Ask question

Ask question

All categories

4 years ago

15

Leap years _____. happen because the Earth revolves around the sun in less than 365 days make up for the extra one-fourth day th

e Earth needs to orbit the sun add an extra day to the end of December, making the month 32 days long subtract a day from February, making the month only 27 days long

1 answer:

anzhelika [568]4 years ago

7 0

The Earth takes (365 and 1/4) days to orbit the sun. That's the length of our 'year'.

Since it's not really possible to make our calendar a quarter of a day longer every year, we make the calendar a <u><em>whole</em></u> day longer every 4 years.

So "Leap years" make up for the extra one-fourth day the Earth takes to orbit the sun.

You might be interested in

A 23.3-kg mass is attached to one end of a horizontal spring, with the other end of the spring fixed to a wall. The mass is pull

sdas [7]

In spring mass system we know that angular frequency is given as

$\omega = 2\pi f$

f = 8.38 Hz

$\omega = 2\pi(8.38)$

$\omega = 52.65 rad/s$

now we know that speed of SHM at its extreme position is given by

$v = A\omega$

here we know that

A = 17.5 cm

$v = 0.175 (52.65)$

$v = 9.21 m/s$

so maximum speed is 9.21 m/s

7 0

3 years ago

A thin rod of length 0.75 m and mass 0.42 kg is suspended

MrRissso [65]

Answer:

a) K = 0.63 J, b) h = 0.153 m

Explanation:

a) In this exercise we have a physical pendulum since the rod is a material object, the angular velocity is

w² = $\frac{m g d}{I}$

where d is the distance from the pivot point to the center of mass and I is the moment of inertia.

The rod is a homogeneous body so its center of mass is at the geometric center of the rod.

d = L / 2

the moment of inertia of the rod is the moment of a rod supported at one end

I = ⅓ m L²

we substitute

w = $\sqrt{\frac{mgL}{2} \ \frac{1}{\frac{1}{3} mL^2} }$

w = $\sqrt{\frac{3}{2} \ \frac{g}{L} }$

w = $\sqrt{ \frac{3}{2} \ \frac{9.8}{0.75} }$

w = 4.427 rad / s

an oscillatory system is described by the expression

θ = θ₀ cos (wt + Φ)

the angular velocity is

w = dθ /dt

w = - θ₀ w sin (wt + Ф)

In this exercise, the kinetic energy is requested in the lowest position, in this position the energy is maximum. For this expression to be maximum, the sine function must be equal to ±1

In the exercise it is indicated that at the lowest point the angular velocity is

w = 4.0 rad / s

the kinetic energy is

K = ½ I w²

K = ½ (⅓ m L²) w²

K = 1/6 m L² w²

K = 1/6 0.42 0.75² 4.0²

K = 0.63 J

b) for this part let's use conservation of energy

starting point. Lowest point

Em₀ = K = ½ I w²

final point. Highest point

Em_f = U = m g h

energy is conserved

Em₀ = Em_f

½ I w² = m g h

½ (⅓ m L²) w² = m g h

h = 1/6 L² w² / g

h = 1/6 0.75² 4.0² / 9.8

h = 0.153 m

5 0

3 years ago

PLS HELP, I AM WILLING TO MARK BRAINLIEST TO FIRST ANSWER AND HAS TOO BE RIGHT

Ann [662]

Weak winds that blow for short periods of time with a short fetch.

6 0

3 years ago

Read 2 more answers

The specific heat of a certain type of metal is 0.128 J/(g⋅∘C).0.128 J/(g⋅∘C). What is the final temperature if 305 J305 J of he

Makovka662 [10]

Answer:

45.3°C

Explanation:

Heat gained = mass × specific heat × increase in temperature

q = mC (T − T₀)

Given C = 0.128 J/g/°C, m = 94.0 g, q = 305 J, and T₀ = 20.0°C:

305 J = (94.0 g) (0.128 J/g/°C) (T − 20.0°C)

T = 45.3°C

6 0

3 years ago

The uniform slender bar AB has a mass of 6.4 kg and swings in a vertical plane about the pivot at A. If θ˙ = 2.7 rad/s when θ =

dolphi86 [110]

Answer:

F=√[(1.5(14.58L+11.96))² + (3.2(2.97L - 157.03) + 62.72)²]

Explanation:

Given data,

The mass of the bar AB, m = 6.4 kg

The angular velocity of the bar, θ˙ = 2.7 rad/s

The angle of the bar at A, θ = 24°

Let the length of the bar be, L = l

The angular moment at point A is,

∑ Mₐ = Iα

Where, Mₐ - the moment about A

α - angular acceleration

I - moment of inertia of the rod AB

$-mg(\frac{lcos\theta}{2})=\frac{1}{3}(ml^{2})\alpha$

$\alpha=\frac{-3gcos\theta}{2l}$

Let G be the center of gravity of the bar AB

The position vector at A with respect to the origin at G is,

$\vec{r_{G}}=[\frac{lcos\theta}{2}\hat{i}-\frac{lcos\theta}{2}\hat{j}]$

The acceleration at the center of the bar

$\vec{a_{G}}=\vec{a_{a}}+\vec{\alpha}X\vec{r_{G}}-\omega^{2}\vec{r_{G}}$

Since the point A is fixed, acceleration is 0

The acceleration with respect to the coordinate axes is,

$(\vec{a_{G}})_{x}\hat{i}+(\vec{a_{G}})_{y}\hat{j}=0+(\frac{-3gcos\theta}{2l})\hat{k}\times[\frac{lcos\theta}{2}\hat{i}-\frac{lcos\theta}{2}\hat{j}]-\omega^{2}[\frac{lcos\theta}{2}\hat{i}-\frac{lcos\theta}{2}\hat{j}]$

$(\vec{a_{G}})_{x}\hat{i}+(\vec{a_{G}})_{y}\hat{j}=[-\frac{cos\theta(2l\omega^{2}+3gsin\theta)}{4}\hat{i}+(\frac{2l\omega^{2}sin\theta-3gcos^{2}\theta}{4})\hat{j}]$

Comparing the coefficients of i

$=-\frac{cos\theta(2l\omega^{2}+3gsin\theta)}{4}$

Comparing coefficients of j

$(\vec{a_{G}})_{y}=\frac{2l\omega^{2}sin\theta-3gcos^{2}\theta}{4}$

Net force on x direction

$F_{x}=(\vec{a_{G}})_{x}$

substituting the values

$F_{x}$ =1.5(14.58L+11.96)

Similarly net force on y direction

$F_{y}=(\vec{a_{G}})_{y}+mg$

= 3.2(2.97L - 157.03) + 62.72

Where L is the length of the bar AB

Therefore the net force,

$F=\sqrt{F_{x}^{2}+F_{y}^{2}}$

F=√[(1.5(14.58L+11.96))² + (3.2(2.97L - 157.03) + 62.72)²]

Substituting the value of L gives the force at pin A

8 0

3 years ago