What coefficient is needed in front of H2O to balance the equation below?

Answer:

The force of the floor on the briefcase

Explanation:

The briefcase hits the floor and stops the falling which is reacting to the briefcase that was falling and hits the floor.

The falling briefcase would be the action force because it is moving downward

and

The reaction force would be the floor because it will react on the briefcase.

Hope this Helps

When the briefcase lands on the floor, the reaction force to the action force of the briefcase on the floor is  force of the floor on the briefcase.

It might be difficult to distinguish between action and reaction forces. This is due to the fact that both forces are always present.

One might argue that there are an equal number of forces in the universe if one were to count them all. However, the definition of the action force is provided by the perspective of the interaction analysis in order to help the argument.

A force that is applied to an object is known as an action force.

An action force with an opposite direction has an effect called a reaction force.

These two forces—also referred to as action and reaction forces—are covered by Newton's third law of motion.

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Answer:

Answer:

Speed of the wave in the string will be 3.2 m/sec

Explanation:

We have given frequency in the string fixed at both ends is 80 Hz

Distance between adjacent antipodes is 20 cm

We know that distance between two adjacent anti nodes is equal to half of the wavelength

So \frac{\lambda }{2}=20cm

2

λ

=20cm

\lambda =40cmλ=40cm

We have to find the speed of the wave in the string

Speed is equal to v=\lambda f=0.04\times 80=3.2m/secv=λf=0.04×80=3.2m/sec

So speed of the wave in the string will be 3.2 m/sec

Answer:

so sorry i don't know the answer so sorry

Answer:

Explanation:

The speed of the wave given only it's frequency and wavelength can be found by using the formula

[tex]c = f \times \lambda[/tex]

where

c is the velocity of the wave in m/s

[tex] \lambda[/tex] is the wavelength in m

f is the frequency in Hz

From the question we have

c = 6 × 1.5 = 9

We have the final answer as

Hope this helps you

Answer:

The car that is hald as massive

Explanation:

We can use the eqation for kinetic energy to solve this problem

KE = 1/2mv^2 (where m is mass and v is velocity/speed)

Lets give the original car a mass of 100kg and use the equation

KE = 1/2(100kg)(16m/s)^2 = 12800J

and now lets find kinetic energy of the half as massive car

KE2 = 1/2(50kg)(32m/s)^2 = 25600J

From this, we find that the hald as massive car will experience a greater kinetic energy while travling at double the velocity

Answer:

D 3, 9, 5, 7, 3, 6

Explanation:

range is the difference between the highest and the lowest value and 9 is the highest while 3 is the lowest, if u subtract them u will get 6. while for the first u will get 4 and the second u will get 2. so the answer is D

The answer is : A₃ = A₁ - A₂ = 3.0 - 2.0 =  1.0A

Ok done. Thank to me :>

you would die.

the core of the earth is as hot as the sun surface so you would be dead

When a log of wood is put over the freshwater, the log start to float. Option B is correct.

Density:

It can be defined as the amount of substance per unit volume. It is usually denoted by D. It is measured in the [tex]\bold { kg/m^3}[/tex].

The less dense object always floats over the more dense substance.

For example- Here, The density of the log is 0.8 [tex]\bold{ cm^3}[/tex] and the density of the of water is 1.0 [tex]\bold{ cm^3}[/tex] .

Since, the density of the log is lesser than the density of the water.

Therefore, when a log of wood is put over the freshwater, the log start to float.

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The distance is equal to 6 m traveled by elevator between 8 s and 24 s.

The distance is described as the total length covered by an object. Distance is a scalar parameter with no direction. The distance covered by an object is always +ve, it can never be zero.

The displacement is described as the shortest length between two points by the object. The displacement is a vector parameter with direction & magnitude. The displacement can be +ve, -ve, or zero and can increase or decrease over time.

Given the position of the elevator at time t = 8 s at d = 9 m

The position of the elevator at time t = 24 s at d = 15 m

The distance covered by the elevator = 15 - (9) =  15 - 9 = 6 m

Therefore, the distance traveled by an elevator between 8 s and 24 s is equal to 6 m.

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Answer:

405

Explanation:

If the car stops at 45m/s in 9 seconds then it took the car 405 meters to stop

Answer:

3.6m/s², 18m/s

Explanation:

So according to h=1/2at², 45=1/2a*25, a=3.6m/s² (which means that it didn't happen on Earth?)

Final velocity = at = 3.6*5 = 18 m/s

Answer:

The answer is true I guess!!!

This is because the more you go up to less air it is and the pressure also gets less!!!!!!!!

Explanation:

I HOPE THAT I WAS HELPFUL TO YOU!

