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
alexdok [17]
3 years ago
8

A cheetah can run at a maximum speed 97.8 km/h and a gazelle can run at a maximum speed of 78.2 km/h. If both animals are runnin

g at full speed, with the gazelle 96.8 m ahead, how long before the cheetah catches its prey? Answer in units of s.
The cheetah can maintain its maximum speed for only 7.5 s. What is the minimum distance the gazelle must be ahead of the cheetah to have a chance of escape? (After 7.5 s the speed of cheetah is less than that of the gazelle.) Answer in units of m.
Physics
2 answers:
emmainna [20.7K]3 years ago
4 0

(1)

Cheetah speed: v_c = 97.8 km/h=27.2 m/s

Its position at time t is given by

S_c (t)= v_c t (1)

Gazelle speed: v_g = 78.2 km/h=21.7 m/s

the gazelle starts S0=96.8 m ahead, therefore its position at time t is given by

S_g(t)=S_0 +v_g t (2)

The cheetah reaches the gazelle when S_c=S_g. Therefore, equalizing (1) and (2) and solving for t, we find the time the cheetah needs to catch the gazelle:

v_c t=S_0 + v_g t

(v_c -v_g t)=S_0

t=\frac{S_0}{v_c-v_t}=\frac{96.8 m}{27.2 m/s-21.7 m/s}=17.6 s


(2) To solve the problem, we have to calculate the distance that the two animals can cover in t=7.5 s.

Cheetah: S_c = v_c t =(27.2 m/s)(7.5 s)=204 m

Gazelle: S_g = v_g t =(21.7 m/s)(7.5 s)=162.8 m

So, the gazelle should be ahead of the cheetah of at least

d=S_c -S_g =204 m-162.8 m=41.2 m

tatuchka [14]3 years ago
3 0

The cheetah catches the prey \fbox{17.78 s} before the gazelle.

The minimum distance the gazelle must be ahead of the cheetah to have a chance of escape is \fbox{40.83 m}.

Further Explanation:

The speed is the rate of change of the distance and the relative speed is the speed of the one object in respect of another object. If two body moves in same direction then the relative speed in respect of the one body is given by subtract of both speed and it they are moving in opposite direction the relative speed in respect of the one body is given by sum of both speeds.

Given:

The maximum speed of the cheetah is 97.8 km/h.

The maximum speed of the gazelle is 78.2 km/h.

The distance between the starting points of cheetah and the gazelle is 96.8 m.

Concept:

The expression for the position of the cheetah at some time t is:

{S_1}={v_c}t                                            …… (1)

Here, {S_1}the position of the cheetah at some time t, {v_c} is the speed of the cheetah and {S_1} is the distance covered at time t.

The expression for the position of the gazelle at some time t is:

{S_2}={S_o}+{v_g}t                                              …… (2)

Here, {S_2} the position of the gazelle at some time t, {v_g} is the speed of the gazelle, {S_o} is the distance between the starting points of cheetah and the gazelle.

At the time when the cheetah reached the gazelle then {S_1}={S_2}.

Equate equation (1) and (2).

\begin{aligned}{v_c}t&={S_o}+{v_g}t\hfill\\{v_c}t-{v_g}t&={S_o}\hfill\\t\left( {{v_c} - {v_g}}\right)&={S_o}\hfill\\t&=\frac{{{S_o}}}{{\left( {{v_c} - {v_g}} \right)}}\hfill\\\end{aligned}

Substitute 97.8 km/h for {v_c}, 78.2 km/h for {v_g} and 96.8 m for {S_o} in the above equation.

\begin{aligned}t&=\frac{{96.8{\text{ m}}}}{{\left( {97.8 - 78.2} \right){\text{km/h}}\left({\frac{{1000{\text{ m}}}}{{1{\text{ km}}}}} \right)\left( {\frac{{1{\text{ h}}}}{{{\text{3600 s}}}}} \right)}}\\&=\frac{{96.8{\text{ m}}}}{{{\text{19}}{\text{.6 m/s}}\left( {\frac{5}{{18}}}\right)}}\\&=17.78{\text{s}}\\\end{aligned}

Therefore, the cheetah catches the prey 17.78 s before the gazelle.

