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bezimeni [28]
3 years ago
6

A freight car moves along a frictionless level railroad track at constant speed. The freight car is open on the top. A large loa

d of sand is suddenly dumped into the freight car. What happens to the speed of the freight car?
A) The speed of the freight car remains the same.
B) The speed of the freight car increases.
C) The speed of the freight car decreases.
D) The speed of the freight car cannot be determined from the information given.
Physics
1 answer:
MakcuM [25]3 years ago
7 0

Answer:

C) The speed of the freight car decreases.

Explanation:

The speed of the freight car decreases, because the sand load increases the mass of the system.

Using the conservation of linear momentum

m_{1}v_{1}=m_{2}v_{2}

The linear momentum m*v is an amount that must be preserved, so this equality must be satisfied before and after the sand is added to the freight car.

Subindices 1 are for before the sand is added, and subindices 2 for after the sand is added.

In this case the mass in the second moment m_{2} is greater due to the mass added by the sand than the original mass m_{1}, so to maintain equality the speed in this situation must decrease.

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In the process of changing a flat tire, a motorist uses ahydraulic jack. She begins by applying a force of 45 N to the inputpist
Verdich [7]

Answer:

3100.05 N

Explanation:

F_1 = Force on piston = 45 N

r_1 = Radius of piston

r_2 = Radius of plunger

\dfrac{r_2}{r_1}=8.3

From Pascal's law we have

\dfrac{F_1}{A_1}=\dfrac{F_2}{A_2}\\\Rightarrow F_2=\dfrac{F_1A_2}{A_1}\\\Rightarrow F_2=\dfrac{F_1\pi r_2^2}{\pi r_1^2}\\\Rightarrow F_2=\dfrac{F_1r_2^2}{r_1^2}\\\Rightarrow F_2=F_1\dfrac{r_2^2}{r_1^2}\\\Rightarrow F_2=45\times 8.3^2\\\Rightarrow F_2=3100.05\ N

The force is 3100.05 N

3 0
3 years ago
If a true heading of 135° results in a ground track of 130° and a true airspeed of 135 knots results in a groundspeed of 140 kno
vladimir2022 [97]

Answer:

245.1° and 13 knots

Explanation:

The given parameters are;

The true heading = 135°

The resultant ground track = 130°

The true airspeed = 135 knots

The ground speed = 140 knots

Given that the true airspeed the ground speed and the wind direction and magnitude form a triangle, we have;

From cosine rule, we have;

a² = b² + c² - 2×b×c×cos(A)

Where

a = The magnitude of the wind speed in knot

b = The true airspeed = 135 knots

c = The ground speed = 140 knots

A = The angle in between the true heading and the resultant ground track heading = 5°

Which gives;

a² = 135² + 140² - 2×135×140×cos(5 degrees) = 168.84 knots²

a = √168.84 = 12.9934 ≈ 13 knots

We have;

135 × sin(135 degrees) - 140× sin(130 degrees) = -11.7868

135 × cos(135 degrees) - 140× cos(130 degrees) = -5.469

Tan(θ) = -11.8/-5.5 = 2.155

θ = tan⁻¹(2.155) = 65.108°

Given that the wind is moving in opposite direction (slowing down the airplane, we add 180°, to get

Therefore, the angle direction = 180 + 65.108 = 245.1

Therefore, we have;

245.1° and 13 knots

3 0
3 years ago
If an airplane is flying at 300 km/h to the east and is facing a headwind of 18.0 km/h, what is the airplane's final velocity?
givi [52]
If an airplane is flying at 300 km/h to the east and is facing a headwind of 18.0 km/h, the final velocity can be calculated using simple vector addition. In this case, the planes velocity is positive (+330 km/h) and head wind has a negative component (-18.0 km/h). Vector addition yields +330 km / h + (-18.0 km /h) = 312 km / h. 
8 0
3 years ago
Which four equations can be used to solve for acceleration
emmainna [20.7K]

The four equations for acceleration are obtained from the three equations of motion and from second law of motion.

Explanation:

Acceleration is defined as the rate of change of velocity with respect to time. So the change in velocity with respect to time can be determined using the three equations of motions.

So from the first equation of motion, v = u + at , we can determine the value of acceleration if time taken, final and initial velocity is known. The equation can be re-written as a = \frac{v-u}{t}

Similarly, from the second equation of motion, s = ut + 1/2 at², we can determine the equation for acceleration as a = 2*\frac{s-ut}{t^{2} }

So this is second equation for acceleration.

Then from the third equation of motion, v^{2}- u^{2} = 2* a *s

the acceleration equation is determined as a = \frac{v^{2}-u^{2}  }{2s}

In addition to these three equation, another equation is present to determine the acceleration with respect to force from the Newton's second law of motion. F = Mass × acceleration. From this, acceleration = Force/mass.

So, these are the four equations for acceleration.

8 0
3 years ago
Which is a diatomic molecule? <br> A. Ar <br> B. CO <br> C. CO2 <br> D. NaCl
snow_tiger [21]

Answer: B. CO

Explanation:

Diatomic molecules are those that are formed by two atoms of the same chemical element (homonuclear diatomic molecule) or different chemical element (heteronuclear diatomic molecule).  

In this sense, oxygen is a homonuclear diatomic molecule because it is formed by two atoms of the same element (O_{2}) and Carbon monoxide (CO) is heteronuclear diatomic molecule.

Sodium Chloride NaCl is not a diatomic molecule because is a product of ionization, but it can be diatomic in its gas phase with a polar covalent bond.

3 0
3 years ago
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