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lara [203]
4 years ago
12

You are creating a carbon cycle poster. What could you include in the poster to demonstrate the cycling

Physics
1 answer:
nalin [4]4 years ago
3 0

Answer:

the answer should be d.

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1. A large ball was let go on a hill and started rolling down with a constant acceleration of 4.2 m/s². What was the velocity of
pochemuha

Answer:

<em>The velocity after 12s is 50.4m/s</em>.

Explanation:

<em>In acceleration formula make velocity the </em><em>subject.</em>

<em> acceleration(a) = velocity(</em>v)÷time(t)

<h3><em> </em><em>velocity</em><em> </em><em>(</em><em>v)</em><em> </em><em>=</em><em> </em><em>acceleration</em><em>(</em><em>a)</em><em>×</em><em>t</em><em>ime</em><em>(</em><em>t)</em></h3>

<em>V </em><em>=</em><em> </em><em>4</em><em>.</em><em>2</em><em>m</em><em>/</em><em>s²</em><em>×</em><em>1</em><em>2</em><em>s</em>

<em>V </em><em>=</em><em> </em><em>5</em><em>0</em><em>.</em><em>4</em><em>m</em><em>/</em><em>s</em>

<em>Therefore</em><em> the</em><em> </em><em>velocity</em><em> </em><em>after</em><em> </em><em>1</em><em>2</em><em>s</em><em> </em><em>is </em><em>5</em><em>0</em><em>.</em><em>4</em><em>m</em><em>/</em><em>s.</em>

8 0
2 years ago
If a body p with a positive charge is placed in contact with a body q (initially uncharged), what will be the nature of the char
uysha [10]

If a body p with a positive charge is placed in contact with a body q (initially uncharged), then the nature of charge gained by q must be positive, because rubbing an uncharged body with a charged body or placed in contact with a positive charged body, helps gain a charge to the uncharged body.

There are a variety of methods to charge an object. One method is known as induction. In the induction process, a charged object is brought near but not touched to a neutral conducting object.

Let's know, how a element gain positive charge?

A positive charge occurs when the number of protons exceeds the number of electrons. A positive charge may be created by adding protons to an atom or object with a neutral charge. A positive charge also can be created by removing electrons from a neutrally charged object.

To learn more about Positive charge here

brainly.com/question/2903220

#SPJ4

5 0
2 years ago
The notes produced by a tuba range in frequency from approximately 45 Hz to 375 Hz. Find the possible range of wavelengths in ai
taurus [48]

Answer:

The possible range of wavelengths in air produced by the instrument is 7.62 m and 0.914 m respectively.

Explanation:

Given that,

The notes produced by a tuba range in frequency from approximately 45 Hz to 375 Hz.

The speed of sound in air is 343 m/s.

To find,

The wavelength range for the corresponding frequency.

Solution,

The speed of sound is given by the following relation as :

v=f_1\lambda_1

Wavelength for f = 45 Hz is,

\lambda_1=\dfrac{v}{f_1}

\lambda_1=\dfrac{343}{45}=7.62\ m

Wavelength for f = 375 Hz is,

\lambda_2=\dfrac{v}{f_2}

\lambda_2=\dfrac{343}{375}=0.914\ m/s

So, the possible range of wavelengths in air produced by the instrument is 7.62 m and 0.914 m respectively.

6 0
4 years ago
A coin with a diameter 3.00 cm rolls up a 30.0 inclined plane. The coin starts out with an initial angular speed of 60.0 rad/s
sweet [91]

This question is in complete.The question is

A coin with a diameter 3.00 cm rolls up a 30.0° inclined plane. The coin starts out with an initial angular speed of 60.0 rad/s and rolls in a straight line without slipping. If the moment of inertia of the coin is(1/2) MR² , how far will the coin roll up the inclined plane (length along the ramp)? Hint: Conservation of mechanical energy.

Answer:

distance=0.124 m

Explanation:

mgh=mglSin\alpha =(1/2)Iw_{i}^{2}+(1/2)mv^{2}\\   v=wR\\Solve for L\\L=((1/2)(1/2)0.015^{2}*60^{2}+(1/2)(60*0.015^{2} ))/9.8Sin30\\   L=0.124m

6 0
4 years ago
A disk-shaped grindstone of mass 3.0 kg and radius 8.0 cm is spinning at 600 rev/min. After the power is shut off, a man continu
kolbaska11 [484]

Answer:

τ=0.060 N.m

Explanation:

By kinematics:

\omega f = \omega o-\alpha*t

Solving for α:

\alpha=\frac{\omega o-\omega f}{t}

where ωo = 600*2*π/60;   ωf = 0;    t=10s

\alpha=6.283rad/s^2

The sum of torque is:

\tau=I*\alpha

\tau=M*R^2/2*\alpha

\tau=0.060 N.m

8 0
3 years ago
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