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Pepsi [2]
3 years ago
15

What is a circumpolar star? A)a star that is close to the north celestial pole a star that is close to the south celestial pole

C) a star that always remains above your horizon and appears to rotate around the celestial pole D) a star that makes a daily circle around the celestial sphere E) a star that is visible from the Arctic or Antarctic circles
Physics
1 answer:
Vedmedyk [2.9K]3 years ago
6 0

Answer:

A star that always remains above your horizon and appears to rotate around the celestial pole.

Explanation:

A) a star that is close to the north celestial pole: a circumpolar star could be close to the north celestial pole, but this answer is omitting the south celestial pole.

B) a star that is close to the south celestial pole: a circumpolar star could be close to the south celestial pole, but this answer is omitting the north celestial pole.

C) a star that always remains above your horizon and appears to rotate around the celestial pole: this is the definition of a circumpolar star.

D) a star that makes a daily circle around the celestial sphere: every star does this.

E) a star that is visible from the Arctic or Antarctic circles : there are many starts visible from there that are not circumpolar.

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a metal bar 70cm long and 4.00 kg in mass supported on two knief -edge placed 10 cm from each end. a 6.00 kg weight is suspended
Klio2033 [76]
First, you make a diagram of all the forces acting on the system. This is shown in the figure. We have to determine F1 and F4. Let's do a momentum balance. Momentum is conserved so the summation of all momentum is equal to zero. Momentum is force*distance.

To determine F1: (reference is F4, so F4=0)

∑Momentum = 0 = -F2 - F3 + F1
0 = (-4 kg)(9.81 m/s2)(0.25m)-(6kg)(9.81 m/s2)(0.5-0.3m)+F1(0.5-0.1m)
F1 = 53.96 N (left knife-edge)

To determine F4: (reference is F1, so F1=0)

∑Momentum = 0 = -F2 - F3 + F4
0 = (-4 kg)(9.81 m/s2)(0.25m)-(6kg)(9.81 m/s2)(0.5-0.2m)+F4(0.5-0.1m)
F4 = 68.67 N (right knife-edge) 


4 0
3 years ago
Read 2 more answers
1. Looking at the planet vs. eccentricity table, which two planets have the greatest eccentricity?
AVprozaik [17]

Answer:

Pluto & Mercury

Explanation:

Pluto's eccentricity is 0.248

Mercury's eccentricity is 0.206

6 0
3 years ago
A 34n force is applied to a 213kg mass how much does the mass accelerate
motikmotik
We Know, F = m*a
Here, F = 34 N
m = 213 Kg

Substitute their values in the equation,
34 = 213 * a
a = 34/213
a = 0.159 m/s²

So, your final answer & the acceleration of the object would be 0.159 m/s²

Hope this helps!
3 0
3 years ago
A 12-kg piece of metal displaces 1.6 L of water when submerged. Part A Find its density. Express your answer to two significant
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Answer:

ρ = 7500 kg/m³

Explanation:

Given that

mass ,m = 12 kg

Displace volume ,V= 1.6 L

We know that

1000 m ³ = 1 L

Therefore V= 0.0016 m ³

When metal piece is fully submerged

We know that

mass = Density x volume

m=\rho \times V

Now by putting the values in the above equation

\rho=\dfrac{12}{0.0016}\ kg/m^3

ρ = 7500 kg/m³

Therefore the density of the metal piece will be  7500 kg/m³.

6 0
3 years ago
A train has a length of 81.1 m and starts from rest with a constant acceleration at time t = 0 s. At this instant, a car just re
arlik [135]

Answer: a) vcar= 7 m/s ; b) a train= 0.65 m/s^2

Explanation: By using the kinematic equation for the car and the train we can determine the above values of the car velocity and the acceletarion of the train, respectively.

We have for the car

distance = v car* t, considering the length of train (81.1 m) travel by the car during the first 11.6 s

the v car =  distance/time= 81.1 m/11.6s= 7 m/s

In order to calculate the acceleration we have to use the kinematic equation for the train from the rest

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distance train : distance travel by the car at constant speed

so distance train= (vcar*36.35)m=421 m

the a traiin= (2* 421 m)/(36s)^2=0.65 m/s^2

4 0
3 years ago
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