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3 years ago

Small, icy bodies that have highly eccentric orbits and can be found in the Oort cloud or the Kuiper belt are called ____.

Physics

1 answer:

irina1246 [14]3 years ago

4 0

Answer:

Small, icy bodies that have highly eccentric orbits and can be found in the Oort cloud or the Kuiper belt are called COMETS.

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A bird watcher meanders through the woods, walking 1.93 km due east, 1.03 km due south, and 3.84 km in a direction 52.8 ° north

Sedaia [141]

Answer:

Magnitude of displacement = 2.07 km

Magnitude of average velocity = 1.17 kmph

Explanation:

Let east represent positive x axis and north represent positive y axis.

A bird watcher meanders through the woods, walking 1.93 km due east, 1.03 km due south, and 3.84 km in a direction 52.8 ° north of west.

1.93 km due wast

s ₁ = 1.93 i km

1.03 km due south

s₂ = -1.03 j km

3.84 km in a direction 52.8 ° north of west

s₃ = -3.84 cos 52.8 i + 3.84 sin 52.8 j = -2.32 i + 3.06 j km

Total displacement

s = s ₁+ s₂+ s₃ = 1.93 i - 1.03 j -2.32 i + 3.06 j = -0.39 i + 2.03 j

Magnitude of displacement, $=\sqrt{(-0.39)^2+2.03^2}=2.07km$

Time taken = 1.771 hour

Magnitude of average velocity, $=\frac{2.07}{1.771}=1.17km/hr$

7 0

3 years ago

3. Specify the wrong sentences.

soldi70 [24.7K]

Answer:

a: false

b: True

c: i dont know

Explanation:

4 0

1 year ago

:What will be the value of the refractive index of the medium? Critical angle of that medium is 30 degree

earnstyle [38]

Answer:

Let the second medium be air (n₁=1)

The refractive index n₂ of the medium where first medium is air is found (a)

(a) n₂ = 2

Explanation:

Critical angle can be defined as the angle of incidence that provides the angle of refraction of 90°.

Refractive index of a medium can be defined as a number that describes that how fast a light will travel through that medium.

Critical angle and Refractive index are related by:

$\theta_{critical}= sin^{-1}(\frac{n_1}{n_2})$

$sin \theta_{critical}=\frac{n_1}{n_2}$

To find refractive index of medium with respect to air, substitute n₁=1 (Refractive index of air is 1)

Also θ(critical)=30°

Find n₂ :

$sin30= \frac{1}{n_2}\\0.5=\frac{1}{n_2}\\n_2=\frac{1}{0.5}\\n_2=2$

8 0

2 years ago

If there is an object in top of another object, why does the upper object exert a downward normal force?

rusak2 [61]

Answer:

This is because normal force is exerted perpendicularly to the point of contact between the upper and lower objects.

Explanation:

This is because the upper object is still subject to gravitational pull. Therefore, the amount of force it exerts on the lower object due to gravity will be equal to the normal force that acts in the negative direction of gravitational force. Additionally, normal force is evident because the upper object will not go into the lower object.

4 0

3 years ago

A 235 kg crate is pulled across a horizontal surface with a force of 760 N applied at an

AnnyKZ [126]

Answer:

658.16N

Explanation:

Step one:

given data

mass m= 235kg

Force F= 760N

angle= 30 degrees

Required

The horizontal component of the force

Step two:

The horizontal component of the force

Fh= 760cos∅

Fh=760cos30

Fh=760*0.8660

Fh=658.16N

3 0

3 years ago