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Lana71 [14]
3 years ago
7

A block of aluminum measures 4.0 cm x 5.0 cm x 2.0 cm is completely immersed in a tank of water.

Physics
1 answer:
Aneli [31]3 years ago
7 0

Answer:

upthrust or BUOYANT FORCE =Vdg

volume=LWH

upthrust=(4cm×5cm×2cm)×1g/cm²×g

upthrust=40cm³×1g/cm³×g

upthrust=40gf or 0.04kg×10m/s²=0.4N

weight of the displaced liquid is upthrust.

so mass=40g or 0.04kg

upthrust=40gf or 0.4Nand mass of the displaced liquid=40g or 0.04kg

please mark brainliest, hope it helped

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If block D weighs 300 lb and block B weighs 275 lb determine the required weight of block C and the angle theta for equilibrium?
adoni [48]
Set up a free body diagram. 

<span>and by reason, Tcd = Tbd </span>

<span>Tbd y = 275 - 300*sinθ </span>
<span>Tcd y = Tc - 300*sin30 </span>

<span>Tbd x = 300*cosθ </span>
<span>Tcdx = 300 * cos30 </span>

<span>Tbd^2 = (275 - 300*sinθ)^2 + (300*cosθ)^2 </span>
<span>Tcd^2 = (300*sin30)^2 + (300 * cos30)^2 </span>

<span>(275 - 300*sinθ)^2 + (300*cosθ)^2 = (300*sin30)^2 + (300 * cos30)^2 </span>
<span>etc.</span>
5 0
3 years ago
A ball rolls down a ramp for 15 seconds. If the initial velocity of the ball was 0.8m/s and the final velocity was 7m/s, what wa
iren [92.7K]
Acceleration = (change in speed) / (time for the change)

Change in speed = (ending speed) - (starting speed)

                            =  (7.0 m/s)  -  (0.8 m/s)  =  6.2 m/s

Time for the change  =  15 seconds

Acceleration  =  (6.2 m/s) / (15 sec)

                      =   (6.2/15)  m/s²

                      =     0.413 m/s²  
6 0
3 years ago
What are the largest and smallest resistances you can obtain by connecting a 36.0-Ω , a 50.0-Ω , and a 700-Ω resistor together?
myrzilka [38]
<h2>Answer:</h2>

786Ω and 20.32Ω respectively.

<h2>Explanation:</h2>

(a) Given a number of resistors each with its own resistance, the largest resistance can be obtained when these resistors are connected in series.

From the question, the resistors have the following resistances;

36.0-Ω, 50.0-Ω , and 700-Ω

Now, when they are connected in series, the total resistance (R) obtainable is given by the sum of these individual resistances as follows;

R = 36.0-Ω + 50.0-Ω + 700-Ω

R = 786Ω

Therefore, the largest resistance that can be obtained by connecting  a 36.0-Ω , a 50.0-Ω , and a 700-Ω resistor together is 786Ω

(b) Similarly, given a number of resistors each with its own resistance, the smallest resistance can be obtained when these resistors are connected in parallel.

From the question, the resistors have the following resistances;

36.0-Ω, 50.0-Ω , and 700-Ω

Now, when they are connected in parallel, the total resistance (R) obtainable is given by using the relation as follows;

\frac{1}{R} = \frac{1}{36.0} + \frac{1}{50.0} + \frac{1}{700.0}

\frac{1}{R} = \frac{35000+25200+1800}{1260000}

\frac{1}{R} = \frac{62000}{1260000}

\frac{1}{R} = \frac{62}{1260}

R = \frac{1260}{62}

R = 20.32Ω

Therefore, the smallest resistance that can be obtained by connecting  a 36.0-Ω , a 50.0-Ω , and a 700-Ω resistor together is 20.32Ω

7 0
3 years ago
You and your little cousin sit on a see-saw. You sit 0.5 m from the fulcrum, and your cousin sits 1.5 m from the fulcrum. You we
svetoff [14.1K]

Answer:

200N

Explanation:

0.5/1.5=x/600N

1/3=x/600

x=200N

7 0
3 years ago
What is a Non-Example of Newton's 1st law?
guajiro [1.7K]

Answer:

Say a 14 year old girl was at a construction site and she was asked to move something like a 10,000 pound brick( one brick). She would be acting on it as the unbalanced force but they would still not change their position.

so to say the girl would be doing everything she could to move that brick but the brick would still be in that same spot so the unbalanced force (the girl) would be acting on the thing that was at rest but it wouldn't move.

so the unbalanced force would not really be acting on the thing at rest; even though the unbalanced force was doing something to the brick.

( just think about it and you will eventually get it...just imagine in your head...)

Explanation:

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