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brilliants [131]
2 years ago
6

Find the odd one out radish,potato, beet,carrot.​

Physics
1 answer:
romanna [79]2 years ago
4 0

Answer:

Potato is not a root whereas the rest of the four items are forms of root.

Potato is a stem.

So, option B is the odd one out

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Exercise combined with a blank can help control weight
SIZIF [17.4K]

Answer:

Exercise combined with a diet can help manage weight, or you can contact a doctor

Explanation:

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Terangkan lokasi zaman prasejarah di Malaysia
Andreas93 [3]

Answer:

Panget ka wala kang alam at Bo bo ka mamatay ka sige patay kana

Done That's the answer

5 0
3 years ago
What is heat energy!? ​
nordsb [41]

Answer:

heat, energy that is transferred from one body to another as the result of a difference in temperature. If two bodies at different temperatures are brought together, energy is transferred—i.e., heat flows—from the hotter body to the colder. example: stove

Explanation:

hope this helps

5 0
3 years ago
Read 2 more answers
An object accelerates 3.0 m/s2 when a force of 6.0 N is applied to it. What is the mass of the object?
makvit [3.9K]

Answer:

<h2>2 kg</h2>

Explanation:

The mass of the object can be found by using the formula

m =  \frac{f}{a}  \\

f is the force

a is the acceleration

From the question we have

m =  \frac{6}{3}  \\

We have the final answer as

<h3>2 kg</h3>

Hope this helps you

5 0
2 years ago
A 6.5 kg rock thrown down from a 120m high cliff with initial velocity 18 m/s down. Calculate
Olegator [25]

Answer:

See the answers below.

Explanation:

In order to solve this problem we must use the principle of energy conservation. Which tells us that the energy of a body will always be the same regardless of where it is located. For this case we have two points, point A and point B. Point A is located at the top at 120 [m] and point B is in the middle of the cliff at 60 [m].

E_{A}=E_{B}

The important thing about this problem is to identify the types of energy at each point. Let's take the reference level of potential energy at a height of zero meters. That is, at this point the potential energy is zero.

So at point A we have potential energy and since a velocity of 18 [m/s] is printed, we additionally have kinetic energy.

E_{A}=E_{pot}+E_{kin}\\E_{A}=m*g*h+\frac{1}{2}*m*v^{2}

At Point B the rock is still moving downward, therefore we have kinetic energy and since it is 60 [m] with respect to the reference level we have potential energy.

E_{B}=m*g*h+\frac{1}{2}*m*v^{2}

Therefore we will have the following equation:

(6.5*9.81*120)+(0.5*6.5*18^{2} )=(6.5*9.81*60)+(0.5*6.5*v_{B}^{2} )\\3.25*v_{B}^{2} =4878.9\\v_{B}=\sqrt{1501.2}\\v_{B}=38.75[m/s]

The kinetic energy can be easily calculated by means of the kinetic energy equation.

KE_{B}=\frac{1}{2} *m*v_{B}^{2}\\KE_{B}=0.5*6.5*(38.75)^{2}\\KE_{B}=4878.9[J]

In order to calculate the velocity at the bottom of the cliff where the reference level of potential energy (potential energy equal to zero) is located, we must pose the same equation, with the exception that at the new point there is only kinetic energy.

E_{A}=E_{C}\\6.5*9.81*120+(0.5*9.81*18^{2} )=0.5*6.5*v_{C}^{2} \\v_{c}^{2} =\sqrt{2843.39}\\v_{c}=53.32[m/s]

5 0
3 years ago
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