago

A certain fuse "blows" if the current in it exceeds 1.0 A, at which instant the fuse melts with a current density of What is the

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Answer:

Explanation:

The complete question is

a certain fuse "blows" if the current in it exceeds 1.0 A, at which instant the fuse melts with a current density of 620 A/ cm^2. What is the diameter of the wire in the fuse?

A) 0.45 mm

B) 0.63 mm

C.) 0.68 mm

D) 0.91 mm

Current in the fuse is 1.0 A

Current density of the fuse when it melts is 620 A/cm^2

Area of the wire in the fuse = I/ρ

Where I is the current through the fuse

ρ is the current density of the fuse

Area = 1/620 = 1.613 x 10^-3 cm^2

We know that 10000 cm^2 = 1 m^2, therefore,

1.613 x 10^-3 cm^2 = 1.613 x 10^-7 m^2

Recall that this area of this wire is gotten as

A = $\frac{\pi d^{2} }{4}$

where d is the diameter of the wire

1.613 x 10^-7 = $\frac{3.142* d^{2} }{4}$

6.448 x 10^-7 = 3.142 x $d^{2}$

$d^{2}$ = $\sqrt{ 2.05*10^-7}$

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3 years ago

A fuzzy bunny sees a butterfly and chases it right off of a 20 m high cliff at 7m/s

bogdanovich [222]

A) 2.02 s

To find the time it takes for the bunny to reach the rocks below the cliff, we just need to analyze its vertical motion. This motion is a free fall motion, which is a uniformly accelerated motion. So we can use the suvat equation:

$s=u_y t+\frac{1}{2}at^2$

where

s is the vertical displacement

u is the initial vertical velocity

a is the acceleration

t is the time

For the bunny here, choosing downward as positive direction,

$u_y = 0$ (initial vertical velocity is zero)

s = 20 m

$a=g=9.8 m/s^2$ (acceleration of gravity)

And solving for t, we find the time of flight:

$t=\sqrt{\frac{2s}{g}}=\sqrt{\frac{2(20)}{9.8}}=2.02 s$

B) 14.1 m

For this part, we need to consider the horizontal motion of the bunny.

The horizontal motion of the bunny is a uniform motion with constant velocity, which is the initial velocity of the bunny:

$v_x = 7 m/s$

Therefore the distance covered after time t is given by

$d=v_x t$

And substituting the time at which the bunny hits the ground,

t = 2.02 s

We find how far the bunny went from the cliff:

$d=(7)(2.02)=14.1 m$

C) 21.0 m/s at $70.5^{\circ}$ below the horizontal

The horizontal component of the velocity is constant during the entire motion, so it will still be the same when the bunny hits the ground:

$v_x = 7 m/s$

Instead, the vertical velocity is given by

$v_y = u_y +at$

And substituting t = 2.02 s, we find the vertical velocity at the moment of impact:

$v_y = 0+(9.8)(2.02)=19.8 m/s$

So, the magnitude of the final velocity is

$v=\sqrt{v_x^2+v_y^2}=\sqrt{7^2+(19.8)^2}=21.0 m/s$

And the angle is given by

$\theta=tan^{-1}(\frac{v_y}{v_x})=tan^{-1}(\frac{19.8}{7})=70.5^{\circ}$

below the horizontal

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3 years ago

A 55 kgkg meteorite buries itself 5.5 mm into soft mud. The force between the meteorite and the mud is given by F(x)F(x) = (630

sashaice [31]

Answer:

W = 1.44 10⁻⁷ J

Explanation:

The expression for the job is

W = ∫ F. dx

Where the point is the scalar product in this case the direction of the meteor and the depth is parallel, whereby the scalar product is reduced to the ordinary product

W = 630 ∫ x³ dx

W = 630 x⁴ / 4

Let's evaluate between the lower limit x = 0, w = 0 to the upper limite the point at x = 5.5 10⁻³ m

W = 157.5 ((5.5 10⁻³)⁴ -0)

W = 1.44 10⁻⁷ J

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3 years ago