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3 years ago

How long does it take for the Earth to make a complete revolution around the sun?

2 answers:

Alex3 years ago

8 0

365 days, 5
Earth revolves around the sun in 365 days, 5 hours, 59 minutes and 16 seconds. The time a planet takes to revolve around the sun is called a year.

timurjin [86]3 years ago

4 0

365 days so a year basically

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Which letter represents the location of the resister in this diagram? A B C D

ololo11 [35]

Answer:

The letter B is the letter that represents the location of the resister in the diagram.

Explanation:

I hope it helped!Please mark brainliest and have a wonderful night! and day!!!!

5 0

4 years ago

Read 2 more answers

Part A Determine the magnitude of the x component of F using scalar notation. Fx F x = nothing lb Request Answer Part B Determin

maksim [4K]

As we know that force F makes an angle of 60 degree with X axis

so the X component is given as

$cos60 = \frac{F_x}{F}$

now we have

$F_x = F cos60$

$F_x = 0.50 F$

Similarly we know that force F makes an angle of 45 degree with Y axis

so the X component is given as

$cos45 = \frac{F_y}{F}$

now we have

$F_y = F cos45$

$F_y = 0.707 F$

Now for the component along z axis we know that

$F_x^2 + F_y^2 + F_z^2 = F^2$

now plug in all components

$(0.707 F)^2 + (0.50 F)^2 + F_z^2 = F^2$

$0.5 F^2 + 0.25 F^2 + F_z^2 = F^2$

$F_z^2 = F^2(1 - 0.75)$

$F_z^2 = 0.25 F^2$

$F_z = 0.5 F$

5 0

3 years ago

Discuss the limitations of using the Doppler shift to determine an object's speed.

pantera1 [17]

Answer and Explanation:

Limitation of Doppler shift :

The Doppler impact is relevant when the speeds of the wellspring of sound and spectator are considerably less than the speed of sound. The movement of both the spectator and the source is along a similar straight line.When movement is not in straight line or velocity is not much less than speed of light then we can not use Doppler shift

This is the limitation of Doppler shift to determine the object distance

3 0

4 years ago

One cycle of the power dissipated by a resistor ( R = 800 Ω R=800 Ω) is given by P ( t ) = 60 W , 0 ≤ t < 5.0 s P(t)=60 W, 0≤

OLga [1]

Answer:

42.5W

Explanation:

To solve this problem we must go back to the calculations of a weighted average based on the time elapsed thus,

$Power_{avg} = \frac{P_1(t_1)+P_2(t_2)}{t_1+t_2}$

We need to calculate the average power dissipated by the 800\Omega resistor.

Our values are given by:

$P(t)=60 W, 0\leq t$

$P(t)=25 W, 5.0\leq t$

Aplying the values to the equation we have:

$Power_{avg} = \frac{P_1(t_1)+P_2(t_2)}{t_1+t_2}$

$Power_{avg} = \frac{60(5-0)+25(10-5)}{(5-0)+(10-5)}$

$Power_{avg} = 42.5W$

5 0

3 years ago

The free length of the spring that is attached to the 0.3-lb slider is 3 in. If the slider is released from rest when x = 6 in.,

hichkok12 [17]

Answer:

a = 64 ft / s²

Explanation:

The force in a spring is given by Hooke's law

F = -k x

Let's use the initial data to calculate the spring constant

k = F / x

Reduscate to the English system

x = 3 in (1foot/12 in) =0.25 foot

k = 0.3 / 0.25

k = 1.2 lb / foot

Now we can use Newton's second law

F = ma

a = F / m

a = -k x / m

m = w / g

m = 0.3 / 32 = 0.009375

x= 6 in (1foot /12 in)= 0.5 foot

a = - 1.2 0.5 / 0.009375

a = 64 ft / s²

5 0

3 years ago