What is the average speed of a cheetah that runs 70 m in 2.5 seconds?

A ski lift is used to transport people from the base of a hill to the top. If the lift leaves the

Liono4ka [1.6K]

The energy of the ski lift at the base is kinetic energy:
$K= \frac{1}{2}mv^2$
where m is the mass of the ski lift+the people carried, and $v=15.5 m/s$ is velocity at the base.
As long as the ski lift goes upward, its velocity decreases and its kinetic energy converts into potential energy. Eventually, when it reaches the top, its final velocity is v=0, so no kinetic energy is left and it has all converted into gravitational potential energy, which is
$U=mgh$
where $g=9.81 m/s^2$ and h is the height at the top of the hill.

So, since the total energy must conserve, we have
$U=K$
and so
$mgh = \frac{1}{2}mv^2$
from which we find the height:
$h= \frac{v^2}{2g}= \frac{(15.5m/s)^2}{2\cdot 9.81 m/s^2}=12 m$

8 0

3 years ago

How many work is done when a force of 33n pulls wagon 13meters

goldenfox [79]

Work = force x distance
$13 \times 33 = 429$
the answer is 429 joules

good luck

3 0

3 years ago

Consider the system consisting of the box and the spring, but not Earth. How does the energy of the system when the spring is fu

BabaBlast [244]

Answer:

the energy when it reaches the ground is equal to the energy when the spring is compressed.

Explanation:

For this comparison let's use the conservation of energy theorem.

Starting point. Compressed spring

Em₀ = K_e = ½ k x²

Final point. When the box hits the ground

Em_f = K = ½ m v²

since friction is zero, energy is conserved

Em₀ = Em_f

1 / 2k x² = ½ m v²

v = $\sqrt{ \frac{k}{m} }$ x

Therefore, the energy when it reaches the ground is equal to the energy when the spring is compressed.

5 0

3 years ago

Car a runs a red light and broadsides car b, which is stopped and waiting to make a left turn. car a has a mass of 1,800kg. car

frez [133]

The law of conservation of momentum tells us that momentum is conserved, therefore total initial momentum should be equal to total final momentum. In this case, we can expressed this mathematically as:

mA vA + mB vB = m v

where, m is the mass in kg, v is the velocity in m/s

since m is the total mass, m = mA + mB, we can write the equation as:

mA vA + mB vB = (mA + mB) v

furthermore, car B was at a stop signal therefore vB = 0, hence

mA vA + 0 = (mA + mB) v

1800 (vA) = (1800 + 1500) (7.1 m/s)

7 0

3 years ago

. A 40.0-kg child standing on a frozen pond throws a 0.500-kg stone to the east with a speed of 5.00 m/s. Neglecting friction be

tamaranim1 [39]

Answer:

Explanation:

A 40kg child throw stone of 0.5kg

At a direction of 5m/s

Recoil can be calculated using recoil of a gun formula

m_1•v_1 + m_2•v_2

m_1•v_1 = -m_2•v_2

The negative sign show that the momentum of the boy is directed oppositely to that of the stone

m_1 Is mass of boy

v_1 is the recoil velocity of the boy

m_2 is mass of stone

v_2 is the velocity of stone

Then,

m_1•v_1 = -m_2•v_2

40•v_1 = -0.5 × 5

40•v_1 = -2.5

v_1 = -2.5 / 40

v_1 = -0.0625 m/s

The recoil velocity of the boy is 0.0625 m/s

6 0

3 years ago