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nadezda [96]
3 years ago
15

Which landform is produced at location E where the Mississippi River enters the Gulf of

Physics
1 answer:
mixer [17]3 years ago
3 0

Answer:

a delta

Explanation:

The landform produced at the location E where the Mississippi River enters the Gulf of Mexico is a delta.

A delta is a depositional landform where a smaller body of water enters into a larger one.

The Gulf of Mexico contains a larger body of water and as the Mississippi river enters into it, it splits up into many distributaries.

So, this feature is a delta.  

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The majority of the time


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The speed is represented by
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<u>Answer:</u>

<em>Instantaneous velocity is equal to speed of the object at that particular instant.</em>

<u>Explanation:</u>

Instantaneous velocity is the velocity of the object at that particular instant. It is also equal to speed of the object at that instant. It can be calculated by drawing a tangent to the position-time graph at that point and finding the tangent’s slope.

The first option ‘The ratio of change in position to the time interval during that change’  gives the average velocity of an object and not speed. Similarly the second option ‘the absolute value of the slope of position time graph’ gives the average speed.

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A 0.40-kg cart with charge 4.0 x 10-5 C starts at rest on a horizontal frictionless surface 0.50 m from a fixed object with char
maw [93]

Answer:

26.82m/s

Explanation:

Given

Mass = m= 0.4kg

Initial Velocity = u = 0

Charge = 4.0E-5C

Distance= d = 0.5m

Object Charge = 2E-4C

First, we'll calculate the initial energy (E)

E = Potential Energy

PE = kQq / d

Where k = coulomb constant = 8.99E9Nm²/C²

Energy is then calculated by;

PE = 8.99E9 * 4E-5 * 2E-4 / 0.5

PE = 143.84J

Energy = Potential Energy = Kinetic Energy

K.E = ½mv² = 143.84J

½mv² = ½ * 0.40 * v² = 143.85

0.2v² = 143.85

v² = 143.85/0.2

v² = 719.25

v = √719.25

v = 26.81883666380777

v = 26.82m/s

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4 years ago
Batteries can supply a steady flow of electrons.<br> True<br> False
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6 0
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Your friend has decided to make some money during the next State Fair by inventing a game of skill. In the game as she has devel
coldgirl [10]

Answer: from the vertical, one should aim  86.6°

Explanation:

height of the center of object = 7.0 m - 0.05 m = 6.95 m

now let the bullet hits centre at point A height x meters from the ground

also let t be the time taken for the bullet to hit the object

so distance travelled by the target will be

d = h - x = 6.95 - x

now using the equation of motion

d = 1/2gt²

so 1/2gt² = 6.95 - x

x = 6.95 - 1/2gt² .........let this be equ 1

let angle of fire be ∅

so v(cos∅) × t = 100

our velocity v is 1200 ft/sec = 365.76 m/s

365.76(cos∅) × t = 100 ........equ 2

also vertical position of the bullet after t is

y = y₀ + c(sin∅)t - 1/2gt²

y = 1 + 365.76(sin∅)t - 1/2gt² ----- equ 3

After time t. the vertical position x and y are same, else the bullet wouldn't have strike target at centre, so;

x = y

we substitute

equ 1 = equ 3

6.95 - 1/2gt² = 1 + 365.76(sin∅)t - 1/2gt²

6.95 - 1 = 365.76(sin∅)t - 1/2gt² +  1/2gt²

5.95 = 365.76(sin∅)t

t = 5.96 / 365.76(sin∅)

now input the above equ  into equ 2

365.76(cos∅) ×  5.96/365.76(sin∅) = 100

5.95(cos∅)/sin∅ = 100

tan∅ = 5.95/100 = 0.0595

∅ =  3.40°

therefore from the vertical, one should aim (90° - 3.40°) = 86.6°

4 0
4 years ago
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