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nikklg [1K]
3 years ago
15

A circular test track for cars has a circumference of 3.5 km . A car travels around the track from the southernmost point to the

northernmost point. what distance does the car travel?
Physics
1 answer:
Marta_Voda [28]3 years ago
8 0
<span>The car would have traveled exactly one-half of the circumference of the track, since it would have gone from one extreme point to its opposite extreme point. This would be equal to (3.5 / 2), or 1.75 km. The northernmost point would be 1.75km away from the southernmost point.</span>
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You lift a stuffed toy up above the ground (over your head).
BigorU [14]

Answer:

Gravitational energy

Explanation:

when over head , when it hits the floor thermal energy

7 0
3 years ago
Need help with this science!!
noname [10]
Lol, I remember I helped my sister with this question but anyways, the card flew because of inertia and the quarter fell in the glass because of gravity :).
3 0
3 years ago
Please help! the mass of an object is measured on a pan balance with a precision of 0.005 g and the recorded value of 128.01 g.
inysia [295]

Yes, these two objects have different masses.

<h3>How can we calculate that this statement is right ?</h3>

To calculate the precision of mass we are using the formula,

Precision(P) = \frac{m_1}{m_1+\triangle m}

Or,\trianglem=\frac{m_1}{P}-m₁

For the first case we are given,

m₁= The recorded value of mass

= 128.01 g

P= Precision of the mass

=0.005g

So, according to the formula, \trianglem will be,

\trianglem=  \frac{128.01}{0.005}-128.01

Or,\trianglem=25,473.99 g

Or,\trianglem=25.47 Kg

For the first case \trianglem is 25.47 Kg..

For the second case we are given,

m₁= The recorded value of mass

= 0.13 Kg

P= Precision of the mass

=0.005 Kg

So, according to the formula, \trianglem will be,

\trianglem= \frac{0.13}{0.005}-0.13

Or,\trianglem= 25.87 kg

For the second case \trianglem is 25.87 Kg.

For the two cases  \trianglem has different values, 25.47 Kg≠25.87 Kg.

Therefore we can conclude that, these two objects have different masses.

Learn more about Mass:

brainly.com/question/3187640

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7 0
2 years ago
You and your friends are having a discussion about weight. He/she claims that he/she weighs less on the 100th floor of a buildin
Viktor [21]

Answer:

if the weight theoretically decreases at this height, but in a fraction of 10⁻⁵, which is not appreciable in any scale, therefore, the reading of the scale in the two places is the same.

Explanation:

The weight of a person in the force with which the Earth attracts the person, therefore can be calculated using the law of universal attraction

          F = G m M / r²

Where m is the mass of the person, M the masses of the earth

Let's call the person's weight at ground level as Wo and suppose the distance to the center of the Earth is Re

            W₀ = G m M / Re²

In the calculation of the weight of the person on the 100th floor the only thing that changes is the distance

          r = Re + 100 r₀

Where r₀ is the distance between the floors, which is approximately 2.5 m, so the distance is

         r = Re + 250

We substitute

     W = G m M / r²

      W = G m M / (Re + 250)²

The value of Re is 6.37 10⁶ m, so we can take it out as a factor and perform a serial expansion of the remaining fraction

      W = G m M / Re² (1+ 250 / Re)²

      (1 + 250 / Re)⁻² = 1 + (-2) 250 / Re + (-2 (-2-1)) / 2 (250 / Re)² +….

The value of the expression is

      (1 + 250 / Re)⁻² = 1 -2 250 / 6.37 10⁶ -30 (250 / 6.37)² 10⁻¹² + ...

We can see that the quadratic term is very small, which is why we despise it, we substitute in the weight equation

      W = G m M / Re² (1 - 78.5 10⁻⁶)

Remains

     W = Wo (1 - 7.85  10⁻⁵)

We can see that if the weight theoretically decreases at this height, but in a fraction of 10⁻⁵, which is not appreciable in any scale, therefore, the reading of the scale in the two places is the same.

4 0
3 years ago
Read 2 more answers
A string attached to an airborne kite was maintained at an angle of 40.0 with the ground. if 120m of string was reeled in to ret
AleksAgata [21]
The hypotenuse is measured at 120 meters of string, and you need to solve for the leg of the triangle that is horizontal. The degree is 40, so use trigonometry to figure it out. 
Cosin (40) is equal to around .766 
Adjacent/Hypotenuse 
x/120 = cos40 
Answer: 91.92533. 
If you use 3 significant figures it should be 91.9 meters.
5 0
3 years ago
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