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2 years ago

Why does a solar and lunar eclips not happen every month

Physics

1 answer:

jasenka [17]2 years ago

4 0

Answer:

"because the Moon's orbit is tilted five degrees from Earth's orbit around the Sun, so most of the time the Moon passes above or below the shadow."

You might be interested in

Which should be done in case of a laboratory accident?

stich3 [128]

Tell your instructor or teacher

4 0

3 years ago

What is the kinetic energy of an object moving 40m/s with a mass of 20kg?

Svet_ta [14]

KE = (1/2) (mass) (speed)²

KE = (1/2) (20 kg) (40 m/s)²

KE = (1/2) (20 kg) (1,600 m²/s²)

KE = (10 kg) (1,600 m²/s²)

KE = 16,000 Joules

7 0

3 years ago

A 9.00 g bullet is fired horizontally into a 1.20 kg wooden block resting on a horizontal surface. The coefficient of kinetic fr

Zinaida [17]

Answer:

The initial speed of bullet is "164 m/s".

Explanation:

The given values are:

mass of bullet,

$m'=9.00 \ g$

or,

$=0.009 \ kg$

mass of wooden block,

$m=1.20 \ kg$

speed,

$s=0.390 \ m$

Coefficient of kinetic friction,

$\mu=0.20$

As we know,

The Kinematic equation is:

⇒ $v^2=u^2+2as$

then,

Initial velocity will be:

⇒ $u=v^2-2as$

$=v^2-2 \mu gs$

On substituting the given values, we get

⇒ $u=\sqrt{0-2\times 0.20\times 9.8\times 0.390}$

$=\sqrt{-1.5288}$

$=1.23 \ m/s$

As we know,

The conservation of momentum is:

⇒ $mu=m'u'$

or,

⇒ Initial speed, $u'=\frac{mu}{m'}$

On substituting the values, we get

⇒ $=\frac{1.20\times 1.23}{0.009}$

⇒ $=\frac{1.476}{0.009}$

⇒ $=164 \ m/s$

3 0

3 years ago

Each of the gears a and b has a mass of 675 g and has a radius of gyration of 40 mm, while gear c has a mass of 3. 6 kg and a ra

navik [9.2K]

9.87 seconds

The time required for this system to come to rest is equal to 9.87 seconds.

We have the following data:

Mass of gear A = 675 g to kg = 0.675 kg.

Radius of gear A = 40 mm to m = 0.04 m.

Mass of gear C = 3.6 kg.

Radius of gear C = 100 mm to m = 0.1 m.

How can I calculate the time needed?

We would need to figure out the moment of inertia for gears A and C in order to compute the time needed for this system to come to rest.

Mathematically, the following formula can be used to determine the moment of inertia for a gear:

I = mr²

Where:

m is the mass.

r is the radius.

We have, For gear A:

I = mr²

I = 0.675 × 0.04²

I = 0.675 × 0.0016

I = 1.08 × 10⁻³ kg·m².

We have, For gear C:

I = mr²

I = 3.6 × 0.1²

I = 3.6 × 0.01

I = 0.036 kg·m².

The initial angular velocity of gear C would therefore be converted as follows from rotations per minute (rpm) to radians per second (rad/s):

ωc₁ = 2000 × 2π/60

ωc₁ = 4000π/60

ωc₁ = 209.44 rad/s.

Also, the initial angular velocity of gears A and B is given by:

ωA₁ = ωB₁ = rc/rA × (ωc₁)

ωA₁ = ωB₁ = 0.15/0.06 × (209.44)

ωA₁ = ωB₁ = 2.5 × (209.44)

ωA₁ = ωB₁ = 523.60 rad/s.

Taking the moment about A, we have:

I_A·ωA₁ + rA∫F_{AC}dt - M(f)_A·t = 0

On Substituting the given parameters into the formula, we have;

(1.08 × 10⁻³)·(523.60) + 0.06∫F_{AC}dt - 0.15t = 0

0.15t - 0.06∫F_{AC}dt = 0.56549 ----->equation 1.

Similarly, the moment about B is given by:

0.15t - 0.06∫F_{BC}dt = 0.56549 ------>equation 2.

Note: Let x = ∫F_{BC}dt + ∫F_{AC}dt

Adding eqn. 1 & eqn. 2, we have:

0.3t - 0.06x = (0.56549) × 2

0.3t - 0.06x = 1.13098 ------>equation 3.

Taking the moment about A, we have:

Ic·ωc₁ - rC∫F_{AC}dt - rC∫F_{BC}dt - Mc(f)_A·t = 0

0.036(209.44) - 0.3t - 0.15(∫F_{BC}dt + ∫F_{AC}dt) = 0

0.3t + 0.15x = 7.5398 ------->equation 4.

Solving eqn. 3 and eqn. 4 simultaneously, we have:

x = 30.5 Ns.

Time, t = 9.87 seconds.

To learn more about moment of inertia visit:

brainly.com/question/15246709

#SPJ4

6 0

2 years ago

In the aerials competition in skiing, the competitors speed down a ramp that slopes sharply upward at the end. The sharp upward

11111nata11111 [884]

Answer:

u = 11.6 m/s

Explanation:

The end of a launch ramp is directed 63° above the horizontal. A skier attains a height of 10.9 m above the end of the ramp.

Maximum height, H = 10.9

Let v is the launch speed of the skier. The maximum height attained by the projectile is given by :

$H=\dfrac{u^2\ sin^2\theta}{g}$

$10.9=\dfrac{u^2\ sin^2(63)}{9.8}$

u = 11.6 m/s

So, the launch speed of the skier is 11.6 m/s. Hence, this is the required solution.

4 0

3 years ago