To calculate the minimum distance the gazelle must be ahead of the cheetah to have a chance of escape we should subtract the distance covered by the cheetah and gazelle in 7.5 s.

The expression for the distance covered is:

S = vt

Here, S is the distance covered, v is the speed and t is the travel time.

Substitute 97.8 km/h for v and 7.5 s for t in the above equation.

\begin{aligned}S&=\left( {97.8{\text{ km/h}}} \right)\left( {\frac{{1000{\text{ m}}}}{{1{\text{ km}}}}} \right)\left( {\frac{{1{\text{ h}}}}{{{\text{3600 s}}}}} \right)\left( {7.5{\text{ s}}} \right)\\&=\left( {97.8{\text{ m/s}}} \right)\left( {\frac{5}{{18}}} \right)\left( {7.5{\text{ s}}} \right)\\&=203.75{\text{m}}\\\end{aligned}

Therefore, the distance covered by the cheetah in 7.5 s is 203.75 m.

Substitute 78.2 km/h for v and 7.5 s for t in the above equation.

\begin{aligned}S&=\left( {78.2{\text{ km/h}}} \right)\left( {\frac{{1000{\text{ m}}}}{{1{\text{ km}}}}} \right)\left( {\frac{{1{\text{ h}}}}{{{\text{3600 s}}}}} \right)\left( {7.5{\text{ s}}} \right)\\&=\left( {78.2{\text{ m/s}}} \right)\left( {\frac{5}{{18}}} \right)\left( {7.5{\text{ s}}} \right)\\&=162.92{\text{ m}}\\\end{aligned}

Therefore, the distance covered by gazelle in 7.5 s is 162.92 m.

The minimum distance the gazelle must be ahead of the cheetah to have a chance of escape is:

\begin{aligned}d&=\left( {203.75 - 162.92} \right){\text{ m}}\\&=40.83{\text{ m}}\\\end{aligned}

Therefore, the minimum distance the gazelle must be ahead of the cheetah to have a chance of escape is \fbox{40.83 m}.

Learn more:

1.  Conservation of momentum brainly.com/question/7031524.

2.  Motion under friction brainly.com/question/11023695.

3. Motion under force brainly.com/question/4033012.

Answer Details:

Grade: High school

Subject: Physics

Chapter: Kinematics

Keywords:

Cheetah, 97.8 km/h, gazelle, 78.2 km/h, full speed, 96.8 m. ahead, catches, prey, 7.5 s, minimum, ahead, escape, answer, unit, s, m, 40.83 m.

You might be interested in
A 25 newton force applied on an object moves it 50 meters. The angle between the force and displacement is 40.0 degrees. What is
soldier1979 [14.2K]
W =F triangle d cosine0. F = 25 Newton’s. Delta d = 50 meters. Theta =40.0 degrees
3 0
3 years ago
What is the definition of permitted orbits ?
givi [52]
.The path of a celestial body or an artificial satellite as it revolves around another body due to their mutual gravitational <span>attraction.</span>
7 0
3 years ago
An electron is placed on a line connecting two fixed point charges of equal charge but the opposite sign. The distance between t
viktelen [127]

Answer:

a)    F_net = 6.48 10⁻¹⁸ ( \frac{1}{x^2} + \frac{1}{(0.300-x)^2} ),   b) x = 0.15 m

Explanation:

a) In this problem we use that the electric force is a vector, that charges of different signs attract and charges of the same sign repel.

The electric force is given by Coulomb's law

         F =k \frac{q_2q_2}{r^2}

         

Since when we have the two negative charges they repel each other and when we fear one negative and the other positive attract each other, the forces point towards the same side, which is why they must be added.

