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Otrada [13]
3 years ago
10

How could two different buses have two different momentum

Physics
1 answer:
REY [17]3 years ago
6 0

Answer:

if the busses mass and velocity are different their momentums will be different

momentum =mass ×velocity

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sonic is sliding down a frictionless 15m tall hill. He starts at the top with a velocity of 10m/s. At the bottom of the hill he
podryga [215]

Answer:

The maximum speed of sonic at the bottom of the hill is equal to 19.85m/s and the spring constant of the spring is equal to (497.4xmass of sonic) N/m

Energy approach has been used to sole the problem.

The points of interest for the analysis of the problem are point 1 the top of the hill and point 2 the bottom of the hill just before hitting the spring

The maximum velocity of sonic is independent of the his mass or the geometry. It is only depends on the vertical distance involved

Explanation:

The step by step solution to the problem can be found in the attachment below. The principle of energy conservation has been applied to solve the problem. This means that if energy disappears in one form it will appear in another.

As in this problem, the potential and kinetic energy at the top of the hill were converted to only kinetic energy at the bottom of the hill. This kinetic energy too got converted into elastic potential energy .

x = compression of the spring = 0.89

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3 years ago
What option lists the types of electromagnetic waves in order of increasing wavelength?
WITCHER [35]

Answer:

I believe its D visible light, infrared, ultraviolet

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3 years ago
8. A 30-kg box is sliding down a frictionless plane that is sloped at 24º. Assuming the object starts at rest,
Katena32 [7]

The net force on the box parallel to the plane is

∑ F[para] = mg sin(24°) = ma

where mg is the weight of the box, so mg sin(24°) is the magnitude of the component of its weight acting parallel to the surface, and a is the box's acceleration.

Solve for a :

g sin(24°) = a ≈ 3.99 m/s²

The box starts at rest, so after 7.0 s it attains a speed of

(3.99 m/s²) (7.0 s) ≈ 28 m/s

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2 years ago
A ladder 10 ft long rests against a vertical wall. If the bottom of the ladder slides away from the wall at a rate of 1.1 ft/s,
IRINA_888 [86]

Explanation:

If the distance between the bottom of the ladder and the wall is x, then:

cos θ = x / 10

Taking derivative with respect to time:

-sin θ dθ/dt = 1/10 dx/dt

Substituting for θ:

-sin (acos(x / 10)) dθ/dt = 1/10 dx/dt

Given that x = 6 and dx/dt = 1.1:

-sin (acos(6/10)) dθ/dt = 1/10 (1.1)

-0.8 dθ/dt = 0.11

dθ/dt = -0.1375

The angle is decreasing at 0.1375 rad/s.

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