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3 years ago

What would be the mass of an object that is moving at 5 m/s, with a total momentum of 1500 kg*m/s southeast?

Physics

1 answer:

USPshnik [31]3 years ago

7 0

Answer:

The mass of the object is 300 kg

Explanation:

The given parameters are;

The velocity of the object = 5 m/s southeast

The total momentum of the object = 1500 kg·m/s

The equation for linear momentum, is given as follows;

Momentum = Mass × Velocity

Therefore, the mass of the object can be found as follows;

$The \ mass \ of \ the \ object= \dfrac{The \ momentum \ of \ the \ object}{The \ velocity \ of \ the \ object}$

Therefore, we have;

$The \ mass \ of \ the \ object= \dfrac{1500 \ kg \cdot m/s}{5 \ m/s} = 300 \ kg$

The mass of the object = 300 kg.

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Which of the following distinguishes the isotope uranium-238 from the isotope uranium-235?

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Answer: A

Explanation:

Isotopes of different elements differ by the number of neutrons inside the nucleus.

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2 years ago

Read 2 more answers

Problem 1: Spherical mirrorConsider a spherical mirror of radius 2 m, and rays which go parallel to the optic axis. What is thep

SIZIF [17.4K]

Answer:

1) iii i= 1m, 2) iii and iv, 3) i = f₂ (L-f₁) / (L - (f₁ + f₂))

Explanation:

Problem 1

For this problem we use two equations the equations of the focal distance in mirrors

f = r / 2

f = 2/2

f = 1 m

The builder's equation

1 / f = 1 / o + 1 / i

Where f is the focal length, "o and i" are the distance to the object and the image respectively.

For a ray to arrive parallel to the surface it must come from infinity, whereby o = ∞ and 1 / o = 0

1 / f = 0 + 1 / i

i = f

i = 1 m

The image is formed at the focal point

The correct answer is iii

Problem 2

For this problem we have two possibilities the lens is convergent or divergent, in both cases the back face (R₂) must be flat

Case 1 Flat lens - convex (convergent)

R₂ = infinity

R₁ > 0

Cas2 Flat-concave (divergent) lens

R₂ = infinity

R₁ <0

Why the correct answers are iii and iv

Problem 3

For a thick lens the rays parallel to the first surface fall in their focal length (f₁), this is the exit point for the second surface whereby the distance to the object is o = L –f₁, let's apply the constructor equation to this second surface

1 / f₂ = 1 / (L-f₁) + 1 / i

1 / i = 1 / f₂ - 1 / (L-f₁)

1 / i = (L-f₁-f₂) / f₂ (L-f₁)

i = f₂ (L-f₁) / (L - (f₁ + f₂))

This is the image of the rays that enter parallel to the first surface

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3 years ago

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3 years ago

A 0.274 kg block is pushed up a frictionless ramp inclined at angle of 32° above the horizon. The push force is a constant 2.55

SpyIntel [72]

Answer:

The answer is 2,416 m/s. Let's jump in.

Explanation:

We do work with the amount of energy we can transfer to objects. According to energy theory:

W = ΔE

Also as we know W = F.x

We choose our reference point as a horizontal line at the block's rest point. At the rest, block doesn't have kinetic energy and since it is on the reference point(as we decided) it also has no potential energy.

Under the force block gains;

W = F.x → $W=2,55.0,71=1,8105\frac{N}{m}$