1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
natulia [17]
3 years ago
5

What is the minimum speed with which he’d need to run off the edge of the cliff to make it safely to the far side of the river?

the answer was 6 m/s the world-record time for the 100 m dash is approximately 10 s. given this, is it reasonable to expect brady to be able to run fast enough to achieve brady's leap?
a.yes, the obtained speed is less than the world-record.
b.yes, the obtained speed is almost equal to the world-record.
c.no, the obtained speed is greater than the world-record.
d.no, the obtained speed is almost equal to the world-record.

Physics
2 answers:
Vladimir [108]3 years ago
7 0

Part (a): The minimum speed of Brady should be \boxed{6\text{ m/s}} to cross the cliff.

Part (b): The speed of Brady’s leap is possible to achieve, as it is less than the speed of world record.

Further Explanation:

(a)

Brady jumps from a cliff to go on other side. He crosses 22\text{ ft} horizontal distance from a height of 20\text{ft}. To jump from cliff, he follows the Newton’s law of motion.

Given:

The horizontal distance is 22\text{ ft}.

The vertical distance is 20\text{ft}.

Concept:

The horizontal distance in meter is 6.71\text{ m}.

The vertical distance in meter is 6.10\text{ m}.

To obtain the time of flight, applying one of the equation of motion given as:

s=ut+\dfrac{1}{2}at^2

Substitute 0\text {m/s} for u and rearrange the above equation for t :

t=\sqrt{\dfrac{2s}{a} }

Substitute 6.10\text{ m} for s and 9.81\text{ m}/\text{s}^2 for a in above equation.

\begin{aligned}t&=\sqrt{\dfrac{2\times6.10}{9.81}}\text{ s}\\&=1.12\text{ s}\end{aligned}

The time of flight and time taken to cover horizontal distance are equal.

Horizontal velocity of Brady given as:

\begin{aligned}v&=\dfrac{6.71}{1.12}\text{ m/s}\\&\approx6\text{ m/s}\end{aligned}

Thus, the minimum speed of Brady should be \boxed{6\text{ m/s}} to cross the cliff.

(b)  

The minimum speed with which Brady is running is 6\text{ m/s}.

The speed of world record given as:

\begin{aligned}V&=\frac{100}{10}\text{ m/s}\\&=10\text{ m/s}\end{aligned}  

Thus, the speed of Brady’s leap is possible to achieve, as it is less than the speed of world record.

Learn more:

1. Projectile motion of a body: brainly.com/question/11023695

2. Body in pure rolling motion: brainly.com/question/9575487

3. Newton’s law of motion: brainly.com/question/6125929

Answer Details:

Grade: High School

Subject: Physics

Chapter: Kinematics

Keywords:

1780, Brady's Leap, Captain, Sam, U.S. Continental Army, horizontally, cliff, Ohio's Cuyahoga, gorge, leap, 22 ft, 20 ft, minimum, speed, river, 100 m, dash, 10 s, jump, time of flight, vertical and distance.

Angelina_Jolie [31]3 years ago
4 0

The answer is

A. Yes, the obtained speed is less than the world record

The explanation:

when the obtained speed is 6 m /s

and the world record speed = distance / time = 100 m / 10 s = 10 m/s

So, Yes, the obtained speed is less than the world record

Not only is it less, its also a reasonable average speed for a somewhat athletic person. Therefore, the leap is entirely possible.

-Samuel Brady gained his lasting notoriety for his leap over the Cuyahoga River around 1780 in what is now Kent, Ohio. After following a band of Indians into the Ohio country, a failed ambush attempt resulted in the band chasing Brady near the Cuyahoga River. To avoid capture, Brady leaped across a 22-foot (6.7 m) wide gorge of the river (which was widened considerably in the 1830s for construction of the Pennsylvania and Ohio Canal) and fled to a nearby lake where he hid in the water under a fallen tree using a reed for air.

You might be interested in
When running a half marathon (13.1 miles), it took Kevin 8 minutes to run from mile marker 1 to mile marker 2, and 18 minutes to
Vsevolod [243]

Answer:

It took Kevin 26 minutes to run from markers 1 to 4

His average speed from mile markers 1 to 4 is 0.154 miles/minute

Kevin must run by average speed 0.1 miles/minute to finish the race

Explanation:

Lets explain how to solve the problem

A half marathon 13.1 miles

Kevin took 8 minutes to run from mile marker 1 to mile marker 2 and

18 minutes to run from mile marker 2 to mile marker 4

→ He took 8 minutes and 18 minutes to run from marker 1 to marker 4

→ The total time of the first 4 marker = 8 + 18 = 26 minutes

<em>It took Kevin 26 minutes to run from markers 1 to 4</em>

<em></em>

Average speed is total distance divided by total time

The average speed of Kevin as he ran from mile marker 1 to mile

marker 4 is the 4 miles divides by 26 minutes

→ Average speed = 4 ÷ 26 = \frac{2}{13} = 0.154 miles/minute

<em>His average speed from mile markers 1 to 4 is 0.154 miles/minute</em>

<em></em>

It took Kevin 71 minutes to pass mile marker 9

Kevin need to complete the race in 112 minutes, then what must

Kevin's average speed be as he travels from mile marker 9 to the

finish line?

