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3 years ago

10

If an object is dropped from a tall building and hits the ground 3.0 s later, what is the velocity of the object when it hits th

e ground? (Remember that the velocity is downward)

1 answer:

dusya [7]3 years ago

6 0

Using the kinematic equation:

Vf=Vi+at then plug the values in.

Vi=0 m/s (Because the ball is dropped)

t=3 s.

a=9.81 m/s=g

Vf=0+3*9.81;

Vf=29.43 m/s

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Select the statement that is not true about the Hubble Telescope:

xxMikexx [17]

Answer:

The option is B is not true for Hubble telescope.

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3 years ago

You and your friend throw balloons filled with water from the roof of a several story apartment house. You simply drop a balloon

Aleks [24]

Answer:

Height = 53.361 m

Explanation:

There are two balloons being thrown down, one with initial speed (u1) = 0 and the other with initial speed (u2) = 43.12

From the given information we make the following summary

$u_{1}$ = 0m/s

$t_{1}$ = t

$u_{2}$ = 43.12m/s

$t_{2}$ = (t-2.2)s

The distance by the first balloon is

$D = u_{1} t_{1} + \frac{1}{2} at_{1}^2$

where

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Inputting the values

$D = (0)t + \frac{1}{2} (9.8)t^2\\ D = 4.9t^2$

The distance traveled by the second balloon

$D = u_{2} t_{2} + \frac{1}{2} at_{2}^2$

Inputting the values

$D = (43.12)(t-2.2) + \frac{1}{2} (9.8)(t-2.2)^2$

simplifying

$D = 4.9t^2 + 21.56t -71.148$

Substituting D of the first balloon into the D of the second balloon and solving

$4.9t^2 = 4.9t^2 + 21.56t -71.148 \\ 21.56t = 71.148\\ t = 3.3s$

Now we know the value of t. We input this into the equation of the first balloon the to get height of the apartment

$D = 4.9(3.3)^2\\ D = 53.361 m$

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3 years ago

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3 years ago

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3 years ago

Read 2 more answers

According to information found in an old hydraulics book, the energy loss per unit weight of fluid flowing through a nozzle conn

suter [353]

This question is incomplete, the complete question is;

According to information found in an old hydraulics book, the energy loss per unit weight of fluid flowing through a nozzle connected to a hose can be estimated by the formula; h= (0.04 to 0.09)(D/d)⁴V²/2g

where h is the energy loss per unit weight, D the hose diameter, d the nozzle tip diameter, V the fluid velocity in the hose, and g the acceleration of gravity.

Do you think this equation is valid in any system of units

Answer:

YES, the equation is a general equation that is valid in any system of units

Explanation:

Given the data in the question;

h = (0.04 to 0.09)(D/d)⁴ × $\frac{V^{2} }{2g}$

so

[ N.m/N ] = (0.04 to 0.09) ( m/m)² × (m²/s²)1/2 × (s²/m)

[ N.L/N ] = (0.04 to 0.09) ( L⁴/L⁴) × (L²/T²)1/2 × (T²/L)

∴ [ L ] = (0.04 to 0.09) [L]

So as each term in the equation must have the same dimensions, the constant term (0.04 to 0.09) must be without dimension.

Therefore, YES, the equation is a general equation that is valid in any system of units

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2 years ago