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3 years ago

Starting from rest and moving in a straight line, a cheetah

Physics

1 answer:

UNO [17]3 years ago

3 0

The average acceleration of the cheetah is 7.75 m/s²

The given parameters;

velocity of the cheetah, v = 31 m/s

time of motion, t = 4 seconds

The average acceleration of the cheetah is calculated as follows;

$Average \ acceleration = \frac{\Delta velocity}{\Delta time } \\\\Average \ acceleration = \frac{v-u}{t}$

where;

v is the final velocity

u is the initial velocity

t is the time of motion

Assuming the cheetah started from rest, the initial velocity, u = 0

The average acceleration is calculated as follows;

$Average \ acceleration = \frac{v-u}{t} = \frac{31-0}{4} = 7.75 \ m/s^2$

Thus, the average acceleration of the cheetah is 7.75 m/s²

Learn more here: brainly.com/question/17280180

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A particle is projected with velocity v0 directly up a slope which makes an angle α with the horizontal. Assume frictionless mot

Nikolay [14]

Answer:

a)T total = 2*Voy/(g*sin( α ))

b)α = 0º , T total≅∞ (the particle, goes away horizontally indefinitely)

α = 90º, T total=2*Voy/g

Explanation:

Velocity in the Y axis:

Voy=Vo*sinα

Time to reach the maximal height :

Kinematics equation: Vfy=Voy-at

a=g*sinα ; g is gravity

if Vfy=0 ⇒ t=T ; time to reach the maximal height

so:

0=Voy-g*sin( α )*T

T=Voy/(g*sin( α ))

Time required to return to the starting point:

After the object reaches its maximum height, the object descends to the starting point, the time it descends is the same as the time it rises.

So T total= 2T = 2*Voy/(g*sin( α ))

α = 0º , sinα=0

The particle goes totally horizontal, goes away indefinitely

T total= 2*Voy/(g*sin( α )) ≅∞

α = 90º, sinα=1

T total=2*Voy/g

6 0

3 years ago

Using this formula Vj = V; + at, If a vehicle starts from rest

Furkat [3]

Answer:

<em>The final speed of the vehicle is 36 m/s</em>

Explanation:

<u>Uniform Acceleration</u>

When an object changes its velocity at the same rate, the acceleration is constant.

The relation between the initial and final speeds is:

$v_f=v_o+a.t$

Where:

vf = Final speed

vo = Initial speed

a = Constant acceleration

t = Elapsed time

The vehicle starts from rest (vo=0) and accelerates at a=4.5 m/s2 for t=8 seconds. The final speed is:

$v_f=0+4.5*8$

$v_f=36\ m/s$

The final speed of the vehicle is 36 m/s

5 0

3 years ago

A body is thrown up with a velocity of 78.4 m per second.How high will it rise and how much time it will take to return to its p

a_sh-v [17]

Answer:

The maximum height reached by the body is 313.6 m

The time to return to its point of projection is 8 s.

Explanation:

Given;

initial velocity of the body, u = 78.4 m/s

at maximum height (h) the final velocity of the body (v) = 0

The following equation is applied to determine the maximum height reached by the body;

v² = u² - 2gh

0 = u² - 2gh

2gh = u²

h = u²/2g

h = (78.4²) / (2 x 9.8)

h = 313.6 m

The time to return to its point of projection is calculated as follows;

at maximum height, the final velocity becomes the initial velocity = 0

h = v + ¹/₂gt²

h = 0 + ¹/₂gt²

h = ¹/₂gt²

2h = gt²

t² = 2h/g

$t = \sqrt{\frac{2h}{g} } \\\\t = \sqrt{\frac{2\times 313.6}{9.8} }\\\\t = 8 \ s$

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3 years ago

The peak luminosity of a white dwarf supernova is around 1010 Lsun, and it remains brighter than 108 Lsun for about 150 days. In

Airida [17]

Answer:

Explanation: find the attached solution below

8 0

3 years ago

Please I need help on Physics ASAP!!!

Dafna11 [192]

Answer:

you cant cheat on test

Explanation:

you will get repoted

7 0

3 years ago