PHYSICS COLLEGE

A can of soup has a mass of 0.35 kg. The can is moved from a shelf that is 1.2 m off the ground to a shelf that is 0.40m off the ground. How does the gravitational potential energy of the can change?

Answers

Answer 1

Answer:

Answer:

2.744 difference

Explanation:

Use Pe=mgh

So when the soup is at a height of 1.2m, its Pe is (.35kg)(9.8m/ $s^(2)$ )(1.2m)=4.116

when the soup is at a height of .40m, its Pe is (.35kg)(9.8m/ $s^(2)$ )(.40m)=1.372

So youre looking at a 2.744 difference in pe

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What happens to a black body radiator as it increases in temperature? A. it gives off a range of electromagnetic radiation of shorter wavelengths. B. It gives off only one wavelength of electromagnetic radiationC. It releases only ultraviolet waves of electromagnetic radiationD. It becomes hotter but gives off less electromagnetic radiation
To see how two traveling waves of the same frequency create a standing wave. Consider a traveling wave described by the formula y1(x,t)=Asin(kx−ωt)This function might represent the lateral displacement of a string, a local electric field, the position of the surface of a body of water, or any of a number of other physical manifestations of waves. 1. Find ye(x) and yt(t). Keep in mind that yt(t) should be a trigonometric function of unit amplitude.2. At the position x=0, what is the displacement of the string (assuming that the standing wave ys(x,t) is present)?3. At certain times, the string will be perfectly straight. Find the first time t1>0 when this is true.4. Which one of the following statements about the wave described in the problem introduction is correct?A. The wave is traveling in the +x direction.B. The wave is traveling in the −x direction.C. The wave is oscillating but not traveling.D. The wave is traveling but not oscillating.Which of the expressions given is a mathematical expression for a wave of the same amplitude that is traveling in the opposite direction? At time t=0this new wave should have the same displacement as y1(x,t), the wave described in the problem introduction.A. Acos(kx−ωt)B. Acos(kx+ωt)C. Asin(kx−ωt)D. Asin(kx+ωt)
A block m1 rests on a surface. A second block m2 sits on top of the first block. A horizontal force F applied to the bottom block pulls both blocks at constant velocity. Here m1 = m2 = m.(a)What is the normal force exerted by the surface on the bottom block? (Use the following as necessary: m and g as necessary.)

One of two nonconducting spherical shells of radius a carries a charge Q uniformly distributed over its surface, the other carries a charge -Q, also uniformly distributed. The spheres are brought together until they touch.A.) What does the electric field look like, both outside and inside the shells?

B.) How much work is needed to move them far apart?

Answers

Answer:

Explanation:

A ) The spheres are non conducting , charge will not move on the surface so neutralization of charge by + ve and - ve charge is not possible. Charges will remain intact on them . The electric field inside them will be zero . Electric field outside shell will not be spherically symmetrical . Lines of force will emanate from the surface of positively charged shell outwardly oriented and end at negatively charged shell .

B )

distance between the centres of spherical shell

= 2 a

potential energy of charges

= k q₁ x q₂ / R

= k x - Q x Q / ( 2a )

= - k Q²/ 2a

So work needed to separate them to infinity will be equal to

= k Q²/ 2a

A softball player swings a bat, accelerating it from rest to 2.6 rev/srev/s in a time of 0.20 ss . Approximate the bat as a 0.90-kgkg uniform rod of length 0.95 mm, and compute the torque the player applies to one end of it.

Answers

Answer:

$\tau=22.13Nm$

Explanation:

information we have:

mass: $m=0.9kg$

lenght: $L=0.95m$

frequency: $f=2.6rev/s$

time: $t=0.2s$

and from the information we have we can calculate the angular velocity $\omega$ . which is defined as

$\omega=2\pi f$

$\omega=2\pi (2.6rev/s)\n\omega=16.336 rev/s$

----------------------------

Now, to calculate the torque

We use the formula

$\tau=I \alpha$

where $I$ is the moment of inertia and $\alpha$ is the angular acceleration

moment of inertia of a uniform rod about the end of it:

$I=(1)/(3)mL^2$

substituting known values:

$I=(1)/(3) (0.9kg)(0.95m)^2\nI=0.271kg/m^2$

for the torque we also need the acceleration $\alpha$ which is defined as:

$\alpha=(\omega)/(t)$

susbtituting known values:

$\alpha=(16.336rev/s)/(0.2s) \n\alpha=81.68rev/s^2$

and finally we substitute $I$ and $\alpha$ into the torque equation $\tau=I \alpha$ : $\tau=(0.271kg/m^2)(81.68rev(s^2)\n\tau=22.13Nm$

Final answer:

To calculate the torque, we need to use the formula: Torque = Moment of Inertia * Angular Acceleration. By approximating the bat as a uniform rod and using its length and mass, we can find the moment of inertia. Then, using the given angular velocity, we can calculate the angular acceleration. Finally, we can determine the torque by multiplying the moment of inertia by the angular acceleration.