          F_net= ∑ F = F₁ + F₂

let's locate a reference system in the load that is on the left side, the distances are

left side - electron       r₁ = x

right side -electron     r₂ = d-x

let's call the charge of the electron (q) and the fixed charge that has equal magnitude Q

we substitute

          F_net = k q Q  ( \frac{1}{r_1^2}+ \frac{1}{r_2^2})

          F _net = kqQ  ( \frac{1}{x^2} + \frac{1}{(d-x)^2} )

         

let's substitute the values

          F_net = 9 10⁹  1.6 10⁻¹⁹ 4.50 10⁻⁹ ( \frac{1}{x^2} + \frac{1}{(0.30-x)^2} )

          F_net = 6.48 10⁻¹⁸ ( \frac{1}{x^2} + \frac{1}{(0.300-x)^2} )

now we can substitute the value of x from 0.05 m to 0.25 m, the easiest way to do this is in a spreadsheet, in the table the values ​​of the distance (x) and the net force are given

x (m)        F (N)

0.05        27.0 10-16

0.10          8.10 10-16

0.15          5.76 10-16

0.20         8.10 10-16

0.25        27.0 10-16

b) in the adjoint we can see a graph of the force against the distance, it can be seen that it has the shape of a parabola with a minimum close to x = 0.15 m

4 0
3 years ago
Singly charged uranium-238 ions are accelerated through a potential difference of 2.90 kV and enter a uniform magnetic field of
Lera25 [3.4K]

Answer: 0.091 m

Explanation:

r = 1/B * √(2mV/e), where

r = radius of their circular path

B = magnitude of magnetic field = 1.29 T

m = mass of Uranium -238 ion = 238 * amu = 238 * 1.6*10^-27 kg

V = potential difference = 2.9 kV

e = charge of the Uranium -238 ion = 1.6*10^-19 C

r = 1/1.29 * √[(2 * 238 * 1.6*10^-27 * 2900) / 1.6*10^-19]

r = 1/1.29 * √(2.21*10^-21 / 1.6*10^-19)

r = 1/1.29 * √0.0138

r = 1/1.29 * 0.117

r = 0.091 m

Therefore, the radius of their circular path is 0.091 m

6 0
3 years ago
A cardinal (Richmondena cardinalis) of mass 3.70×10−2 kg and a baseball of mass 0.144 kg have the same kinetic energy. What is t
Radda [10]

Answer:

\frac{p_{c}}{p_{b}}\approx 0.507

Explanation:

Since the cardinal and ball have the same kinetic energy, it is possible to determine the ratio between speeds. (c for cardinal, b for baseball)

K_{c} = K_{b}

\frac{1}{2}\cdot m_{c}\cdot v_{c}^{2}= \frac{1}{2}\cdot m_{b}\cdot v_{b}^{2}

\frac{v_{c}}{v_{b}}=\sqrt{\frac{m_{b}}{m_{c}} }

The ratio is obtained by multiplying each side by \frac{m_{c}}{m_{b}}:

\frac{p_{c}}{p_{b}}=\frac{m_{c}}{m_{b}}\cdot \sqrt{\frac{m_{b}}{m_{c}} }

\frac{p_{c}}{p_{b}}= \sqrt{\frac{m_{c}}{m_{b}} }

The value of this ratio is:

\frac{p_{c}}{p_{b}}\approx 0.507

3 0
3 years ago
Read 2 more answers
Other questions:
  • Is this statement true or false?<br><br> The littoral zone is exposed to air.
    13·1 answer
  • What is the acceleration of a car that starts from rest and attains a final speed of 20 m
    15·1 answer
  • A bar magnet moves through a loop of wire with constant velocity, and the north pole enters the loop first the induced current w
    5·2 answers
  • If you change the magnetic field in a closed loop of wire You induce in the loop a(n) A. current. B. voltage. C. electric field.
    13·2 answers
  • What elements are bridges exposed to yearly?
    9·1 answer
  • Select the correct answer.
    15·1 answer
  • If a proton and an electron are released when they are 2.50×10^-10m apart (typical atomic distances), find the initial accelerat
    15·1 answer
  • g Two cars, car 1 and car 2 are traveling in opposite directions, car 1 with a magnitude of velocity v1=13.0 m/s and car 2 v2= 7
    5·1 answer
  • A box is being moved with a velocity v by a force p (parallel to v) along a level horizontal floor. The normal force is FN, the
    15·1 answer
  • Guys please help me on the rest of the numbers
    8·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!