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prohojiy [21]
2 years ago
11

Complete the sentence with the word "element" or "compound." O is a(n) and H2O2 is a(n) .

Physics
2 answers:
Debora [2.8K]2 years ago
9 0
O (oxygen) is an element, whereas H2O2 (hydrogen peroxide) is a compound.
jeka942 years ago
8 0
O is an element (Oxygen) and H2O2 is a compound (Hydrogen Peroxide)
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A steel rod with a length of l = 1.55 m and a cross section of A = 4.45 cm2 is held fixed at the end points of the rod. What is
Blababa [14]

To solve this problem it is necessary to apply the concepts related to thermal stress. Said stress is defined as the amount of deformation caused by the change in temperature, based on the parameters of the coefficient of thermal expansion of the material, Young's module and the Area or area of the area.

F = AY\alpha \Delta T

Where

A = Cross-sectional Area

Y = Young's modulus

\alpha= Coefficient of linear expansion for steel

\Delta T= Temperature Raise

Our values are given as,

A = 4.45cm^2

T = 37K

\alpha = 1.17*10^{-5}K^{-1}

Y = 200*10^9Gpa

Replacing we have,

F = (4.45*10^{-4})(200*10^9)(1.17*10^{-5})(37)

F = 38526.1N

Therefore the size of the force developing inside the steel rod when its temperature is raised by 37K is 38526.1N

7 0
2 years ago
Tyler stands at rest on a skateboard. He has a mass of 120 kg. His friend (m = 60 kg) jumps into his arms at a speed of 2 m/s. I
Andrews [41]
Momentum question. This is an inelastic collision, so 

m1v1+m2v2=Vf(m1+m2)
Vf=(m1v1+m2v2)/(m1+m2)=[(120kg)(0m/s)+(60kg)(2m/s)] / (120kg+60kg)
Vf=120kg m/s  /   180kg
Vf=0.67m/s

0.67m/s
5 0
2 years ago
A particle in the first excited state of a one-dimensional infinite potential energy well (with U = 0 inside the well) has an en
nataly862011 [7]

Answer:

The energy of this particle in the ground state is E₁=1.5 eV.

Explanation:

The energy E_{n} of a particle of mass <em>m</em> in the <em>n</em>th energy state of an infinite square well potential with width <em>L </em>is:

                                                    E_{n}=\frac{n^{2}h^{2}}{8mL^{2}}

In the ground state (n=1). In the first excited state (n=2) we are told the energy is E₂= 6.0 eV. If we replace in the above equation we get that:

                                                    E_{1}=\frac{h^{2}}{8mL^{2}}            

                                                    E_{2}=\frac{h^{2}}{2mL^{2}}

So we can rewrite the energy in the ground state as:

                                                   E_{1}=\frac{1}{4}(\frac{h^{2}}{2mL^{2}})

                                                      E_{1}=\frac{1}{4} E_{2}

                                                   E_{1}=\frac{1}{4} ( 6.0\ eV)

Finally

                                                    E_{1}=1.5\ eV

                                                   

                                                   

6 0
2 years ago
A balloon is at a height of 81m and is ascending upwards with a velocity of 12m/s. A body of 2kg weight is dropped from it. If g
kap26 [50]
I know you are Indian by your question, HC Verma class 9 or 11 !!

if you got any problem, comment !!

7 0
2 years ago
A 35 g steel ball is held by a ceiling-mounted electromagnet 4.0 m above the floor. A compressed-air cannon sits on the floor, 4
HACTEHA [7]

Answer:

7.9 m/s

Explanation:

When both balls collide, they have spent the same time for their motions.

Motion of steel ball

This is purely under gravity. It is vertical.

Initial velocity, <em>u </em>= 0 m/s

Distance, <em>s</em> = 4.0 m - 1.2 m = 2.8 m

Acceleration, <em>a</em> = g

Using the equation of motion

s = ut+\frac{1}{2}at^2

2.8 \text{ m} = 0+\dfrac{gt^2}{2}

t = \sqrt{\dfrac{5.6}{g}}

Motion of plastic ball

This has two components: a vertical and a horizontal.

The vertical motion is under gravity.

Considering the vertical motion,

Initial velocity, <em>u </em>= ?

Distance, <em>s</em> = 1.2 m

Acceleration, <em>a</em> = -<em>g                   </em> (It is going up)

Using the equation of motion

s = ut+\frac{1}{2}at^2

1.2\text{ m} = ut-\frac{1}{2}gt^2

Substituting the value of <em>t</em> from the previous equation,

1.2\text{ m} = u\sqrt{\dfrac{5.6}{g}}-\dfrac{1}{2}\times g\times\dfrac{5.6}{g}

u\sqrt{\dfrac{5.6}{g}} = 4.0

Taking <em>g</em> = 9.8 m/s²,

u = \dfrac{4.0}{0.756} = 5.29 \text{ m/s}

This is the vertical component of the initial velocity

Considering the horizontal motion which is not accelerated,

horizontal component of the initial velocity is horizontal distance ÷ time.

u_h = \dfrac{4.4\text{ m}}{0.756\text{ s}} = 5.82\text{ m/s}

The initial velocity is

v_i = \sqrt{u^2+u_h^2} = \sqrt{(5.29\text{ m/s})^2+(5.82\text{ m/s})^2} = 7.9 \text{ m/s}

4 0
2 years ago
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