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3 years ago

A force or an event that changes the earth's surface is known as

Physics

1 answer:

Natasha2012 [34]3 years ago

3 0

The answer is Geological Process

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2) Um gás ideal sofre uma determinada transformação, conforme mostra o gráfico abaixo. Considere

Roman55 [17]

Answer:

yes

Explanation:

5 0

3 years ago

Spiderman, whose mass is 70.0 kg, is dangling on the free end of a 12.2-m-long rope, the other end of which is fixed to a tree l

Lana71 [14]

Answer:

U = -3978.8 J

Explanation:

The work of the gravitational force U just depends of the heigth and is calculated as:

U = -mgh

Where m is the mass, g is the gravitational acceleration and h the alture.

for calculate the alture we will use the following equation:

h = L-Lcos(θ)

Where L is the large of the rope and θ is the angle.

Replacing data:

h = 12.2-12.2cos(58.4)

h = 5.8 m

Finally U is equal to:

U = -70(9.8)(5.8)

U = -3,978.8 J

5 0

3 years ago

What is the density of the block if it has a mass of 45.5 grams and a volume of 2.4 cm³?

rjkz [21]

Explanation:

density= mass/volume

therefore;45.5/2.4

=19.0 g/cm³

5 0

2 years ago

A huge tank of glycerine with a density of 1.260 g/cm3 is vertically stationed on a platform which is 15 m above the ground. The

EleoNora [17]

Answer:

The tank is losing $4.976*10^{-4} m^3/s$

$v_g = 19.81 \ m/s$

Explanation:

According to the Bernoulli’s equation:

$P_1 + 1 \frac{1}{2} \rho v_1^2 + \rho gh_1 = P_2 + \frac{1}{2} \rho v_2^2 + \rho gh_2$

We are being informed that both the tank and the hole is being exposed to air :

∴ P₁ = P₂

Also as the tank is voluminous ; we take the initial volume $v_1$ ≅ 0 ;

then $v_2$ can be determined as: $\sqrt{[2g (h_1- h_2)]$

h₁ = 5 + 15 = 20 m;

h₂ = 15 m

$v_2 = \sqrt{[2*9.81*(20 - 15)]$

$v_2 = \sqrt{[2*9.81*(5)]$

$v_2= 9.9 \ m/s$ as it leaves the hole at the base.

radius r = d/2 = 4/2 = 2.0 mm

(a) From the law of continuity; its equation can be expressed as:

J = $A_1v_2$

J = πr² $v_2$

J = $\pi *(2*10^{-3})^{2}*9.9$

J = $1.244*10^{-4} m^3/s$

How fast is the water from the hole moving just as it reaches the ground?

In order to determine that; we use the relation of the velocity from the equation of motion which says:

v² = u² + 2gh ₂

v² = 9.9² + 2×9.81×15

v² = 392.31

The velocity of how fast the water from the hole is moving just as it reaches the ground is : $v_g = \sqrt{392.31}$

$v_g = 19.81 \ m/s$

4 0

3 years ago

Can someone explain how the weight of the block is 10.26N, with reference to an appropriate law of motion?

spayn [35]

this process is called parellelogram method of resolving vectors.

5 0

3 years ago