Answer:
The distance away the center of the earthquake is 1083.24 km.
Explanation:
Given that,
Speed of transverse wave = 9.1\ km/s
Speed of longitudinal wave = 5.7 km/s
Time = 71 sec
We need to calculate the distance of transverse wave
Using formula of distance

....(I)
The distance of longitudinal wave
....(II)
From the first equation

Put the value of t in equation (II)




Hence, The distance away the center of the earthquake is 1083.24 km.
In order to decrease the friction on the slide,
we could try some of these:
-- Install a drippy pipe across the top that keeps continuously
dripping olive oil on the top end of the slide. The oil oozes
down the slide and keeps the whole slide greased.
-- Hire a man to spread a coat of butter on the whole slide,
every 30 minutes.
-- Spray the whole slide with soapy sudsy water, every 30 minutes.
-- Drill a million holes in the slide,and pump high-pressure air
through the holes. Make the slide like an air hockey table.
-- Keep the slide very cold, and keep spraying it with a fine mist
of water. The water freezes, and a thin coating of ice stays on
the slide.
-- Ask a local auto mechanic to please, every time he changes
the oil in somebody's car, to keep all the old oil, and once a week
to bring his old oil to the park, to spread on the slide. If it keeps
the inside of a hot car engine slippery, it should do a great job
keeping a simple park slide slippery.
-- Keep a thousand pairs of teflon pants near the bottom of the ladder
at the beginning of the slide. Anybody who wants to slide faster can
borrow a set of teflon pants, put them on before he uses the slide, and
return them when he's ready to go home from the park.
Answer:
magnitude: 21.6; direction: 33.7 degrees
Explanation:
When we multiply a vector by a scalar, we have to multiply each component of the vector by the scalar number. In this case, we have
vector: (-3,-2)
Scalar: -6
so the vector multiplied by the scalar will have components

The magnitude is given by Pythagorean's theorem:

and the direction is given by the arctan of the ratio between the y-component and the x-component:

Answer:
option (c) is correct
Explanation:
Half life of a substance is the time in which the element becomes half of is initial value.
half life, T = 8 days
Amount remaining, N = 10 % of original value
Let the original value is No.
N = 10% of No
N = 0.1 No
Let the time taken is t and the decay constant is λ.
The relation between the decay constant and the half life is given by

Us the equation of radioactivity



Taking natural log on both the sides, we get
0.08664 t = 2.303
t = 26.6 days
<h2>5.3 km</h2>
Explanation:
This question involves continuous displacement in various directions. When it becomes difficult to imagine, vector analysis becomes handy.
Let us denote each of the individual displacements by a vector. Consider the unit vectors
as the unit vectors in the direction of East and North respectively.
By simple calculations, we can derive the unit vectors
in the directions North,
South of West and
North of West respectively.
So Total displacement vector = Sum of individual displacement vectors.
Displacement vector = 
Magnitude of Displacement = 
∴ Total displacement = 