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Nataly_w [17]
4 years ago
10

A mass M suspended by a spring with force constant k has aperiod T when set into oscillations on Earth. Its period on Mars,whose

mass is about 1/9 and radius 1/2 that of Earth, is mostnearly
A) 1/3 T
B) 2/3T
C)T
D) 3/2T
E) 3 T
Physics
1 answer:
olchik [2.2K]4 years ago
6 0

Answer:

C)T

Explanation:

The period of a mass-spring system is:

T=2\pi\sqrt\frac{m}{k}

As can be seen, the period of this simple harmonic motion, does not depend at all on the gravitational acceleration (g), neither the mass nor the spring constant depends on this value.

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3- (a) What is the period of rotation of Earth in seconds? (b) What is the angular velocity of Earth? (c) Given that Earth has a
ASHA 777 [7]

Explanation:

a) The Earth makes 1 rotation in 24 hours.  In seconds:

24 hr × (3600 s / hr) = 86400 s

b) 1 rotation is 2π radians.  So the angular velocity is:

2π rad / 86400 s = 7.27×10⁻⁵ rad/s

c) The earth's linear velocity is the angular velocity times the radius:

40075 km × 7.27×10⁻⁵ rad/s = 2.91 km/s

7 0
4 years ago
The part of the building structure, typically below grade, upon which all other construction is built is known as:________
Degger [83]

Answer:

horizontal structural member that supports a floor. Beams are typically wood, cold formed metal framing or steel.

Joists

Horizontal timbers, beams or bars supporting a floor.

8 0
4 years ago
A motorist is driving at 20m/s when she sees that a traffic light 200m ahead has just turned red. She knows that this light stay
yuradex [85]

Answer:

5.71428571422 m/s

Explanation:

u = Initial velocity = 20 m/s

v = Final velocity

s = Displacement

a = Acceleration

Time taken = 15-1 = 14 s

Distance traveled in 1 second = 20\times 1=20\ m

s=ut+\frac{1}{2}at^2\\\Rightarrow 200-20=20\times 14+\frac{1}{2}\times a\times 14^2\\\Rightarrow a=\frac{2(180-20\times14)}{14^2}\\\Rightarrow a=-1.02040816327\ m/s^2

v=u+at\\\Rightarrow v=20-1.02040816327\times 14\\\Rightarrow v=5.71428571422\ m/s

The speed as she reaches the light at the instant it turns green is 5.71428571422 m/s

4 0
4 years ago
Honeybees accumulate charge as they fly, and they transfer charge to the flowers they visit. Honeybees are able to sense electri
Vilka [71]

Answer:

ΔE> E_minimo

We see that the field difference between these two flowers is greater than the minimum field, so the bee knows if it has been recently visited, so the answer is if it can detect the difference

Explanation:

For this exercise let's use the electric field expression

         E = k q / r²

where k is the Coulomb constant that is equal to 9 109 N m² /C², q the charge and r the distance to the point of interest positive test charge, in this case the distance to the bee

let's calculate the field for each charge

 

Q = 24 pC = 24 10⁻¹² C

         E₁ = 9 10⁹ 24 10⁻¹² / 0.20²

         E₁ = 5.4 N / C

Q = 32 pC = 32 10⁻¹² C

         E₂ = 9 10⁹ 32 10⁻¹² / 0.2²

         E₂ = 7.2 N / C

let's find the difference between these two fields

         ΔE = E₂ -E₁

         ΔE = 7.2 - 5.4

         ΔE = 1.8 N / C

the minimum detection field is

         E_minimum = 0.77 N / C

        ΔE> E_minimo

We see that the field difference between these two flowers is greater than the minimum field, so the bee knows if it has been recently visited, so the answer is if it can detect the difference

8 0
3 years ago
"In a Young’s double-slit experiment, the separation between slits is d and the screen is a distance D from the slits. D is much
san4es73 [151]

Answer:

The number of bright fringes per unit width on the screen is, x=\dfrac{\lambda D}{d}      

Explanation:

If d is the separation between slits, D is the distance between the slit and the screen and \lambda is the wavelength of the light. Let x is the  number of bright fringes per unit width on the screen is given by :

x=\dfrac{n\lambda D}{d}

\lambda is the wavelength

n is the order

If n = 1,

x=\dfrac{\lambda D}{d}

So, the the number of bright fringes per unit width on the screen is \dfrac{\lambda D}{d}. Hence, the correct option is (B).

6 0
3 years ago
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