1. On this graph, what is the change in velocity between 5 seconds and 8 seconds?

How can we predict how high a roller coaster can go knowing its velocity?

KatRina [158]

“to predict the speed that a coaster will reach before it is ever placed on the track. ... When coaster goes up by height h then its gravitational potential energy increases by the amount m.g.h where m = mass of coaster, h = height of coaster and g = gravitational acceleration due to Earth”

6 0

3 years ago

A hot iron ball of mass 200 g is cooled to a temperature of 22°C. 6.9 kJ of heat is lost to the surroundings during the process.

qaws [65]

Heat lost or gained, H = mc(θ₂ - θ₁)
Where m = mass, c = Specific heat capacity, θ₂= final temperature, θ₁ = initial temperature

m = 200g, c = 0.444 J/g°C, θ₁ = 22 °C (Since it was cooled).

H = 6.9 kj = 6.9 *1000J = 6900 J

6900 = 200*0.444* (θ₂ - 22)

6900/(200*0.444) = θ₂ - 22

77.70 = θ₂ - 22

θ₂ - 22 = 77.7

θ₂ = 77.7 + 22 = 99.7

So initial temperature before cooling ≈ 100°C . Option C.

5 0

3 years ago

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In a double slit experiment, the distance between the slits is 0.2 mm and the distance to the screen is 100 cm. What is the phas

Anni [7]

Answer:

The phase difference is $\Delta \phi = 180^o$

Explanation:

From the question we are told that

The distance between the slits is $d = 0.2 \ mm = \frac{0.2}{1000} = 0.2 *10^{-3} \ m$

The distance to the screen is $D = 100 cm = \frac{100}{100} = 1 \ m$

The wavelength is $\lambda = 400nm$

The distance of the wave from the central maximum is $L = 5mm = 5*10^{-3} m$

Generally the path difference of this waves is mathematically represented as

$y = d sin \theta$

Here $\theta$ is the angle between the the line connecting the mid-point of the slits with the screen and the line connecting the mid-point of the slits to the central maximum