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2 years ago

6

why do this I'm trying to cheat and you put a question omgggggggggggggggggggggg!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

1 answer:

Dahasolnce [82]2 years ago

5 0

Answer:

yeah

Explanation:

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A 34-kg child runs with a speed of 2.8 m/s tangential to the rim of a stationary merrygo-round. The merry-go-round has a momentu

tekilochka [14]

Answer: $0.43\ rad/s$

Explanation:

Given

Mass of child $m=34\ kg$

speed of child is $v=2.8\ m/s$

Moment of inertia of merry go round is $I=510\ kg.m^2$

radius $r=2.31\ m$

Conserving the angular momentum

$\Rightarrow mvr=I\omega \\\Rightarrow 34\times 2.8\times 2.31=510\times \omega\\\\\Rightarrow \omega=\dfrac{219.912}{510}\\\Rightarrow \omega=0.43\ rad/s$

7 0

2 years ago

White light viewed through a prism is an example of a(n) _______________ spectrum.

hoa [83]

Answer:

White light viewed through a prism is an example of a visible spectrum.

3 0

3 years ago

Which nucleus completes the following equation?<br> A. 299 Np<br> B. 20Pa<br> C. 2 Pa<br> D. - Np

blondinia [14]

Answer:

Option D. ²³⁹₉₃Np

Explanation:

Let the unknown be ʸₓA.

Thus, the equation becomes:

²³⁹₉₂U —> ⁰₋₁e + ʸₓA

Next, we shall determine the x, y and A. This can be obtained as follow:

92 = –1 + x

Collect like terms

92 + 1 = x

93 = x

x = 93

239 = 0 + y

239 = y

y = 239

ʸₓA => ²³⁹₉₃A => ²³⁹₉₃Np

Thus, the complete equation is:

²³⁹₉₂U —> ⁰₋₁e + ²³⁹₉₃Np

7 0

3 years ago

A wheel with moment of inertia 25 kg. m2 and angular velocity 10 rad/s begins to speed up, with angular acceleration 15 rad/sec2

Pani-rosa [81]

Answer:

(A) Angular speed 40 rad/sec

Rotation = 50 rad

(b) 37812.5 J

Explanation:

We have given moment of inertia of the wheel $I=25kgm^2$

Initial angular velocity of the wheel $\omega _0=10rad/sec$

Angular acceleration $\alpha =15rad/sec^2$

(a) We know that $\omega =\omega _0+\alpha t$

We have given t = 2 sec

So $\omega =10+15\times 2=40rad/sec$

Now $\Theta =\omega _0t+\frac{1}{2}\alpha t^2=10\times 2+\frac{1}{2}\times 15\times 2^2=50rad$

(b) After 3 sec $\omega =10+15\times 3=55rad/sec$

We know that kinetic energy is given by $Ke=\frac{1}{2}I\omega ^2=\frac{1}{2}\times 25\times 55^2=37812.5J$

7 0

3 years ago

The fastest pitched baseball was clocked at 46 m/s. assume that the pitcher exerted his force (assumed to be horizontal and cons

Zielflug [23.3K]

Using the Equation:
v² = vi² + 2 · a · s → Eq.1
where,
v = final velocity
vi = initial velocity
a = acceleration
s = distance

<span><span>We know that vi = 0 because the ball was at rest initially.
</span><span>
Therefore,

Solving Eq.1 for acceleration,

</span></span> v² = vi² + 2 · a · s
v² = 0 + 2 · a · s
v² = 2 · a · s
Rearranging for a,
a = v ²/2·<span>s
Substituting the values,
a = 46</span>²/2×1<span>
a = 1058 m/s</span>²

<span>Now applying Newton's 2nd law of motion,
</span>
<span>F = ma
= 0.145</span>×<span>1058

F = 153.4 N</span>

8 0

3 years ago