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zhenek [66]
3 years ago
15

Physical Science Lesson 4 unit 2 questions, I need the answers

Physics
2 answers:
neonofarm [45]3 years ago
7 0
I know I'm a bit late but just in case you still need some of the answers.

1: Oxygen
2: Change in shape
3: Water breaking down into hydrogen and oxygen
4: I think it's Milk but I might be wrong.

I might be wrong about some of them, sorry it's not much.

lubasha [3.4K]3 years ago
5 0

1- Oxygen

2- Change in shape

3- Water breaking down into hydrogen and oxygen

4- Milk

5-Liquid

6- Solid

7-Physical

8- Chemical property

9- Volume

10- Kinetic Energy

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Archy [21]
If you dont wear protection while shooting a rifle it will damge your hear
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3 years ago
Read 2 more answers
The Great Sandini is a 60 kg circus performer who is shotfrom a cannon (actually a spring gun). You don't find many men ofhis ca
d1i1m1o1n [39]

Answer:

V=15.3 m/s

Explanation:

To solve this problem, we have to use the energy conservation theorem:

U_e+K_i+U_{gi}+W_{friction}=K_f+U_{gf}

the elastic potencial energy is given by:

U_e=\frac{1}{2}*k*x^2\\U_e=\frac{1}{2}*1100N/m*(4m)^2\\U_e=8800J

The work is defined as:

W_{friction}=F_f*d*cos(\theta)\\W_{friction}=40N*2.5m*cos(180)\\W_{friction}=-100J

this work is negative because is opposite to the movement.

The gravitational potencial energy at 2.5 m aboves is given by:

U_{gf}=m*g*h\\U_{gf}=60kg*9.8*2.5\\U_{gf}=1470J

the gravitational potential energy at the ground and the kinetic energy at the begining are 0.

8800J+0+0+-100J=\frac{1}{2}*62kg*v^2+1470J\\v=\sqrt{\frac{2(8800J-100J-1470J)}{62kg}}\\v=15.3m/s

3 0
3 years ago
If the given wave has a frequency of 100 Hz, what frequency will the sixth harmonic have?
alukav5142 [94]

Answer:

600Hz

Explanation:

In electrical systems of alternating current, the harmonics are, as in acoustics, frequencies multiples of the fundamental working frequency of the system and whose amplitude decreases as the multiple increases. For example, if we have systems fed by the 50 Hz network, harmonics of 100, 150, 200, etc. may appear.

In our case having a fundamental wave of 100Hz, I can have harmonics of 200,300,400, ..., 600Hz

4 0
3 years ago
A vw beetle goes from 0 to 60 mi/h with an acceleration of 2.35 m/s^2.
Vlad1618 [11]
Part a.
u = 0, the initial velocity
v = 60 mi/h, the final velocity
a = 2.35 m/s², the acceleration.

Note that
1 m = 1609.34 m.
Therefore
v = (60 mi/h)*(1609.34 m/mi)*(1/3600 h/s) = 26.822 m/s
Use the formula
v = u + at
(26.822 m/s) = (2.35 m/s²)*(t s)
t = 26.822/2.35 = 11.4 s

Answer: 11.4 s

Part b.
We already determined that v = 60 mi/h = 26.822 m/s.
t = 0.6 s
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6 0
3 years ago
Determine the components of reaction at the fixed support
denis-greek [22]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The components of reaction at the fixed support are  

    A_{(x)}  = 400  \ N ,  A_{(y)}  = -500  \ N ,  A_{(z)}  = 600  \ N ,  M_x  = 1225 \  N\cdot m , M_y  = 750 \  N\cdot m ,  M_z  = 0 \  N\cdot m

Explanation:

Looking at  the diagram uploaded we see that there are two  forces acting along the x-axis on the fixed support    

   These force are  400 N  and  A_{(x)} [ i.e the reactive force of  400 N  ]

Hence the sum of forces along the x axis is mathematically represented as

        A_{(x)}  - 400  = 0

=>     A_{(x)}  = 400  \ N

Looking at  the diagram uploaded we see that there are two  forces acting along the y-axis on the fixed support  

   These force are  500 N  and  A_{(y)} [ i.e the force acting along the same direction with 500 N   ]

Hence the sum of forces along the x axis is mathematically represented as

        A_{(y)}  + 500  = 0

=>     A_{(y)}  = -500  \ N

Looking at  the diagram uploaded we see that there are two  forces acting along the z-axis on the fixed support  

       These force are  600 N  and  A_{(z)} [ i.e the reactive force of  600 N  ]

Hence the sum of forces along the x axis is mathematically represented as

        A_{(z)}  - 600  = 0

=>     A_{(z)}  = 600  \ N

Generally taking moment about A along the x-axis we have that

    \sum M_x  = M_x  - 500 (0.75 + 0.5) + 600 ( 1 ) = 0

=>   M_x  = 1225 \  N\cdot m

Generally taking moment about A along the y-axis we have that

    \sum M_y  = M_y  - 400 (0.75 ) + 600 ( 0.75 ) = 0

=>   M_y  = 750 \  N\cdot m

Generally taking moment about A along the z-axis we have that

    \sum M_z  = M_z = 0

=>   M_z  = 0 \  N\cdot m

8 0
3 years ago
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