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MaRussiya [10]
3 years ago
6

If the price of gasoline at a particular station in Europe is 5 euros per liter. An American student in Europe is allowed to use

40 euros to buy gasoline. Knowing that 4 quarts make a gallon, and that 1 liter is about 1 US liquid quart, about how many gallons can the student buy?
Physics
1 answer:
Mashcka [7]3 years ago
5 0

Answer:

Number of gallons =2 gallon

Explanation:

given data:

rate of gasoline ineurope = 5 euro per liter

total money to buy gasoline =  40 euro

total gasoline an american can buy in europe = \frac{40}{5}

= 8 litres of gasoline

As given in the question 1 ltr is 1 quarts therefore  

Total no. of quarts is 8 quarts

As from question 4 quarts is equal to one gallon, hence

Number of gallons= \frac{8}{4} = 2 gallon

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A satellite of mass m is in a circular orbit of radius R2 around a spherical planet of radius R1 made of a material with density
Anit [1.1K]

Answer: The gravitational force Fg exerted on the orbit by the planet is Fg = G 4/3πr3rhom/ (R1 + d+ R2)^2

Explanation:

Gravitational Force Fg = GMm/r2----1

Where G is gravitational constant

M Mass of the planet, m mass of the orbit and r is the distance between the masses.

Since the circular orbit move around the planet, it means they do not touch each other.

The distance between two points on the circumference of the two massesb is given by d, while the distance from the radius of each mass to the circumferences are R1 and R2 from the question.

Total distance r= (R1 + d + R2)^2---2

Recall, density rho =

Mass M/Volume V

Hence, mass of planet = rho × V

But volume of a sphere is 4/3πr3

Therefore,

Mass M of planet = rho × 4/3πr3

=4/3πr3rho in kg

From equation 1 and 2

Fg = G 4/3πr3rhom/ (R1 + d+ R2)^2

6 0
3 years ago
The objects in the inner part of our solar system are composed mostly of
N76 [4]

“Inner Planets” – Mercury, Venus, Earth, and Mars – which are so named because they orbit closest to the Sun. ... For starters, the inner planets are rocky and terrestrial, composed mostly of silicates and metals, whereas the outer planets are gas giants.

4 0
3 years ago
You find a piece of iron with a density of 10.4g/cm3. Which layer of the earth might it come from
topjm [15]

Answer:

the inner core if I believe. I apologize if I am incorrect

3 0
3 years ago
A light platform is suspended from the ceiling by a spring. A student with a mass of 90 kg climbs onto the platform. When it sto
Ilya [14]
Refer to the diagram shown.

When the student climbs onto the platform, the spring stretches by 0.82 m to reach the equilibrium position.
The mass of the student is m = 90 kg, so his weight is
mg = (90 kg)*(9.8 m/s²) = 882 N

By definition, the spring constant is
k = (882 N)/(0.82 m) = 1075.6 N/m

When the spring is stretched by x from the equilibrium position, the restoring force is
F = - k*x.

If damping is ignored, the equation of motion is
F = m * acceleration
or
m \frac{d^{2}x}{dt^{2}} = -kx \\ \frac{d^{2}x}{dt^{2}} + \frac{k}{m} x = 0

Define ω² = k/m = 11.751 => ω = 3.457.
Then the solution of the ODE is
x(t) = c₁ cos(ωt) + c₂ sin(ωt)

x'(t) = -c₁ω sin(ωwt) + c₂ω cos(ωt)
When t=0, x' =0, therefore c₂ = 0

The solution is of the form
x(t) = c₁ cos(ωt)
When t = 0, x = 0.32 m. Therefore c₁ = 0.32

The motion is
x(t) = 0.32 cos(3.457t)
The single amplitude is 0.32 m, and the double amplitude is 0.64 m.

Answer: 
0.32 m (single amplitude), or
0.64 m (double amplitude)

6 0
3 years ago
A(n) 1.3 kg mass sliding on a frictionless surface has a velocity of 7.1 m/s east when it undergoes a one-dimensional elastic co
Oxana [17]

Answer: 2.12 kg

Explanation:

Since the 1.3 kg object moves to the west after the collision, the other object will move to the east after the collision.

In an elastic collision, the relative velocity after the collision is the opposite of the relative velocity before the collision. Since the 1.3 kg object’s velocity before the collision is 6.7 m/s greater than the other object, after the collision, its velocity will be 6.7 m/s less than the other object. To determine the other object’s velocity, use the following equation.

v = 1.7 – 7.1 = -5.4 m/s

The negative sign means it is moving eastward. Let’s use this number is a momentum equation to determine its mass.

Initial momentum = 1.3 * 7.1 = 9.23 east

For the 1.3 object, final momentum = 1.3 * 1.7 = 2.21 west

To determine the final momentum of the other object, add these two numbers.

Final momentum = 11.44 east

To determine its mass, use the following equation.

m * 5.4 = 11.44

m = 11.44 ÷ 5.4 = 2.12 kg

To make sure that kinetic energy is conserved, let’s round this number to 2 kg and determine the final kinetic energies.

For the 1.3 object, KE = 1/2 * 1/3* 1.7^2 = 0.48

For the 2 kg object, KE = 1/2* 2 * 5.4^2 = 29.64

Total final KE = 29.64

Initial KE = 0.5* 1.3 * 7.1^2 = 32.77

Since I rounded the mass up to 2kg, this proves that kinetic energy is conserved and the mass is correct!

3 0
3 years ago
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