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Mandarinka [93]
3 years ago
13

A 1500 kg car traveling east at 40 km/hr turns a corner and speeds up to a velocity of 50 km/hr due north. What is the change in

the car's momentum?
Physics
1 answer:
pantera1 [17]3 years ago
5 0

Answer:

96046  Ns.

Explanation:

We shall represent velocity in vector form considering east direction as + ve x axis and north as + y direction.

40 km/h in the east

V₁ = 40 i

V₂ = 50j

momentum p₁ = mV₁

= 1500 X 40 i

= 60000 i

Momentum p₂ = mV₂

= 1500 X 50j

= 75000 j

Change in momentum

p₂ - p₁

75000j - 60000i

Magnitude of change

= \sqrt{(750000)^2 +(60000)^2

= 96046  Ns.

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Explanation:

It is given that,

Mass of the ball, m = 0.06 kg

Initial speed of the ball, u = 50.4 m/s

Final speed of the ball, v = -37 m/s (As it returns)  

(a) Let J is the magnitude of the impulse delivered to the ball by the racket. It can be calculated as the change in momentum as :

J=m(v-u)

J=0.06\times (-37-(50.4))  

J = -5.24 kg-m/s

(b) Let W is the work done by the racket on the ball. It can be calculated as the change in kinetic energy of the object.

W=\dfrac{1}{2}m(v^2-u^2)

W=\dfrac{1}{2}\times 0.06\times ((-37)^2-(50.4)^2)

W = -35.1348 Joules

Hence, this is the required solution.

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Answer:

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Explanation:

4 0
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The diagram shows a stone suspended under the surface of a liquid from a string. The stone experiences a pressure caused by the
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An object in the shape of a thin ring has radius a and mass M. A uniform sphere with mass m and radius R is placed with its cent
madreJ [45]

Answer:

F = GMmx/[√(a² + x²)]³

Explanation:

The force dF on the mass element dm of the ring due to the sphere of mass, m at a distance L from the mass element is

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Since the ring is symmetrical, the vertical components of this force cancel out leaving the horizontal components to add.

So, the horizontal components add from two symmetrically opposite mass elements dM,

Thus, the horizontal component of the force is

dF' = dFcosФ where Ф is the angle between L and the x axis

dF' = GmdMcosФ/L²

L² = a² + x² where a = radius of ring and x = distance of axis of ring from sphere.

L = √(a² + x²)

cosФ = x/L

dF' = GmdMcosФ/L²

dF' = GmdMx/L³

dF' = GmdMx/[√(a² + x²)]³

Integrating both sides we have

∫dF' = ∫GmdMx/[√(a² + x²)]³

∫dF' = Gm∫dMx/[√(a² + x²)]³    ∫dM = M

F = GmMx/[√(a² + x²)]³  

F = GMmx/[√(a² + x²)]³

So, the force due to the sphere of mass m is

F = GMmx/[√(a² + x²)]³

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Eduardwww [97]

Answer:

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