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2 years ago

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A 4.0 kg box is pressed into a spring. The spring constant is k=220 n/m. The box is released at the bottom of a frictionless ram

p and rises to a height of y=0.75m

1 answer:

VikaD [51]2 years ago

4 0

What’s the question, exactly? I’d be happy to help :)

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This chemical equation represents a ______________ reaction.

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When a gun is fired at the shooting range, the gun recoils (moves backward). Explain this using the law of conservation of momen

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4 years ago

Which type of macromolecules helps a cell brake down food?

velikii [3]

Answer:

Hope this helps you find the answer

Explanation:

Chemical Digestion of Carbohydrates, Proteins, Lipids, and Nucleic Acids. The chemical breakdown of the macromolecules contained in food is completed by various enzymes produced in the digestive system.

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3 years ago

Dario, a prep cook at an Italian restaurant, spins a salad spinner and observes that it rotates 20.0 times in 5.00 seconds and t

BabaBlast [244]

Answer:

$-\frac{8\pi}{3}rad/s^2$

Explanation:

To solve this problem we need to apply the concept related to Angular Acceleration. We can find it through the equation

$\omega_f^2-\omega_i^2=2\alpha\theta$

Where for definition,

$\omega_i = \frac{\theta}{t}$

The number of revolution $(\theta)$ was given by 20 times, then

$\omega_i = \frac{20*2pi}{5}$

$\omega = 8\pi rad/s$

We know as well that the salad rotates 6 more times, therefore in angle measurements that is

$\theta = 6*2\pi rad = 12\pi rad$

The cook at the end stop to spin, then using our first equation,

$0-8\pi = 2\alpha (12\pi)$

re-arrange to solve $\alpha$ ,

$\alpha = \frac{-8\pi}{2*12\pi}$

$\alpha = -\frac{8\pi}{3}rad/s^2$

We can know find the required time,

$\omega_f-\omega_i = \alpha t$

Re-arrange to find t, and considering that $\omega_f=0$

$t= \frac{\omega_i}{\alpha}$

$t=\frac{-8\pi}{-8\pi/3}$

$t=3s$

Therefore take for the salad spinner to come to rest is 3 seconds with acceleration of $-\frac{8\pi}{3}rad/s^2$

6 0

3 years ago

Read 2 more answers

pantera1 [17]

Answer:

(a) The time the ball stays in the air is approximately 4.66 seconds

(b) The distance from the cannon at which the ball will hit the ground is approximately 83.18 meters

(c) The maximum altitude of the ball is approximately 26.62 m

Explanation:

The given parameters of the question are;

The initial velocity at which a baseball cannon fires balls, u = 29 m/s

The angle at which the cannon is tilted, θ = 52°

(a) The time duration the ball stays in the air is given by the time of flight of the projected ball as follows;

$2 \cdot t = \dfrac{2 \cdot u \cdot sin (\theta)}{g}$

Where;

t = The time it takes the baseball to reach maximum height

2·t = The total time of flight = The time the ball stays in the air

θ = The angle at which the ball is tilted = 52°

g = The acceleration due to gravity ≈ 9.81 m/s²

u = The initial velocity = 29 m/s

Therefore, we have;

$2 \cdot t = \dfrac{2 \times 29 \ m/s\times sin (52^{\circ})}{9.81 \ m/s^2} \approx 4.66 \ s$

The time the ball stays in the air, 2·t ≈ 4.66 seconds

(b) The distance from the cannon at which the ball will hit the ground is given by the horizontal range, 'R', of the projectile as follows;

$Horizontal \ range, R = \dfrac{u^2 \cdot sin(2 \cdot \theta) }{g}$

∴ The distance from the cannon at which the ball will hit the ground = R

$Horizontal \ range, R = \dfrac{(29 \ (m/s))^2 \cdot sin(2 \times 52^{\circ}) }{9.81 \ m/s^2} \approx 83.18 \, m$

The distance from the cannon at which the ball will hit the ground = R ≈ 83.18 meters

(c) The maximum altitude of the ball is equal to the maximum height reached by the ball, H, which is given as follows;

$H = \dfrac{u^2 \cdot sin^2 (\theta)}{2 \cdot g}$

Therefore, we have;

$H = \dfrac{(29 \ m/s)^2 \times sin^2 (52^{\circ})}{2 \times 9.81 \ m/s^2} \approx 26.62 \ m$

The maximum altitude of the ball ≈ 26.62 m

3 0

3 years ago