All categories

3 years ago

A Lemon falls off a blimp and hits the ground 14 seconds later. How fast was it moving before it hits the ground?

1 answer:

5 0

Answer: Lemon blossoms falling off a potted lemon tree may also be caused by cool drafts, as well as under or over watering. A lemon tree dropping flowers may be a sign of drought or other changes in watering. When water is scarce, a lemon tree will drop flowers or fruit to conserve energy. Flooding, waterlogged soil or over-watering can also cause ...

Explanation:

You might be interested in

We wrap a light, nonstretching cable around a 8.00 kg solid cylinder with diameter of 30.0 cm. The cylinder rotates with negligi

antoniya [11.8K]

Answer:

h = 16.67m

Explanation:

If the kinetic energy of the cylinder is 510J:

$Kc=510=1/2*Ic*\omega c^2$

$\omega c=\sqrt{510*2/Ic}$

Where the inertia is given by:

$Ic=1/2*m_c*R_c^2=1/2*(8)*(0.15)^2=0.0225kg.m^2$

Replacing this value:

$\omega c=106.46rad/s$

Speed of the block will therefore be:

$V_b=\omega_c*R_c=106.46*0.15=15.969m/s$

By conservation of energy:

Eo = Ef

Eo = 0

$Ef = 510+1/2*m_b*V_b^2-m_b*g*h$

So,

$0 = 510+1/2*m_b*V_b^2-m_b*g*h$

Solving for h we get:

h=16.67m

3 0

3 years ago

A 125kg bumper car going 12m/s hits a 235kg bumper car going -13m/s.if the first car bounces back at -12.5m/s what is the veloci

vovikov84 [41]

According to the law of conservation of momentum:

$m_{1}v_{1}+m_{2}v_{2}=m_{1}v_{1}'+m_{2}v_{2}'$

m1 = mass of first object
m2 = mass of second object
v1 = Velocity of the first object before the collision
v2 = Velocity of the second object before the collision
v'1 = Velocity of the first object after the collision
v'2 = Velocity of the second object after the collision

Now how do you solve for the velocity of the second car after the collision? First thing you do is get your given and fill in what you know in the equation and solve for what you do not know.

m1 = 125 kg v1 = 12m/s v'1 = -12.5m/s
m2 = 235kg v2 = -13m/s v'2 = ?

$m_{1}v_{1}+m_{2}v_{2}=m_{1}v_{1}'+m_{2}v_{2}'$
$(125kg)(12m/s)+(235kg)(-13m/s)=(125kg)(-12.5m/s)+(235kg)(v_{2}'$
$1,500kg.m/s+(-3055kg.m/s)=(-1562.5kg.m/s)+(235kg)(v_{2}')$
$-1,555kg.m/s=(-1562.5kg.m/s)+(235kg)(v_{2}')$

Transpose everything on the side of the unknown to isolate the unknown. Do not forget to do the opposite operation.

$-1,555kg.m/s + 1562.5kg.m/s=(235kg)(v_{2}')$
$7.5kg.m/s=(235kg)(v_{2}')$
$(7.5kg.m/s)/(235kg)=(v_{2}')$
$0.03m/s=(v_{2}')$

The velocity of the 2nd car after the collision is 0.03m/s.

5 0

3 years ago

If a rock were dropped from a building it would hit the ground with a certain kinetic energy and velocity. If it fell from a roo

STALIN [3.7K]

Answer:

Explanation:

because if you test it they hit the ground at the same time.

5 0

4 years ago

An idea is being proposed. The steps that lead to the idea are listed below.

gavmur [86]

there were different outcomes each time.