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3 years ago

What is the final speed of a lion initially running at a speed of 30 m/s, if it accelerates at

Physics

1 answer:

cricket20 [7]3 years ago

8 0

The final speed of a lion running 30 m/s accelerates at a rate of 3 m/s3 for 5 seconds it’s 3.2

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When reading the printout from a laser printer, you are actually looking at an array of tiny dots.

solniwko [45]

Answer:

The value is $y = 3.097 * 10^{-5} \ m$

Explanation:

From the question we are told that

The diameter of the pupil is $d_p = 4.2 \ mm = 4.2 *10^{-3} \ m$

The distance of the page from the eye $d = 29 \ cm = 0.29 \ m$

The wavelength is $\lambda = 500 \ nm = 500 *10^{-9} \ m$

The refractive index is $n_r = 1.36$

Generally the minimum separation of adjacent dots that can be resolved is mathematically represented as

$y = [ \frac{1.22 * \lambda }{d_p * n_r } ]* d$

$y = [ \frac{1.22 * 500 *10^{-9} }{4.2 *10^{-3} * 1.36} ]* 0.29$

$y = 3.097 * 10^{-5} \ m$

7 0

3 years ago

If the jet is moving at a speed of 1300 km/h at the lowest point of the loop, determine the minimum radius of the circle so that

Natalka [10]

Answer:

R = 2216m and The normal force of the seat on the pilot is 5008N

Explanation:

See attachment below please.

5 0

3 years ago

A 50.0 kg woman climbs a flight of stairs 6.00 m high in 15.0 s. How much power does she use.

WITCHER [35]

Her weight = (mass) · (gravity) = (50kg) · (9.8 m/s²)

Work = (weight) · (height) = (50kg) · (9.8 m/s²) · (6 m)

Power = (work) / (time) = (50kg) · (9.8 m/s²) · (6 m) / (15 s)

Power = (50 · 9.8 · 6 / 15) · (kg · m² / s³)

Power = 196 (kg · m / s²) · (m) / s

Power = 196 Newton-meter/second

<em>Power = 196 watts</em>

6 0

3 years ago

video for A door 1 m wide, of mass 15 kg, is hinged at one side so that it can rotate without friction about a vertical axis. It

natka813 [3]

Answer:

$\omega_f = 0.4\ rad/s$

Explanation:

given,

width of door dimension = 1 m

mass of door = 15 Kg

mass of bullet = 10 g = 0.001 Kg

speed of bullet = 400 m/s

$I_{total} =I_{door} + I_{bullet}$

$I_{total} =\dfrac{1}{3}MW^2 + m(\dfrac{W}{2})^2$

a) from conservation of angular momentum

$L_i = L_f$

$mv\dfrac{W}{2} = I_{total}\omega_f$

$mv\dfrac{W}{2}= (\dfrac{1}{3}MW^2 + m(\dfrac{W}{2})^2)\omega_f$

$\omega_f= \dfrac{mv\dfrac{W}{2}}{\dfrac{1}{3}MW^2 + m(\dfrac{W}{2})^2}$

$\omega_f= \dfrac{\dfrac{mv}{2}}{\dfrac{MW }{3}+(\dfrac{mW}{4})}$

$\omega_f= \dfrac{\dfrac{0.01\times 400}{2}}{\dfrac{15\times 1 }{3}+(\dfrac{0.01\times 1}{4})}$

$\omega_f = 0.4\ rad/s$

8 0

3 years ago

How does an atom of potassium-41 become a potassium ion with a +1 charge? 19 K 39.10

stiv31 [10]

It is very difficult for an atom to accept a proton. It can only be done under very special circumstances. So A and C are both incorrect. I don't see how D is possible. The atom does lose 1 electron, but how it gets 21 is think air.

The answer is B which is exactly what happens.

5 0

3 years ago

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