What acceleration is produced when a 12-N force is exerted on a 3-kg object?

Determine the average distance between the Earth and the Sun. Then calculate the average speed of the Earth in its orbit in kilo

FrozenT [24]

Answer:

The average speed of the Earth in its orbit is 4.74 km/s.

Explanation:

Distance between the Earth and the Sun ,r= 149.60 million km = $149.60\times 10^6 km$

Considering the movement of Earth around the Sun follows circular path.

The circumference of circular path will be give by :

$D=2\pi r=2\times 3.14\times 149.60\times 10^6 km$

Time taken by Earth to complete revolution around earth ,T= 1 year

1 year = 365 days

1 year = 31,536,000 seconds

T = 31,536,000 seconds

Speed of the Earth around the Sun :S

$S=\frac{D}{T}=\frac{149.60\times 10^6 km}{31,536,000 s}$

$S=4.74 km/s$

The average speed of the Earth in its orbit is 4.74 km/s.

8 0

3 years ago

The L-ft ladder has a uniform weight of W lb and rests against the smooth wall at B. θ = 60. If the coefficient of static fricti

Colt1911 [192]

This question is incomplete, the complete question;

The L-ft ladder has a uniform weight of W lb and rests against the smooth wall at B. θ = 60. If the coefficient of static friction at A is μ = 0.4.

Determine the magnitude of force at point A and determine if the ladder will slip. given the following; L = 10 FT, W = 76 lb

Answer:

- the magnitude of force at point A is 79.1033 lb

- since FA < FA_max; Ladder WILL NOT slip

Explanation:

Given that;

∑'MA = 0

⇒ NB [Lsin∅] - W[L/2.cos∅] = 0

NB = W / 2tan∅ -------let this be equation 1

∑Fx = 0

⇒ FA - NB = 0

FA = NB

therefore from equation 1

FA = NB = W / 2tan∅

we substitute in our values

FA = NB = 76 / 2tan(60°) = 21.9393 lb

Now ∑Fy = 0

NA - W = 0

NA = W = 76 lb

Net force at A will be

FA' = √( NA² + FA²)

= √( (W)² + (W / 2tan∅)²)

we substitute in our values

FA' = √( (76)² + (21.9393)²)

= √( 5776 + 481.3328)

= √ 6257.3328

FA' = 79.1033 lb

Therefore the magnitude of force at point A is 79.1033 lb

Now maximum possible frictional force at A

FA_max = μ × NA

so, FA_max = 0.4 × 76

FA_max = 30.4 lb

So by comparing, we can easily see that the actual friction force required for keeping the the ladder stationary i.e (FA) is less than the maximum possible friction available at point A.

Therefore since FA < FA_max; Ladder WILL NOT slip

5 0

3 years ago

Figure 2. sketch of bar and magnetic field lines observations PLEASE HELPPPP

labwork [276]

Hope this sketch helps

5 0

3 years ago

Now consider a wave which is paired with seven other waves into seven pairs. The two waves in each pairing are identical, except

Nikitich [7]

Answer:

Explanation:

Between 2 waves, if path difference is mλ, m is an integer and they interfere constructively.

Between 2 waves, if path difference is (m+1/2)λ, m is an integer and they interfere destructively.

8 0

3 years ago

Unpolarized light passes through a combination of two ideal polarizers. The transmission axes of the first polarizer and the sec

Yuliya22 [10]

Answer:

62.5 %

Explanation:

Let the initial intensity of unpolarized light is Io.

After first polariser the intensity of light becomes I'.

So, $I' = \frac{I_{0}}{2}$

Now it passes through another polariser. The angle between the first polariser and the second polariser is given by Ф. The intensity is I''.

According to the law of Malus

$I'' = I' Cos^{2}\phi$

Here, Ф = 30 degree

$I'' = \frac{I_{0}}{2} Cos^{2}30=0.375I_{0}$

The percentage change in the intensity is given by

$\frac{I_{0}-I''}{I_{0}}\times 100=\frac{I_{0}-0.375I_{0}}{I_{0}}\times100$

= 62.5 %

7 0

3 years ago