Please solve this problem!​

1 kilo = 1000 gms

Therefore, 5 kilo = 5000 gms

Answer:

The hydrogen peroxide decomposes as: [tex]H_2O_2\rightarrow 2H_2O+O_2[/tex]

Explanation:

Hydrogen peroxide antiseptic although tightly capped but left on the shelf for a long time leads to swollen bottle and opens with a hissing sound because hydrogen peroxide reacts with air in presence of light to dissociate into water and oxygen molecules that leads to high volume of content in the tightly closed bottle and formation of water molecules from hydrogen peroxide.

Once the hydrogen peroxide has turned into water molecules it is no longer effective as an antiseptic.

Answer:

66.83 meters

Explanation:

After a quick online search, it seems that scientists calculate the average reaction time of individuals as 2.3 seconds between seeing an obstacle and putting their foot on the brakes. Now that we have this reaction time we need to turn the miles/hour into meters/second.

1 mile = 1609.34 meters  (multiply these meters by 65)

65 miles = 104,607 meters

1 hour = 3600 seconds

Therefore the car was going 104,607 meters every 3600 seconds. Let's divide these to find the meters per second.

[tex]\frac{104,607}{3600} = \frac{29.0575 meters}{1 second}[/tex]

Now we simply multiply these meters by 2.3 seconds to find out the distance covered before the driver puts his/her foot on the brakes...

29.0575m * 2.3s = 66.83 meters

Answer:

Por lo tanto, la presión del ozono es:

[tex]P=0.011\: atm[/tex]  

Explanation:

Podemos usar la ecuacion de los gases ideales;

[tex]PV=nRT[/tex] (1)

Tenemos:

El volumen V = 224 dm³ = 224 L

La temperatura T = 51.09 C = 324.09 K

La masa es m = 4.561 kg

Lo necesitamos ahora es calvular n que es el numero de moles;

recordemos que el peso molecular del ozono M = 48 g/mol.

[tex]n=\frac{m}{M}=\frac{4.561}{48}=0.095\: mol[/tex]

Finalmente, usando la ecuacion 1 despejamos la presion P

[tex]P=\frac{nRT}{V}[/tex]

[tex]P=\frac{0.095*0.082*324.09}{224}[/tex]  

Por lo tanto, la presion del ozono es:

[tex]P=0.011\: atm[/tex]  

Espero te haya ayudado!

Answer:

pretty sure its A

Explanation:

please give brainliest if i'm correct

Answer:

A.

Explanation:

I have done this and that was correct

hope this helps

Answer: The period of a spring if it has a mass of 5 kg and a spring constant of 6 N/m is 5.73 sec.

Explanation:

Given: Mass = 5 kg

Spring constant = 6 N/m

Formula used to calculate period is as follows.

[tex]T = 2 \pi \sqrt\frac{m}{k}[/tex]

where,

T = period

m = mass

k = spring constant

Substitute the values into above formula as follows.

[tex]T = 2 \pi \sqrt\frac{m}{k}\\= 2 \times 3.14 \times \sqrt\frac{5}{6}\\= 5.73 s[/tex]

Thus, we can conclude that the period of a spring if it has a mass of 5 kg and a spring constant of 6 N/m is 5.73 sec.

Answer:

Initially...

Object A is at rest(v=0)

With Mass =8kg

Object B moving at 10ms- and Mass of 16kg.

Now After collision(Elastic)... B Came to rest...Meaning its final vel =0

But its mass remains the same

What velocity would A be moving with?

From Conversation of Momentum

M'U' + M"U"= M'V' + M"V"

8(0) + (16)(10) = 8V' + 16(0)

0 + 160 = 8V' + 0

8V'= 160

V'= 160/8

V'=20ms-¹.

A would be Moving at 20ms-¹ after collision.

Answer:

Because weight W = M g, the ratio of weights equals the ratio of masses.

