Are small molecular units joined together in large molecules

A girl of mass m1=60 kilograms springs from a trampoline with an initial upward velocity of v1=8.0 meters per second. At height

AleksandrR [38]

a) 5.0 m/s

This first part of the problem can be solved by using the conservation of energy. In fact, the mechanical energy of the girl just after she jumps is equal to her kinetic energy:

$E_i=\frac{1}{2}m_1v_1^2$

where m1 = 60 kg is the girl's mass and v1 = 8.0 m/s is her initial velocity.

When she reaches the height of h = 2.0 m, her mechanical energy is sum of kinetic energy and potential energy:

$E_f = \frac{1}{2}m_1 v_2 ^2 + m_1 gh$

where v2 is the new speed of the girl (before grabbing the box), and h = 2.0m. Equalizing the two equations (because the mechanical energy is conserved), we find

$\frac{1}{2}m_1 v_1^2 = \frac{1}{2}m_1 v_2 ^2 + m_1 gh\\v_1^2 = v_2^2 +2gh\\v_2 = \sqrt{v_1^2 -2gh}=\sqrt{(8.0 m/s)^2-(2)(9.8 m/s^2)(2.0 m)}=5.0 m/s$

b) 4.0 m/s

After the girl grab the box, the total momentum of the system must be conserved. This means that the initial momentum of the girl must be equal to the total momentum of the girl+box after the girl catches the box:

$p_i = p_f\\m_1 v_2 = (m_1 + m_2) v_3$

where m2 = 15 kg is the mass of the box. Solving the equation for v3, the combined velocity of the girl+box, we find

$v_3 = \frac{m_1 v_2}{m_1 + m_2}=\frac{(60 kg)(5.0 m/s)}{60 kg+15 kg}=4 m/s$

c) 2.8 m

We can use again the law of conservation of energy. The total mechanical energy of the girl after she catches the box is sum of kinetic energy and potential energy:

$E_i = \frac{1}{2}(m_1+m_2) v_3^2 + (m_1+m_2)gh=\frac{1}{2}(75 kg)(4 m/s)^2+(75 kg)(9.8 m/s^2)(2.0m)=2070 J$

While at the maximum height, the speed is zero, so all the mechanical energy is just potential energy:

$E_f = (m_1 +m_2)gh_{max}$

where h_max is the maximum height. Equalizing the two expressions (because the mechanical energy must be conserved) and solving for h_max, we find

$E_i = (m_1+m_2)gh_{max}\\h_{max}=\frac{E_i}{(m_1+m_2)g}=\frac{2070 J}{(75 kg)(9.8 m/s^2)}=2.8 m$

4 0

3 years ago

What is the transfer of energy by electromagnetic waves

Ann [662]

The transfer of energy by electromagnetic waves is called electromagnetic radiation

3 0

3 years ago

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The following three questions refer to a situation in which a driver is in a car that crashes into a solid wall. The car comes t

Andreas93 [3]

Answer:

1) p₀ = 45000 N / s , p₀ '= 1800 , b) I = -45000 N s , I = 1800 Ns

Explanation:

Impulse equals the change in momentum

I = Δp

1) the initial moment of the car

p₀ = M v

p₀ = 1500 30

p₀ = 45000 N / s

the change at the moment is

Δp = 45000

because the end the car is stopped

moment of the person

P₀ ’= m v

p₀ '= 60 30

p₀ '= 1800

D₀ '= 1800

2) of the momentum change impulse ratio

car

I = Δp

I = -45000 N s

person

I = Δpo '

I = 1800 Ns

3) the object that give the momentum to stop the wall motoring

The person is stopped by the impulse given by the car

a) This area is the one that absorbs most of the vehicle impulse

be) If using a safety painter, the time during which the greater force will act, therefore the lessons decrease

c) The air bag helps reduction in the speed of the person relatively quickly.

6 0

3 years ago

A fold is a _ in a rock , and a fault is a _in a rock?

777dan777 [17]

A geological fold occurs when one or a stack of originally flat and planar surfaces, such as sedimentary strata, are bent or curved as a result of permanent deformation.

So A fold is a Bend? in a rock. Maybe.

A fault is a planar fracture or discontinuity in a volume of rock, across which there has been significant displacement as a result of rock-mass movement.

3 0

3 years ago

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It takes a minimum distance of 57.46 m to stop a car moving at 13.0 m/s by applying the brakes (without locking the wheels). Ass

vivado [14]

Answer:

The minimum stopping distance when the car is moving at

29.0 m/sec = 285.94 m

Explanation:

We know by equation of motion that,

$v^{2}=u^{2}+2\cdot a \cdot s$