The calculation for kinetic energy is this
KE = 1/2mv^2
KE = 1/2(50)(7^2)
KE = 1/2(49•50)
KE = 1225 kgm^2/s^2.
Or simply 1225 J.
She possess this much energy when she runs.
Question:
How do mountain glaciers and continental glaciers differ in terms of dimensions, thickness and patterns of movement?
Answer:
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Continental glaciers are thicker, much more expansive sheets. Mountain glaciers flow downhill as a result of gravity acting on the mass of ice. Continental glaciers move in response to pressure from the weight of material in their thick midsections.
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Hope this helped!
~Shane :}
The tension in the first and second rope are; 147 Newton and 98 Newton respectively.
Given the data in the question
- Mass of first block;

- Mass of second block,

- Tension on first rope;

- Tension on second rope;
To find the Tension in each of the ropes, we make use of the equation from Newton's Second Laws of Motion:

Where F is the force, m is the mass of the object and a is the acceleration ( In this case the block is under gravity. Hence ''a" becomes acceleration due to gravity
)
For the First Rope
Total mass hanging on it; 
So Tension of the rope;

Therefore, the tension in the first rope is 147 Newton
For the Second Rope
Since only the block of mass 10kg is hang from the second, the tension in the second rope will be;

Therefore, the tension in the second rope is 98 Newton
Learn More, brainly.com/question/18288215
Answer:
a)906.5 Nm^2/C
b) 0
c) 742.56132 N•m^2/C
Explanation:
a) The plane is parallel to the yz-plane.
We know that
flux ∅= EAcosθ
3.7×1000×0.350×0.700=906.5 N•m^2/C
(b) The plane is parallel to the xy-plane.
here theta = 90 degree
therefore,
0 N•m^2/C
(c) The plane contains the y-axis, and its normal makes an angle of 35.0° with the x-axis.
therefore, applying the flux formula we get
3.7×1000×0.3500×0.700×cos35°= 742.56132 N•m^2/C
Answer:
Moment of the force is 20 N-m.
Explanation:
Given:
Force exerted by the person is, 
Distance of application of force from the point about which moment is needed is, 
Now, we know that, moment of a force 'F' about a point at a perpendicular distance of 'd' from the same point is given as the product of the force and the perpendicular distance.
Therefore, the moment of the force about the end of the claw hammer is given as:

Hence, the moment of the force exerted by the person about the end of the claw hammer is 20 N-m.