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ziro4ka [17]
4 years ago
9

Question 2 (2 points)

Physics
1 answer:
tankabanditka [31]4 years ago
5 0

Answer:

Testing

Explanation:

This is basically asking you what the next step of the scientific method is. First, you would ask a question. Next, you would research that question to get some background information. Once you have that information, you would create a hypothesis (typically "If..., then..., because..." format). Then you would conduct an experiment to test your hypothesis. Then you would gather data, draw conclusions, and see if your hypothesis was correct.

Hope this helps!

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What happens when charged object is brought near uncharged object?<br> Attract or Repel ?
Makovka662 [10]
When a charged object is brought near to but does not touch a neutral object, it causes the side of the neutral object that the charged object is near to become the other charge. It causes charge migration within the neutral object so the two charges (positive and negative) move to opposite sides of the object. Because the two objects do not touch, they do not repel each other, but rather have a slight attraction because of charge migration. If the two object were to touch then they would repel.
5 0
3 years ago
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What is the velocity of 360?
9966 [12]

Answer:

3/360 is 120. your velocity is 120

Explanation:

8 0
3 years ago
An object 16.8 cm tall is placed in front of a converging lens. A real image, 46 cm tall, is formed on the other side of the len
worty [1.4K]

Answer:

2.74

Explanation:

Magnification = image distance/object distance

Mag = v/u

Given

v = 46cm

u = 16.8

Magnification = 46/16.8

Magnification = 2.74

Hence the magnification is 2.74

7 0
3 years ago
A 1300-N crate rests on the floor. How much work is required to move it at constant speed (a)
kherson [118]

a) The work done is 920 J

b) The work done is 5200 J

Explanation:

a)

In this first part of the problem, the crate is moved horizontally at constant speed.

The work required in this case is given by

W=Fd cos \theta

where

F is the magnitude of the force applied

d is the displacement of the crate

\theta is the angle between the direction of the force and of the displacement

Here the crate is moved at constant speed: this means that the acceleration of the crate is zero, and so according to Newton's second law, the net force on the crate is zero: this means that the force applied, F, must be equal to the force of friction (but in opposite direction), so

F = 230 N

The displacement is

d = 4.0 m

And the angle is \theta=0^{\circ}, since the force is applied horizontally. Therefore, the work done is

W=(230)(4.0)(cos 0^{\circ})=920 J

b)

In this case, the crate is moved vertically. The force that must be applied to lift the crate must be equal to the weight of the crate (in order to move it a constant speed), therefore

F = W = 1300 N

The displacement this time is again

d = 4.0 m

And the angle is \theta=0^{\circ}, since the force is applied vertically, and the crate is moved also vertically. Therefore, the work done on the crate this time is

W=(1300)(4.0)(cos 0^{\circ})=5200 J

Learn more about work:

brainly.com/question/6763771

brainly.com/question/6443626

#LearnwithBrainly

4 0
3 years ago
A missile is moving 1350 m/s at 25.0° angle. It needs to hit in a 55.0° direction in 10.20 s. What is the direction of its final
77julia77 [94]

Answer:

final velocity = 3504 m/s  

Explanation:

<em>Given data:</em>

velocity of missile = Vi = 1350m/s

angle at which missile is moving = 25degree

distance between missile and targets = 23500m

angle between target and missile=55degree

time=10.2s

<em>To find:</em>

Final velocity: ?

<em>Formula:</em>

x = Vx*t + ½*ax*t²  

Let x be the horizontal component of distance

x = ertical component of distance

t-time

ax = horizontal component of acceleration

ay = Vertical component of acceleration

Vx = horizontal component of velocity

Vy = Vertical component of velocity

<em>Solution:</em>

x = Vx*t + ½*ax*t²

23500m * cos55.0º = 1350m/s * cos25.0º * 10.20s + ½ * ax * (10.20s)²  

ax = 19.2 m/s²  

V'x = Vx + ax*t = 1350m/s * cos25.0º + 19.2m/s² * 10.20s = 1419 m/s  

<em>similarly vertically:</em>

y = Vy*t + ½*ay*t² 

23500m * sin55.0º = 1350m/s * sin25.0º * 10.20s + ½ * ay * (10.20s)²  

ay = 258 m/s²  

V'y = Vy + ay*t

     = 1350m/s * sin25.0º + 258m/s² * 10.20s = 3204 m/s  

V = √(V'x² + V'y²)

   = 3504 m/s  

8 0
3 years ago
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