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3 years ago

10

A small boat is moving at a velocity of 3.35m/s when it is accelerated by a river current perpendicular to the initial direction

of motion. The current accelerates the boat at 0.750m/s^2. what will the new velocity (magnitude and direction) of the boat be after 5 s?

1 answer:

mr_godi [17]3 years ago

7 0

Answer:

v =7.1 m/s

Explanation:

Given that

u = 3.35 m/s

t= 5 s

a= 0.75 m/s²

The final velocity = v

We know v = u +at

v=final velocity

u=initial velocity

Now by putting the values in the above equation

v = 3.35 + 0.75 x 5 m/s

v =7.1 m/s

Therefore the final velocity will be 7.1 m/s

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A 730-keV gamma ray Compton-scatters from an electron. Find the energy of the photon scattered at 120°, the kinetic energy of th

romanna [79]

Answer:

Energy of scattered photon is 232.27 keV.

Kinetic energy of recoil electron is 497.73 keV.

The recoil angle of electron is 13.40°

Explanation:

The energy of scattered photon is given by the relation :

$E_{2}=\frac{E_{1} }{1+(\frac{E_{1} }{m_{e}c^{2} })(1-\cos\theta) }$ .....(1)

Here E₁ is the energy of incident photon, E₂ is the energy of scattered photon, $m_{e}$ is mass of electron and θ is scattered angle.

Substitute 730 keV for E₁, 511 keV for $m_{e}$ and 120° for θ in equation (1).

$E_{2}=\frac{730 }{1+(\frac{730 }{511 })(1-\cos120) }$

E₂ = 232.27 keV

Kinetic energy of recoil electron is given by the relation :

K.E. = E₁ - E₂ = (730 - 232.27 ) keV = 497.73 keV

The recoil angle of electron is given by :

$\cot\phi=(1+\frac{E_{1} }{m_{e}c^{2} })\tan\frac{\theta}{2}$

Substitute the suitable values in above equation.

$\cot\phi=(1+\frac{730 }{511 })\tan\frac{120}{2}$

$\cot\phi=4.20$

$\phi$ = 13.40°

8 0

3 years ago

A capacitor has a charge of 4.6 Î¼C. An E-field of 1.8 kV/mm is desired between the plates. There's no dielectric. What must be

Scrat [10]

Answer:

$A = 0.2875 m^2$

Explanation:

As we know that

$Q = 4.6 \mu C$

E = 1.8 kV/mm

now we know that electric field between the plated of capacitor is given as

$E = \frac{\sigma}{\epsilon_0}$

now we will have

$1.8 \times 10^6 = \frac{\sigma}{\epsilon_0}$

$\sigma = (1.8 \times 10^6)(8.85 \times 10^{-12})$

$\sigma = 1.6 \times 10^{-5} C/m^2$

now we have

$\sigma = \frac{Q}{A}$

now we have area of the plates of capacitor

$A = \frac{Q}{\sigma}$

$A = \frac{4.6 \times 10^{-6}}{1.6 \times 10^{-5}}$

$A = 0.2875 m^2$

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3 years ago

What force does it take to accelerate a 7.2 kg object 3.0 m/s^2.

grin007 [14]

Answer:

<h2>21.6 N</h2>

Explanation:

The force acting on an object given it's mass and acceleration can be found by using the formula

force = mass × acceleration

From the question we have

force = 7.2 × 3 = 21.6

We have the final answer as

<h3>21.6 N</h3>

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2 years ago

The cornea behaves as a thin lens of focal lengthapproximately 1.80 {\rm cm}, although this varies a bit. The material of whichi

Keith_Richards [23]

Answer:

Explanation:

a )

from lens makers formula

$\frac{1}{f} =(\mu-1)(\frac{1}{r_1} -\frac{1}{r_2})$

f is focal length , r₁ is radius of curvature of one face and r₂ is radius of curvature of second face

putting the values

$\frac{1}{1.8} =(1.38-1)(\frac{1}{.5} -\frac{1}{r_2})$

1.462 = 2 - 1 / r₂

1 / r₂ = .538

r₂ = 1.86 cm .

= 18.6 mm .

b )

object distance u = 25 cm

focal length of convex lens f = 1.8 cm

image distance v = ?

lens formula

$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$

$\frac{1}{v} - \frac{1}{-25} = \frac{1}{1.8}$

$\frac{1}{v} = \frac{1}{1.8} -\frac{1}{25}$

.5555 - .04

= .515

v = 1.94 cm

c )

magnification = v / u

= 1.94 / 25

= .0776

size of image = .0776 x size of object

= .0776 x 10 mm

= .776 mm

It will be a real image and it will be inverted.

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A. Divergent (pretty sure)

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3 years ago

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