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An object propelled upwards with an acceleration of 2.0 m / s ^ 2 is launched from rest. After 6 seconds the fuel runs out. Determine the speed at this time and the maximum height at which it reaches.

Answers

Answer 1

Answer:

Answer:43.34 m

Explanation:

Given

acceleration(a) $=2 m/s^2$

Initial Velocity(u)=0 m/s

After 6 s fuel runs out

Velocity after 6 s

v=u+at

$v=0+2* 6=12 m/s$

After this object will start moving under gravity

height reached in first 6 s

$s=ut+(at^2)/(2)$

$s=0+(2* 6^2)/(2)$

s=36 m

After fuel run out distance traveled in upward direction is

$v^2-u^2=2as_0$

here v=0

u=12 m/s

$a=9.8 m/s^2$

$0-12^2=2(-9.8)(s)$

$s_0=(144)/(2* 9.8)=7.34 m$

$s+s_0=36+7.34=43.34 m$

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A square coil (length of side = 24 cm) of wire consisting of two turns is placed in a uniform magnetic field that makes an angle of 60° with the plane of the coil. If the magnitude of this field increases by 6.0 mT every 10 ms, what is the magnitude of the emf induced in the coil?

Answers

Explanation:

It is given that,

Length of side of a square, l = 24 cm = 0.24 m

The uniform magnetic field makes an angle of 60° with the plane of the coil.

The magnetic field increases by 6.0 mT every 10 ms. We need to find the magnitude of the emf induced in the coil. The induced emf is given by :

$\epsilon=N(d\phi)/(dt)$

$(d\phi)/(dt)$ is the rate of change if magnetic flux.

$\phi=BA\ cos\theta$

$\theta$ is the angle between the magnetic field and the normal to area vector.

$\theta=90-60=30$

$\epsilon=NA(dB)/(dt)* cos30$

$\epsilon=2* (0.24\ m)^2* (6\ mT)/(10\ mT)* cos(30)$

$\epsilon=0.0598\ T$

$\epsilon=59.8\ mT$

EMF = 60 mT

So, the magnitude of emf induced in the coil is 60 mT. Hence, this is the required solution.

At what partial pressure are argon atoms expected to have a free travel of approximately 5 µm, if the gas is at a temperature of 400 K? The cross section of collision, σ, or Argon is 0.28 nm2Ar molar mass is 39.9 g/mole

Answers

Answer:

2790 Pa

Explanation:

Given wavelength λ= 5μm

temperature T= 400 K

cross section of collision σ= 0.28 nm^2

molar mass = 39.9 g/mole

pressure = $P= (RT)/(√(2)N_A\sigma\lambda )$

putting values we get

= $(8.314*400)/(√(2)*6.022*10^(23)*0.28*10^(-18)*5*10^(-6) )$

⇒P = 2790 J/m^3

the partial pressure are argon atoms expected= 2790 Pa

Energy is the capacity to do work, but not to produce heat

Answers

Energy and Work Energy is the capacity to do work or to produce heat. Internal energy is the sum of kinetic energy and the potential energy. ... The KE would increase because heating something causes an increase in temperature.

Final answer:

Energy is the capacity to do work but not to produce heat. In physics, energy can exist in various forms, including mechanical and thermal energy.

Explanation:

Energy is the capacity to do work and is an important concept in physics. In the context of this question, it is stated that energy is the capacity to do work but not to produce heat. This highlights the distinction between the two forms of energy. For example, mechanical energy can be used to perform work on an object and cause it to move, while thermal energy is associated with heat and not directly related to work. However, it's important to note that energy can be converted from one form to another, such as converting mechanical energy to thermal energy in a friction process.

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A 580-mm long tungsten wire, with a 0.046-mm-diameter circular cross section, is wrapped around in the shape of a coil and used as a filament in an incandescent light bulb. When the light bulb is connected to a battery, a current of 0.526 A is measured through the filament. (Note: tungsten has a resistivity of 4.9 × 10-8 Ω • m.). How many electrons pass through this filament in 5 seconds?

Answers

Answer: 1.64 *10^19 electrons

Explanation: In order to the explain this problem we have to consider the following:

The current= charge/time; so

as the electrons move in the tungsten wire we have:

0.526 C/s= N electrons per second* charge of electron=

N electrons/s= 0.526/1.6*10^-19= 3.28 *10^18 electrons/s

Then, during 5 seconds will pass:

3.28 *10^18 electrons/s*5 5s= 1.64 *10^19 electrons

Answer:

1.64 x 10^19 electrons

Explanation:

The current is defined as I=ΔQ/Δt where ∆Q is the amount of charge flowing past a point in the filament. This charge is comprised of electrons that each carry charge of e = 1.602 × 10^-19 C. So ΔQ=Ne=IΔt and the number of electrons flowing through the filament in 5 s is N=IΔte=(0.526 A)(5 s)1.602×10^−19 C=1.64×10^19 electrons.

What best describes the bromide ion that forms

Answers

Answer:

it A

Explanation:

Its a negative ion that hss one less valence electron than a netural bromine atom

A volumetric flask made of Pyrex is calibrated at 20.0°C. It is filled to the 150-mL mark with 34.5°C acetone. After the flask is filled, the acetone cools and the flask warms so that the combination of acetone and flask reaches a uniform temperature of 32.0°C. The combination is then cooled back to 20.0°C. (The average volume expansion coefficient of acetone is 1.50 10-4(°C)−1.) (a) What is the volume of the acetone when it cools to 20.0°C?

Answers

Answer:149.73 ml

Explanation:

Given

$\beta \ of\ acetone=1.50* 10^(-4) ^(\circ)C^(-1)$

change in volume is given by

$\Delta V=V_(final)-V_(initial)$

$\Delta V=\nu_(initial)\beta _(acetone)\left [ T_f-T_i\right ]$

$V_(final)=\nu_(initial)+\nu_(initial)\beta _(acetone)\left [ T_f-T_i\right ]$

$V_(final)=150+150* 1.50* 10^(-4)\left [ 20-32\right ]$

$V_(final)=149.73 ml$

Final answer:

The volume of the acetone when it cools to 20.0°C is approximately 142.39 mL.

Explanation:

In order to determine the volume of the acetone when it cools to 20.0°C, we can use the equation for the volume change caused by a temperature change at constant pressure, known as Charles's law. Charles's law states that the volume of a gas is directly proportional to its temperature in Kelvin. We can use the formula V2 = V1 * (T2 / T1) to calculate the volume of the acetone at the lower temperature.

Given that the initial volume of the acetone is 150 mL at a temperature of 34.5°C, we need to convert this temperature to Kelvin by adding 273.15. Therefore, T1 = 34.5°C + 273.15 = 307.65 K.

Since the final temperature is 20.0°C, the final temperature in Kelvin will be T2 = 20.0°C + 273.15 = 293.15 K. We can now plug these values into the equation to find the volume of the acetone at the lower temperature: V2 = 150 mL * (293.15 K / 307.65 K) = 142.39 mL.

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