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Liono4ka [1.6K]
3 years ago
13

. Write the following in the scientific notation of 0.0000002400m ​

Physics
1 answer:
aleksandr82 [10.1K]3 years ago
7 0

Explanation:

<h2><u>Steps </u><u>:</u></h2>
  1. <u>Move </u><u>decimal</u><u> </u><u>from</u><u> </u><u>left </u><u>to </u><u>right</u><u> </u><u>=</u><u>0</u><u> </u><u>0</u><u>0</u><u>0</u><u>0</u><u>0</u><u>0</u><u>2</u><u>4</u><u>0</u><u>.</u><u>0</u>
  2. <u>Then </u><u>count </u><u>the</u><u> </u><u>numbers</u><u> </u><u>before</u><u> </u><u>decimal </u><u>and </u><u>w</u><u>rite </u><u>it </u><u>like</u><u> </u><u>this </u><u>=</u><u>2</u><u>4</u><u>0</u><u>.</u><u>0</u><u>x</u><u>1</u><u>0</u><u> </u><u>power-</u><u>9</u><u> </u>
  3. <u>That's</u><u> </u><u>all </u>

<u>hope</u><u> it</u><u> </u><u>help</u>

<h2><u>#</u><u>H</u><u>o</u><u>p</u><u>e</u></h2>
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A scientist measures the growth of a bamboo plant over time. The table above shows the results. What is the best interference fo
lisabon 2012 [21]

(B) 2.25cm

<u>Explanation:</u>

Given:

At 40 hours, the height of the bamboo plant is 2.1cm

At 50 hours, the height of the bamboo plant is 2.4cm

Height of the bamboo plant after 45 hours = ?

The difference in length from 40 to 50 hours = 2.4 - 2.1cm

                                                                      = 0.3 cm

Mean of 40 and 50 is 45.

Thus,

At 45 hours, the height will increase by 0.3/2

                                                         = 0.15 cm

Height at 45 hour = 2.1 + 0.15cm

                           = 2.25cm

Therefore, the height of the plant after 45 hours is 2.25cm

7 0
3 years ago
2. How long must a 400 W electrical engine work in order to produce 300 kJ of work?
yanalaym [24]

Answer:

Explanation:

400 W = 400 J/s

300000 J / 400 J/s = 750 s or 12.5 minutes

7 0
3 years ago
The particles that make up protons and neutrons are called
Sphinxa [80]

Answer:

Quark

Explanation:

In the past the proton,neutron and electron were believed to be indivisible. Quarks were later discovered to be particles that make up the proton and neutron. Quarks are known as fast moving points of energy and are usually joined to one another through gluons. Neutron has two down quarks and one up quark while proton has two up quarks and one down quark.

6 0
3 years ago
Does a rigid object in uniform rotation about a fixed axis satisfy the first and second conditions for equilibrium?
Vladimir [108]

Answer:

Explanation:

a rigid object in uniform rotation about a fixed axis does not satisfy both the condition of equilibrium .

First condition of equilibrium is that net force on the body should be zero.

or F net = 0

A body under uniform rotation is experiencing a centripetal force all the time so F net ≠ 0

So first condition of equilibrium is not satisfied.

Second condition is that , net torque acting on the body  must be zero.

In case of a rigid object in uniform rotation , centripetal force is applied towards the centre ie  towards the  line joining the body under rotation with the axis .

F is along r

torque = r x F

= r F sinθ

θ = 0 degree

torque = 0

Hence 2nd condition is fulfilled.

7 0
3 years ago
A 0.58 kg mass is moving horizontally with a speed of 6.0 m/s when it strikes a vertical wall. The mass rebounds with a speed of
Sunny_sXe [5.5K]

Answer:

5.8\; {\rm kg\cdot m \cdot s^{-1}}.

Explanation:

If the mass of an object is m and the velocity of that object is v, the linear momentum of that object would be m\, v.

Assume that the initial velocity of the mass is positive (6.0\; {\rm m\cdot s^{-1}}.) However, the direction of the velocity is reversed after the impact. Thus, the sign of the new velocity of the object would be negative- the opposite of that of the initial velocity. The new velocity would be (-4.0\; {\rm m\cdot s^{-1}}).

Thus, the change in the velocity of the mass would be:

\begin{aligned}& (\text{Change in Velocity}) \\ =\; & (\text{Final Velocity}) - (\text{Initial Velocity}) \\ =\; & (-4.0\; {\rm m\cdot s^{-1}}) - (6.0\; {\rm m\cdot s^{-1}}) \\ =\; & (-10\; {\rm m\cdot s^{-1})\end{aligned}.

The change in the linear momentum of the mass would be:

\begin{aligned} & \text{change in momentum} \\ =\; & (\text{mass}) \times (\text{change in velocity}) \\ =\; & 0.58\; {\rm kg} \times (-10\; {\rm m\cdot s^{-1}}) \\  =\; & (-5.8\; {\rm kg \cdot m \cdot s^{-1}})\end{aligned}.

Thus, the magnitude of the change of the linear momentum would be 5.8\; {\rm kg \cdot m \cdot s^{-1}}.

7 0
3 years ago
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