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qwelly [4]
3 years ago
10

What force does a trampoline have to apply to a gymnast to accelerate her straight up at ? Note that the answer is independent o

f the velocity of the gymnastâshe can be moving either up or down, or be stationary.
Physics
1 answer:
Andrew [12]3 years ago
5 0

Answer: Force applied by trampoline = 778.5 N

<em>Note: The question is incomplete.</em>

<em>The complete question is : What force does a trampoline have to apply to a 45.0 kg gymnast to accelerate her straight up at 7.50 m/s^2? note that the answer is independent of the velocity of the gymnast. She can be moving either up or down or be stationary. </em>

Explanation:

The total required the trampoline by the trampoline = net force accelerating the gymnast upwards + force of gravity on her.

= (m * a) + (m * g)

= m ( a + g)

= 45 kg ( 7.50 *  9.80) m/s²

Force applied by trampoline = 778.5 N

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7 0
3 years ago
Consider the displacement vectors Ā=(i +6j)m, B = (3i– 7j)m,
andrezito [222]

Answer:

Dx = -0.5

Dy = -0.25

Explanation:

Two vectors are given in rectangular components form as follows:

A = i + 6j

B = 3i - 7j

It is also given that:

A - B - 4D = 0

so, we solve this to find D vector:

(i + 6j) - (3i - 7j) - 4D = 0

- 2i - j = 4D

D = - (2/4)i - (1/4)j

D = - (1/2)i - (1/4)j

<u>D = - 0.5i - 0.25j</u>

Therefore,

<u>Dx = -0.5</u>

<u>Dy = -0.25</u>

8 0
3 years ago
What are (a) the kinetic energy, (b) the rest energy, and (c) the total energy of a 1.50 g particle with a speed of 0.600 c ?
MAVERICK [17]
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3 years ago
Read 2 more answers
The 2.50 kg cube in the figure has edge lengths d = 6.50 cm and is mounted on an axle
kozerog [31]

Answer:

0.191 s

Explanation:

The distance from the center of the cube to the upper corner is r = d/√2.

When the cube is rotated an angle θ, the spring is stretched a distance of r sin θ.  The new vertical distance from the center to the corner is r cos θ.

Sum of the torques:

∑τ = Iα

Fr cos θ = Iα

(k r sin θ) r cos θ = Iα

kr² sin θ cos θ = Iα

k (d²/2) sin θ cos θ = Iα

For a cube rotating about its center, I = ⅙ md².

k (d²/2) sin θ cos θ = ⅙ md² α

3k sin θ cos θ = mα

3/2 k sin(2θ) = mα

For small values of θ, sin θ ≈ θ.

3/2 k (2θ) = mα

α = (3k/m) θ

d²θ/dt² = (3k/m) θ

For this differential equation, the coefficient is the square of the angular frequency, ω².

ω² = 3k/m

ω = √(3k/m)

The period is:

T = 2π / ω

T = 2π √(m/(3k))

Given m = 2.50 kg and k = 900 N/m:

T = 2π √(2.50 kg / (3 × 900 N/m))

T = 0.191 s

The period is 0.191 seconds.

7 0
3 years ago
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