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3 years ago

Which car has the greatest potential energy? What type of potential energy does the

Physics

1 answer:

REY [17]3 years ago

8 0

Explanation:

*car A has the greatest PE

*gravitational potential energy

*mass and height

*car D has the least PE

You might be interested in

If a battery causes a wire to carry a current of 4 Amps how many coulombs of charge flow past any point in the wire in 3 seconds

BabaBlast [244]

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According to above question ~

Current (I) = 4 Amperes

Time (t) = 3 seconds

Charge (q) = ?

Let's find the charge (q) by using formula ~

$I = \dfrac{q}{t}$

$4 = \dfrac{q}{3}$

$q = 4 \times 3$

$q = 12 \: \: coulombs$

Hence, 12 coulombs of charge flow past any point in the wire in 3 seconds

7 0

3 years ago

1. If 800 C of electric charge passed through a light bulb in 4 min; what is the mag

9966 [12]

The electric current passing through the bulb would be 3.3A

<u>Explanation:</u>

Given:

Electric charge, q = 800C

Time, t = 4 min

= 4 X 60 sec

= 240 sec

Electric current, I = ?

We know,

$Current = \frac{Charge}{Time}$

On substituting the value we get:

$I = \frac{800}{4 X 60} \\\\I = 3.3A$

Thus, the electric current passing through the bulb would be 3.3A

5 0

3 years ago

A 34.0 kg sled is accelerated by a constant force of 45.0 N from 2.00 m/s to 16.00 m/s. For how long was the force acting on the

goblinko [34]

Answer: 10.6 sec

Explanation:

Because I got it right on my quiz :D

also because you can use the impulse momentum formula, Ft=m(triangle)v

so basically u do 16m/s-2m/s=14m/s and thats your triangle v (change in velocity) then multiply 14 times the mass, which is 34. Thats 476, so now you have ft=476, and you know F, the force, so all you have to do is divide 476 by 45, and you get like 10.5777777 which rounds up to 10.6!

4 0

3 years ago

The maximum allowable resistance for an underwater cable is one hundredth of an ohm per

Studentka2010 [4]

Answer:

A = 1.54 x 10⁻⁵ m² = 15.4 mm²

Explanation:

The resistance of a wire can be given by the following formula:

$R = \frac{\rho L}{A}\\\\A = \frac{\rho L}{R} = \frac{\rho}{\frac{R}{L} }$

where,

A = smallest cross-sectional area = ?

ρ = resistivity of copper = 1.54 x 10⁻⁸ Ωm

$\frac{R}{L}$ = resistance per unit length of wire = 0.001 Ω/m

Therefore,

$A = \frac{1.54\ x\ 10^{-8}\ \Omega m}{0.001\ \Omega/m}$

6 0

3 years ago

Water, with a density of 1000 kg/m3, flows out of a spigot, through a hose, and out a nozzle into the air. The hose has an inner

stepan [7]

Answer:

P₁ = 2.3506 10⁵ Pa

Explanation:

For this exercise we use Bernoulli's equation and continuity, where point 1 is in the hose and point 2 in the nozzle

P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂

A₁ v₁ = A₂ v₂

Let's look for the areas

r₁ = d₁ / 2 = 2.25 / 2 = 1,125 cm

r₂ = d₂ / 2 = 0.2 / 2 = 0.100 cm

A₁ = π r₁²

A₁ = π 1.125²

A₁ = 3,976 cm²

A₂ = π r₂²

A₂ = π 0.1²

A₂ = 0.0452 cm²

Now with the continuity equation we can look for the speed of water inside the hose

v₁ = v₂ A₂ / A₁

v₁ = 11.2 0.0452 / 3.976

v₁ = 0.1273 m / s

Now we can use Bernoulli's equation, pa pressure at the nozzle is the air pressure (P₂ = Patm) the hose must be on the floor so the height is zero (y₁ = 0)

P₁ + ½ ρ v₁² = Patm + ½ ρ v₂² + ρ g y₂

P₁ = Patm + ½ ρ (v₂² - v₁²) + ρ g y₂

Let's calculate

P₁ = 1.013 10⁵ + ½ 1000 (11.2² - 0.1273²) + 1000 9.8 7.25

P₁ = 1.013 10⁵ + 6.271 10⁴ + 7.105 10⁴

P₁ = 2.3506 10⁵ Pa

7 0

3 years ago