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sdas [7]
3 years ago
9

A cylinder of mass 8.0 kg rolls without slipping on a horizontal surface. At a certain instant its center of mass has a speed of

13.0 m/s.
(a) Determine the translational kinetic energy of its center of mass.
J
(b) Determine the rotational kinetic energy about its center of mass.
J
(c) Determine its total energy.
J
Physics
1 answer:
olganol [36]3 years ago
8 0

Answer:

b

Explanation:

determine the rotational kinetic energy about it's center of mass

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Please help. I don’t understand this
skad [1K]

The short answer is that the displacement is equal tothe area under the curve in the velocity-time graph. The region under the curve in the first 4.0 s is a triangle with height 10.0 m/s and length 4.0 s, so its area - and hence the displacement - is

1/2 • (10.0 m/s) • (4.0 s) = 20.00 m

Another way to derive this: since velocity is linear over the first 4.0 s, that means acceleration is constant. Recall that average velocity is defined as

<em>v</em> (ave) = ∆<em>x</em> / ∆<em>t</em>

and under constant acceleration,

<em>v</em> (ave) = (<em>v</em> (final) + <em>v</em> (initial)) / 2

According to the plot, with ∆<em>t</em> = 4.0 s, we have <em>v</em> (initial) = 0 and <em>v</em> (final) = 10.0 m/s, so

∆<em>x</em> / (4.0 s) = (10.0 m/s) / 2

∆<em>x</em> = ((4.0 s) • (10.0 m/s)) / 2

∆<em>x</em> = 20.00 m

5 0
2 years ago
O que é cena fone de luz na visão da fisica
yulyashka [42]
Me don’t speak spanish
4 0
3 years ago
A gray kangaroo can bound across level ground with each jump carrying it 9.1 m from the takeoff point. Typically the kangaroo le
Alona [7]

Answer:

u = 10.63 m/s

h = 1.10 m

Explanation:

For Take-off speed ..

by using the standard range equation we have

R = u² sin2θ/g

R = 9.1 m

θ = 26º,

Initial velocity = u

solving for u

u² = \frac{Rg}{sin2\theta}

u^2 = \frac{9.1 x 9.80}{sin26}

u^2 = 113.17

u = 10.63 m/s

for Max height

using the standard h(max) equation ..

v^2 = (v_osin\theta)^2 -2gh

h =\frac{(v_o^2sin\theta)^2}{2g}

h  =  \frac{(113.17)(sin26)^2}{(2 x 9.80)}}

h = 1.10 m

7 0
3 years ago
To calibrate the calorimeter electrically, a constant voltage of 3.6 V is applied and a current of 2.6 A flows for a period of 3
iren [92.7K]

Answer:

372.3 J/^{\circ}C

Explanation:

First of all, we need to calculate the total energy supplied to the calorimeter.

We know that:

V = 3.6 V is the voltage applied

I = 2.6 A is the current

So, the power delivered is

P=VI=(3.6)(2.6)=9.36 W

Then, this power is delivered for a time of

t = 350 s

Therefore, the energy supplied is

E=Pt=(9.36)(350)=3276 J

Finally, the change in temperature of an object is related to the energy supplied by

E=C\Delta T

where in this problem:

E = 3276 J is the energy supplied

C is the heat capacity of the object

\Delta T =29.1^{\circ}-20.3^{\circ}=8.8^{\circ}C is the change in temperature

Solving for C, we find:

C=\frac{E}{\Delta T}=\frac{3276}{8.8}=372.3 J/^{\circ}C

5 0
2 years ago
Please help on this one?
Nonamiya [84]

30 or c because 60-30=30

3 0
3 years ago
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