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artcher [175]
3 years ago
8

Imagine that a tank is filled with water. The height of the liquid column is 7 meters and the area is 1.5 square meters (m2). Wh

at's the force of gravity acting on the column of water?
Physics
2 answers:
Korolek [52]3 years ago
4 0

Answer: 1.03 × 10⁵ N

Explanation:

Force of gravity due to liquid column in a tank is given by the product of pressure due to the liquid column and area on which this pressure is acting.

Force, F = P × A

Pressure, P = ρ g h

where, ρ is the density of the liquid

g is the acceleration due to gravity

h is the height of the liquid column

It is given that, h = 7 m

Area, A = 1.5 m²

g = 9.81 m/s²

density of water, ρ = 1000 kg/m³

⇒ F =  ρ g h × A = 1000 kg/m³ × 9.81 m/s² × 7 m × 1.5 m² = 103005 N

Hence, force of gravity acting on the column of water is 1.03 × 10⁵ N

Marta_Voda [28]3 years ago
3 0
The answer to your question is 102,900N
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A projectile is launched from the ground at an angle of 60o above the horizontal. At what point in its trajectory does it have t
Elodia [21]

Answer:

It's constant everywhere in its trajectory.

Explanation:

the projectile was launched with an initial velocity, the only acceleration that is affecting the projectile's velocity is gravity.

The acceleration of gravity is practically equal everywhere on earth, so during its trajectory, we have to take into consideration only the acceleration because of gravity.

This is only correct because the projectile was launched with an initial velocity and it's not accelerating from rest and then falls.

5 0
3 years ago
A block of mass 12.2 kg is sliding at an initial velocity of 3.9 m/s in the positive x-direction. The surface has a coefficient
Studentka2010 [4]

Answer:

Explanation:

a) Force of friction = μ R where μ is coefficient of kinetic friction and R is reaction force

R = mg where m is mass of the block

Force of friction F = μ x mg

= .173 x 12.2 x 9.8

= 20.68 N

b ) Only force of friction is acting on the body so

deceleration = force / mass = 20.68 / 12.2 = 1.7 m /s²

acceleration = - 1.7 m /s²

c )

v² = u² - 2 a s

v = 0 , u = 3.9 m /s

a = 1.7 m /s

0 = 3.9² - 2 x 1.7 x s

s = 4.47  m

5 0
3 years ago
A sample of helium behaves as an ideal gas as energy is
alisha [4.7K]

Answer:

0.0321 g

Explanation:

Let helium specific heat c_h = 5.193 J/g K

Assuming no energy is lost in the process, by the law of energy conservation we can state that the 20J work done is from the heat transfer to heat it up from 273K to 393K, which is a difference of ΔT = 393 - 273 = 120 K. We have the following heat transfer equation:

E_h = m_hc_h \Delta T = 20 J

where m_h is the mass of helium, which we are looking for:

m_h = \frac{20}{c_h \Delta T} = \frac{20}{5.193 * 120} \approx 0.0321 g

4 0
3 years ago
A dog of mass 4 kg runs up a hill of height 8 m. How much gravitational potential energy does the dog gain?
Genrish500 [490]
A. 314 because when you use the formula for the GPE ; GPE=MGH or means mass times gravity time height (4x8x9.8) and thats equivalent to 313.6 which rounds up to 314. Hope it helps 
8 0
3 years ago
Read 2 more answers
In the diagram, q1 = -6.39*10^-9 C and q2 = +3.22*10^-9 C. What is the electric field at point P? pls help
Alexxx [7]

Answer:

Below

Explanation:

First draw the vectors that represent both electric fields.

E1 is the elictric field created by q1, E2 is the one created by q2.

● q1 is negative so E1 will point from P.

● q2 is positive so E2 will point out of P

(Picture below)

■■■■■■■■■■■■■■■■■■■■■■■■■■

The resulting electric field is equal to the sum of the two fields since both vectors are colinear.

Let E be the total field.

● E = E1 + E2

The formula of the electric field intensity is:

● E = K ×(q/d^2)

-K is Coulomb's constant

-d is the distance between the charge and the object ( here P)

-q is the charge

■■■■■■■■■■■■■■■■■■■■■■■■■■

● E1 = K × (q1/d1^2)

The distance between q1 and P is the qum of 0.15 m 0.25 m. (0.4 m)

Coulombs constant is 9×10^9 m^2/C^2

● E1 = 9×10^9 ×[-6.39 × 10^(-9)/ 0.4^2]

● E1 = -359.43 N/C

■■■■■■■■■■■■■■■■■■■■■■■■■■

● E2 = K ×(q2/d^2)

The distance between q2 and P is 0.25 m.

● E2 = 9×10^9×[3.22×10^(-9) /0.25^2]

● E2 = 463.68 N/C

■■■■■■■■■■■■■■■■■■■■■■■■■■

● E = E1 + E2

● E = -359.43+463.68

● E = 105.25 N/C

4 0
3 years ago
Read 2 more answers
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