A bodybuilder lifts a 10 N weight a distance of 2.5 m. How much energy has the weight gained?

A split highway has a number of lanes for traffic. For traffic going in one direction, the radius for the inside of the curve is

Arte-miy333 [17]

Cuál es tu pregunta

7 0

1 year ago

Charge q1 is distance s from the negative plate of a parallel-plate capacitor. Charge q2=q1/3 is distance 2s from the negative p

Svetlanka [38]

Answer:

The ratio (U₁/U₂) = 6

Explanation:

U, the potential energy is given as

U = kqQ/r

k = Coulomb's constant

q = charge we're concerned about

Q = charge of the negative plate of the capacitor

r = distance of q from the negative plate of the capacitor.

For charge q₁

U₁ = kq₁Q/s

U₂ = kq₂Q/2s

But q₂ = q₁/3

U₂ becomes U₂ = kq₁Q/6s

U₁ = kq₁Q/s

U₂ = kq₁Q/6s

(U₁/U₂) = 6

5 0

2 years ago

Suppose that now you want to make a scale model of the solar system using the same ball bearing to represent the sun. How far fr

Dahasolnce [82]

Answer:

d = 0.645 m (assuming a radius of the ball bearing of 3 mm)

Explanation:

The given information is:

The distance from the center of the sun to the center of the earth is 1.496x10¹¹m = $d_{e}$
The radius of the sun is 6.96x10⁸m = $r_{s}$

We need to assume a radius for the ball bearing, so suppose that the radius is 3 mm = $r_{b}$ .

First, we need to find how many times the radius of the sun is bigger respect to the radius of the ball bearing, which is given by the following equation:

$\frac{r_{s}}{r_{b}} = \frac{6.96\cdot 10^{8}m}{3\cdot 10^{-3}m} = 2.32\cdot 10^{11}$

Now, we can calculate the distance from the center of the sun to the center of the sphere representing the earth, $d_{s}$ :

[tex] d_{s} = \frac{d_{e}}{r_{s}/r_{b}} = \frac{1.496 \cdot 10^{11} m}{2.32\cdot 10^{11}} = 0.645 m

I hope it helps you!

7 0

1 year ago

If the humidity in a room of volume 450 m3 at 30 ∘C is 75%, what mass of water can still evaporate from an open pan?

Yuliya22 [10]

From tables,

SVP at 30°C = 4.24 kPa

From ideal gas expressions;
n = PV/RT = (4.24*1000*450)/(8.314*303) = 757.4 moles

Now, 75% of 757.4 moles will evaporate leaving 20%. Then, 25% of 757.5 moles...
25% of 757.4 moles = 25/100*757.4 = 189.35 moles
Mass of 189.35 moles = 189.35 moles*18 g/mol = 3408.3 g ≈ 3.4 kg

5 0

2 years ago

A coat rack weighs 65.0 lbs when it is filled with winter coats and 40.0 lbs when it is empty. The base of the coat rack has an

Whitepunk [10]

Answer:

0.056 psi more pressure is exerted by filled coat rack than an empty coat rack.

Explanation:

First we find the pressure exerted by the rack without coat. So, for that purpose, we use formula:

P₁ = F/A

where,

P₁ = Pressure exerted by empty rack = ?

F = Force exerted by empty rack = Weight of Empty Rack = 40 lb

A = Base Area = 452.4 in²

Therefore,

P₁ = 40 lb/452.4 in²

P₁ = 0.088 psi

Now, we calculate the pressure exerted by the rack along with the coat.

P₂ = F/A

where,

P₂ = Pressure exerted by rack filled with coats= ?

F = Force exerted by filled rack = Weight of Filled Rack = 65 lb

A = Base Area = 452.4 in²

Therefore,

P₂ = 65 lb/452.4 in²

P₂ = 0.144 psi

Now, the difference between both pressures is:

ΔP = P₂ - P₁

ΔP = 0.144 psi - 0.088 psi

ΔP = 0.056 psi

8 0

2 years ago