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3 years ago

Define the problem. Look around your house or outside. Is there anything that you're wondering about or have questions about?

Physics

1 answer:

bekas [8.4K]3 years ago

6 0

Answer:

Water

Explanation:

How do we have water at our disposal, like that's crazy.

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For incompressible fluids, density changes

mr_godi [17]

Just to make it easier on you It’s C.Pressure

7 0

3 years ago

An 800-N billboard worker stands on a 4.0-m scaffold supported by vertical ropes at each end. If the scaffold weighs 500 N and t

Arturiano [62]

Answer:

T = 850 N

Explanation:

given,

mass of billboard worker = 800 N

length of scaffold = 4 m

weight of the scaffold = 500 N

worker is standing at 1 m from one end.

Tension in the rope = ?

To calculate the tension in the string we have to balance the clockwise and counterclockwise moment of the system.

Weight of the worker and the weight of the scaffold will be in clockwise direction where as the tension will be in counterclockwise direction

now,

800 x 3 + 500 x 2 = T x 4

4T = 3400

T = 850 N

hence, tension in the rope is equal to 850 N

5 0

4 years ago

49 POINTS!!!

Sladkaya [172]

Answer:

Explanation:

chemical kinetic

electric potenitial

Electromagnetic (light, x Ray's, microwaves, etc) kinetic

Gravitational potential

heat kinetic

Movement of mass kentic

8 0

3 years ago

Read 2 more answers

A vessel at rest at the origin of an xy coordinate system explodes into three pieces. Just after the explosion, one piece, of ma

tresset_1 [31]

Answer:

Explanation:

a. Given that:

$m-$ mass of first & second piece, $3m-$ mass of 3rd piece, $\bar v_1$ -velocity of first piece( $-23\dot i \ m/s$ ) and $\bar v_2$ as velocity of 2nd piece ( $-23\dot j \ m/s$ )

Let $\bar v_3$ be velocity of 3rd piece=?

#Vessel is at rest before explosion. Considering conservation of linear momentum:

$m\bar v_1 +m\bar v_2 +3m \bar v_3=0$ #Dividing both sides by $m, m\neq 0$

$\bar v_1+ \bar v_2 +3\bar v_3=0\\3\bar v_3=-(\bar v_1 +\bar v_2)\\\bar v_3=-\frac{1}{3}(\bar v_1 +\bar v_2)$

#Plug the $\bar v_1 ,\bar v_2$ values:

$\bar v_3=-\frac{1}{3}(-23\dot i -23\dot j)=\frac{1}{3}(23\dot i +23\dot j)$

#So the magnitude of the third piece is:

$|\bar v_3|=\sqrt{(23/3)^2+(23/3)^2}\\=10.84m/s$

Magnitude of the 3rd piece is 10.84 m/s

b. To find direction of the magnitude (as an angle relative to the $x$ -axis), we find $\angle \theta$ . The angle is obtained by getting the tan inverse as: