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Indigenous people sometimes cook in watertight baskets by placing hot rocks into water to bring it to a boil. What mass of 500ºC

scoray [572]

Answer:

m = 4.65 kg

Explanation:

As we know that the mass of the water that evaporated out is given as

$m = 0.0250 kg$

so the energy released in form of vapor is given as

$Q = mL$

$Q = (0.0250)(2.25 \times 10^6)$

$Q = 56511 J$

now the heat required by remaining water to bring it from 15 degree to 100 degree

$Q_2 = ms\Delta T$

$Q_2 = (4 - 0.025)(4186)(100 - 15)$

$Q_2 = 1.41\times 10^6J$

total heat required for above conversion

$Q = 56511 + 1.41 \times 10^6 = 1.47 \times 10^6 J$

now by heat energy balance

heat given by granite = heat absorbed by water

$m(790)(500 - 100) = 1.47 \times 10^6$

$m = 4.65 kg$

4 0

4 years ago

a 55.0 kg cheerleader uses an oil-filled hydraulic lift to hold four 130 kg football players at a height of 0.800 m . the diamet

suter [353]

The diameter of the football player's piston is 0.55 m

Given that the mass of the cheerleader(m) is 55 kg, mass of football player to be hold (M) is 130 kg, height of the players (h) is 1.30 m, radius of the piston corresponding to the diameter (r) is 0.09 m, Diameter of football player's piston (R), P1 is Pressure on the cheerleader's side, P2 is Pressure on the football player's side

Using Pascal's law,

This law states that if there is a change at a point of a body immersed in a fluid then that change will spread thoroughly to each and every point of the body.

The formula of hydraulic system is,

P1= P2

F1/A1 = F2/A2

mg/πr^2 = 4Mg/πR^2

m/r^2=4M/R^2

R^2=4M×r^2/m

By plugging the values, we get.,

R^2=(4×130×0.09^2)/55

R^2=4.21/55=0.076

R=√0.076 = 0.275 m

Hence, diameter of football player's piston is 0.55 m

Learn more about Hydraulic lift here

brainly.com/question/19052774

#SPJ4

8 0

1 year ago

A bicyclist passing through a city accelerates after he passes the sign post marking the city limits at x=0. His acceleration is

Anni [7]

Answer:

A. 24 m, 14 m/s

B. 8.0 m

Explanation:

Given:

x₀ = 6.0 m

v₀ = 4.0 m/s

a = 5.0 m/s²

t = 2.0 s

A. Find: x and v

x = x₀ + v₀ t + ½ at²

x = (6.0 m) + (4.0 m/s) (2.0 s) + ½ (5.0 m/s²) (2.0 m/s)²

x = 24 m

v = at + v₀

v = (5.0 m/s²) (2.0 s) + (4.0 m/s)

v = 14 m/s

B. Find x when v = 6.0 m/s.

v² = v₀² + 2a (x − x₀)

(6.0 m/s)² = (4.0 m/s)² + 2 (5.0 m/s²) (x − 6.0 m)

x = 8.0 m

8 0

3 years ago

A 2-kg object is moving at 3 m/s. A 4-N force is applied in the direction of motion and if the object has a final velocity 7m/s.

aniked [119]

Intial velocity u=3m/s
final velocity v
2
=u
2
+2as=3
2
+(2×2×5)=29 ⟹v=5.3m/s

KE=
2
1

m(v
2
−u
2
)=
2
1

×2×((5.3)
2
−3
2
)=20J

7 0

3 years ago

A warehouse worker is pushing a 90.0-kg crate with a horizontal force of 282 N at a speed of v = 0.850 m/s across the warehouse

Elanso [62]

Answer:

$v_{f} = 0.51 \frac{m}{s}$

Explanation:

We apply Newton's second law at the crate :

∑F = m*a (Formula 1)

∑F : algebraic sum of the forces in Newton (N)

m : mass in kilograms (kg)

a : acceleration in meters over second square (m/s²)

Data:

m=90kg : crate mass

F= 282 N

μk =0.351 :coefficient of kinetic friction

g = 9.8 m/s² : acceleration due to gravity

Crate weight (W)

W= m*g

W= 90kg*9.8 m/s²

W= 882 N

Friction force : Ff

Ff= μk*N Formula (2)

μk: coefficient of kinetic friction

N : Normal force (N)

Problem development

We apply the formula (1)

∑Fy = m*ay , ay=0

N-W = 0

N = W

N = 882 N

We replace the data in the formula (2)

Ff= μk*N = 0.351* 882 N

Ff= 309.58 N

We apply the formula (1) in x direction:

∑Fx = m*ax , ax=0

282 N - 309.58 N = 90*a

a= (282 N - 309.58 N ) / (90)

a= - 0.306 m/s²

Kinematics of the crate

Because the crate moves with uniformly accelerated movement we apply the following formula :

vf²=v₀²+2*a*d Formula (3)

Where:

d:displacement in meters (m)

v₀: initial speed in m/s

vf: final speed in m/s

a: acceleration in m/s²

Data

v₀ = 0.850 m/s

d = 0.75 m

a= - 0.306 m/s²

We replace the data in the formula (3)

vf²=(0.850)²+(2)( - 0.306 )(0.75 )

$v_{f} = \sqrt{(0.850)^{2} +(2)( - 0.306 )(0.75 )}$

$v_{f} = 0.51 \frac{m}{s}$

8 0

3 years ago