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Masja [62]
3 years ago
12

describe how acceleration and velocity are related and specify if these are scalar and vector quantities

Physics
1 answer:
vitfil [10]3 years ago
8 0

Velocity is the rate of change in distance over change in time, this can be written as:

v = Δd / Δt

While acceleration is the rate of change in velocity over change in time, this is written as:

a = Δv / Δt

 

<span>Both quantities are vector quantities because negative values means that the acceleration or velocity is acting on the opposite direction.</span>

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4. AN OBJECT INCREASES ITS VELOCITY
Cloud [144]

Answer:

it is accelerating 14 m/s

Explanation:

7 0
2 years ago
The wheels of a wagon can be approximated as the combination of a thin outer hoop, of radius ????h=0.209 m and mass 4.32 kg , an
Stolb23 [73]

Answer:

r= 98.3 mm

Explanation:

For rim

R= 0.209 m

M= 4.32 kg

For rods

m= 7.37 kg

L= 2 R= 2 x 0.209 = 0.418 m

The Total moment of inertia of the  wagon

I=MR²+2 x 1/12 m L²

Now by putting the values

I=4.32\times 0.209^2+2\times \dfrac{1}{12}\times 7.73\times 0.418^2\ kg.m^2

I=0.413 kg.m²

For disk:

t= 0.0462 m

Density ρ = 5990 kg/m³

Lets take r is the radius of disk

So the mass of the disc

m'=ρ πr² t

The moment of inertia of disc

I'=1/2 m'r²

I'=1/2 x r² x ρ πr² t

Given that

I = I'

1/2 x r² x ρ πr² t = 0.413 kg.m²

1/2 x r³ x ρ π t = 0.413

r³ x ρ π t = 0.826

r^3=\dfrac{0.826}{\pi \times 5990 \times 0.0462}

r³=0.00095

r=0.0983 m

r= 98.3 mm

8 0
3 years ago
What is the minimum force require to move a 5kg wooden crate on a wooden floor?
kolbaska11 [484]

You need to know the coefficient of static friction between a wooden object and a wooden surface. I'll denote it with <em>µ</em>. If you're given a specific value you should obviously use that.

By Newton's second law, the horizontal and vertical net forces are

• net horizontal:

∑ <em>F</em> = <em>p</em> - <em>f</em> = 0

• net vertical:

∑ <em>F</em> = <em>n</em> - <em>w</em> = 0

where

<em>p</em> = magnitude of the <u>p</u>ushing force

<em>f</em> = mag. of <u>f</u>riction

<em>n</em> = mag. of the <u>n</u>ormal force

<em>w</em> = <u>w</u>eight of the crate

The second equation gives

<em>n</em> = <em>w</em> = (5 kg) (9.8 m/s²) = 49 N

Friction is proportional to the normal force by a factor of <em>µ</em>, so

<em>f</em> = <em>µ</em> (49 N) = 49<em>µ</em> N

To overcome static friction, the push has to exceed this in magnitude, so that

<em>p</em> > 49<em>µ</em> N

For instance, if <em>p</em> = 0.25, then <em>p</em> would need to greater than 12.25 N. (This example isn't particularly helpful, though, since both possibly correct options are larger than 12.25 N...)

7 0
3 years ago
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3 years ago
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3 years ago
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