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2 years ago

Under the Big Top elephant, Ella (2500 kg), is attracted to Phant, the 3,000 kg

Physics

1 answer:

Vladimir [108]2 years ago

4 0

Under the Big Top elephant, Ella (2500 kg), is attracted to Phant, the 3,000 kg elephant. They are separated by 8

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A 0.3 mm long invertebrate larva moves through 20oC water at 1.0 mm/s. You are creating an enlarged physical model of this larva

AleksandrR [38]

Answer:

Explanation:

For the problem, we should have same reynolds number

ρvd/mu = constant

1000×1×10⁻³×0.3×10⁻³/1.002×10⁻³ = 1400×0.5×d/600

d = 25.66 cm

5 0

2 years ago

In a semiclassical model of the hydrogen atom, the electron orbits the proton at a distance of 0.053 nm. Part A What is the elec

Bezzdna [24]

Answer with Explanation:

We are given that

$r=0.053 nm=0.053\times 10^{-9} m$

$1 nm=10^{-9} m$

Charge on proton,q= $1.6\times 10^{-19} C$

a.We have to find the electric potential of the proton at the position of the electron.

We know that the electric potential

$V=\frac{kq}{r}$

Where $k=9\times 10^9$

$V=\frac{9\times 10^9\times 1.6\times 10^{-19}}{0.053\times 10^{-9}}$

$V=27.17 V$

B.Potential energy of electron,U= $\frac{kq_e q_p}{r}$

Where

$q_e=-1.6\times 10^{-19} c=$ Charge on electron

$q_p=q=1.6\times 10^{-19} C$ =Charge on proton

Using the formula

$U=\frac{9\times 10^9\times (-1.6\times 10^{-19}\times 1.6\times 10^{-19}}{0.053\times 10^{-9}}$

$U=-4.35\times 10^{-18} J$

8 0

2 years ago

A ball was kicked upward at a speed of 64.2 m/s. how fast was the ball going 1.5 seconds later

UNO [17]

Anything that's not supported and doesn't hit anything, and
doesn't have any air resistance, gains 9.8 m/s of downward
speed every second, on account of gravity. If it happens to
be moving up, then it loses 9.8 m/s of its upward speed every
second, on account of gravity.

(64.2 m/s) - [ (9.8 m/s² ) x (1.5 sec) ]

= (64.2 m/s) - [ 14.7 m/s ]

= 49.5 m/s . (upward)

7 0

2 years ago

A cliff diver running 3.60 m/s dives out horizontally from the edge of a vertical cliff and reaches the water below 2.00 s later

mart [117]

Explanation:

It is given that,

The horizontal speed of a cliff diver, $v_x=3.6\ m/s$

It reaches the water below 2.00 s later, t = 2 s

Let $d_x$ is the distance where the diver hit the water. It can be calculated as follows :

$d_x=v_x\times t\\\\=3.6\times 2\\\\=7.2\ m$

Let $d_y$ is the height of the cliff. It can be calculated using second equation of motion as follows :

$d_y=u_yt+\dfrac{1}{2}gt^2\\\\d_y=\dfrac{1}{2}\times 9.8\times 2^2\\\\=19.6\ m$

So, the cliff is 19.6 m high and it will hit the water at a distance of 19.6 m.

8 0

1 year ago

Light with a wavelength of 495 nm is falling on a surface and electrons with a maximum kinetic energy of 0.5 eV are ejected. Wha

devlian [24]

Answer:

To increase the maximum kinetic energy of electrons to 1.5 eV, it is necessary that ultraviolet radiation of 354 nm falls on the surface.

Explanation:

First, we have to calculate the work function of the element. The maximum kinetic energy as a function of the wavelength is given by:

$K_{max}=\frac{hc}{\lambda}-W$

Here h is the Planck's constant, c is the speed of light, $\lambda$ is the wavelength of the light and W the work function of the element:

$W=\frac{hc}{\lambda}-K_{max}\\W=\frac{(4.14*10^{-15}eV\cdot s)(3*10^8\frac{m}{s})}{495*10^{-9}m}-0.5eV\\W=2.01eV$

Now, we calculate the wavelength for the new maximum kinetic energy:

$W+K_{max}=\frac{hc}{\lambda}\\\lambda=\frac{hc}{W+K_{max}}\\\lambda=\frac{(4.14*10^{-15}eV\cdot s)(3*10^8\frac{m}{s})}{2.01eV+1.5eV}\\\lambda=3.54*10^{-7}m=354*10^{-9}m=354nm$

This wavelength corresponds to ultraviolet radiation. So, to increase the maximum kinetic energy of electrons to 1.5 eV, it is necessary that ultraviolet radiation of 354 nm falls on the surface.

8 0

2 years ago