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3 years ago

Where a subducting plate slides beneath the lithosphere, melting takes place and a(n) ____ is created?

Physics

1 answer:

11111nata11111 [884]3 years ago

3 0

under ground cave will be created

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An ion in a mass spectrometer follows a semicircular path of radius 14.8. What is the distance it travels?

aleksklad [387]

The circumference of a circle is (2π · the circle's radius).

The length of a semi-circle is (1π · the circle's radius) =

(π · 14.8) = 46.5 (rounded)

(The unit is the same as whatever the unit of the 14.8 is.)

7 0

3 years ago

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Difference between derived quantity and physical quantity

Lerok [7]

Answer:

<h2>Derived quantities are based on fundamental quantities, and they can be given in terms of fundamental quantities.</h2>

<h3>Fundamental quantities are the base quantities of a unit system, and they are defined independent of the other quantities. </h3>

Explanation:

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2 years ago

Gravity on Earth is 9.8 m/s^2, and gravity on Jupiter is 23.1 m/s^2. So, if the mass of a rock is 70 kilograms, it's weight on E

Levart [38]

~686newtons on earth and
~1617 newtons on jupiter
the formula is weight = gravitational acceleration * mass of the object

3 0

3 years ago

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Please help true or false

NISA [10]

Answer:

the answer is true.......

8 0

2 years ago

A force F = (2.75 N)i + (7.50 N)j + (6.75 N)k acts on a 2.00 kg mobile object that moves from an initial position of d = (2.75 m

Natasha2012 [34]

Answer:

The work done on the object by the force in the 5.60 s interval is 40.93 J.

Explanation:

Given that,

Force $F=(2.75\ N)i+(7.50\ N)j+(6.75\ N)k$

Mass of object = 2.00 kg

Initial position $d=(2.75)i-(2.00)j+(5.00)k$

Final position $d=-(5.00)i+(4.50)j+(7.00)k$

Time = 4.00 sec

We need to calculate the work done on the object by the force in the 5.60 s interval.

Using formula of work done

$W=F\cdot d$

$W=F\cdot(d_{f}-d_{i})$

Put the value into the formula

$W=(2.75)i+(7.50)j+(6.75)k\cdot (-(5.00)i+(4.50)j+(7.00)k-(2.75)i+(2.00)j-(5.00)k)$

$W=(2.75)i+(7.50)j+(6.75)k\cdot((-7.75)i+6.50j+2.00k)$

$W=2.75\times-7.75+7.50\times6.50+6.75\times2.00$

$W=40.93\ J$

Hence, The work done on the object by the force in the 5.60 s interval is 40.93 J.

4 0

3 years ago