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3 years ago

What is the net force of this object? A. 400 newtons B. 200 newtons C. 0 newtons D. 600 newtons @Dexteright02

Physics

1 answer:

Katena32 [7]3 years ago

6 0

In this case, as the meanings of the vectors are opposite, one of the forces receives the negative signal and thus a piece of the other vector is drawn, as shown in the figure. And to know the module of the resultant force, if it subtracts the modules of the forces.

$F_{Net} = F_{grav} - F_{air}$
$F_{Net} = 600 - 400$
$\boxed{F_{Net} = 200N}
$
Answer:
B. 200N

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Electron grops could be considered which of the following

Maurinko [17]

Answer:

Electron groups could be considered as Lone pair electrons and bonded pairs of electrons.

Answer: Option D & B

Explanation:

The two or more electrons can be bonded by single bond, double bond, covalent bond of electrons can simply be lone pair of electrons. Unshared pair of electrons are generally termed as lone pair of electrons in an atom which are generally present in the outermost shell of atoms. Hence electron groups can be determined by bonded pairs and lone pairs of electrons.

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2 years ago

A parachutist falls 50.0 m without friction. When the parachute opens, he slows down at a rate of 67 m/s*2. If he reaches the gr

KIM [24]

Answer:

3.49 seconds

3.75 seconds

-43200 ft/s²

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

$s=ut+\frac{1}{2}at^2\\\Rightarrow 50=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{50\times 2}{9.81}}\\\Rightarrow t=3.19\ s$

Time the parachutist falls without friction is 3.19 seconds

$v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 9.81\times 50+0^2}\\\Rightarrow v=31.32\ m/s$

Speed of the parachutist when he opens the parachute 31.32 m/s. Now, this will be considered as the initial velocity

$v=u+at\\\Rightarrow 11=31.32+9.81t\\\Rightarrow t=\frac{11-31.32}{-67}=0.3\ s$

So, time the parachutist stayed in the air was 3.19+0.3 = 3.49 seconds

$s=ut+\frac{1}{2}at^2\\\Rightarrow \frac{s}{2}=0t+\frac{1}{2}\times a\times t^2\\\Rightarrow \frac{s}{2}=\frac{1}{2}at^2$

$s=ut+\frac{1}{2}at^2\\\Rightarrow \frac{s}{2}=u1.1+\frac{1}{2}\times a\times 1.1^2$

Now the initial velocity of the last half height will be the final velocity of the first half height.

$v=u+at\\\Rightarrow v=at$

Since the height are equal

$\frac{1}{2}at^2=u1.1+\frac{1}{2}\times a\times 1.1^2\\\Rightarrow \frac{1}{2}at^2=at1.1+\frac{1}{2}\times a\times 1.1^2\\\Rightarrow 0.5t^2-1.1t-0.605=0\\\Rightarrow 500t^2-1100t-605=0$

$t=\frac{11\left(1+\sqrt{2}\right)}{10},\:t=\frac{11\left(1-\sqrt{2}\right)}{10}\\\Rightarrow t=2.65, -0.45$

Time taken to fall the first half is 2.65 seconds

Total time taken to fall is 2.65+1.1 = 3.75 seconds.

When an object is thrown with a velocity upwards then the velocity of the object at the point to where it was thrown becomes equal to the initial velocity.

$v^2-u^2=2as\\\Rightarrow a=\frac{v^2-u^2}{2s}\\\Rightarrow a=\frac{0^2-240^2}{2\times \frac{8}{12}}\\\Rightarrow a=-43200\ ft/s^2$

Magnitude of acceleration is -43200 ft/s²

5 0

3 years ago

In practice, if a voltmeter was connected across any combination of the terminals, the potential difference would be less than w

VladimirAG [237]

Answer:

This difference is kept to a minimum because the resistance in transformers is a few tens of ohms and the resistance of modern voltmeters is of the order of MΩ.

Explanation:

A voltmeter is built by a galvanometer and a resistance in series, this set is connected in parallel to the resistance where the voltage is to be measured, therefore the voltage is divided between the voltmeter and the element to be measured, consequently the measured voltage It is less than the calculated one, since for them the resistance of the voltmeter is assumed infinite.

This difference is kept to a minimum because the resistance in transformers is a few tens of ohms and the resistance of modern voltmeters is of the order of MΩ.

8 0

3 years ago

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Minchanka [31]

Infrared, visible light, then ultraviolet. Infrared is light that the human eye can not see and visible light is clearly light we can see then ultraviolet is has such a high frequency we can't see it either.

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3 years ago

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A 5 kg ball rolling at 1.0 m/s hits a 15 kg ball at rest. The balls stick together after the collision What is the

Reika [66]

Answer:

0.25m/s

Explanation:

Given parameters

m₁ = 5kg

v₁ = 1.0m/s

m₂ = 15kg

v₂ = 0m/s

Unknown:

velocity after collision = ?

Solution:

Momentum before collision and after collision will be the same. For inelastic collision;

m₁v₁ + m₂v₂ = v(m₁ + m₂)

Insert parameters and solve for v;

5 x 1 + 15 x 0 = v (5 + 15 )

5 = 20v

v = $\frac{5}{20}$ = 0.25m/s

5 0

3 years ago