Subtract 7.987 m - 0.54 m and the final answer must be in decimal form

A lion has a mass of 45 kg. Answer the following questions about it, using correct units. a. The lion runs at a speed of 14.2 m/

Eva8 [605]

A) The kinetic energy of an object is given by:
$K= \frac{1}{2}mv^2$
where m is the mass of the object, and v its speed. For the lion in our problem, m=45 kg and v=14.2 m/s, so its kinetic energy is
$K= \frac{1}{2}mv^2= \frac{1}{2}(45 kg)(14.2 m/s)^2=4537 J$

b) the increase in gravitational potential energy of the lion is given by:
$\Delta U = mg \Delta h$
where g is the gravitational acceleration, and $\Delta h$ is the increase in altitude of the lion. In this problem, $\Delta h=28 m$ , so the increase in gravitational potential energy is
$\Delta U=mg \Delta h=(45 kg)(9.81 m/s^2)(28 m)=12361 J$

c) When the fox reaches the top of the tree, its gravitational potential energy is
$U=mgh=(1.8 kg)(9.81 m/s^2)(3.8 m)=67 J$
As it jumps, its kinetic energy is
$K= \frac{1}{2}mv^2= \frac{1}{2}(1.8 kg)(8.1 m/s)^2=59 J$
So the total mechanical energy of the fox as it jumps is
$E=U+K=67 J + 59 J =126 J$

6 0

3 years ago

A pirate ship shoots a cannon towards an oncoming ship. The cannon ball has a mass of 25 kg and the ship has a mass of 2500 kg.

mars1129 [50]

Answer:

25 m/s in the opposite direction with the ship recoil velocity.

Explanation:

Assume the ship recoil velocity and velocity of the cannon ball aligns. By the law of momentum conservation, the momentum is conserved before and after the shooting. Before the shooting, the total momentum is 0 due to system is at rest. Therefore, the total momentum after the shooting must also be 0:

$m_sv_s + m_bv_b = 0$

where $m_s = 2500 kg, m_b = 25 kg$ are masses of the ship and ball respectively. $v_s = 0.25 m/s, v_b$ are the velocities of the ship and ball respectively, after the shooting.

$2500*0.25 + 25*v_b = 0$

$25v_b = -2500*0.25$

$v_b = -2500*0.25/25 = -25 m/s$

So the cannon ball has a velocity of 25 m/s in the opposite direction with the ship recoil velocity.

3 0

3 years ago

an airplane releases a ball as it flies parallel to the ground at a height of 235m. if the ball lands on the ground exactly at 2

Oksanka [162]

<span>When the question says the ball lands a distance of 235 meters from the release point, we can assume this means the horizontal distance is 235 meters. Let's calculate the time for the ball to fall 235 meters to the ground. y = (1/2)gt^2 t^2 = 2y / g t = sqrt{ 2y / g } t = sqrt{ (2) (235 m) / (9.81 m/s^2) } t = 6.9217 s We can use the time t to find the horizontal speed. v = d / t v = 235 m / 6.9217 s v = 33.95 m/s Since the horizontal speed is the speed of the plane, the speed of the plane is 33.95 m/s</span>

7 0

3 years ago

A vertically hung spring has a spring constant of 150. newtons per meter. A

stiks02 [169]

What fraction of stipend triangle is a shaded triangle? What fraction of the spotted triangle is a shaded triangle? Use >,

5 0

4 years ago

The __________-second rule applies to any speed in ideal weather and road conditions.

SOVA2 [1]

The two-second rule.

It is a common guideline to follow while driving.

It means that any given driver should be AT LEAST two seconds behind any vehicle that is driving in front of his vehicle. It might apply for any kind of vehicle.

8 0

3 years ago