When 93 is written in simplest radical form, which value remains under the radical?

Sound travels at about 360 meters/sec. What is the wavelength of sound with a

ollegr [7]

<u>Answer:</u>

0.24 m

<u>Explanation:</u>

Given:

Wave velocity ( v ) = 360 m / sec

Frequency ( f ) = 1500 Hz

We have to calculate wavelength ( λ ):

We know:

v = λ / t [ f = 1 / t ]

v = λ f

= > λ = v / f

Putting values here we get:

= > λ = 360 / 1500 m

= > λ = 36 / 150 m

= > λ = 0.24 m

Hence, wavelength of sound is 0.24 m.

6 0

3 years ago

An electron is initially at rest in a uniform electric field having a strength of 1.85 × 106 V/m. It is then released and accele

kirza4 [7]

Answer:

$W = 462.5 keV$

Explanation:

As we know that when electron moved in electric field then work done by electric field must be equal to the change in kinetic energy of the electron

So here we have to find the work done by electric field on moving electron

So we have

$F = qE$

$F = (1.6 \times 10^{-19})(1.85 \times 10^6)$

$F = 2.96 \times 10^{-13} N$

now the distance moved by the electron is given as

$d = 0.25 m$

so we have

$W = F.d$

$W = (1.6 \times 10^{-19})(1.85 \times 10^6)(0.25)$

$W = 7.4 \times 10^{-14} J$

now we have to convert it into keV units

so we have

$1 keV = 1.6 \times 10^{-16} J$

$W = 462.5 keV$

5 0

2 years ago

Which of these has the greatest influence over how big a baby elephant will grow?

lina2011 [118]

Is this a multiple choice question? Or would you just like me to state generally the greatest influence over how big a baby elephant will grow ?

4 0

2 years ago

A loaded ore car has a mass of 950 kg. and rolls on rails ofnegligible friction. It starts from rest ans is pulled up a mineshaf

stiks02 [169]

(a) 10241 W

In this situation, the car is moving at constant speed: this means that its acceleration along the direction parallel to the slope is zero, and so the net force along this direction is also zero.

The equation of the forces along the parallel direction is:

$F - mg sin \theta = 0$

where

F is the force applied to pull the car

m = 950 kg is the mass of the car

$g=9.8 m/s^2$ is the acceleration of gravity

$\theta=30.0^{\circ}$ is the angle of the incline

Solving for F,

$F=mg sin \theta = (950)(9.8)(sin 30.0^{\circ})=4655 N$

Now we know that the car is moving at constant velocity of

v = 2.20 m/s

So we can find the power done by the motor during the constant speed phase as

$P=Fv = (4655)(2.20)=10241 W$

(b) 10624 W

The maximum power is provided during the phase of acceleration, because during this phase the force applied is maximum. The acceleration of the car can be found with the equation

$v=u+at$

where

v = 2.20 m/s is the final velocity

a is the acceleration

u = 0 is the initial velocity

t = 12.0 s is the time

Solving for a,

$a=\frac{v-u}{t}=\frac{2.20-0}{12.0}=0.183 m/s^2$

So now the equation of the forces along the direction parallel to the incline is

$F - mg sin \theta = ma$

And solving for F, we find the maximum force applied by the motor:

$F=ma+mgsin \theta =(950)(0.183)+(950)(9.8)(sin 30^{\circ})=4829 N$

The maximum power will be applied when the velocity is maximum, v = 2.20 m/s, and so it is:

$P=Fv=(4829)(2.20)=10624 W$

(c) $5.82\cdot 10^6 J$

Due to the law of conservation of energy, the total energy transferred out of the motor by work must be equal to the gravitational potential energy gained by the car.

The change in potential energy of the car is:

$\Delta U = mg \Delta h$