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exis [7]
3 years ago
8

Find change of velocity for: 0-1.7 s 1.7-6s

Physics
1 answer:
tekilochka [14]3 years ago
3 0
The integral of acceleration is velocity. The area under the curve is an integral. You can see the relation here. So just take the area under those time intervals. I’m assuming 1.7 is where it crosses the x-axis

For the first interval:
Base = 1.7
Height = -2
[1.7*(-2)]/2 = -1.7 cm/s

For the second interval:
There are two triangles and a rectangle here.
Base #1 = 0.3
Height #1 = 1

Base #2 = 1
Height #2 = 1

Length = 3
Width = 1

Apply the area formulas to get an answer of 3.65 cm/s
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A 6 kg block initially at rest is pulled to East along a horizontal surface with coefficient of kinetic friction μk=0.15 by a co
ozzi

Answer:

1.8 m/s

Explanation:

Draw a free body diagram of the block.  There are four forces:

Normal force Fn up.

Weight force mg down.

Applied force F to the east.

Friction force Fn μ to the west.

Sum the forces in the y direction:

∑F = ma

Fn − mg = 0

Fn = mg

Sum the forces in the x direction:

F − Fn μ = ma

F − mg μ = ma

a = (F − mg μ) / m

a = (12 N − 6 kg × 9.8 m/s² × 0.15) / 6 kg

a = 0.53 m/s²

Given:

Δx = 3 m

v₀ = 0 m/s

a = 0.53 m/s²

Find: v

v² = v₀² + 2aΔx

v² = (0 m/s)² + 2 (0.53 m/s²) (3 m)

v = 1.8 m/s

8 0
3 years ago
Dòng điện dịch là:
yarga [219]

Answer: the answer is a

Explanation:

8 0
3 years ago
Which is one-dimensional
Tema [17]

Answer:

B

Explanation:

They are all one-dimensional

6 0
3 years ago
Why don't atoms get too close?
Lera25 [3.4K]

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3 0
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Read 2 more answers
A lawn roller in the form of a thin-walled, hollow cylinder with mass M is pulled horizontally with a constant horizontal force
olga nikolaevna [1]

Answer:

a_{cm}=\frac{F}{2M}\\\\F_{fr}=\frac{F}{2}

Explanation:

Given the mass as M, the rotational inertia of the mower is;

I_{cm}=MR^2

-The roller doesn't slip while rolling;

v_{cm}=wR, a_{cm}=\alpha R

\sum F_x=Ma_x, -F+F_{fr}=-Ma_{cm}\\\\a_{cm}=\frac{F-Fr}{M}    \ \ \ \ \ \ \ \ eqtn1\\\\\sum \tau_{cm}=I_{cm}\alpha, F_{fr}(R)}=(MR^2)(\frac{a_{cm}}{R}), ->F_{fr}=Ma_c_m\ \ \  \ \ \ \ eqtn2\\\\a_{cm}=\frac{F-Ma_{cm}}{M}, ->F=2Ma_{cm}\\\\a_{cm}=\frac{F}{2M}\\\\\\\therefore F_{fr}=M(\frac{F}{2M})\\\\F_{fr}=\frac{F}{2}

6 0
3 years ago
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