(M_m g)/ (M_w g) = [ (p^2 Man )/ (2 K_man)] / [ (p^2 Woman )/ (2 K_woman)

but p's are equal, so

K_m/K_m = (M_w g)/(M_m g) = W_woman / W_man = 450/680 = 0.662Explanation:

Explanation:

the efficiency of a machine can be increased by reducing the friction

please mark the brainliest

Answer:

[tex]f_n=3.75N[/tex]

Explanation:

From the question we are told that:

Frictional force [tex]F=0.150N[/tex]

Coefficient of kinetic friction [tex]\mu=0.04[/tex]

Generally the equation for Normal for is mathematically given by

 [tex]f_n=\frac{F}{\mu}[/tex]

Therefore

 [tex]f_n=\frac{0.150}{0.04}[/tex]

 [tex]f_n=3.75N[/tex]

Answer:

The amount of heat required to boil away 1 kg of water is 10 times the amount of heat required to melt 1 kg of ice

Explanation:

let the latent heat of fusion of ice  = L

then, the latent heat of vaporization of water = 10L

The heat of fusion of 1 kg of ice = 1 x L = L

The heat of vaporization 1 kg of water = 1 x 10L = 10L

Therefore, the amount of heat required to boil away 1 kg of water is 10 times the amount of heat required to melt 1 kg of ice

Explanation:

the unit of area is called a derived unit because it is made of two fundamental unit metre and metre.

Answer: hello your question is incomplete below is the complete question

Water stands at a depth H in a large open tank whose side walls are vertical  . A hole is made in one of the walls at a depth h below the water surface. Part B How far above the bottom of the tank could a second hole be cut so that the stream emerging from it could have the same range as for the first hole

answer :

At Height ( h ) from the bottom of Tank

Explanation:

Determine how far above the bottom of the tank a second hole be cut

For the second hole to have the same range as the first hole

Range of first hole = Velocity of efflux of water * time of fall of water

                               =  √ (2gh) * √( 2g (H - h) / g)

                               = √ ( 4(H-h) h)

Hence the Height at which the second hole should be placed to exercise same range of stream emerging = h from the bottom of the Tank

Answer:

Explanation:

one electron has [tex]1.60217662*10^{-19}~coulombs~then\\\\1.02*10^{20}~electrons------->1.02*10^{20}*1.60217662*10^{-19}~coulombs= 16.3422~coulombs[/tex]

Answer:

[tex]|P_3|=21.242m/s[/tex]

Explanation:

From the question we are told that:

Mass [tex]m=200g=0.2[/tex]

Speed [tex]v=82.0m/s[/tex]

Mass [tex]M_2=300g=0.3kg[/tex]

Speed [tex]V_2=45.0m/s[/tex]

Generally the equation for Magnitude of the Third piece is mathematically given by

[tex]P_1+P_2+P_3=0[/tex]

[tex]|P_3|=|P_1+P_2|[/tex]

Where

[tex]P_1=m_1v_1[/tex]

[tex]P_1=0.2*82[/tex]

[tex]P_1=16.4kgm/s[/tex]

And

[tex]P_2=0.3*45[/tex]

[tex]P_2=13.5kgm/s[/tex]

Therefore

[tex]|P_3|=\sqrt{16.4^2+13.5^2}[/tex]

[tex]|P_3|=21.242m/s[/tex]

Answer:

8.37 Joules

Explanation:

The amount of energy that a substance, such as animal tissue or rubber can take and yet return to its original condition is known as resilience. When we stress such a substance and then let it restore to its previous state, we are referring to the substance to maintain its elastic area of the stress-strain curve.

From the given information:

the resilence is 0.93

The amount of work done during stretching of the tendon = 9.00 J

Thus,

the work done when relaxing = 0.93 × 9.00 J

= 8.37 Joules

Answer:

when we get images from the telescope, we're not seeing those object like they are right now. we're seeing them like they looked when that light left those distant objects...... the animation cuts to a visualization of distant galaxies imaged by Hubble.

Answer:

d.it is used in electronic devices

Answer:

Light-year is the distance light travels in one year. Light zips through interstellar space at 186,000 miles (300,000 kilometers) per second and 5.88 trillion miles (9.46 trillion kilometers) per year.

