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tresset_1 [31]
4 years ago
12

27. Calculate the impulse when an average force of 10 N acts on a cart for 5.0 s.

Physics
1 answer:
wlad13 [49]4 years ago
6 0

Answer:

50 Ns

Explanation:

Impulse = force × time

J = F Δt

J = (10 N) (5.0 s)

J = 50 Ns

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A 3.00 kg block moving 2.09 m/s
Talja [164]

Answer:11.64kgm/s

Explanation:

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3 years ago
A scuba diver is sitting on a boat while waiting to go on a dive and sees light reflected from the water's surface. At what angl
LUCKY_DIMON [66]

Answer:

θ_p = 53.0º

Explanation:

For reflection polarization occurs when a beam is reflected at the interface between two means, the polarization in total when the angle between the reflected and the transmitted beam is 90º

Let's write the transmission equation

     n1 sin θ₁ = ne sin θ₂

The angle to normal (vertcal) is

    180 = θ2 + 90 + θ_p

    θ₂ = 90 - θ_p

Where θ₂ is the angle of the transmitted ray θ_p is the angle of the reflected polarized ray

We replace

     n1 sin θ_p = n2 sin (90 - θ_p)

Let's use the trigonometry relationship

    Sin (90- θ_p) = sin 90 cos θ_p - cos 90 sin θ_p = cos θ_p

In the law of reflection  incident angle equals reflected angle,  

    ni sin θ_p = ns cos θ_p

    n₂ / n₁ = sin θ_p / cos θ_p

    n₂ / n₁ = tan θ_p

    θ_p = tan⁻¹ (n₂ / n₁)

Now we can calculate it

The refractive index of air is 1 (n1 = 1) the refractive index of seawater varies between 1.33 and 1.40 depending on the amount of salts dissolved in the water

n₂ = 1.33

      θ_p = tan⁻¹ (1.33 / 1)

      θ_p = 53.0º

n₂ = 1.40

      θ_p = tan⁻¹ (1.40 / 1)

      Tep = 54.5º

4 0
3 years ago
Does anyone know how to solve this?
FromTheMoon [43]
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7 0
3 years ago
What’s something that is 50cm
Firlakuza [10]

Answer:

50 cm is equivalent to 19,6850393701 inches.

Explanation:

A meter has 100 centimeters. 100 millimeters make one centimeter. The centimeter can be written as cm. While calculating the surface area of an object, the unit of measurement becomes cm2.

5 0
3 years ago
Read 2 more answers
Astronomers discover an exoplanet, a planet obriting a star other than the Sun, that has an orbital period of 3.27 Earth years i
Naddik [55]

Answer:

  r = 3.787 10¹¹ m

Explanation:

We can solve this exercise using Newton's second law, where force is the force of universal attraction and centripetal acceleration

    F = ma

    G m M / r² = m a

The centripetal acceleration is given by

    a = v² / r

For the case of an orbit the speed circulates (velocity module is constant), let's use the relationship

    v = d / t

The distance traveled Esla orbits, in a circle the distance is

    d = 2 π r

Time in time to complete the orbit, called period

     v = 2π r / T

Let's replace

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    G M / r² = (2π r / T)² / r

    G M / r² = 4π² r / T²

    G M T² = 4π² r3

     r = ∛ (G M T² / 4π²)

Let's reduce the magnitudes to the SI system

     T = 3.27 and (365 d / 1 y) (24 h / 1 day) (3600s / 1h)

     T = 1.03 10⁸ s

Let's calculate

      r = ∛[6.67 10⁻¹¹ 3.03 10³⁰ (1.03 10⁸) 2) / 4π²2]

      r = ∛ (21.44 10³⁵ / 39.478)

      r = ∛(0.0543087 10 36)

      r = 0.3787 10¹² m

      r = 3.787 10¹¹ m

7 0
3 years ago
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