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Bumek [7]
3 years ago
13

In Europe, gasoline efficiency is measured in km/L. If your car's gas mileage is 36.0mi/gal , how many liters of gasoline would

you need to buy to complete a 142-km trip in Europe? Use the following conversions: 1km=0.6214mi and 1gal=3.78L.
Physics
1 answer:
Zigmanuir [339]3 years ago
3 0
36 \frac{mi}{gal} * \frac{1 km}{0.6214 mi}* \frac{1 gal}{3.78 L} = 15.3 \frac{km}{L}

142/15.3=9.27L
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What is the magnitude of the magnetic dipole moment of the bar magnet
Annette [7]

The magnitude of the magnetic dipole moment of the bar magnet is 1.2 Am²

<h3> Magnetic dipole moment of the bar magnet</h3>

The magnitude of the magnetic dipole moment of the bar magnet at distance from its axis is calculated as follows;

B = \frac{2\mu_0m}{4\pi r^3} \\\\m = \frac{4\pi r^3 B}{2\mu_0}

where;

  • B is magnetic field
  • m is dipole moment
  • μ is permeability of free space

m = (4π x 0.1³ x 2.4 x 10⁻⁴)/(2 x 4π x 10⁻⁷)

m = 1.2 Am²

The complete question is below:

What is the magnitude of the magnetic dipole moment of the bar magnet from 0.1 m of its axis and magnetic field strength of 2.4 x 10⁻⁴ T.

Learn more about dipole moment here: brainly.com/question/27590192

#SPJ11

6 0
1 year ago
21/33
postnew [5]

Answer:

400N

Explanation:

5 0
3 years ago
The lightest car in the world was built in London and had a mass of less than 10 kg. it's maximum speed was 25.0 km/h. Suppose t
Softa [21]
25km/h = 6.94 m/s
suvat
s=16
u=6.94
v=0
a=a
v^2=u^2+2as
(v^2-u^2)/2s = a =1.5ms^-2
6 0
3 years ago
A 5 kg ball moving to the right at a speed of 6 m/s strikes another 4 kg
Dahasolnce [82]

Answer:

The 5 kg ball moves 3.78 m/s to the left, and the 4 kg ball moves 7.22 m/s to the right.

Explanation:

Momentum before = momentum after

m₁ u₁ + m₂ u₂ = m₁ v₁ + m₂ v₂

(5 kg) (6 m/s) + (4 kg) (-5 m/s) = (5 kg) v₁ + (4 kg) v₂

10 m/s = 5 v₁ + 4 v₂

Assuming an elastic collision, kinetic energy is conserved.

½ m₁ u₁² + ½ m₂ u₂² = ½ m₁ v₁² + ½ m₂ v₂²

m₁ u₁² + m₂ u₂² = m₁ v₁² + m₂ v₂²

(5 kg) (6 m/s)² + (4 kg) (-5 m/s)² = (5 kg) v₁² + (4 kg) v₂²

280 m²/s² = 5 v₁² + 4 v₂²

Substituting:

v₂ = (10 − 5 v₁) / 4

280 = 5 v₁² + 4 [(10 − 5 v₁) / 4]²

280 = 5 v₁² + (10 − 5 v₁)² / 4

1120 = 20 v₁² + (10 − 5 v₁)²

1120 = 20 v₁² + 100 − 100 v₁ + 25 v₁²

0 = 45 v₁² − 100 v₁ − 1020

0 = 9 v₁² − 20 v₁ − 204

0 = (9 v₁ + 34) (v₁ − 6)

v₁ = -3.78 m/s or 6 m/s

u₁ = 6 m/s, so v₁ = -3.78 m/s.  Solving for v₂:

v₂ = (10 − 5 v₁) / 4

v₂ = 7.22 m/s

The 5 kg ball moves 3.78 m/s to the left, and the 4 kg ball moves 7.22 m/s to the right.

6 0
3 years ago
You toss a rock up vertically at an initial speed of 39 feet per second and release it at an initial height of 6 feet. The rock
3241004551 [841]

Answer:

2.583 s, 29.77 ft and 1.219 s

Explanation:

Using equation of motion and taken the motion upward as positive, also a = g ( acceleration due to gravity) = - 32 fts⁻², V= 39 fts⁻¹ V₁ is final velocity, y is the distance in ft from the ground

H = 6 ft, the height from which it is tossed

V₁ = V + gt = V - gt

at maximum height the body came to rest momentarily V₁ = 0

0 = V - gt

-V = -gt

- 39 / -32 = t

t time to reach maximum height = 1.219 s

To Maximum height reached can be calculated with the formula

V₁² = V² + 2g( y - H) where H is the initial height reached by the tossed rock

where V₁ is the final velocity at maximum height which = 0

0 = V² - 2g(y-H) where y is the distance traveled from the ground

-V² = -2g(y-H)

₋V² / -2g = y-H

(V²/2g) + H = y in ft

(39² / (2 × 32)) + 6

y = 29.77 ft

The total time it will be in air can be calculated with the formula below

y = H + Vt - 0.5gt² from y-H = ut + 0.5at²

0.5gt² - Vt - H = 0 since the body returned to the ground ( y = 0)

0.5gt² - Vt - H = 0

using quadratic formula

- (-V)² ± √ ((-V²) - 4 × 0.5g × -H) / (2 × 0.5 × g)

(V ± √ (V² + 2gH)) ÷ g

substitute the values into the expression

t = (39 + √(39² + (2×-32× 6)))/ 32 or (39 - √ (39² + (2 × -32×6))/ 32

t = (39 + √(1521 +384))/32 = (39 + √1905) / 32  = 2.583 s

t = (39 - √1905) / 32 =  -0.15 s

The will remain in air (V ± √ (V² + 2gH)) / g seconds. It will reach a maximum height of (V²/2g) + H feet after V/g seconds

8 0
2 years ago
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