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2 years ago

Scientific notation. jst 11a

Physics

1 answer:

coldgirl [10]2 years ago

5 0

Answer:

2 * 10^4

Explanation:

(3 * 10^4)(4 * 10^4) / (6 * 10^4)

= (12 * 10^8)/(6 * 10^4)

= 2 * 10^4

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When magnesium fluoride reacts with iodine, the products are fluorine and magnesium iodide. What would the equation be?

likoan [24]

Answer:

Explanation:

Every chemical reaction can be represented by simple chemical equations that shows how compounds are combining to give some products.

Such reactions are usually made up of:

Reactants on the left hand side

Products on the right hand side

Reactants → Products

In this given problem;

Reactants are : Magnesium fluoride = MgF₂

Iodine in form of Iodide = I₂

Product : Magnesium iodide = MgI₂

MgF₂ + I₂ → MgI₂ + F₂

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3 years ago

What is the IMA of the following pulley system?

Furkat [3]

Answer: $IMA= \frac {F_r}{F_e}$

Explanation:

IMA stands for Ideal Mechanical advantage.

The IMA of pulley system can be defined as the ratio of output force to input force.

From the given pulley system,

The input force = $F_e$

The output force = $F_r$

Hence, IMA of the given pulley system,

$IMA= \frac {F_r}{F_e}$

5 0

3 years ago

Read 2 more answers

The period of a simple pendulum is 3.5 s. The length of the pendulum is doubled. What is the period T of the longer pendulum?

Rudik [331]

Answer:

Explanation:

T = 2π√(L/g)

If you increase L to 2L, the period is increased by a factor of √2

T = 3.5√2 ≈ 4.9 s

6 0

3 years ago

Convert 2 kg to cg give you answer in SI

BARSIC [14]

1 kg=100000 cg

2 kg=200000 cq

If mass is the quantity then kg is the S.I

2 kg=2kg

6 0

3 years ago

The quantity of charge Q in coulombs (C) that has passed through a point in a wire up to time t (measured in seconds) is given b

Mnenie [13.5K]

Explanation:

The quantity of charge Q in coulombs (C) that has passed through a point in a wire up to time t (measured in seconds) is given by :

$Q=t^3-2t^2+4t+4$

We need to find the current flowing. We know that the rate of change of electric charge is called electric current. It is given by :

$I=\dfrac{dQ}{dt}\\\\I=\dfrac{d(t^3-2t^2+4t+4)}{dt}\\\\I=3t^2-4t+4$

At t = 1 s,

Current,

$I=3(1)^2-4(1)+4\\\\I=3\ A$

So, the current at t = 1 s is 3 A.

For lowest current,

$\dfrac{dI}{dt}=0\\\\\dfrac{d(3t^2-4t+4)}{dt}=0\\\\6t-4=0\\\\t=0.67\ s$

Hence, this is the required solution.

7 0

3 years ago