Newton's second law and universal gravitation law allow to find the results for questions about the motion of satellites in orbit are:

1) The correct answer is D:  v = 8.17 10³ m/s

2) The correct answer is B

    The velocity of Satellite A is greater than the velocity of Satellite B.

3) The correct answer is B

   Toward the center of Earth

4) The correct answer is B

   The gravitational force increases and the velocity increases.

Newton's second law establishes a relationship between force, mass, and the acceleration of bodies.

           F = m a

In the case of the satellite the force is given by the law of universal gravitation.

           [tex]F = - G \frac{Mm}{r^2 }[/tex]

Where G is the constant of universal gravitation. M and m the mass of each object and r the distance between them.

In this case the satellite is in a circular orbit, therefore the acceleration is centripetal.

          [tex]a = \frac{v^2}{r}[/tex]  

We substitute.

      [tex]G \frac{Mm}{r^2} = m \frac{v^2}{r}[/tex]  

       [tex]\frac{GM}{r} = v^2[/tex]  

Let's analyze the answers to find the correct one.

1) They indicate the height of the space station r = 4.08 10⁵ m and ask the speed.

          [tex]v= \sqrt{ \frac{6.67 \ 10 ^{-11} 5.9 \ 10^{24}}{4.08 \ 10^6 } }[/tex]  

          v = 9.82 10³ m / s

The correct answer is D.

2) Satellite A has an orbit of hₐ = 500 km and satellite b an orbit of

    h_b = 800 km

The distance from the center of the earth to each satellite is:

          rₐ = R + hₐ

          r_b = R + h_b

          rₐ = 6.37 106 + 500 10³ = 6.87 10⁶ m

          r_b = 6.37 10⁶ + 800 10³ = 7.17 10⁶ m

Let's find the ratio of the speeds

         [tex]\frac{v_a}{v_b} = \sqrt{ \frac{r_b}{r_a} } \\\frac{v_a}{v_b} = \sqrt{ \frac{7.17}{6.87} }[/tex]

        [tex]\frac{v_a}{v_b}[/tex]  = 1,022

we see that the speed of satellite a is slightly greater than the speed of satellite b.

Let's analyze the claims.

A) False. The speed does not depend on the mass of the satellites.

B) True. The velocity of a is slightly greater than the velocity of b.

C) False. The speed of a is greater.

D) False. The speed of a is greater.

3) As the orbit is circular, the force must be radial, that is, it points towards the center of the earth.

Let's analyze the claims.

A) False. The speed modulus does not change, therefore there is no acceleration in the direction of the satellite.

B) True. Aim for the center of the Earth, change the direction of the velocity.

C) False. Aim for the scepter of the earth.

D) false. The modulus of velocity is constant and the direction changes towards the center of the earth, therefore the force must go towards the center of the earth.

4) The force in the law of universal gravitation increased as the distance decreased.

When a satellite approaches the earth its speed must increase since the speed is proportional in inverse of the square root of the distance.

Let's examine the claims.

A) False. The attraction force increases.

B) True. You agree with the explanation.

C) False. The gravitational force increases.

D) False. Speed ​​increases.

In conclusion, using Newton's second law and the universal law of gravitation we can find the results for the questions about the movement of satellites in orbit are:

    1) The correct answer is D:  v = 8.17 10³ m/s  

    2) The correct answer is B

    The velocity of Satellite A is greater than the velocity of Satellite B.

    3) The correct answer is B

    Toward the center of Earth

    4) The correct answer is B

   The gravitational force increases and the velocity increases.

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Answer:

1. 7.66 × 10^3 m/s

2. The velocity of Satellite A is greater than the velocity of Satellite B.

3. toward the center of Earth

4. The gravitational force increases and the velocity increases.

Explanation:

[tex]\\ \sf\Rrightarrow kE=\dfrac{1}{2}mv^2[/tex]

[tex]\\ \sf\Rrightarrow 14=\dfrac{1}{2}17v^2[/tex]

[tex]\\ \sf\Rrightarrow 17v^2=28[/tex]

[tex]\\ \sf\Rrightarrow v^2=28/17[/tex]

[tex]\\ \sf\Rrightarrow v^2=1.64[/tex]

[tex]\\ \sf\Rrightarrow v\approx 0.13m/s[/tex]

Answer:

The magnitude of acceleration is 9.8 m/s

Explanation:

Its not always in m/s^2 but in m/s due to the unit of distance and time

Answer:

Like other emissions resulting from fossil fuel combustion, aircraft engines produce gases, noise, and particulates, raising environmental concerns over their global effects and their effects on local air quality.[2]Jet airliners contribute to climate change by emitting carbon dioxide (CO2), the best understood greenhouse gas, and, with less scientific understanding, nitrogen oxides, contrails and particulates. Their radiative forcing is estimated at 1.3–1.4 that of CO2 alone, excluding induced cirrus cloud with a very low level of scientific understanding. In 2018, global commercial operations generated 2.4% of all CO2 emissions.

