A train accelerates from 23m/s to 190m/s in 54 seconds. What was its acceleration?

Physics

1 answer:

Ierofanga [76]4 years ago

3 0

Answer:

Use the method on the image and solve it.

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When you look at yourself in the mirror, what is the approximate angle of incidence of the light rays?

Kitty [74]

46 degrees
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A pan hangs from a 50 cm spring. When a 10 kg mass is placed in the pan, it stretches the spring 6 cm. What is a function rule l

MrRissso [65]

Answer:

Explanation:

A Spring stretches / compresses when force is applied on them and they are governed by the Hookes Law which states that the force required to stretch or compress a spring is directly proportional to the distance it is stretched.

$F = -kx$

F is the force applied and x is the elongation of the spring

k is the spring constant.

negative sign indicates the change in direction from equilibrium position.

In the given question, we dont have force but we know that the pan is hanging. We also know from the Newton's second law of motion that

$F=mg$

Inserting this into Hooke's Law

$mg=-kx$

computing it for x,

$-x=mg/k$

This is the model which will tell the length of the spring against change in the mass located in the pan.

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Witch planet takes 84 earth year's to orbit the sun just once

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Uranus takes 84 earth years to make a full rotation around the sun<span />

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3 years ago

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An electron in the n = 5 level of an h atom emits a photon of wavelength 1282.17 nm. to what energy level does the electron move

lions [1.4K]

This is an interesting (read tricky!) variation of Rydberg Eqn calculation.
Rydberg Eqn: 1/λ = R [1/n1^2 - 1/n2^2]
Where λ is the wavelength of the light; 1282.17 nm = 1282.17×10^-9 m
R is the Rydberg constant: R = 1.09737×10^7 m-1
n2 = 5 (emission)
Hence 1/(1282.17 ×10^-9) = 1.09737× 10^7 [1/n1^2 – 1/25^2]
Some rearranging and collecting up terms:
1 = (1282.17 ×10^-9) (1.09737× 10^7)[1/n2 -1/25]
1= 14.07[1/n^2 – 1/25]
1 =14.07/n^2 – (14.07/25)
14.07n^2 = 1 + 0.5628
n = √(14.07/1.5628) = 3

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3 years ago

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A shell is fired from the ground with an initial speed of 1.60 × 103 m/s (approximately five times the speed of sound) at an ini

Volgvan

Answer:

The horizontal range will be $2.55\times 10^5m$