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love history [14]
3 years ago
10

A hawk flew 600 meters in 60 seconds. A sparrow flew 400 meters in 30 seconds. Which bird flew faster? How fast did each bird fl

y?
Physics
1 answer:
Andru [333]3 years ago
4 0

Answer:

A sparrow flew faster. the sparrow flew 10 meters per second. The sparow flew 13.(3) meters per second

Explanation:

You might be interested in
A student fills a tank of radius r with water to a height of h1 and pokes a small, 1.0 cm diameter hole at a distance h2 from th
Alik [6]

when a hole is made at the bottom of the container then water will flow out of it

The speed of ejected water can be calculated by help of Bernuolli's equation and Equation of continuity.

By Bernoulli's equation we can write

Po + \frac{1}{2}\rho v_1^2 + \rho g h = Po + \frac{1}{2}\rho v_2^2 + \rho g *0

Now by equation of continuity

A_1v_1 = A_2v_2

\pi (0.2)^2 v_1 = \pi (0.01)^2 v_2

from above equation we can say that speed at the top layer is almost negligible.

v_1 = 0

now again by equation of continuity

\rho g h = \frac{1}{2} \rho v^2

v = \sqrt{2 g h}

here we have

h = h_1 - h_2

h = 0.50 - 0.03 = 0.47m

now speed is given by

v = \sqrt{2* 9.8 * 0.47}

v = 3.03 m/s

7 0
3 years ago
A person jogs eight complete laps around a quarter-mile track in a total time of 12.5 min. Calculate (a) the average speed and (
Margarita [4]

\large\displaystyle\text{$\begin{gathered}\sf \huge \bf{\underline{Data:}} \end{gathered}$}

  • \large\displaystyle\text{$\begin{gathered}\sf 1\ mile = 1609.34 \ m \end{gathered}$}
  • \large\displaystyle\text{$\begin{gathered}\sf  1/4 \ mile = 402.33 \ m \end{gathered}$}

                           \large\displaystyle\text{$\begin{gathered}\sf 12.5 \not{min}*\frac{60 \ s}{1\not{min}}=750 \ s \end{gathered}$}

                   \large\displaystyle\text{$\begin{gathered}\sf \bf{A) \ Calculate \ the \ average \ speed: } \end{gathered}$}

                         \large\displaystyle\text{$\begin{gathered}\sf 402.33 \ m*8 \ laps = 3218.64 \ m \end{gathered}$}

                         \large\displaystyle\text{$\begin{gathered}\sf d=3218.64 \ m \end{gathered}$}

                         \large\displaystyle\text{$\begin{gathered}\sf t=750 \ s \end{gathered}$}

                         \large\displaystyle\text{$\begin{gathered}\sf V=\frac{d}{t} \ \ \ \ \ \  V= \frac{3218.64 \ m }{750 \ s} \end{gathered}$}\\\\\\\large\displaystyle\text{$\begin{gathered}\sf V=4.29 \ m/s \end{gathered}$}

                  \large\displaystyle\text{$\begin{gathered}\sf \bf{B) \ Calculate \ the \ average \ speed \  in \ m/s} \end{gathered}$}

                          \large\displaystyle\text{$\begin{gathered}\sf V=402.33 \ m \end{gathered}$}  

                          \large\displaystyle\text{$\begin{gathered}\sf t=750 \ s \end{gathered}$}

                          \large\displaystyle\text{$\begin{gathered}\sf V=\frac{D}{T} \ \ \ \ \ V=\frac{402.33 \ m}{750 \ s}   \end{gathered}$}\\\\\\\large\displaystyle\text{$\begin{gathered}\sf V= 0.53 \ m/s \end{gathered}$}

4 0
1 year ago
Acceleration is defined as the rate of change for which of the following? Question 4 options: time position velocity displacemen
vovangra [49]

Acceleration can be defined as the rate of change in the velocity of an object. Option C is correct.

<h3>What is Acceleration?</h3>
  • It is defined as the rate of change in velocity.
  • It can also be defined as the rate of change in position in a particular direction.

a =  \dfrac {v-u}t\ \ \ \ \rmor}\\\\a = \dfrac {\Delta v }t

Where,

a - acceleration

\Delta v - change in velocity

t - time

Therefore, acceleration can be defined as the rate of change in the velocity of an object.

Learn more about Velocity:

brainly.com/question/2239252

4 0
2 years ago
A train, traveling at a constant speed of 22.0 m/s, comes to an incline with a constant slope. While going up the incline, the t
Charra [1.4K]

Answer:

123.30 m

Explanation:

Given

Speed, u = 22 m/s

acceleration, a = 1.40 m/s²

time, t = 7.30 s

From equation of motion,

                       v = u + at

where,

v is the final velocity

u is the initial velocity

a is the acceleration

t is time  

                       V = at + U

using equation  v - u = at to get line equation for the graph of the motion of the train on the incline plane

                       V_{x} = mt + V_{o}      where m is the slope

Comparing equation (1) and (2)

V = V_{x}

a = m    

U = V_{o}

Since the train slows down with a constant acceleration of magnitude 1.40 m/s² when going up the incline plane. This implies the train is decelerating. Therefore, the train is experiencing negative acceleration.

          a = -  1.40 m/s²

Sunstituting a = -  1.40 m/s² and  u = 22 m/s

                        V_{x} = -1.40t + 22

                            V_{x} = -1.40(7.30) + 22

                             V_{x} = -10.22 + 22

                             V_{x} = 11. 78 m/s

The speed of the train at 7.30 s is 11.78 m/s.

The distance traveled after 7.30 sec on the incline is the area cover on the incline under the specific interval.

           Area of triangle +  Area of rectangle

          [\frac{1}{2} * (22 - 11.78) * (7.30)]  + [(11.78 - 0) * (7.30)]

                           = 37.303 + 85.994

                           = 123. 297 m

                           ≈ 123. 30 m

                 

4 0
2 years ago
Read 2 more answers
For a steady two-dimensional flow, identify the boundary layer approximations.
Georgia [21]

Answer:

  • The velocity component in the flow direction is much larger than that in the normal direction ( A )
  • The temperature and velocity gradients normal to the flow are much greater than those along the flow direction ( b )

Explanation:

For a steady two-dimensional flow the boundary layer approximations are The velocity component in the flow direction is much larger than that in the normal direction and The temperature and velocity gradients normal to the flow are much greater than those along the flow direction

assuming Vx ⇒ V∞ ⇒ U and Vy ⇒ u from continuity equation we know that

Vy << Vx

4 0
3 years ago
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