Between 1940 and 2018, aviation CO2 emissions grew from 0.7% to 2.65% of all CO2 emissions.[1]

Jet airliners have become 70% more fuel efficient between 1967 and 2007, and CO2 emissions per Revenue Ton-kilometer (RTK) in 2018 were 47% of those in 1990. In 2018, CO2 emissions averaged 88 grams of CO2 per revenue passenger per km. While the aviation industry is more fuel efficient, overall emissions have risen as the volume of air travel has increased. By 2020, aviation emissions were 70% higher than in 2005 and they could grow by 300% by 2050.

Aircraft noise pollution disrupts sleep, children's education and could increase cardiovascular risk. Airports can generate water pollution due to their extensive handling of jet fuel and deicing chemicals if not contained, contaminating nearby water bodies. Aviation activities emit ozone and ultrafine particles, both of which are health hazards. Piston engines used in general aviation burn Avgas, releasing toxic lead.

Aviation's environmental footprint can be reduced by better fuel economy in aircraft or Air Traffic Control and flight routes can be optimised to lower non-CO2 effects on climate from NO

x, particulates or contrails. Aviation biofuel, emissions trading and carbon offsetting, part of the ICAO's CORSIA, can lower CO2 emissions. Aviation usage can be lowered by short-haul flight bans, train connections, personal choices and aviation taxation and subsidies. Fuel-powered aircraft may be replaced by hybrid electric aircraft and electric aircraft or by hydrogen-powered aircraft.

Answer:

8 seconds

Explanation:

Since the carspeed is in km/h, we need equal units, so we will make 100.0m 0.1000km.

Then we need to find how long it takes the car to travel 0.1km

We can use the formula distance=speed * time and get

0.1=45 * time

Therefore we get .002222... hours

Multiplying this by 3600 (to get seconds, 60x60), we get 8 seconds

Answer:

8m/s

Explanation:

speed = 45km/h

Distance = 100m

we have to find time = ?

Formula for speed is = Distance/ Time

Here Distance is given in 'm' so we need to convert speed value in 'm'

So to convert km/h in m formula is divide 45km/h by 3.6 then the km/h value gets converted in m so now the value is 12.5 m/s

So now,

speed = Distance/Time

we have to find Time

Then,

Time = Distance/ Speed

= 100/12.5

= 8m/s

These Milky Way companion galaxies are easily visible from dark locations in the Southern Hemisphere. Prime examples of erratic galaxies are the Large and Small Magellanic clouds (left and right, respectively).

In comparison to a rich cluster, the poor cluster typically has a slightly more erratic shape. A number of smaller galaxies orbit each major spiral. The Small and Large Magellanic clouds are the two most well-known examples of atypical galaxies.

When two galaxies collide, irregular galaxies frequently result. This unusual Cartwheel Galaxy was created when a tiny galaxy slid through the centre of a massive spiral galaxy.

Therefore, Rich clusters are other clusters that include hundreds to thousands of galaxies. A weak cluster can't cling to its members strongly because of its low bulk.

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Newton's second equation of motion :-

S=ut+1/2at^2 [where, u is the initial velocity, a is the acceleration and t is the time interval]

This Equation simply finds a relation between distance travelled by a particle (classically) under uniform acceleration.

So let's see what pieces of information (bundles of equations) do we have with us, initially.

We have, a very primary equation with us,

dS/dt = v… (I)

(Considering motion in a straight line only)

And we also have the equation

dv/dt = a…(II)

Simply replacing the v in eqn (II) by eqn (I), we find

d2S/dt^2 = a…(III)

This is what we need to solve. It's easy.

You know,

d2S/dt^2 = d/dt(dS/dt) = a

⟹ dS/dt = ∫adt = at+c1

Since, dS/dt is the velocity of the particle,

Therefore, at t = 0, dS/dt|t = 0 = u

⟹ u = a∗0 + c1 = c1

⟹ c1 = u

Therefore, dS/dt = u + at

Thus, S = ∫(udt + atdt)

⟹ S = ut + 1/2at^2 +c^2

If say, the particle is already having a displacement S0 the moment you start measuring it's motion. Then, at t = 0, S = S0

This makes S = S0 +ut + 1/2at^2

Since, in most of the practical cases, we start measuring a motion when the particle starts displacing (i.e., when S0=0 ),

We get

S = ut + 1/2at^2

Hope it helps :)

The conservation of energy and moment allows to find the results for the velocity of the pendulum and the bullet in the system are:

          m (kg)    v (m / s)

            0.05      41.58

            0.10       21.78

            0.15       15.18

            0.20      11.88

           0.25       9.90

The conservation of mechanical energy is one of the most important principles of physics, it establishes that if there is no friction force, the energy is constant at all points. Mechanical energy is the sum of the kinetic energy plus the potential energies.

