Which two types of energy does a book have as it falls to the floor

Answers

Answer 1

Answer:

Answer:

kinetic and potential energy

Explanation:

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A 140 W lightbulb emits 4% of its energy as electromagnetic radiation. What is the radiation pressure (in N/m2) on a perfectly absorbing sphere of radius 14 m that surrounds the bulb

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.

In a cyclotron, the orbital radius of protons with energy 300 keV is 16.0 cm . You are redesigning the cyclotron to be used instead for alpha particles with energy 300 keV . An alpha particle has charge q=+2e and massm=6.64×10−27kg .If the magnetic field isnt changed, what will be the orbital radius of the alpha particles? Express your answer to three significant figures and include the appropriate units.

Answers

Answer:

16 cm

Explanation:

For protons:

Energy, E = 300 keV

radius of orbit, r1 = 16 cm

the relation for the energy and velocity is given by

$E = (1)/(2)mv^(2)$

So, $v = \sqrt{(2E)/(m)}$ .... (1)

Now,

$r = (mv)/(Bq)$

Substitute the value of v from equation (1), we get

$r = (√(2mE))/(Bq)$

Let the radius of the alpha particle is r2.

For proton

So, $r_(1) = \frac{\sqrt{2m_(1)E}}{Bq_(1)}$ ... (2)

Where, m1 is the mass of proton, q1 is the charge of proton

For alpha particle

So, $r_(2) = \frac{\sqrt{2m_(2)E}}{Bq_(2)}$ ... (3)

Where, m2 is the mass of alpha particle, q2 is the charge of alpha particle

Divide equation (2) by equation (3), we get

$(r_(1))/(r_(2))=(q_(2))/(q_(1))* \sqrt{(m_(1))/(m_(2))}$

q1 = q

q2 = 2q

m1 = m

m2 = 4m

By substituting the values

$(r_(1))/(r_(2))=\frac{2q}}{q}}* \sqrt{\frac{m}}{4m}}=1$

So, r2 = r1 = 16 cm

Thus, the radius of the alpha particle is 16 cm.

Answer:15.95 cm

Explanation:

Given

Energy=300 kev

radius of Proton=16 cm

mass of alpha particle $=6.64* 10^(-27) kg$

mass of proton $=1.67* 10^(-27) kg$

charge on alpha particle is twice of proton

radius of Proton is given by

$r=(mv)/(|q|B)$

and Kinetic energy $K=(P^2)/(2m)$

where P=momentum

$P=√(2Km)$

$r=(√(2km))/(qB)$ ---1

Radius for Alpha particle is

$r_(alpha)=\frac{\sqrt{2k\cdot m_(alpha)}}{2qB}$ -----2

Divide 1 & 2 we get

$(r)/(r_(alpha))=(√(m))/(q)* \frac{2q}{\sqrt{m_(alpha)}}$

$r_(alpha)=\sqrt{(6.64* 10^(-27))/(1.67* 10^(-27))}* 0.5$

$r_(alpha)=0.997* 16$

$r_(alpha)=15.95 cm$

Suppose that a particle accelerator is used to move two beams of particles in opposite directions. In a particular region, electrons move to the right at 6020 m/s and protons move to the left at 1681 m/s. The particles are evenly spaced with 0.0476 m between electrons and 0.0662 m between protons. Assuming that there are no collisions and that the interactions between the particles are negligible, what is the magnitude of the average current in this region

Answers

Answer:

$2.429783984* 10^(-14)\ A$

Explanation:

Velocity of electron = 6020 m/s

Velocity of proton = 1681 m/s

Electron space = 0.0476 m

Proton space = 0.0662 m

e = Charge of particle = $1.6* 10^(-19)\ C$

Number of electrons passing per second

$n_e=(6020)/(0.0476)\n\Rightarrow n_e=126470.588$

Number of protons passing per second

$n_p=(1681)/(0.0662)\n\Rightarrow n_p=25392.749$

Current due to electrons

$I_e=n_ee\n\Rightarrow I_e=126470.588* 1.6* 10^(-19)\n\Rightarrow I_e=2.0235* 10^(-14)\ A$

Current due to protons

$I_p=n_pe\n\Rightarrow I_p=25392.749* 1.6* 10^(-19)\n\Rightarrow I_p=4.06283984* 10^(-15)\ A$

Total current

$I=2.0235* 10^(-14)+4.06283984* 10^(-15)\n\Rightarrow I=2.429783984* 10^(-14)\ A$

The average current is $2.429783984* 10^(-14)\ A$

The index of refraction for red light in water is 1.331, and that for blue light is 1.340. If a ray of white light enters the water at an angle of incidence of 83.00o , what are the underwater angles of refraction for the blue and red components of the light

Answers

Answer:

The underwater angles of refraction for the blue and red components of the light is 47.8° and 48.2°

Explanation:

Using the Snell's law

n1 * sin Ф1 = n2 sin Ф2

1 * sin 83 = n2 sin Ф2

Ф2 = $sin^(-1) ((1)/(n2) * sin 83)$

Red light

n2 = 1.331

Ф2 = $sin^(-1) ((1)/(1.331) * sin 83) = 48.2$ °

Blue light

n2 = 1.340

Ф2 = $sin^(-1) ((1)/(1.340) * sin 83) = 47.8$ °

Suppose a fast-pitch softball player does a windmill pitch, moving her hand through a circular arc with her arm straight. She releases the ball at a speed of 25.5 m/s (about 57.0 mph ). Just before the ball leaves her hand, the ball's radial acceleration is 1060 m/s2 . What is the length of her arm from the pivot point at her shoulder

Answers

Answer:

61.3 cm

Explanation:

Radial acceleration of the object in circular motion is given by formula

$a = (v^2)/(R)\n$

Given:

$a = 1060 m/s^2\nv = 25.5 m/s$

Plugging in the values in the formula

$1060 = (25.5^2)/(R)\nR = 0.613 m$

so length of his arm is 61.3 cm

A piece of soft iron is placed in a solenoid increasing the magnetic field in an arrangement that can be switched on and off. Such a device is calleda. a hysteresis loop.b. an electrosolenoid.c. a permanent magnet.d. a ferromagnet.

Answers

Question:

A piece of soft iron is placed in a solenoid increasing the magnetic field in an arrangement that can be switched on and off. Such a device is called

Answer

a hysteresis loop.

an electrosolenoid.

a permanent magnet.

a ferromagnet.

an electromagnet

Answer:

A piece of soft iron is placed in a solenoid increasing the magnetic field in an arrangement that can be switched on and off. Such a device is called an electromagnet.

Explanation:

When a magnet is processed by supplying electricity than it is called as electromagnet and therefore its strength can be fluctuated as dependent on amount of electricity supplied.

Here, when a iron piece is moved closer to a long coil of wire named solenoid physical interaction named electromagnetic force take place between two electrically charged particles.

This force is supplied by electromagnetic fields constructed from electric and magnetic field. This is also called as Lorentz force constituted from both electricity and magnetism. This property itself gives evidence of characteristics of material used for electromagnetism.

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