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2 years ago

Where is energy stored in a compound?

1 answer:

5 0

Chemical energy is energy stored in the bonds of chemical compounds, like atoms and molecules. This energy is released when a chemical reaction takes place.

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An elevator motor provides 45.0 kW of power while lifting an elevator 35.0 m. If the elevator contains seven passengers each wit

mylen [45]

Find how much work ∆W is done by the motor in lifting the elevator:

P = ∆W / ∆t

where

• P = 45.0 kW = power provided by the motor

• ∆W = work done

• ∆t = 20.0 s = duration of time

Solve for ∆W :

∆W = P ∆t = (45.0 kW) (20.0 s) = 900 kJ

In other words, it requires 900 kJ of energy to lift the elevator and its passengers. The combined mass of the system is M = (m + 490.0) kg, where m is the mass of the elevator alone. Then

∆W = M g h

where

• g = 9.80 m/s² = acceleration due to gravity

• h = 35.0 m = distance covered by the elevator

Solve for M, then for m :

M = ∆W / (g h) = (900 kJ) / ((9.80 m/s²) (35.0 m)) ≈ 2623.91 kg

m = M - 490.0 kg ≈ 2133.91 kg ≈ 2130 kg

4 0

3 years ago

you find a rock in which the grains are arrenged in wavy, parallel bands of white and black crystals. What kind of rock have you

dezoksy [38]

I think your in the wrong section kid. This should be in Earth and Space Science.

7 0

2 years ago

Imagine a 10kg block moving with a speed of 20m/s. calculate the kinetic energy of this block

avanturin [10]

$E_{c}=\frac{mv^2}{2}=2000J$

7 0

2 years ago

A 1300 kg car traveling at 35 mph rear-ends a 1000 kg car traveling 25 mph. Just after the collision (but before the driver’s sl

guajiro [1.7K]

Answer

given,

before collision

mass of car A = m_a = 1300 kg

velocity of car A = v_a = 35 mph

mass of car B = m_b= 1000 kg

velocity of car B = v_b = 25 mph

after collision

V_a = 30 mph

V_b = 31.5 mph

Initial momentum

$P_1 = m_av_a + m_b v_b$

$P_1 = 1300 \times 35+ 1000 \times 25$

$P_1 =70500 Kg.m/s$

final momentum

$P_2 = m_aV_a + m_b V_b$

$P_2 = 1300 \times 30+ 1000 \times 31.5$

$P_2 =70500 Kg.m/s$

here initial momentum is equal to the final momentum of the car.

hence, momentum is conserved in the collision.

6 0

3 years ago

A 0.58 kg mass is moving horizontally with a speed of 6.0 m/s when it strikes a vertical wall. The mass rebounds with a speed of

Sunny_sXe [5.5K]

Answer:

$5.8\; {\rm kg\cdot m \cdot s^{-1}}$ .

Explanation:

If the mass of an object is $m$ and the velocity of that object is $v$ , the linear momentum of that object would be $m\, v$ .

Assume that the initial velocity of the mass is positive ( $6.0\; {\rm m\cdot s^{-1}}$ .) However, the direction of the velocity is reversed after the impact. Thus, the sign of the new velocity of the object would be negative- the opposite of that of the initial velocity. The new velocity would be $(-4.0\; {\rm m\cdot s^{-1}})$ .

Thus, the change in the velocity of the mass would be:

$\begin{aligned}& (\text{Change in Velocity}) \\ =\; & (\text{Final Velocity}) - (\text{Initial Velocity}) \\ =\; & (-4.0\; {\rm m\cdot s^{-1}}) - (6.0\; {\rm m\cdot s^{-1}}) \\ =\; & (-10\; {\rm m\cdot s^{-1})\end{aligned}$ .

The change in the linear momentum of the mass would be:

$\begin{aligned} & \text{change in momentum} \\ =\; & (\text{mass}) \times (\text{change in velocity}) \\ =\; & 0.58\; {\rm kg} \times (-10\; {\rm m\cdot s^{-1}}) \\ =\; & (-5.8\; {\rm kg \cdot m \cdot s^{-1}})\end{aligned}$ .

Thus, the magnitude of the change of the linear momentum would be $5.8\; {\rm kg \cdot m \cdot s^{-1}}$ .

7 0

2 years ago