A baseball flying through the air has what kind of energy?

Why do high and low tides happen 1 hour later each day? just answer please and it is a written response

8_murik_8 [283]

With each passing day, the high tides occur about an hour later. The moon rises about an hour later each day, too (actually, 54 minutes later). Since the moon pulls up the tides, these two delays are connected. As the earth rotates through one day, the moon moves in its orbit.

6 0

2 years ago

If the distance d (in meters) traveled by an object in time t (in seconds) is given by the formula d=a+bt2, the si units of a an

ss7ja [257]

Distance , d = a+b $t^2$

The unit of d is in meter and t is in seconds.

So the unit of a a must be meter.

Now we have unit of b $t^2$ is meter.

So unit of b* $second^2$ = meter

Unit of b = meter/ $second^2$

So unit of a = m and unit of b = m/ $s^2$ .

8 0

2 years ago

Describe how the number of photoelectrons emitted from a metal plate in the photoelectric effect would change if the following o

andrew-mc [135]

Answer:

a) True. The number of photoelectrons is proportional to the amount (intensity) of the incident beam. From the expression above we see that threshold frequency cannot emit electrons.

b) λ = c / f

Therefore, as the wavelength increases, the frequency decreases and therefore the energy of the photoelectrons emitted,

c) threshold energy

h f =Ф

Explanation:

It's photoelectric effect was fully explained by Einstein by the expression

Knox = h f - fi

Where K is the kinetic energy of the photoelectrons, f the frequency of the incident radiation and fi the work function of the metal

a) True. The number of photoelectrons is proportional to the amount (intensity) of the incident beam. From the expression above we see that threshold frequency cannot emit electrons.

b) wavelength is related to frequency

λ = c / f

Therefore, as the wavelength increases, the frequency decreases and therefore the energy of the photoelectrons emitted, so there is a wavelength from which electrons cannot be removed from the metal.

c) As the work increases, more frequency radiation is needed to remove the electrons, because there is a threshold energy

h f =Ф

7 0

2 years ago

A circular ring with area 4.45 cm2 is carrying a current of 13.5 A. The ring, initially at rest, is immersed in a region of unif

Gwar [14]

Answer:

a) ( 0.0015139 i^ + 0.0020185 j^ + 0.00060556 k^ ) N.m

b) ΔU = -0.000747871 J

c) w = 47.97 rad / s

Explanation:

Given:-

- The area of the circular ring, A = 4.45 cm^2

- The current carried by circular ring, I = 13.5 Amps

- The magnetic field strength, vec ( B ) = (1.05×10−2T).(12i^+3j^−4k^)

- The magnetic moment initial orientation, vec ( μi ) = μ.(−0.8i^+0.6j^)

- The magnetic moment final orientation, vec ( μf ) = -μ k^

- The inertia of ring, T = 6.50×10^−7 kg⋅m2

Solution:-

- First we will determine the magnitude of magnetic moment ( μ ) from the following relation:

μ = N*I*A

Where,

N: The number of turns

I : Current in coil

A: the cross sectional area of coil

- Use the given values and determine the magnitude ( μ ) for a single coil i.e ( N = 1 ):

μ = 1*( 13.5 ) * ( 4.45 / 100^2 )

μ = 0.0060075 A-m^2

- From definition the torque on the ring is the determined from cross product of the magnetic moment vec ( μ ) and magnetic field strength vec ( B ). The torque on the ring in initial position:

vec ( τi ) = vec ( μi ) x vec ( B )

= 0.0060075*( -0.8 i^ + 0.6 j^ ) x 0.0105*( 12 i^ + 3 j^ -4 k^ )

= ( -0.004806 i^ + 0.0036045 j^ ) x ( 0.126 i^ + 0.0315 j^ -0.042 k^ )

- Perform cross product:

$\left[\begin{array}{ccc}i&j&k\\-0.004806&0.0036045&0\\0.126&0.0315&-0.042\end{array}\right] = \left[\begin{array}{ccc}-0.00015139\\-0.00020185\\-0.00060556\end{array}\right] \\\\$

- The initial torque ( τi ) is written as follows:

vec ( τi ) = ( 0.0015139 i^ + 0.0020185 j^ + 0.00060556 k^ )

- The magnetic potential energy ( U ) is the dot product of magnetic moment vec ( μ ) and magnetic field strength vec ( B ):

- The initial potential energy stored in the circular ring ( Ui ) is:

Ui = - vec ( μi ) . vec ( B )

Ui =- ( -0.004806 i^ + 0.0036045 j^ ) . ( 0.126 i^ + 0.0315 j^ -0.042 k^ )

Ui = -[( -0.004806*0.126 ) + ( 0.0036045*0.0315 ) + ( 0*-0.042 )]

Ui = - [(-0.000605556 + 0.00011)]

Ui = 0.000495556 J

- The final potential energy stored in the circular ring ( Uf ) is determined in the similar manner after the ring is rotated by 90 degrees with a new magnetic moment orientation ( μf ) :

Uf = - vec ( μf ) . vec ( B )

Uf = - ( -0.0060075 k^ ) . ( 0.126 i^ + 0.0315 j^ -0.042 k^ )

Uf = - [( 0*0.126 ) + ( 0*0.0315 ) + ( -0.0060075*-0.042 ) ]

Uf = -0.000252315 J

- The decrease in magnetic potential energy of the ring is arithmetically determined:

ΔU = Uf - Ui

ΔU = -0.000252315 - 0.000495556

ΔU = -0.000747871 J

Answer: There was a decrease of ΔU = -0.000747871 J of potential energy stored in the ring.

- We will consider the system to be isolated from any fictitious forces and gravitational effects are negligible on the current carrying ring.

- The conservation of magnetic potential ( U ) energy in the form of Kinetic energy ( Ek ) is valid for the given application:

Ui + Eki = Uf + Ekf

Where,

Eki : The initial kinetic energy ( initially at rest ) = 0

Ekf : The final kinetic energy at second position

- The loss in potential energy stored is due to the conversion of potential energy into rotational kinetic energy of current carrying ring.

-ΔU = Ekf

0.5*T*w^2 = -ΔU

w^2 = -ΔU*2 / T

Where,

w: The angular speed at second position

w = √(0.000747871*2 / 6.50×10^−7)

w = 47.97 rad / s

6 0

3 years ago

What kind of weather is signaled by cumulus clouds

Novay_Z [31]

Cumulonimbus clouds are associated by severe weather. They are thunderstorm clouds that bring heavy rain, snow, hail, lightning and even tornadoes.

5 0

3 years ago