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kiruha
7 days ago
7

4. In a closed system consisting of a cannon and a cannonball, the kinetic energy of a cannon is 72,000 J. If the cannonball is

18 kg it has a velocity of 23 m/s when fired from the cannon, what is the total kinetic energy in the system?
A. 72, 207 J
B. 71, 793 J
C. 67, 239 J
D. 76, 761 J



5. In a computational model of the energy loss of a house, if the energy flux increases and the total area insulated increases, what happens to the energy loss of the house?

A. The energy loss must decrease.
B. The energy loss must increase.
C. The energy loss can increase or decrease.
D. The energy loss will neither increase nor decrease.
Physics
2 answers:
Yuliya22 [3.3K]7 days ago
4 0

Respuesta:

1. La velocidad se reduce y la energía cinética disminuye.

2. un incremento en la diferencia de temperatura entre el interior y el exterior del edificio

3. La energía cinética total se mantiene constante.

4. 76,761 J

5. La pérdida de energía debe aumentar.

Explicación:

Agradece al sonógrafo y no a mí, solo quería que él/ella recibiera el agradecimiento que merece. #FÍSICASAPESTAN

serg [3.5K]7 days ago
3 0

Respuesta:

D y B

Explicación:

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A superhero swings a magic hammer over her head in a horizontal plane. The end of the hammer moves around a circular path of rad
Keith_Richards [3271]

Answer:

9.21954 m/s

54 m/s²

The angle is zero

Explanation:

r = Length of the arm = 1.5 m

\omega = Angular velocity = 6 rad/s

The speed's horizontal component can be calculated using

v_h=\omega r\\\Rightarrow v_h=6\times 1.5\\\Rightarrow v_h=9\ m/s

The vertical speed is determined by

v_v=2\ m/s

The combination of these two components yields the hammer’s speed in relation to the ground

v=\sqrt{v_h^2+v_v^2}\\\Rightarrow v=\sqrt{9^2+2^2}\\\Rightarrow v=9.21954\ m/s

The hammer’s speed relative to the ground measures 9.21954 m/s

Acceleration in the vertical direction is zero

The total acceleration is expressed as

a_n=a_h=\omega^2r\\\Rightarrow a_n=6^2\times 1.5\\\Rightarrow a_n=54\ m/s^2

Total acceleration is 54 m/s²

Since acceleration points towards the center, the angle remains zero.

3 0
1 month ago
Bradley gets an x-ray at a radiology clinic that employs its own technologists and radiologists. Would the coder at the clinic r
Maru [3345]

Answer:

Explanation:

If Bradley's examination was done and analyzed in the same facility, the radiologist code is utilized as shown for example- procedure code 72100- Radiologic examination, spine, lumbosacral, 2 or 3 views is reported.

if the X-ray was conducted by Dr. X but he doesn't interpret the results and instead passes it on to the radiologist for initial assessment, then a 26-modifier is applied. For instance, a report from the technologist would be procedure code 72050-Radiologic examination, spine, cervical, 2 or 3

views or under specific circumstances, 72050-TC and the consulting radiologist could report 72050-26.

if Bradley’s x-ray were referred to an independent radiologist for interpretation, then procedure code 76140 would be used in the reporting.

8 0
25 days ago
Which contributions did Galileo make to the model of the solar system? Select two options.a mathematical model for the orbits of
Sav [3153]
Galileo's contributions to the solar system model include: Data indicating that planets reflect sunlight like the Moon, and his observations of Jupiter's moons orbiting the gas giant. With the assistance of an early telescope that he constructed, Galileo made these two significant discoveries.
7 0
1 month ago
A large box of mass m sits on a horizontal floor. You attach a lightweight rope to this box, hold the rope at an angle θ above t
inna [3103]

Answer:

The answer to the specified question will be "\mu_{s}=\frac{T_{m}Cos\theta}{M_{g}-T_{m}Sin\theta}".

Explanation:

Referring to the question,

\sum F_{x}

⇒  TCos \theta-F_{s}=0

⇒  T_{m}Cos \theta =F_{s}...(equation 1)

\sum F_{y}

⇒  TSin \theta+F_{N}=m_{g}

⇒  M_{g}-TSin \theta=F_{N}...(equation 2)

Now,

From equation 1 and equation 2, we conclude

⇒  T_{m} Cos \theta = \mu_{s}F_{N}

By substituting the value of F_{N}, we derive

⇒  T_{m} Cos\theta = \mu_{s}(M_{g}-T_{m}Sin \theta)

⇒  \mu_{s}=\frac{T_{m}Cos\theta}{M_{g}-T_{m}Sin\theta}

4 0
1 month ago
If a 50.0-kg mass weighs 554 n on the planet saturn, calculate saturn’s radius
ValentinkaMS [3465]

Answer:

17.35 × 10^(-6) m

Explanation:

Mass; m = 50 kg

Weight; W = 554 N

From the formula:

W = mg

This simplifies to; 554 = 50g

g = 554/50

g = 11.08 m/s²

Also, using the formula;

mg = GMm/r²

hence; g = GM/r²

Rearranging gives;

r = √(GM/g)

With G as a known constant of 6.67 × 10^(-11) Nm²/kg²

r = √(6.67 × 10^(-11) × 50/11.08)

r = 17.35 × 10^(-6) m

8 0
1 month ago
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