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LenaWriter [7]
4 years ago
14

What is the control center of the bacteria cell

Physics
1 answer:
iren2701 [21]4 years ago
7 0
The control center of the bacteria cell is RNA is "free floating".

Till then. -ayeitswesley
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How much force is needed to accelerate a 3kg book in 15m/s
kotegsom [21]
F=M×A
F=3 ×15
F=45N
so the force is 45 N
7 0
3 years ago
A small dog is trained to jump straight up a distance of 1.2 m. How much kinetic energy does the 7.2-kg dog need to jump this hi
LUCKY_DIMON [66]
Answer:84.672 joules.

Explanation:

1) Data:

m = 7.2 kg
h = 1.2 m
g = 9.8 m / s²

2) Physical principle

Using the law of mechanical energy conservation principle, you have that the kinetic energy of the dog, when it jumps, must be equal to the final gravitational potential energy.

3) Calculations:

The gravitational potential energy, PE, is equal to m × g × h

So, PE = m × g × h = 7.2 kg × 9.8 m/s² × 1.2 m = 84.672 joules.

And that is the kinetic energy that the dog needs.
8 0
4 years ago
A student performs a lab measuring the velocities of toy cars of different masses
stiv31 [10]
The answer is Car 1 and Car 2.
5 0
4 years ago
Read 2 more answers
What is a voltage ? What is potential Difference ?
ANEK [815]

Answer:

A voltage is an electromotive force or potential difference expressed in volts. A potential difference is the difference of electrical potential between two points

8 0
3 years ago
Use Stefan's law to find the intensity of the cosmic background radiation emitted by the fireball of the Big Bang at a temperatu
Rus_ich [418]

Complete Question

Use Stefan's law to find the intensity of the cosmic background radiation emitted by the fireball of the Big Bang at a temperature of 2.81 K. Remember that Stefan's Law gives the Power (Watts) and Intensity is Power per unit Area (W/m2).

Answer:

The intensity is I  = 3.535 *10^{-6} \  W/m^2

Explanation:

From the question we are told that

    The temperature is  T = 2.81 \ K

Now  According to Stefan's law

        Power(P) =  \sigma  *  A  * T^4

Where  \sigma is the Stefan Boltzmann constant with value  \sigma  =  5.67*10^{-8} m^2 \cdot kg \cdot s^{-2} K^{-1}

  Now the intensity of the cosmic background radiation emitted according to the unit from the question is mathematically evaluated as

        I  =  \frac{P}{A}

=>      I  =  \frac{\sigma *  A  * T^4}{A}

=>      I  =  \sigma  *  T^4

substituting values

      I  = 5.67 *10^{-8}  *  (2.81)^4

       I  = 3.535 *10^{-6} \  W/m^2

       

4 0
3 years ago
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