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Otrada [13]
2 years ago
11

A 241 kg mass is lifted 1.8 m. What is the potential energy of the mass (in J)?

Physics
2 answers:
Korolek [52]2 years ago
6 0

Answer:

mass is lifted 1.8 m. What is the potential energy of the mass 4. A 100 kg

Nitella [24]2 years ago
4 0
4251.24 Joules
Using acceleration as 9.8m/s^2,241*1.8*9.8
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5) A person holds a 1.7 kg bucket and lets it move down at
Elena L [17]

Answer:

12.5J

Explanation:

Given parameters:

Mass of bucket  = 1.7kg

Height  = 75cm  = 0.75m

Unknown;

Work done on the bucket by the person  = ?

Solution:

To solve this problem, we use the work done equation;

  Work done  = force x distance  = mgh

 m is the mass  

  g is the acceleration due to gravity

  h is the height

Now, insert parameters and solve ;

     Work done  = 1.7 x 9.8 x 0.75  = 12.5J

7 0
3 years ago
The energy stored in a wooden log transforms when the log is burned. Which of the following explanations BEST describes how the
Makovka662 [10]

Answer:

D. the amount of chemical energy equals the amount of heat and light energy.

Explanation:

Given that the first law of thermodynamics affirmed that energy is neither created nor destroyed however, it can be transformed from one form to another. In other words, while, during the transformation of energy, no energy is lost, the input energy is also equal to output energy.

Hence, the chemical energy stored in the log is EQUAL to the heat and light energy produced by burning.

5 0
3 years ago
If The density of this stainless steel is7.85 g/cm3,specific heatis 0.5 J/g.K, melting pointis 1673K, heat of fusion s0.260J/kg.
kodGreya [7K]

Answer:

\Delta H=687.4 J

Explanation:

Hello!

In this case, for this melting process, we can identify two sub-processes in order to take the stainless steel from solid to liquid:

1. Heat up from 298.15 K to 1673 K.

2. Undergo the phase transition.

Both process have an associated enthalpy as shown below:

\Delta H_1=1g*0.5\frac{J}{g*K} (1673K-298.15K)=687.4J

\Delta H_2=0.001kg*\frac{0.260J}{kg} =0.00026J

Therefore, the required heat is:

\Delta H=\Delta H_1+\Delta H_2\\\\\Delta H=687.4J+0.00026J\\\\\Delta H=687.4J

Notice the problem is not providing neither the mass or volume, that is why we assumed the mass is 1 g; however, it can be changed to the mass you are given.

Best regards!

4 0
2 years ago
Solve the inequality 2(n+3) – 4&lt;6. Then graph<br> the solution.
Aloiza [94]
The solution is 22 2(n+3)-4&6
6 0
3 years ago
Early black-and-white television sets used an electron beam to draw a picture on the screen. The electrons in the beam were acce
lutik1710 [3]

Answer:

speed of electrons = 3.25 × 10^{7} m/s

acceleration in term g is 3.9 × 10^{17} g.

radius of circular orbit is 2.76 × 10^{-4} m

Explanation:

given data

voltage = 3 kV

magnetic field = 0.66 T

solution

law of conservation of energy

PE = KE

qV = 0.5 × m × v²

v = \sqrt{\frac{2qV}{m}}

v = \sqrt{\frac{2\times 1.6 \times 10^{-19}\times 3}{9.1\times 10^{-31}}

v = 3.25 × 10^{7} m/s

and

magnetic force on particle movie in magnetic field

F = Bqv

ma = Bqv

a = \frac{Bqv}{m}  

a =  \frac{0.67\times 1.6\times 10^{-19}\times 3.25\times 10^7}{9.1\times 10^{-31}}

a = 3.82 × 10^{18} m/s²

and acceleration in term g

a = \frac{3.82\times 10^{18}}{9.81}  

a = 3.9 × 10^{17} g

acceleration in term g is 3.9 × 10^{17} g.

and

electron moving in circular orbit has centripetal force

F = \frac{mv^2}{r}  

Bqv = \frac{mv^2}{r}  

r = \frac{mv}{Bq}  

r = \frac{9.1\times 10^{-31}\times 3.25\times 10^7}{0.67\times 1.6\times 10^{-19}}  

r = 2.76 × 10^{-4} m

radius of circular orbit is 2.76 × 10^{-4} m

8 0
3 years ago
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