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3 years ago

Which ocean had the greatest average depth

Physics

2 answers:

OLga [1]3 years ago

6 0

Yes, it would be the Pacific Ocean.

agasfer [191]3 years ago

4 0

The Pacific ocean has the greatest average depth of 4280 meters.

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Un bloque de 2.5kg de masa es empujado 2.2m a lo largo de una mesa horizontal sin fricción por una fuerza constante de 16.0 N di

Tanzania [10]

Answer:

Explanation:

(a) The applied force has two components Fx and Fy. The Fx component is the only one that does work

$W_{x}=F_{x}x=(16N)cos(25)(2.2m)=31.9J$

(b) There in no net force in the vertical component

$F_{N}-F_{y}-F_{g}=0\\F_{N}=F_{y}+F_{g}=(16N)sin(25)+(2.5kg)(9.8\frac{m}{s^{2}})=31.26N$

(c)

$F_{g}=Mg=(2.5kg)(9.8\frac{m}{s^{2}})=24.5N$

(d)

$F_{T}=F_{x}=(16N)cos(25)=14.5N$

I attached an scheme of the force diagram

8 0

4 years ago

An object of mass 8kg is attached to massless string of length 2m and swum with a tangential velocity of 3 what is the tension o

Paul [167]

Answer:

36 N

Explanation:

If the object of mass, m = 8 kg is swung in a horizontal circle of radius, r = 2m = length of string with tangential velocity v = 3 m/s, the tension in the string is the centripetal force which is T = mv²/r

= 8 kg × (3 m/s)²/2 m

= 4 kg × 9 m/s²

= 36 N

6 0

3 years ago

Give a real-world example of how energy is transformed from electrical energy to thermal energy. Describe how the heat can be tr

timofeeve [1]

Answer:

A microwave

Explanation:

Car

Lightbulb is a really good one

Same with the sun. That one has Chemical as well

3 0

2 years ago

What is the gravitational force between a 45 kg person, and the Earth at 5.98 x 1024 kg, with a distance of

Assoli18 [71]

Answer:

C. 441 N

Explanation:

Gravitational force between two objects can by calculated by the formula

= G m₁m₂ / r² , m₁ and m₂ are masses at distance r

= ( 6.67 x 10⁻¹¹ x 45 x 5.98 x 10²⁴) / ( 6.38 x 10⁶ )²

= 44.09 x 10

= 440.9 N

= 441 N .

7 0

3 years ago

The components of vector A are:

Korvikt [17]

here as it is given that x component of the vector is positive while y component of the vector is negative so we can say the vector must inclined in Fourth quadrant.

So angle must be more than 270 degree and less than 360 degree

Now in order to find the value we can say that

$tan\theta = \frac{opposite\: side}{adjacent\: side}$