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postnew [5]
2 years ago
14

How much potintial energy is in a closed system where the height of the object is 2 m, and the mass is 10 kg?

Physics
2 answers:
Dovator [93]2 years ago
8 0

Answer:

<u>Answer</u><u>:</u><u> </u><u>B</u><u>.</u><u> </u><u>1</u><u>9</u><u>6</u><u> </u><u>J</u>

• potential energy is calculated from the formula:

{ \tt{p.e = mgh}}

• m = 10 kg

• g = 9.81 m/s²

• h = 2m

→ Substitute:

{ \tt{p.e = (10) \times (2) \times (9.81)}} \\  \\ → \: { \boxed{ \tt{potential \: energy = 196.2 \:  \: joules}}}

Lorico [155]2 years ago
6 0

Answer:

D. 196 J

Explanation:

PE=mgh=10×9.8×2=196J

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Firdavs [7]

The answer is C. Click C.

3 0
3 years ago
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Water initially at 200 kPa and 300°C is contained in a piston–cylinder device fitted with stops. The water is allowed to cool at
netineya [11]

Answer:

Δu=1300kJ/kg  

Explanation:

Energy at the initial state

p_{1}=200kPa\\t_{1}=300^{o}\\u_{1}=2808.8kJ/kg(tableA-5)

Is saturated vapor at initial pressure we have

p_{2}=200kPa\\x_{2}=1(stat.vapor)\\v_{2}=0.8858m^3/kg(tableA-5)

Process 2-3 is a constant volume process

p_{3}=100kPa\\v_{3}=v_{2}=0.8858m^{3}/kg\\u_{3}=1508.6kJ/kg(tableA-5)

The overall in internal energy

Δu=u₁-u₃

We replace the values in equation

Δu=u₁-u₃

=2808.8kJ/kg-1508.6kJ/kg\\=1300kJ/kg

Δu=1300kJ/kg  

3 0
3 years ago
A box of weight 74 N is sliding on a horizontal surface at a constant velocity due to an external force F-&gt; of magnitude 4.8
n200080 [17]

Answer

4.8 N

If the box is moving with a constant velocity, then we can say that the system is in equilibrium. This is because if the external force (F->) was greater than other forces the box would be accelerating. This tells us that this force (F->) is just enough to overcome friction and so it must be equal to 4.8 N.

The normal force has no effect to the horizontal velocities or forces. It is equal to -Weight. That is -74 N. The negative sign shows that the force is in opposite direction.

6 0
3 years ago
The kinetic energy of a body of mass 15 kg is 30 joule. What is its momentum?
lys-0071 [83]

This problem is a piece o' cake, IF you know the formulas for both kinetic energy and momentum.  So here they are:

Kinetic energy = (1/2) · (mass) · (speed²)

Momentum = (mass) · (speed)

So, now ... We know that

==> mass = 15 kg,  and

==> kinetic energy = 30 Joules

Take those pieces of info and pluggum into the formula for kinetic energy:

Kinetic energy = (1/2) · (mass) · (speed²)

30 Joules = (1/2) · (15 kg) · (speed²)

60 Joules = (15 kg) · (speed²)

4 m²/s² = speed²

Speed = 2 m/s

THAT's all you need !  Now you can find momentum:

Momentum = (mass) · (speed)

Momentum = (15 kg) · (2 m/s)

<em>Momentum = 30 kg·m/s</em>

<em>(Notice that in this problem, although their units are different, the magnitude of the KE is equal to the magnitude of the momentum.  When I saw this, I wondered whether that's always true.  So I did a little more work, and I found out that it isn't ... it's a coincidence that's true for this problem and some others, but it's usually not true.)</em>

8 0
3 years ago
The downward force produced when air flows over the winglike spoiler on a race car is an example of ___ principle
Murljashka [212]

As per bernoulli's principle

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here

P_1 = pressure upwards

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v_1 = velocity of air upwards

v_2 = velocity of air downwards

now from this equation we can say that the pressure difference will be

P_1 - P_2 = \frac{1}{2}\rho v_2^2 - \frac{1}{2}\rho v_1^2

now the force due to this pressure difference will be

F = (P_1 - P_2)A

so this is the above force which is given above

5 0
3 years ago
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