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mel-nik [20]
2 years ago
8

PLEASE ANSWER ASAP BEFORE MY TEACHER AND MY MOM KILLES ME PLEASE ASAP

Physics
2 answers:
almond37 [142]2 years ago
8 0

That is not meant to be red, it‘s the bottom of the beaker. The star is at the very bottom of the beaker. it’s just the base of the beaker.

yawa3891 [41]2 years ago
6 0
The red at the bottom is just the holder for the beaker. It’s not apart of the density.
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A 30 kg mass and a 20 kg mass are joined by a light rigid rod and this system is free to rotate in the plane of the page about a
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55 J

Explanation:

Kinetic energy is given as: 0.5MV^2 where M is the mass and V is the speed of rotation. Since the masses are point masses, we calculate the point mass for each mass.

M1 = 30*0.2^2 = 1.2kgm^2

M2 = 20*0.4^2 = 3.2kgm^2

V = 5 rad/s

Calculating using the formula above, we obtain :

0.5(1.2+3.2)5^2 =0.5*4.4*25 = 55 J

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3 years ago
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On a sunny day, a family decided to take sail on a nearby lake. During sailing the wind applies a force of 25N north on the sail
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its n

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Energy that does not involve the large-scale motion or position of objects in a system is called:
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I believe the answer is C.
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You turn on the faucet for your garden hose and let go immediately the end search flying around and spraying water everywhere wh
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It’s Newton’s third law
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Two workers pull horizontally on a heavy box. but one pulls twice as hard as the other. The larger pull is directed at 21.0° wes
pantera1 [17]

Answer:

The  magnitude of F1 is

|F1|=358.74 \ N

The magnitude of F2 is

|F2|=179.37\ N

And the direction of F2 is

\alpha = 44.214^o

Explanation:

<u>Net Force </u>

Forces are represented as vectors since they have magnitude and direction. The diagram of forces is shown in the figure below.  

The larger pull F1 is directed 21° west of north and is represented with the blue arrow. The other pull F2 is directed to an unspecified direction (red arrow). Since the resultant Ft (black arrow) is pointed North, the second force must be in the first quadrant. We must find out the magnitude and angle of this force.  

Following the diagram, the sum of the vector components in the x-axis of F1 and F2 must be zero:

\displaystyle -2F\ sin21^o+F\ cos\alpha =0

The sum of the vertical components of F1 and F2 must equal the total force Ft

\displaystyle -2F\ cos21^o+F\ sin\alpha =460

Solving for \alpha in the first equation

\displaystyle cos\alpha =\frac{2F\ sin21^o}{F}=2sin21^o

\displaystyle cos\alpha =0.717=>\alpha =44.214^o

\displaystyle F(2cos21^o+sin\alpha)=460

\displaystyle F=\frac{460}{2cos21^o+sin\alpha}

\displaystyle F=\frac{460}{2cos21^o+sin44.214^o}

\displaystyle F=179.37\ N

The  magnitude of F1 is

|F1|=2*F=358.74 \ N

The magnitude of F2 is

|F2|=179.37\ N

And the direction of F2 is

\alpha = 44.214^o

4 0
3 years ago
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