3 years ago

A point from which the position of other objects can be described is called what?

Physics

1 answer:

Alisiya [41]3 years ago

3 0

Answer:

Reference Point

Explanation:

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Combine these three velocity vectors into a resultant: 3.0 m/s north, 4.0 m/s east 1.0 m/s west. Identify the resultant vector

Slav-nsk [51]

Answer:

The resultant vector is 1 m/s

Explanation:

The resultant vector is 1 m/s west based on triangle law of vector addition, when two sides of a triangle is represented by two vectors, the resultant vector is the third side of the triangle.

5 0

3 years ago

The components of a 10.8-meters-per-second velocity at an angle of 34.° above the horizontal are

Kisachek [45]

Answer:

6.0 m/s vertical and 9.0 m/s horizontal

Explanation:

For the vertical component, we use the formula:

Sin(34°) = y / 10.8

Then we solve for y:

0.559 = y / 10.8

y = 6.0

And for the horizontal component, we use the formula:

Cos(34°) = x / 10.8

Then we solve for x:

0.829 = x / 10.8

y = 9.0

So the answer is " 6.0 m/s vertical and 9.0 m/s horizontal".

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2 years ago

Comparativo entre a máquina a vapor de Neu carmen e James Watt!

Bas_tet [7]

Answer:

jame watt

Explanation:

James Watt tiña máis importancia que outro home

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3 years ago

PLEASE HELP ME A lens with a surface that curves outward like the exterior of a sphere is __________. (Points : 1) reflected ref

yanalaym [24]

Convex.

Concave curved inward (like how a cave foes in) and convex curves outward. Reflected and refracted do not apply to a lens.

3 0

3 years ago

Read 2 more answers

It has been suggested, and not facetiously, that life might have originated on Mars and been carried to Earth when a meteor hit

damaskus [11]

Answer:

$a=3125000 m/s^2\\a=3.125*10^6 m/s^2$

Acceleration, in m/s, of such a rock fragment = $3.125*10^6m/s^2$

Explanation:

According to Newton's Third Equation of motion

$V_f^2-V_i^2=2as$

Where:

$V_f$ is the final velocity

$V_i$ is the initial velocity

a is the acceleration

s is the distance

In our case:

$V_f=V_{escape}, V_i=0,s=4 m$

So Equation will become:

$V_{escape}^2-V_i^2=2as\\V_{escape}^2-0=2as\\V_{escape}^2=2as\\a=\frac{V_{escape}^2}{2s}\\a=\frac{(5*10^3m)^2}{2*4}\\a=3125000 m/s^2\\a=3.125*10^6 m/s^2$

Acceleration, in m/s, of such a rock fragment = $3.125*10^6m/s^2$

5 0

2 years ago