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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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Please helps me with this question..
Alenkasestr [34]
1) S.I. Unit for electric current = "Ampere"

2) S.I. Unit for resistance = "Ohm"

3) S.I. Unit for potential difference = "Volt"

Hope this helps!
8 0
3 years ago
A bicyclist starts at rest and speeds up to 30 m/s while accelerating at 4 m/s^2. Determine the distance traveled.
raketka [301]

Answer:

Distance, d = 112.5 meters

Explanation:

Initially, the bicyclist is at rest, u = 0

Final speed of the bicyclist, v = 30 m/s

Acceleration of the bicycle, a=4\ m/s^2

Let s is the distance travelled by the bicyclist. The third equation of motion is given as :

v^2-u^2=2as

s=\dfrac{v^2-u^2}{2a}

s=\dfrac{(30)^2}{2\times 4}

s = 112.5 meters

So, the distance travelled by the bicyclist is 112.5 meters. Hence, this is the required solution.

6 0
3 years ago
A jet liner must reach a speed of 82 m/s for takeoff. If the
SIZIF [17.4K]

Answer:

The acceleration that the jet liner that must have is 2.241 meters per square second.

Explanation:

Let suppose that the jet liner accelerates uniformly. From statement we know the initial (v_{o}) and final speeds (v_{f}), measured in meters per second, of the aircraft and likewise the runway length (d), measured in meters. The following kinematic equation is used to calculate the minimum acceleration needed (a), measured in meters per square second:

a = \frac{v_{f}^{2}-v_{o}^{2}}{2\cdot d}

If we know that v_{o} = 0\,\frac{m}{s}, v_{f} = 82\,\frac{m}{s} and d = 1500\,m, then the acceleration that the jet must have is:

a = \frac{\left(82\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}}{2\cdot (1500\,m)}

a = 2.241\,\frac{m}{s^{2}}

The acceleration that the jet liner that must have is 2.241 meters per square second.

3 0
2 years ago
How to find the gradient of a velocity time graph
Sidana [21]
The area-

The area under the line in a velocity-time graph represents the distance travelled. To find the distance travelled in the graph above, we need to find the area of the light-blue triangle and the dark-blue rectangle.

<span><span>Area of light-blue triangle -
<span>The width of the triangle is 4 seconds and the height is 8 meters per second. To find the area, you use the equation: <span>area of triangle = 1⁄2 × base × height </span><span>so the area of the light-blue triangle is 1⁄2 × 8 × 4 = 16m. </span></span></span><span> Area of dark-blue rectangle The width of the rectangle is 6 seconds and the height is 8 meters per second. So the area is 8 × 6 = 48m.</span><span> Area under the whole graph <span>The area of the light-blue triangle plus the area of the dark-blue rectangle is:16 + 48 = 64m.<span>This is the total area under the distance-time graph. This area represents the distance covered.</span></span></span></span>
7 0
3 years ago
A bullet leaves a gun with a horizontal velocity of 100 m/s. Calculate the distance it would travel in 1.3 seconds.
stiks02 [169]

Ignoring air resistance, the bullet's horizontal velocity is constant:

v_x=v_{0x}=100\,\dfrac{\mathrm m}{\mathrm s}

In 1.3 seconds, we can expect it to travel

v_xt=\left(100\,\dfrac{\mathrm m}{\mathrm s}\right)(1.3\,\mathrm s)=130\,\mathrm m

4 0
3 years ago
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