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Oksana_A [137]
3 years ago
14

A river is flowing south at a rate of 3 m/s. Steven can roe directly across the river if he aims the raft 30 degrees. What rate

is Steven rowing at
Physics
1 answer:
levacccp [35]3 years ago
3 0

Answer:

Steven has to row at a speed to reach the same horizontal spot at the other side of the river is, V = 6 m/s

Explanation:

Given data,

The river flowing south at the rate, v = 3 m/s

To reach the other side directly across the river, he aims the raft, Ф = 30°

The speed of his raft across the river is given by the formula,

                                          V = v / Sin Ф

                                             = 3 / Sin 30°

                                              = 6 m/s

Steven has to row at a speed to reach the same horizontal spot at the other side of the river is, V = 6 m/s

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List and explain briefly similarities and differences between the electric force between two charges and the gravitational force
bazaltina [42]

Answer:

Please see below as the answers are self-explanatory.

Explanation:

  • Similarities

1) The resultant force is along the line that joins both charges or both masses (assuming both objects can be represented as points)

2) Both type of forces obey  Newton's 3rd law.

3) Both are proportional to the product of the property that is affected by the force (charges and masses)

4) Both obey an inverse - square law (consequence of our universe being three-dimensional)

  • Differences

1) Main difference, is that while the gravitational force is always attractive, the electrostatic force can be attractive or repulsive, as there are two types of charges, which attract each other being of different type, and repel each other if they are of the same type.

2) It  is possible, artificially, to block the influence of the electrostatic force, shielding a room, for instance, which is not possible for the gravitational force.

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3 years ago
Which of the following is a category of mechanical wave?
givi [52]

Answer:

a

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Light propagate faster through medium “a” than medium “b”
dangina [55]

1) Medium "b" has more optical density

2) Light must hit the interface between the two mediums perpendicularly

Explanation:

1)

Refraction occurs when light propagates from a medium into a second medium.

The optical density of a medium is given by its index of refraction, which is defined as:

n=\frac{c}{v}

where

c is the speed of light in a vacuum

v is the speed of light in a medium

Higher index of refraction means higher optical density, and light propagater slower into a medium with higher optical density.

In this problem, light propagates faster through medium "a" than medium "b": this means that medium "a" has lower refractive index of medium "b", and so "b" has more optical density.

2)

We can answer this part by referring to Snell's law, which gives the relationship between the direction of the incident ray and of the refracted ray when light passes through the interface between two media:

n_1 sin \theta_1 = n_2 sin \theta_2

where

n_1, n_2 are the index of refraction of the two mediums

\theta_1, \theta_2 are the angle of incidence and of refraction (the angle that light makes with the normal to the surface in medium 1 and medium 2)

Here we want the direction of propagation of the light ray not to change: this means that it must be

sin \theta_1 = sin \theta_2 (1)

However, here we have two mediums "a" and "b" with different index of refraction, so

n_1\neq n_2

Therefore the only angle that can satisfy eq.(1) is

\theta_1 = \theta_2 = 0

So, the light must hit the surface perpendicular to the interface between the two mediums.

Learn more about refraction:

brainly.com/question/3183125

brainly.com/question/12370040

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3 0
3 years ago
A car is traveling at a constant speed of 33 m/s on a highway. At the instant this car passes an entrance ramp, a second car ent
Paha777 [63]

Answer:

0.8712 m/s²

Explanation:

We are given;

Velocity of first car; v1 = 33 m/s

Distance; d = 2.5 km = 2500 m

Acceleration of first car; a1 = 0 m/s² (constant acceleration)

Velocity of second car; v2 = 0 m/s (since the second car starts from rest)

From Newton's equation of motion, we know that;

d = ut + ½at²

Thus,for first car, we have;

d = v1•t + ½(a1)t²

Plugging in the relevant values, we have;

d = 33t + 0

d = 33t

For second car, we have;

d = v2•t + ½(a2)•t²

Plugging in the relevant values, we have;

d = 0 + ½(a2)t²

d = ½(a2)t²

Since they meet at the next exit, then;

33t = ½(a2)t²

simplifying to get;

33 = ½(a2)t

Now, we also know that;

t = distance/speed = d/v1 = 2500/33

Thus;

33 = ½ × (a2) × (2500/33)

Rearranging, we have;

a2 = (33 × 33 × 2)/2500

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