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The space shuttle travels at a speed of about 7.6times10^3 m/s. The blink of an astronaut's eye lasts about 110 ms. How many foo

sveta [45]

Answer:

It covers distance of 9.15 football fields in the said time.

Explanation:

We know that

$Distance=Speed\times Time$

Thus distance covered in blinking of eye =

$Distance=7.6\times 10^{3}m/s\times 110\times 10^{-3}s\\\\Distance=836 meters$

Thus no of football fields= $\frac{936}{91.4}=9.15Fields$

7 0

2 years ago

Suppose a firm is producing 2,475 units of output by hiring 50 workers (W = $20 per hour) and 25 units of capital (R = $10 per h

Neko [114]

Answer

given,

firm is producing = 2,475 units

output by hiring 50 workers W = $20 per hour

25 units of capital R = $10 per hour

marginal product of labor = 40

marginal product of capital = 25

$\dfrac{MP_l}{MP_c}=\dfrac{40}{25}$

$\dfrac{MP_l}{MP_c}=\dfrac{8}{5}$

$\dfrac{W}{R}=\dfrac{20}{10}$

$\dfrac{W}{R}=2$

$\dfrac{MP_l}{MP_c} < \dfrac{W}{R}$

Firm is not minimizing the cost because the firm use more capital and less labor.

3 0

3 years ago

Which situation best describes the act of reducing? paying bills online instead of sending paper through the mail donating gentl

Ugo [173]

In my opinion, i think the first one is the best one since we're cutting down on paper wasted in mails when its better to do it online.

"paying bills online instead of sending paper through mail"

3 0

3 years ago

Read 2 more answers

What is the speed of a wave that has a wavelength of 0.4 m and a frequency of 10 Hz

Shkiper50 [21]

<span>speed = wavelength x frequency
speed = 0.4m X 10 Hz
speed = 4 m/s</span>

4 0

2 years ago

A physics student stands on a cliff overlooking a lake and decides to throw a softball to her friends in the water below. She th

Andre45 [30]

The horizontal distance covered by the ball before hitting the water is 70.4 m

Explanation:

The motion of the ball is the motion of a projectile, so it consists of two independent motions:

A uniform motion along the horizontal (x) direction
A uniformly accelerated motion along the vertical (y) direction

We start by calculating the time of flight of the ball. This can be done by analyzing the vertical motion. We can use the following suvat equation:

$s=u_y t + \frac{1}{2}at^2$

where:

s = -16.5 m is the vertical displacement of the ball (it is negative because we take upward as positive direction)

$u_y$ is the initial vertical velocity of the ball, which is given by

$u_y = u sin \theta$

where

u = 23.5 m/s is the initial velocity

$\theta=33.5^{\circ}$ is the angle of projection

Substituting,

$u_y=(23.5)(sin 33.5^{\circ})=13.0 m/s$

$a=g=-9.8 m/s^2$ is the acceleration of gravity, downward

Substituting everything into the equation we get:

$-16.5=13.0t-4.9t^2\\4.9t^2-13.0t-16.5=0$

Solving the equation for t, we find the time of flight of the ball:

t = -0.94 s

t = 3.59 s

We ignore the 1st solution since it is negative, so the ball reaches the water after 3.59 seconds.

Now we analyze the horizontal motion of the ball. The horizontal velocity is constant and it is:

$v_x=u cos \theta=(23.5)(cos 33.5^{\circ})=19.6 m/s$

Therefore, the horizontal distance covered in a time t is

$d=v_x t$

And substituting t = 3.59 s, we find

$d=(19.6)(3.59)=70.4 m$

So, the horizontal distance covered by the ball before hitting the water is 70.4 m.

Learn more about projectile motion:

brainly.com/question/8751410

#LearnwithBrainly

4 0

3 years ago