The will dog catch up with the rabbit in 6 minutes assuming both their velocities remain constant during the chase.
<h3>What time will the dog catch the rabbit?</h3>
The time that the dog will catch up with the rabbit is given as follows:
Let the distance covered by the rabbit be x.
Distance covered by dog = x + 30
- Time taken = distance/speed
The time taken will be the same T
- Time taken by dog, T = (x + 30)/10
- Time taken by rabbit, T = x/5
Equating both times.
(x + 30)/10 = x/5
x = 30 m
Solving for T in equation (ii);
T = 30/5 = 6 minutes
In conclusion, time is obtained as a ratio of distance and speed.
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Answer:
9.8m/s²
Explanation:
The acceleration of the ball thrown after leaving my hand is 9.8m/s². This will be the acceleration due to gravity on the body.
- Acceleration due to gravity is caused by the pull of the earth on a massive object.
- The value of this acceleration is 9.8m/s².
- As the ball nears the surface, it comes near zero.
Sedimentary rock I believe
<span>The ball clears by 11.79 meters
Let's first determine the horizontal and vertical velocities of the ball.
h = cos(50.0)*23.4 m/s = 0.642788 * 23.4 m/s = 15.04 m/s
v = sin(50.0)*23.4 m/s = 0.766044 * 23.4 m/s = 17.93 m/s
Now determine how many seconds it will take for the ball to get to the goal.
t = 36.0 m / 15.04 m/s = 2.394 s
The height the ball will be at time T is
h = vT - 1/2 A T^2
where
h = height of ball
v = initial vertical velocity
T = time
A = acceleration due to gravity
So plugging into the formula the known values
h = vT - 1/2 A T^2
h = 17.93 m/s * 2.394 s - 1/2 9.8 m/s^2 (2.394 s)^2
h = 42.92 m - 4.9 m/s^2 * 5.731 s^2
h = 42.92 m - 28.0819 m
h = 14.84 m
Since 14.84 m is well above the crossbar's height of 3.05 m, the ball clears. It clears by 14.84 - 3.05 = 11.79 m</span>
The sun is a clear example of objects releasing radiation in nature
