PLEASE HELP !!!!!!!!!!

A mass of 4.8 kg is dropped from a height of 4.84 meters above a vertical spring anchored at its lower end to the floor. If the

krek1111 [17]

Answer:

45.6 cm

Explanation:

Let x (m) be the length that the spring is compressed. We know that when we drop the mass from 4.84 m above and compress the springi, ts gravitational energy shall be converted to spring potential energy due to the law of energy conservation

$E_g = E_p$

$mgh = 0.5kx^2$

where h = 4.84 + x is the distance from the dropping point the the compressed point, and k = 24N/cm = 2400N/m is the spring constant, g = 9.81 m/s2 is the gravitational acceleration constant. And m = 4.8 kg is the object mass.

$4.8*9.81(4.84 + x) = 0.5 * 2400 * x^2$

$47.088x + 227.906 = 1200x^2$

$1200x^2 - 47.088x - 227.906 = 0$

$x = 0.456m$ or 45.6 cm

5 0

3 years ago

A mass of 80 g of KNO3 is dissolved in 100 g of water at 50 ºC. The solution is heated to 70ºC. How many more grams of potassium

slava [35]

Hi um i think its At 30 ° C your welcome

7 0

3 years ago

At a given point on a horizontal streamline in flowing air, the static pressure is â2.0 psi (i.e., a vacuum) and the velocity is

Nastasia [14]

At a point on the streamline, Bernoulli's equation is
p/ρ + v²/(2g) = constant
where
p = pressure
v = velocity
ρ = density of air, 0.075 lb/ft³ (standard conditions)
g = 32 ft/s²

Point 1:
p₁ = 2.0 lb/in² = 2*144 = 288 lb/ft²
v₁ = 150 ft/s

Point 2 (stagnation):
At the stagnation point, the velocity is zero.

The density remains constant.
Let p₂ = pressure at the stagnation point.
Then,
p₂ = ρ(p₁/ρ + v₁²/(2g))
p₂ = (288 lb/ft²) + [(0.075 lb/ft³)*(150 ft/s)²]/[2*(32 ft/s²)
= 314.37 lb/ft²
= 314.37/144 = 2.18 lb/in²

Answer: 2.2 psi

5 0

3 years ago

On a safari, a team of naturalists sets out toward a research station located 4.63 km away in a direction 38.7 ° north of east.

Talja [164]

This is a vector subraction problem.

5 0

3 years ago

Read 2 more answers

8. A 40.0kg block of metal is suspended from a scale and is immersed in water. The dimensions of the block are 12.0cm x 10.0cm x

Alja [10]

Answer:

a.1017.9N

b.1029.7N

c.T=380.6N

d.11.8 N

Explanation:

(a) The absolute pressure at the level of the top of the block is

Po=atmospheric pressure

g=gravity

h1=height

rh0=density of water

P1=Po+rho*g*h1

1.013*10^5+1000(9.81)(0.05)

1.0179*10^5Pa

at the level of the bottom of the block we have

P2=Po+rho*g*h2

1.013*10^5+1000(9.81)(0.17)

1.0297* 10^5 Pa

the downward force exerted on the top by the water is

Ftop=P*A== × = 1.0179* 10^5* 0.100

= 1017.9 N

and the upward force the water exerts on the bottom of the block , which is the buoyant force

Fbot== × = 1.0297 10^5 * 0.100 m

= 1029.7 N

(b) The scale reading is the tension, T, in the cord supporting the block.

if the block is at equilibrium, then sum of vertical forces

EFy=T+Fbot-Ftop-mg=0

T=mg+Ftop-Fbot

T=40*9.81-(1029.7-1017.9)

T=380.6N

(c) Archimedes principle state that, the buoyant force on the block equals the weight of the displaced water. Thus,

Buoyant force=rho *g*h

= = 1000* 0.100^2 * 0.120 m *9.80 m s=11.8 N

from the answer a Ftop-Fbot

1029.7-1017.9=11.8N, the same as the buoyant force

5 0

3 years ago