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pshichka [43]
3 years ago
15

This is my first question don’t really know how to use this app yet lol but somebody answer it for me pls!! Seded the corredare

Physics
1 answer:
miskamm [114]3 years ago
7 0
3x6



(This just for extras )
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2. This diagram represents a top-down view of an experiment on a table. The 250 g and 100 g masses are falling and are pulling t
Rasek [7]

Answer:

According to the data given in the question, experiment on table two pulling and falling masses are arranged in the fig. 250  g is pulling right side and   100 g pulling down. The gravitational force is common to both the masses, so we cannot say that the block moves towards heavier mass, also the block  does not move towards the lighter mass.

Obviously, the effect of heavier mass of 250 g is more on the block, so the block moves towards right bottom corner. i.e., diagonally between two masses


please find the attachment.

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3 years ago
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What is a substance?
Alika [10]

Answer:

A single component that can’t be separated

brainliest please ;)

5 0
3 years ago
Rusting metal is an example of a _________ change. A) phase B) state C) chemical D) physical
sergeinik [125]

Answer:

C. Chemical Change

Explanation:

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3 years ago
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A certain resistor dissipates 0.5 W when connected to a 3 V potential difference. When connected to a 1 V potential difference,
Stels [109]

Answer:

<h2>0.056 W</h2>

Explanation:

Power = IV

From ohms law we know that

V= IR\\\\I= \frac{V}{R} \\\\Power= \frac{V}{R}*V\\\\Power= \frac{V^2}{R}

Given data

P1 = 0.5 Watt

P2 = ?

V1= 3 Volts

V2= 1 Volt

Thus we can solve for the power dissipated as follows

P1= \frac{V1^2}{R1}\\\\P2= \frac{V2^2}{R2}

\frac{P1}{P2} = \frac{V1^2}{V2^2}\\\\ P2=\frac{ V2^2}{ V1^2} *P1\\\\ P2=\frac{ 1^2}{ 3^2} *0.5= 0.055= 0.056 W

<em>The  resistor will dissipate 0.056 Watt</em>

7 0
3 years ago
uniform ladder of length 6.0 m and weight 300 N leans against a frictionless vertical wall. The foot of the ladder isplaced 3.0
olganol [36]

Answer:

Fx1 (6 m) sin 60 = 300 (3 m) cos 60  balancing torques about floor

Fx1 = 900 * 1/2 / 5.20 = 86.6 N  this is the horizontal force that must be supplied by the wall to balance torques about the floor

This is also equal to the static force of friction that must be applied at the point of contact with the floor to balance forces in the x-direction.

Fx1 = Fx2 = 86.6 N

3 0
3 years ago
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