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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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In the long jump, an athlete launches herself at an angle above the ground and lands at the same height, trying to travel the gr

NikAS [45]

A) 2.64t

B) 2.64h

C) 2.64D

Explanation:

The motion of the athlete is equivalent to the motion of a projectile, which consists of two independent motions:

- A uniform motion (constant velocity) along the horizontal direction

- A uniformly accelerated motion (constant acceleration) along the vertical direction

The time of flight of a projectile can be found from the equations of motion, and it is found to be

$t=\frac{2u sin \theta}{g}$

where

u is the initial speed

$\theta$ is the angle of projection

g is the acceleration due to gravity

In this problem, when the athlete is on the Earth, the time of flight is $t$ .

When she is on Mars, the acceleration due to gravity is:

$g'=0.379 g$

where g is the acceleration due to gravity on Earth. Therefore, the time of flight on Mars will be:

$t'=\frac{2usin \theta}{g'}=\frac{2u sin \theta}{0.379g}=\frac{1}{0.379}t=2.64t$

The maximum height reached by a projectile can be also found using the equations of motion, and it is given by

$h=\frac{u^2 sin^2\theta}{2g}$

where

u is the initial speed

$\theta$ is the angle of projection

g is the acceleration due to gravity

In this problem, when the athlete is on the Earth, the maximum height is $h$ .

When she is on Mars, the acceleration due to gravity is:

$g'=0.379 g$

where g is the acceleration due to gravity on Earth. So, the maximum height reached on Mars will be:

$h'=\frac{u^2 sin^2\theta}{2g'}=\frac{u^2 sin^2\theta}{(0.379)2g}=\frac{1}{0.379}h=2.64h$

The horizontal distance covered by a projectile is also found from the equations of motion, and it is given by

$D=\frac{u^2 sin(2\theta)}{g}$

where:

u is the initial speed

$\theta$ is the angle of projection

g is the acceleration due to gravity

In this problem, when the athlete is on the Earth, the horizontal distance covered is D.

When she is on Mars, the acceleration due to gravity is:

$g'=0.379 g$

where g is the acceleration due to gravity on Earth. Therefore, the horizontal distance reached on Mars will be:

$D'=\frac{u^2 sin(2\theta)}{g'}=\frac{u^2 sin(2\theta)}{(0.379)g}=\frac{1}{0.379}D=2.64D$

7 0

3 years ago

Do the math: How many seconds would it take an echo sounder’s ping to make the trip from a ship to the Challenger Deep (10,994 m

Lera25 [3.4K]

Answer:

Explanation:

Given the formula for calculating the depth in metres expressed as

depth in meters = ½ (1500 m/sec × Echo travel time in seconds)

Given depth of the challenger = 10, 994 meters, we will substitute this given value into the formula given to calculate the time take for the echo to travel.

10, 994 = depth in meters = ½ * 1500 m/sec × Echo travel time in seconds

10,994 = 750 * Echo travel time in seconds

Dividing both sides by 750;

Echo travel time in seconds = 10,994 /750

Echo travel time in seconds ≈ 14.66secs (to two decimal places)

Therefore, it would take an echo sounder’s ping 14.66secs to make the trip from a ship to the Challenger Deep and back

6 0

3 years ago

PEDALING A BIKE : ACCELERATION:: PULLING A DOGS LEASH: _____.

Alex_Xolod [135]

Pulling a dogs leash: inertia

6 0

3 years ago

What factor affects weight( what force causes weight to change)

Alex Ar [27]

Answer:

Gravitational force affects weight, weight changes with change in gravity

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3 0

3 years ago

The gravitational force between two asteroids is 6. 2 × 108 N. Asteroid Y has three times the mass of asteroid Z. If the distanc

Sliva [168]

Answer:Let m = mass of asteroid y.Because asteroid y has three times the mass of asteroid z, the mass of asteroid z is m/3.Given:F = 6.2x10⁸ Nd = 2100 km = 2.1x10⁶ mNote thatG = 6.67408x10⁻¹¹ m³/(kg-s²)The gravitational force between the asteroids isF = (G*m*(m/3))/d² = (Gm²)/(3d²)orm² = (3Fd²)/G = [(3*(6.2x10⁸ N)*(2.1x10⁶ m)²]/(6.67408x10⁻¹¹ m³/(kg-s²)) = 1.229x10³² kg²m = 1.1086x10¹⁶ kg = 1.1x10¹⁶ kg (approx)Answer: 1.1x10¹⁶ kg

Explanation:

7 0

2 years ago