In the attachment we can see a diagram of the system, let's start by finding the speed of the pendulum when it leaves, for this we use the  conservation of energy.

Starting point. In the lowest part of the movement.

      Em₀ = K = ½ (m + M) v₀²

Final point. At the top of the movement.

      Em_f = U = (m + M) g y

Energy is conserved because there is no friction.

      Em₀ = Em_f

      ½ (m + M) v₀² = (m + M) g y

      [tex]v_o = \sqrt{2gy}[/tex]  

They indicate a table with several measurements of the masses and the period, let's use the relationship of the simple harmonic motion.

        y = y₀ cos wt

The period and the angular velocity are related.

        w = 2π / T

we substitute

          y = y₀ cos ( [tex]2 \pi \frac{t}{T}[/tex] )

Let's analyze how long it takes to reach the point of maximum height, the period is the time of a complete oscillation, therefore from the lowest point to the highest point we have ¼ of oscillation, consequently the time to the highest point.

        t = T / 4

        y = y₀ cos ( [tex]2 \pi 4[/tex])  

        y = y₀

Therefore the point of maximum amplitude coincides with the maximum height and must be average by the student, suppose that the height is

          y₀ = 20 cm = 0.20 m

Let's calculate the initial velocity.

     v₀ = [tex]\sqrt{2 \ 9.8 \ 0.20 }[/tex]  

     v₀ = 1.98 m / s

They ask for the speed of the bullet before striking the basket of mass

       M = 1 Kg.

The conservation of the momentum that for an isolated system the momentum is constant in all the instants. Let's form the system by the bullet and the basket.

Initial instant. Before the crash

       p₀ = m v

Final moment. Right after the crash.

      P_f = (m + M) v₀

The momentum is conserved because the system is isolated.

      p₀ = p_f

      m v = (m + M) v₀

      v = m + M / m v₀

they have tabulated various mass for the bullet, we calculate the speed of each bullet.

m = 0.05 kg

    v = [tex]\frac{0.05+1}{0.05} \ 1.98[/tex]  

    v = 41.58 m / s

m = 0.10 kg

     v = [tex]\frac{0.10+1}{0.10} \ 1.98[/tex]  

     v = 21.78 m / s

m = 0.15 kg

     v = [tex]\frac{0.15+1}{0.15y}[/tex]  

     v = 15.18 m / s

m = 0.20 kg

     v = 11.88 m / s

m = 0.25 kg

     v = 9.9 m / s

In conclusion with the conservation of energy and the momentum we can find the results for the speed of the pendulum and the bullet in the system are:

          m (kg)    v (m / s)

            0.05      41.58

            0.10       21.78

            0.15       15.18

            0.20      11.88

           0.25       9.90

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Answer:

the meaning of conduction is the process by which heat or electricity is directly transmitted through a substance when there is a difference of temperature or of electrical potential between adjoining regions, without movement of the material.

Explanation:

Answer:

the process by which heat or electricity is directly transmitted through a substance when there is a difference of temperature or of electrical potential between adjoining regions, without movement of the material.

Explanation:

Answer:

The solutions to this exercise is B. 400N

Answer:

The mass will stay the same throughout time

Answer:

147 J

Explanation:

PE = mgh

PE = (1.0)(9.8)(15)

PE = 147 joules

The force on the truck is the same in magnitude as the force on the car.

The impulse experienced by an object during collision is directly proportional to the applied force and time of collision of the objects.

J = Ft

where;

The increase in the force applied to an object causes an increase in the impulse experienced by the object.

Also, according to Newton's third law of motion, the force exerted on the loaded truck and the car are equal in magnitude but opposite in direction.

Thus, we can conclude that, the force on the truck is the same in magnitude as the force on the car.

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Answer:

0.667; 4.965

Explanation:

Look at the picture I attached for the force analysis.

a) The coefficient=Friction/Normal Force. Because it's at constant speed, the force of friction + mgsin58.1° (because it's on an inclined plane and has split forces) is equal to the applied force (70.4N). Normal force is not equal to weight force though, because the box is on an inclined plane; it's equal to mgcos58.1°.

b) If it were to break, then the box no longer has an applied force, and the direction of friction has changed to up the inclined plane. F=m/a, so acceleration = mgsin58.1°- Friction/mass