Explanation:

To compute the torque the player applies to one end of the bat, we need to use the formula:

Torque = Moment of Inertia * Angular Acceleration

Given that the bat is approximated as a uniform rod and we know its length and mass, we can calculate the moment of inertia. Then, using the given angular velocity, we can compute the angular acceleration. Finally, we can find the torque by multiplying the moment of inertia by the angular acceleration.

Learn more about Torque here:

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Suppose that a barometer was made using oil with rho=900 kg/m3. What is the height of the barometer at atmospheric pressure?

Answers

Hey there!

The pressure under a liquid column can be , calculated using the following formula :

P = p x g x h

P atm = 1.013 x 10⁵ Pa

g = 9.8 m/s²

h = ?

h = P / ( p x g ) =

h= ( 1.013 x 10⁵ Pa ) / ( 900 x 9.8 ) =

h = ( 1.013 x 10⁵ ) / ( 8820 ) =

h = 11.48 m ≈ 11.50 m

Hope this helps!

What is the wavelength of the photons emitted by hydrogen atoms when they undergo n =5 to n =3 transitions?

Answers

Answer:

$\lambda=1282nm$

Explanation:

The wavelength of the photons emitted due to an atomic electron transition in a hydrogen atom, is given by the Rydberg formula:

$(1)/(\lambda)=R_H((1)/(n_1^2)-(1)/(n_2^2)})$

Here $R_H$ is the Rydberg constant for hydrogen and $n_1,n_2$ are the lower and higher quantum number for the energy levels of the atomic electron transition, respectively. Replacing the given values and solving for $\lambda$

$(1)/(\lambda)=1.097*10^7m^(-1)((1)/(3^2)-(1)/(5^2)})\n(1)/(\lambda)=7.81*10^5m^(-1)\n\lambda=(1)/(7.81*10^5m^(-1))\n\lambda=1.282*10^(-6)m\n\lambda=1.282*10^(-6)m*(1nm)/(10^(-9)m)\n\lambda=1282nm$

An object of mass 10.0kg is released at point A, slidesto the bottom of the 30° incline, then collides with ahorizontal massless spring, compressing it a maximumdistance of 0.750m. (See below.) The spring constant is 500N/m, the height of the incline is 2.0 m, and the horizontalsurface is frictionless. (a) What is the speed of the object atthe bottom of the incline? (b) What is the work of frictionon the object while it is on the incline? (c) The springrecoils and sends the object back toward the incline. Whatis the speed of the object when it reaches the base of theincline? (d) What vertical distance does it move back up theincline?

Answers

Final answer:

The final speed at the bottom of the incline can be calculated using the conservation of energy principle. There is no work done against friction as the object is moving on a frictionless surface. The speed does not change when the spring pushes it back towards the base of the incline due to lack of friction and it moves to a certain height given the angle of the incline and the initial speed.

Explanation:

"Speed at the bottom of the incline:" This can be calculated using conservation of energy. The potential energy at the top (m*g*h) will convert into kinetic energy at the bottom (1/2*m*v^2). Here, m is the mass, g is acceleration due to gravity, h is the height, and v is the velocity. Using this, we can solve for v.
Work of friction on the incline: As per the question, the surface is frictionless. Therefore, the work done by friction is automatically 0 as there is no force of friction.
Speed of the object when it reaches the base of the incline again: As the surface is frictionless, the object reaches the incline with the same speed with which it left as there are no opposing forces to reduce its momentum.
Vertical distance it moves back up the incline: This can be calculated using the principles of conservation of energy and kinematic equations, taking into account the angle of the incline and the velocity of the object.

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What does a planet need in order to retain an atmosphere? How does an atmosphere affect the surface of a planet and the ability of life to exist?

Answers

Answer:

Explained

Explanation:

In order to retain atmosphere a planet needs to have gravity. A gravity sufficient enough to create a dense atmosphere around it, so that it can retain heat coming from sun. Mars has shallow atmosphere as its gravity is only 40% of the Earth's gravity. Venus is somewhat similar to Earth but due to green house effect its temperature is very high. Atmosphere has a huge impact on the planets ability to sustain life. Presence of certain kind gases make the atmosphere poisnous for life. The atmosphere should be such that it allows water to remain in liquid form and maintain an optimum temperature suitable for life.

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