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3 years ago

An 850.0 kg car wants to accelerate from rest to a speed of 18.0 m/s.

Physics

1 answer:

strojnjashka [21]3 years ago

8 0

Answer:

$1.07*10^5 Joules$

Explanation:

From Kinetic energy K.E = m* $v^2$ /2

Where mass of the car, m = 850 kg

velocity of the car, v = 18 m/s

Hence K.E = $850*18^2/2 = 1.07*10^5 Joules$

Also, we can evaluate the given time

time = K.E / Power

= $1.07*10^5 / (50*746)$

= 2.87 secs

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Answer:

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Explanation:

Given that,

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We need to find the wavelength of these photons.

$1.2\times 10^{-3}\ eV=1.2\times 10^{-3}\times 1.6\times 10^{-19}\\\\=1.92\times 10^{-22}\ J$

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Put all the values,

$\lambda=\dfrac{6.63\times 10^{-34}\times 3\times 10^8}{1.92\times 10^{-22}}\\\\\lambda=1.03\times 10^{-3}\ m$

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2 years ago

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2 years ago

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Consider one such cell where the magnitude of the potential difference is 65 mV, and the inner surface of the membrane is at a h

Gelneren [198K]

Answer: W = $1.04.10^{-20}$ J

Explanation: Since the potassium ion is at the outside membrane of a cell and the potential here is lower than the potential inside the cell, the transport will need work to happen.

The work to transport an ion from a lower potential side to a higher potential side is calculated by

$W=q.\Delta V$

q is charge;

ΔV is the potential difference;

Potassium ion has +1 charge, which means:

p = $1.6.10^{-19}$ C

To determine work in joules, potential has to be in Volts, so:

$\Delta V=65.10^{-3}V$

Then, work is

$W=1.6.10^{-19}.65.10^{-3}$

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To move a potassium ion from the exterior to the interior of the cell, it is required $W=1.04.10^{-20}$ J of energy.

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3 years ago

7. A beam of light converges at a point P. Now a lens is placed in the path of the convergent

Rudiy27

Answer:

Lens at a distance = 7.5 cm

Lens at a distance = 6.86 cm (Approx)

Explanation:

Given:

Object distance u = 12 cm

a) Focal length = 20 cm

b) Focal length = 16 cm

Computation:

a. 1/v = 1/u + 1/f

1/v = 1/20 + 1/12

v = 7.5 cm

Lens at a distance = 7.5 cm

b. 1/v = 1/u + 1/f

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2 years ago

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