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3 years ago

1.

Physics

2 answers:

ivolga24 [154]3 years ago

7 0

Answer:

Explanation:

You must park at lest 15 Feet away from a fire hydrant.

If you disobey this law, you could be fined up to $115 dollars.

Agata [3.3K]3 years ago

6 0

Answer:

15 feet

Explanation:

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If the archerfish spits its water 30 degrees from the horizontal aiming at an insect 1.2 m above the surface of the water, how f

Burka [1]

Answer:

The speed is $v = 9.8 \ m/s$

Explanation:

From the question w are told that

The angle made is $\theta = 30^o$

The distance above the surface of the water is $h_{max} = 1.2 \ m$

The value of $g = 10 \ m/s^2$

The maximum height attained by the fish is mathematically evaluate as

$h_{max} = \frac{v^2 sin ^2 \theta }{2g }$

Making v which is the speed of the fish the subject of the formula

$v = \sqrt{ \frac{2gh_{max}}{ sin^2 \theta } }$

substituting values

$v = \sqrt{ \frac{2*10 *1.2 }{ [sin (30)]^2 } }$

$v = 9.8 \ m/s$

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3 years ago

Please need help ASAP 15 points????I need a,b,c

Anettt [7]

Answer:

The acceleration is negative and very close to zero.

Very Small, because it slowly comes to a stop, which means there is nothing to stop it any faster

It would keep traveling in that direction without slowing down

Explanation:

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3 years ago

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Please Help! If you send a sound wave of the same wavelength (_ = 2.00 m) through air, helium, and carbon dioxide, describe how

Andreas93 [3]

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An object moves in one dimensional motion with constant acceleration a = 5 m/s^2. At time t = 0 s, the object is at x0 = 2.7 m a

andreyandreev [35.5K]

Answer:

x = 5 m ;approximate

Explanation:

Kinematic equation for the object

Because the object moves with uniformly accelerated motion we apply the following equation:

$(v_{f})^{2} =(v_{o})^{2} + 2*a*d$

$v_{f}$ : final speed ( m/s)

$v_{o}$ : initial speed ( m/s)

a: acceleration: ( m/s²)

d : distance (m)

Initial Conditions:

t₀=0 , x₀= 2.7 m, v₀= 4.3 m/s

x₀= inicial position

Final Conditions:

a= 5 m/s² , vf= 6.4 m/s

Calculation of the distance traveled by the object in final condition

We apply the formula (1)

$(v_{f})^{2} = (v_{o} )^{2} +2*a*d$

$d = \frac{6.4^{2}- 4.3^{2} }{2*5}$

d= 2.247 m

Calculation of the final position : $x_{f}$

$x_{f} =x_{o} +d$

$x_{f} = 2.7 m + 2.247 m$

$x_{f} =4.947 m$

$x_{f} = 5m$ : approximate

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3 years ago

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