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erastova [34]
3 years ago
14

Which of these planets is the odd one out based on their physical features?

Physics
1 answer:
Maurinko [17]3 years ago
7 0

Answer:

mars

Explanation:

the planet that is the odd is Mars

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A dry cell is better than simple cell​
777dan777 [17]

Answer:

umm this is not a question

Explanation:

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2 years ago
A 1.55-kg object hangs in equilibrium at the end of a rope (taken as massless) while a wind pushes the object with a 13.3-n hori
yarga [219]
<span>To answer this problem, we use balancing of forces: x and y components to determine the tension of the rope.


First, the vertical component of tension (Tsin theta) is equal to the weight of the object. 
 T * sin θ = mg =</span> 1.55 * 9.81 <span>
 T * sin θ = 15.2055

Second, the horizontal component of tension (t cos theta) is equal to the force of the wind. 
 T * cos θ =  13.3 

Tan θ = sin  </span>θ / cos θ = 15.2055/13.3 = 1.143
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T then is equal to 20.20 N
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3 years ago
Efficiency is the ratio of output work to input work, expressed as a percentage. Light bulbs put out less light energy than the
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5 0
3 years ago
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a boy looks at the reflection of his digital watch in a plane mirror and thinks the time is 10:11. what is the correct time?
Gekata [30.6K]

Answer:

11:10 will be the time. reflection causes the object to be flipped when you see its image at the mirror

6 0
3 years ago
A spring on Earth has a 0.500 kg mass suspended from one end and the mass is displaced by 0.3 m. What will the displacement of t
julia-pushkina [17]

To solve this problem we will apply the concepts related to the Force of gravity given by Newton's second law (which defines the weight of an object) and at the same time we will apply the Hooke relation that talks about the strength of a body in a system with spring.

The extension of the spring due to the weight of the object on Earth is 0.3m, then

F_k = F_{W,E}

kx_1 = mg

The extension of the spring due to the weight of the object on Moon is a value of x_2, then

kx_2 = mg_m

Recall that gravity on the moon is a sixth of Earth's gravity.

kx_2 = m\frac{g}{6}

kx_2 = \frac{1}{6} mg

kx_2 = \frac{1}{6} kx_1

x_2 = \frac{1}{6} x_1

We have that the displacement at the earth was x_1 = 0.3m, then

x_2 = \frac{1}{6} 0.3

x_2 = 0.05m

Therefore the displacement of the mass on the spring on Moon is 0.05m

6 0
3 years ago
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