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3 years ago

Que es una ondaaaa?? no me deja escribiiiirrrrrr

Physics

1 answer:

UkoKoshka [18]3 years ago

7 0

Answer:

yo queiro escribir mi hermano

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A dolphin in an aquatic show jumps straight up out of the water at a velocity of 13.0 m/s.

SVEN [57.7K]

Answer:

a) Knowns, initial speed $v_{i}=13.0 m/s$ , final speed $v_{f}=0 m/s$ and gravity due it is a constant $g=9.8m/s^{2}$

b) The maximum high reached by the dolphin is $y_{max}=8.62 m$

c) Total time is $t=2.65s$

Explanation:

a) First of all the initial speed is given at the start of the problem, gravity is constant and final speed is known any object thrown straight up reaches its max high at $0m/s$ speed.

b) Second, now that we know final speed we use $v_{f} =v_{i}-gt$ , as we clear for $t=\frac{v_{i}-v_{f} }{g}=\frac{20.0m/s}{9.8m/s^{2} }=1.32 s$ .

Then we use $y=v_{i}t-\frac{1}{2} gt^{2}=(20.0m/s)(1.32s)-\frac{1}{2} (9.8m/s^{2} ) (1.32s)^{2} =8.62m$

c)Third, finally we can use $y=v_{i}t-\frac{1}{2} gt^{2}$ , as we know $y=0m$ when the dolphin fall into the water again and $v_{i} =13.0m/s$ , then we have $0=(13m/s)t-\frac{1}{2} (9.8m/s^{2} )t^{2}$ is a quadratic form $0=t(13.0-4.9t)$ so we have $t_{i}=0s$ and $t_{f}=\frac{13}{4.90} =2.65s$

6 0

3 years ago

A) Two workers are trying to move a heavy crate. One pushes onthe crate with a force A, which has amagnitude of 445 newtons and

Reptile [31]

Answer:

Divide then multiply or multiply then divide

Explanation:

to get the answer of a and b

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3 years ago

Which statement best explains acceleration?

Alexxx [7]

Acceleration is the ratio of a change in velocity to the time over which the change happend

5 0

3 years ago

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If the magnets are brought close together, they will

murzikaleks [220]

It depends on which side. Opposites attract, so north and south would attract to each other and collide, while north and north or south and south would go away from eachother.

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3 years ago

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A hockey puck slides across the ice with an initial velocity of 7.2 m/s. It has a deceleration of 1.1 m/s2 and is traveling towa

dezoksy [38]

Answer:

0.74 s

Explanation:

We can solve the problem by using the following SUVAT equation:

$d = ut + \frac{1}{2}at^2$

where

d = 5.0 m is the displacement

u = 7.2 m/s is the initial velocity

a = -1.1 m/s^2 is the acceleration (which is negative since it is a deceleration)

t is the time

Substituting numbers into the equation, we find:

$5.0 = 7.2 t -0.55t^2\\0.55t^2 -7.2t + 5.0 = 0$

This is a second-order equation, whose solutions are given by:

$t=\frac{-(-7.2) \pm \sqrt{(-7.2)^2-4(0.55)(5.0)}}{2(0.55)}$

And the solutions are

t = 0.74 s

t = 12.36 s

The solution we are looking for is the first one, because it corresponds to the first time at which the hockey puck has travelled the distance of 5.0 m, reaching the goal.

8 0

3 years ago