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solniwko [45]
3 years ago
9

Ice sheet movement rates have varied from about 50 to 320 meters per year for the margins of the ice sheet advancing from the hu

dson bay region during the ice age. if an ice sheet moved from the southern end of hudson bay to the south shore of present-day lake erie, a distance of 1600 kilometers, what would be the maximum amount of time required?
Physics
1 answer:
harina [27]3 years ago
5 0

First let us convert the distance into meters.

distance = 1600 km = 1,600,000 m

 

Then we get the maximum time by dividing the distance with the smallest movement rate possible, that is:

maximum time = 1,600,000 m / (50 m / year)

<span>maximum time = 32,000 years</span>

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A cannon with a muzzle speed of 1 000 m/s is used to start an avalanche on a mountain slope. The target is 2 000 m from the cann
Nataliya [291]

Answer:

∅ = 89.44°

Explanation:

In situations like this air resistance are usually been neglected thereby making g= 9.81 m/s^{2}

Bring out the given parameters from the question:

Initial Velocity (V_{1}) = 1000 m/s

Target distance (d) = 2000 m

Target height (h) =  800 m

Projection angle ∅ = ?

Horizontal distance = V_{1x}tcos ∅     .......................... Equation 1

where V_{1x} = velocity in the X - direction

           t = Time taken

Vertical Distance = y = V_{1y} t - \frac{1}{2}gt^{2}        ................... Equation 2

Where   V_{1y} = Velocity in the Y- direction

              t  = Time taken

V_{1y} = V_{1}sin∅

Making time (t) subject of the formula in Equation 1

                    t = d/(V_{1x}cos ∅)

                      t = \frac{2000}{1000coso} = \frac{2}{cos0}  =    \frac{d}{cos o}             ...................Equation 3

substituting equation 3 into equation 2

Vertical Distance = d = V_{1y} \frac{d}{cos o} - \frac{1}{2}g\frac{2}{cos0}   ^{2}

                                  Vertical Distance = h = sin∅ \frac{d}{cos o} - \frac{1}{2}g\frac{2}{cos0}   ^{2}

  Vertical Distance = h = dtan∅   - \frac{1}{2}g\frac{2}{cos0}   ^{2}

  Applying geometry

                              \frac{1}{cos o} = tan^{2} o + 1

  Vertical Distance = h = d tan∅   - 2 g (tan^{2} o + 1)

               substituting the given parameters

               800 = 2000 tan ∅ - 2 (9.81)( tan^{2} o + 1)

              800 = 2000 tan ∅ - 19.6( tan^{2} o + 1)  Equation 4

Replacing tan ∅ = Q     .....................Equation 5

In order to get a quadratic equation that can be easily solve.

            800 = 2000 Q - 19.6Q^{2} + 19.6

Rearranging 19.6Q^{2} - 2000 Q + 780.4 = 0

                    Q_{1} = 101.6291

                      Q_{2} = 0.411

    Inserting the value of Q Into Equation 5

                 tan ∅ = 101.63    or tan ∅ = 0.4114

Taking the Tan inverse of each value of Q

                  ∅ = 89.44°     ∅ = 22.37°

             

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Which explains the information needed to calculate speed and velocity?
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Explanation:

Speed and velocity are two different quantities. In fact:

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s=\frac{distance}{time}

Therefore, speed does not take into account the direction of motion.

- Velocity, on the other hand, is a vector quantity, so it has magnitude and a direction.

The magnitude of the velocity is given by

v=\frac{displacement}{time}

where displacement is a vector connecting the initial point with the final point of motion of an object.

The direction of the velocity corresponds to the direction of the displacement.

It must be noted also that in certain situations, the average velocity is zero, while the speed is not zero: for example, for an object completing a circle in a certain time interval, its speed is not zero, while its velocity is zero (because the object returns to the starting point, so the displacement is zero.

Therefore, the correct answer is

Both require time, but velocity requires displacement and speed requires distance.

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brainly.com/question/8893949

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The force needed to accelerate a vehicle with a mass of 1000kg at a rate of 5m/s2 would be 5000
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Answer:

through direct contact

Explanation:

took the test!

3 0
3 years ago
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