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3 years ago

The tropic of cancer is to the tropic of capricorn as the arctic circle is to the

Physics

1 answer:

Y_Kistochka [10]3 years ago

7 0

Answer:

Antarctic Circle

Explanation:

The Tropic of Cancer, which is also referred to as the Northern Tropic, is the most northerly circle of latitude on Earth at which the Sun can be directly overhead. This occurs on the June solstice, when the Northern Hemisphere is tilted toward the Sun to its maximum extent.

Tropic of Capricorn Is it Southern Hemisphere counterpart, marking the most southerly position at which the Sun can be directly overhead.

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When a 3.0 kg mass is hung from a vertical massless spring, the spring is stretched 40 cm. What is the spring constant of the sp

Dmitry_Shevchenko [17]

Answer:

0.74 N/cm

Explanation:

The following data were obtained from the question:

Mass (m) = 3 Kg

Extention (e) = 40 cm

Spring constant (K) =?

Next, we shall determine the force exerted on the spring.

This can be obtained as follow:

Mass (m) = 3 Kg

Acceleration due to gravity (g) = 9.8 m/s²

Force (F) =?

F = mg

F = 3 × 9.8

F = 29.4 N

Finally, we shall determine the spring constant of the spring. This can be obtained as follow:

Extention (e) = 40 cm

Force (F) = 29.4 N

Spring constant (K) =?

F = Ke

29.4 = K × 40

Divide both side by 40

K = 29.4 / 40

K = 0.74 N/cm

Therefore, the spring constant of the spring is 0.74 N/cm

5 0

3 years ago

Siven an element's atomic number and mass number, how can you tell the number of protons and neutrons in its nucleus?

sukhopar [10]

It would be the first option.

Explanation-
The number of protons is equal to the atomic number the number of neutrons is the mass minus the atomic number.

3 0

3 years ago

Read 2 more answers

Damping is negligible for a 0.135-kg object hanging from a light, 6.30-N/m spring. A sinusoidal force with an amplitude of 1.70

IceJOKER [234]

Answer:

0.72 Hz minimum frequency

Explanation:

When the damping is negligible,Amplitude is given as

$A = (F/m)/[\sqrt{(\omega ^{^{2}}-\omega _{o}^{2}})^2$

here $\omega _{o}^{2}= k/m$ = (6.30)/(0.135) = 46.67 N/m kg

$F / mA$ = 1.70/(0.135)(0.480) = 26.2 N/m kg

From the above equation , rearranging for ω,

$\omega ^{2}= \omega _{o}^{2}\pm F/m$

⇒ ω² =46.67 ± 26.2 = 72.87 or 20.47

⇒ ω = 8.53 or 4.52 rad/s

Frequency = f

ω=2 π f

⇒ f = ω / 2π = 8.53 /6.28 or 4.52 / 6.28 = 1.36 Hz or 0.72 Hz

The lower frequency is 0.72 Hz and higher is 1.36 Hz

8 0

3 years ago

A 50 N object is on Earth. What is its mass?

faust18 [17]

Answer: 5.10 KG

Explanation: W = 50 N

BY FORMULA

W = MG

M = W/G

M = 50 /9.8

M= 5.10 KG

7 0

2 years ago

A man claims that he can hold onto a 15.0-kg child in a head-on collision as long as he has his seat belt on. Consider this man

suter [353]

Answer:

F= 5195.625 N

Explanation:

To obtain the force needed to hold the child, we need to know the aceleration in which the car is breaking.

Aceleration is equal to velocity divided by the time of breaking

In international system, velocity [m/s] is

v= (62 mi/h)*(1609 m/mi)*(1 h/3600 s)

v= 27.71 m/s

Now, we part the velocity by the time that is 0.08 seconds

a= v/t= (27.71 m/s)/(0.08 s)

a= 346.375 m/ $s^{2}$

The force in agreement with the Newton's second law is

F=m*a = 15 Kg*346.375 m/ $s^{2}$

F= 5195.625 N

(Note: 1 N = 1 Kg*m/ $s^{2}$ )

3 0

3 years ago