Explanation:

Answer:

In "Young Goodman Brown," Hawthorne reveals his feelings about his Puritan ancestors when the dark man reveals that he helped Brown's forebears persecute others.

Explanation:

hey mate i know it is right so don't worry this is the correct

Answer:

In "Young Goodman Brown," Hawthorne reveals his feelings about his Puritan ancestors when the dark man reveals that he helped Brown's forebears persecute others. so that means the answer is D.

Explanation:

A. brown strives to resist his dark mission

B. faith expresses her anxieties about young brown's departure

C. brown discovers his catechism teacher is on speaking terms with the devil

D. the dark man reveals that he helped brown's forebears persecute others(correct answer)

Answer:

cause they give u milk

Explanation:

Answer:

Cows are important as they provide humans many things for survival. They provide milk, meat, and leather, all of these are important resources.

Answer:

1.2 x 10^-12

Explanation:

3/2.5 x 10^-6/10^6

1.2 x 10^-6 x 10^-6

1.2 x 10^-12

Answer:

Explanation:

When an object's speed increases at a constant rate, we say it has constant/uniform acceleration.

Answer:

ok

Explanation:

nskakkskdnsksmskzkksnsnxksjsjos

Answer:

the particles vibrate inside the tyre

Explanation:

as the car moves kinetic energy is transfered in the tyres which causes the particles to vibrate inside the tyre so the kinetic store is. transferred into thermal

Answer:

400

Explanation:

Answer:

A. bends towards the Equator.

Explanation:

Rossby waves are inertial waves that are naturally occurring in a rotating fluids. These waves are also called as the planetary waves.

The Rossby waves are undulated that occur in the polar front jet stream when there is a significant differences in the temperatures between the polar and the tropical air masses.

It occurs when the polar air masses moves towards the equator and when the tropical air masses moves towards the pole.  It is formed when the air bends away from the poles and bends towards the equator.

Hence the correct option is (A).

Answer: (d)

Explanation:

Given

Mass of object [tex]m=2\ kg[/tex]

Speed of object [tex]u=5\ m/s[/tex]

Mass of object at rest [tex]M=3\ kg[/tex]

Suppose after collision, speed is v

conserving momentum

[tex]\Rightarrow mu+0=(m+M)v\\\\\Rightarrow v=\dfrac{2\times 5}{2+3}\\\\\Rightarrow v=2\ m/s[/tex]

Initial kinetic energy

[tex]k_1=\dfrac{1}{2}\times 2\times 5^2\\\\k_1=25\ J[/tex]

Final kinetic energy

[tex]k_2=\dfrac{1}{2}\times (2+3)\times 2^2\\\\k_2=10\ J[/tex]

So, it is clear there is decrease in kinetic energy . Thus, energy decreases and velocity becomes 2 m/s.

Answer:

the required speed with which the missile move relative to the Earth is -0.727c

Explanation:

Given the data in the question;

relative velocity relation;

u' = u-v / 1 - [tex]\frac{uv}{c^2}[/tex]

so let V[tex]_B[/tex] represent the velocity as seen by an external reference frame; u=V[tex]_B[/tex]

and let V[tex]_A[/tex] represent the speed of the secondary reference frame; v=V[tex]_A[/tex]

hence, u' is the speed of B as seen by A

so

u' = V[tex]_B[/tex]-V[tex]_A[/tex] / 1 - [tex]\frac{V_BV_A}{c^2}[/tex]

now, given that; V[tex]_A[/tex] = 0.9c  and V[tex]_B[/tex]  = 0.5c

we substitute

u' = ( 0.5c - 0.9c ) / 1 - [tex]\frac{(0.5c)(0.9c)}{c^2}[/tex]

u' = ( 0.5c - 0.9c ) / 1 - [tex]\frac{c^2(0.5)(0.9)}{c^2}[/tex]

u' = ( 0.5c - 0.9c ) / 1 - (0.5 × 0.9)

u' = ( -0.4c ) / 1 - 0.45

u' = -0.4c / 0.55

u' = -0.727c

Therefore, the required speed with which the missile move relative to the Earth is -0.727c