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Roman55 [17]
2 years ago
14

ElloNY1 ZINDA HERE xDWHAT IS MOLE ??​

Physics
1 answer:
Julli [10]2 years ago
8 0

Answer:

yes I am

how are you

have a great day

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Compound name: for Al2O3
Deffense [45]

Answer:

Aluminum oxide

6 0
3 years ago
A 0.10-kilogram piece of modeling clay is tossed
viva [34]

a) The initial speed of the clay is 30 m/s

b) The final velocity of the block increases (40 m/s)

Explanation:

a)

We can solve this problem by using the law of conservation of momentum. In fact, in absence of external forces (=no friction), the total momentum of the clay + block system is conserved before and after the collision. Therefore, we can write:

p_i = p_f\\m_1 u_1 + m_2 u_2 = (m_1+m_2)v  

where:  

m_1 = 0.10 kg is the mass of the clay

u_1 is the initial velocity of the clay

m_2 = 0.10 kg is the mass of the wood block

u_2 = 0 is the initial velocity of the wood block (at rest)

v = 15 m/s is the final combined velocity of the clay+block after the collision

Re-arranging the equation, we can find u_1, the speed at which the clay was tossed:

u_1 = \frac{(m_1+m_2)v-m_2 u_2}{m_1}=\frac{(0.10+0.10)(15)-0}{0.10}=30 m/s

b)

In this second case, the clay is replaced by a bouncy ball, which rebounds back after the collision, instead of sticking with the block.

In this second case, the law of conservation of momentum becomes:

p_i = p_f\\m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2  

where

m_1 = 0.10 kg is the mass of the ball

u_1 = 30 m/s is the initial velocity of the ball

m_2 = 0.10 kg is the mass of the wood block

u_2 = 0 is the initial velocity of the wood block (at rest)

v_1 = -10 m/s is the velocity of the bouncing ball after the collision (negative because it goes backward)

v_2 is the velocity of the block after the collision

Solving for v_2, we find the final velocity of the block:

v_2 = \frac{m_1 u_1 - m_1 v_1}{m_2}=\frac{(0.10)(30)-(0.10)(-10)}{0.10}=40 m/s

As we can see, the final velocity of the block has increased. The reason for that is that, as the ball bounces back, part of the total momentum is "carried away" by the ball in the backward direction, and since the total momentum must remain constant, this means that the momentum carried by the block in the forward direction must be larger than the previous situation.

Learn more about conservation of momentum:

brainly.com/question/7973509

brainly.com/question/6573742

brainly.com/question/2370982

brainly.com/question/9484203

#LearnwithBrainly

7 0
3 years ago
A 220 kg crate hangs from the end of a rope of length L = 14.0 m. You push horizontally on the crate with a varying force F to m
kifflom [539]

<u>Answer</u>:

(a) magnitude of F = 797 N

(b)the total work done  W = 0

(c)work done by the gravitational force =  -1.55 kJ

(d)the work done by the pull  = 0

(e) work your force F does on the crate = 1.55 kJ

<u>Explanation</u>:

<u>Given</u>:

Mass of the crate, m =  220 kg

Length of the rope, L = 14.0m

Distance, d =  4.00m

<u>(a) What is the magnitude of F when the crate is in this final position</u>

Let us first determine vertical angle as follows

=>Sin \theta = \frac{d }{L}

=> \theta = Sin^{-1} \frac{d}{L} =

Now substituting thje values

=> \theta = Sin^{-1} \frac{4}{12} =

=> \theta = Sin^{-1} \frac{1}{3}

=> \theta = Sin^{-1}(0.333)

=> \theta = 19.5^{\circ}

Now the tension in the string resolve into components

The vertical component supports the weight

=>Tcos\theta = mg

=>T = \frac{mg}{cos\theta}

=>T = \frac{230 \times 9.8 }{cos(19.5)}

=>T = \frac{2254 }{cos(19.5)}

=>T = \frac{2254 }{0.9426}

=>T =2391N

Therefore the horizontal force

F = TSin(19.5)

F = 797 N

b) The total work done on it

As there is no change in Kinetic energy

The total work done W = 0

<u>c) The work done by the gravitational force on the crate</u>

The work done by gravity

Wg = Fs.d = - mgh

Wg = - mgL ( 1 - Cosθ )

Substituting the values                                                            

= -230 \times 9.8\times 12 ( 1 - cos(19.5) )

= -230 \times 9.8\times 12 ( 1 - 0.9426) )

= -230 \times 9.8\times 12 (0.0574)

= -230 \times 9.8\times 0.6888

=  -230 \times 6.750

= -1552.55 J

The work done by gravity = -1.55 kJ

<u>d) the work done by the pull on the crate from the rope</u>

Since the pull  is perpendicular to the direction of motion,

The work done = 0

e)Find the work your force F does on the crate.

Work done by the Force on the crate

WF = - Wg  

WF = -(-1.55)

WF = 1.55 kJ

<u>(f) Why is the work of your force not equal to the product of the horizontal displacement and the answer to (a)</u>

Here the work done by force is not equal to F*d  

and it is equal to product of the cos angle and F*d

So, it is not equal to the product of the horizontal displacement and the answer to (a)      

7 0
3 years ago
Compressional forces within the crust can produce:
brilliants [131]
<span>tension, compression, and shearing and can i get brainliest plz</span>
4 0
3 years ago
How do you transfer kinetic energy to electrical energy using a generator system
san4es73 [151]

Answer: so when a turbine converts the K.E and the potential of any moving fluid (more likely liquid or gas) to energy. once the proc is started the turbine generato, the fluid such as water, steam, combus gasses, or air pushes s big series of blades that have mounted on a shaft, which then will rotate the shaft that’s conn to the generator

Explanation: hope this helped plz mark brainest

3 0
3 years ago
Read 2 more answers
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