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shepuryov [24]
3 years ago
6

A horizontal spring with spring constant 750 N/m is attached to a wall. An athlete presses against the free end of the spring, c

ompressing it 5.0 cm. How hard is the athlete pushing?
Physics
2 answers:
riadik2000 [5.3K]3 years ago
7 0

Answer:

37.5N

Explanation:

According to Hooke's law, the load or force, F, applied on an elastic material (e.g a spring) is directly proportional to the extension or compression, e, caused by the load. i.e

F ∝ e

F = k x e         -------------------------(i)

where;

k = proportionality constant known as the spring constant.

From the question;

k = 750N/m

e = 5.0cm = 0.05m

Substitute these values into equation (i) as follows;

F = 750 x 0.05

F = 37.5N

Therefore, the load applied which is a measure of how hard the athlete is pushing is 37.5N

ANTONII [103]3 years ago
4 0

Answer:

37.5 N Hard

Explanation:

Hook's law: The force applied to an elastic material is directly proportional to the extension provided the elastic limit of the material is not exceeded.

Using the expression for hook's law,

F = ke.............. Equation 1

F = Force of the athlete, k = force constant of the spring, e = extension/compression of the spring.

Given: k = 750 N/m, e = 5.0 cm = 0.05 m

Substitute into equation 1

F = 750(0.05)

F = 37.5 N

Hence the athlete is pushing 37.5 N hard

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Explanation:

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The voltage V just before action of potential is -70mV,

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A tiger leaps horizontally out of a tree that is 6.00 m high. If he lands 2.00 m from the base of the tree, calculate his initia
Musya8 [376]

Answer:

The initial speed of the tiger is 1.80 m/s

Explanation:

Hi there!

The equation of the position vector of the tiger is the following:

r = (x0 + v0 · t, y0 + 1/2 · g · t²)

Where:

r = position vector at a time t.

x0 = initial horizontal position.

v0 = initial horizontal velocity.

t = time.

y0 = initial vertical position,

g = acceleration due to gravity.

Let´s place the origin of the frame of reference on the ground at the point where the tree is located so that the initial position vector will be:

r0 = (0.00, 6.00) m

We can use the equation of the vertical component of the position vector to obtain the time it takes the tiger to reach the ground.

y = y0 + 1/2 · g · t²

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2 · (-6.00 m) / -9.81 m/s² = t²

t = 1.11 s

We know that in 1.11 s the tiger travels 2.00 m in the horizontal direction. Then, using the equation of the horizontal component of the position vector we can find the initial speed:

x = x0 + v0 · t

At t = 1.11 s, x = 2.00 m

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2.00 m = v0 · 1.11 s

2.00 m / 1.11 s = v0

v0 = 1.80 m/s

The initial speed of the tiger is 1.80 m/s

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3 years ago
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