The total distance of the race is 13.1 miles, he ran 9 miles

→ The remaining distance = 13.1 - 9 = 4.1 miles

He must run 4.1 miles to complete the race

The total time is 112 minutes, he used 71 minutes to run the first 9 miles

→ The remaining time = 112 - 71 = 41 minutes

He must finish the 4.1 miles in 41 minutes

→ His average speed for the last part of the race = 4.1 ÷ 41 = 0.1 mi/min

<em>Kevin must run by average speed 0.1 miles/minute to finish the race</em>

7 0
3 years ago
Read 2 more answers
Physics question help
S_A_V [24]

Answer:

not know sorry sorry

Explanation:

sorry sorry

8 0
3 years ago
HALP me!! This is a physics question.
emmasim [6.3K]
<h2>Answer:</h2>

<u>Distance covered is 6.9 meters</u>


<h2>Explanation:</h2>

Data given:

Work Done = 345 kJ = 345000 J

Force = 5 x 10 ^ 4 =  50000 N

Distance = ?


Solution:

As we know that

Work Done = Force applied x Distance covered

By arranging the equation we get

Work / Force = Distance covered

By putting the values

345000 / 50000 = 6.9

So distance covered is 6.9 meters

4 0
3 years ago
Read 2 more answers
An electron is projected with an initial speed of 5 3.2 10 / ⋅ m s directly toward a proton that is fixed in place. If the elect
Dvinal [7]

Answer:

The distance is d =1.66*10^{-9}m

Explanation:

From the question we are told that

         The initial speed of the  electron is v_i  = 3.2 *10^5 m/s

         The mass of electron is m = 9.11*10^{-31}kg

         Let d be the distance between the electron and the proton when the speed of the electron instantaneously equal to twice the initial value

         Let KE_i be the initial kinetic energy of the electron \

          Let KE_d be the kinetic energy of the electron at the distance d from the proton

  Considering that energy is conserved,

  The energy at the initial position of the electron = The energy at the final position of the electron

      i.e

             KE_i +U_1 = KE_d + U_2

U_1 \ and \ U_2 are the potential energy at the initial  position of the electron and at distance d of the electron to the proton

                Here U_1 = 0

So the equation becomes

                   \frac{1}{2} mv_i^2 = \frac{1}{2} mv_d^2  + \frac{kq_1 q_2}{d}

Here q_1 \ and  \ q_2 are the charge on the electron and the proton and their are the same since a charge on an electron is equal to charge on a proton

 k is electrostatic constant with value 8.99*10^9 N \cdot m^2 /C^2

i.e q = 1.602 *10^{-19}C

           v_d is the velocity at distance d from the proton = 2v_i

  So the equation becomes

             \frac{1}{2}mv_i^2 = \frac{1}{2} m (2v_i)^2 -\frac{k(q)^2}{d}

            \frac{1}{2} mv_i^2  = 4 [\frac{1}{2}mv_i^2 ]- \frac{k(q)^2}{d}

           3[\frac{1}{2}mv_i^2 ] = \frac{k(q)^2}{d}

Making d the subject of the formula

           d = \frac{2k(q)^2}{3mv_i^2}

              = \frac{2* 8.99*10^9 *(1.602*10^{-19}^2)}{3 * 9.11*10^{-31} *(3.2*10^5)^2}

              =1.66*10^{-9}m

             

           

         

                 

   

6 0
3 years ago
what features of stars are plotted on the Hertzsprung-Russell diagram? density and mass B. physical structure and composition ap
aliina [53]

Answer:

its luminosity (brightness) and temperature

Explanation:

3 0
3 years ago
Read 2 more answers
Other questions:
  • How many orbitals are allowed in a subshell if the angular momentum quantum number for electrons in that subshell is 3?
    14·1 answer
  • It takes 15 min to drive 6.0 mi in a straight line to the local hospital. It takes 10 min to go the last 3.0 mi, 2.0 min to go t
    11·1 answer
  • Mixtures occur when two or more materials are combined and each material keeps its own properties. All of the following are mixt
    11·2 answers
  • Which of the following should NOT be discussed during safety training
    7·1 answer
  • 12. What is the net force on a 1-kg apple held at rest in your hand?
    11·1 answer
  • What would happen if the moon was hit by an asteroid
    6·1 answer
  • The crew of a cargo plane wishes to drop a crate of supplies on a target below. To hit the target, when should the crew drop the
    12·2 answers
  • In your everyday life you come across a range of motions in which
    14·1 answer
  • Light of the same wavelength passes through two diffraction gratings. One grating has 4000 lines/cm, and the other one has 6000
    7·1 answer
  • A train is traveling at a speed of 80\,\dfrac{\text{km}}{\text{h}}80 h km ​ 80, start fraction, start text, k, m, end text, divi
    9·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!