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borishaifa [10]
3 years ago
11

I need help!!! I need to have pictures for Absolute and Apparent Magnitude! Someone please help?!!!

Physics
1 answer:
valkas [14]3 years ago
7 0
Pictures you asked, I hope you like the pictures

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A defibrillator is used during a heart attack to restore the heart to its normal beating pattern. A defibrillator passes 18 A of
miskamm [114]

Answer:

q = 0.036 C

Explanation:

Given that,

Current passes through a defibrillator, I = 18 A

Time, t = 2 ms

We need to find the charge moved during this time. We know that,

Electric current = charge/time

q=It

Put all the values,

q=18\times 0.002\\\\q=0.036\ C

So, 0.036 C of charge moves during this time.

8 0
3 years ago
Write any two different between work and power?​
attashe74 [19]

Answer:

1. Work Is a type of Physical Activity  

2. Power is Basically Having control of Society

Explanation:

3 0
3 years ago
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statuscvo [17]

Answer:

huh ?ion understand

Explanation:

have a bad day <3

7 0
3 years ago
Martije has made a slight error in naming a compound monocarbon tetrabromide. What compound is she most likely naming, and what
Alik [6]

Answer: CBr_4 : carbon tetrabromide

Explanation:

CBr_4 is a covalent compound because in this compound the sharing of electrons takes place between carbon and bromine. Both the elements are non-metals. Hence, it will form covalent bond.

The naming of covalent compound is given by:

1. The less electronegative element is written first.

2. The more electronegative element is written second. Then a suffix is added with it. The suffix added is '-ide'.

3. If atoms of an element is greater than 1, then prefixes are added which are 'mono' for 1 atom, 'di' for 2 atoms, 'tri' for 3 atoms and so on.

Hence, the correct name for CBr_4 is carbon tetrabromide.

7 0
3 years ago
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Two forces F1 and F2 act on an object at a point P in the indicated directions. The magnitude of F1 is 15 lb and the magnitude o
adoni [48]

Answer

given,

F₁ = 15 lb

F₂ = 8 lb

θ₁ = 45°

θ₂ = 25°

Assuming the question's diagram is attached below.

now,

computing the horizontal component of the forces.

F_h = F₁ cos θ₁ - F₂ cos θ₂

F_h = 15 cos 45° - 8 cos 25°

F_h = 3.36 lb

now, vertical component of the forces

F_v = F₁ sin θ₁ + F₂ sin θ₂

F_v = 15 sin 45° + 8 sin 25°

F_v = 13.98 lb

resultant force would be equal to

F = \sqrt{F_h^2+F_v^2}

F = \sqrt{3.36^2+13.98^2}

F = 14.38 lb

the magnitude of resultant force is equal to 14.38 lb

direction of forces

\theta =tan^{-1}(\dfrac{F_v}{F_h})

\theta =tan^{-1}(\dfrac{13.98}{3.36})

   θ = 76.48°

3 0
3 years ago
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