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Leya [2.2K]
4 years ago
6

Calculate the Potential Energy of an object that has a mass of 14-kg and is located at a height of 24-m?

Physics
1 answer:
inessss [21]4 years ago
4 0
Potential Energy= 24m * 14kg * 9.8N/kg = 3292.8J
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Help! Will Mark Brainliest!
boyakko [2]

Answer:

D 5m/s

Explanation:

6 0
3 years ago
What would occur if there was a gain-of-function mutation in the promoter for the cyclin E gene such that cyclin E protein was a
rjkz [21]

Answer:

b) Cells will pass through the G1/S checkpoint even if conditions are not ideal for cell division.

Explanation:

In the given problem, if there exists a gain-of-function mutation for the given cell, there would not be the formation of cyclin E when there is the possibility of cells movement via the checkpoint of the G1/S, even when there are non-deal conditions for the division of cell. Thus, the correct option in the lists of options is the option b.

7 0
4 years ago
Convert <img src="https://tex.z-dn.net/?f=%5Cfrac%7B%280.779mg%29%28min%29%7D%7BL%7D" id="TexFormula1" title="\frac{(0.779mg)(mi
Orlov [11]

The number converted is 0.0467 \frac{(kg)(s)}{m^3}

Explanation:

In order to convert from the original units to the final units, we have to keep in mind the following conversion factors:

1 kg = 1000 g = 10^6 mg

1 min = 60 s

1 m^3 = 1000 L

The original unit that we have is

\frac{mg\cdot min}{L}

Therefore, it can be rewritten as:

=\frac{mg \frac{1}{10^6 mg/kg}\cdot min\cdot  60 s/min}{L\frac{1}{1000L/m^3}}=0.06 \frac{(kg)(s)}{m^3}

Therefore, since the initial number was 0.779, the final value is

0.779\cdot 0.06 \frac{(kg)(s)}{m^3}=0.0467 \frac{(kg)(s)}{m^3}

#LearnwithBrainly

5 0
4 years ago
Write the dimension of a / b in the x = at + bt2. Where x is the distance and t is the time?​
Luba_88 [7]

The dimension of a/b where x is the distance and t is the time is T

Given the expression

x = at + bt²

where

x is the distance

t is the time

Based on the homogeneity principle, the expression on the left-hand side must be equal to that on the right. Hence;

x = at

a = \frac{x}{t}

Since x is the distance and distance is measured in metres, the dimension equivalent will be the length 'L'

Since t is the time and time is measured in seconds, the dimension equivalent will be the seconds 'T'

a=\frac{L}{T}

Similarly;

x  = bt²

b=\frac{x}{t^2}\\b=\frac{L}{T^2}

Next is to get a/b;

\frac{a}{b} =  \frac{L}{T} \div \frac{L}{T^2}\\\frac{a}{b} = \frac{L}{T}*\frac{T^2}{L}  \\\frac{a}{b} =\frac{T^2}{T}\\\frac{a}{b} =T

Hence the dimension of a/b is T

4 0
3 years ago
What term is used to denote the change of velocity with time
Westkost [7]
The answers is acceleration.
6 0
3 years ago
Read 2 more answers
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