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SashulF [63]
3 years ago
5

Suppose a student takes a study break from astronomy homework and steps out onto his or her balcony on a summer evening.

Physics
1 answer:
Korvikt [17]3 years ago
6 0

Answer:3,1 and 2

Explanation:

When the students step out for another break three hour later, how has the location of Cygnus changed. The constellations Cygnus is low in the eastern part of the sky and the big dipper to locate Polaris is fairly high in the sky to the north

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What are 3 ways to say velocity is decreasing
inn [45]

Answer: cessation,cease,slowing momentum

Explanation:

6 0
3 years ago
A person of mass m will bungee-jump from a bridge over a river where the height from the bridge to the river is h. The bungee co
Likurg_2 [28]

Answer:k_{min}=\frac{18mg}{h}

Explanation:

Given

mass of person is m

Distance between bridge and river is h

chord has an un-stretched length of \frac{2h}{3}

Let spring constant be k

Person will just stop before hitting the river

Conserve energy i.e. Potential Energy of Person is converted in to elastic energy of chord

mgh=\frac{kx^2}{2}

x=h-\frac{2h}{3}=\frac{h}{3}

mgh=\frac{kh^2}{18}

k=\frac{18mg}{h}

Thus k_{min}=\frac{18mg}{h}

5 0
4 years ago
A conveyer belt moves a 40 kg box at a velocity of 2 m/s. What is the kinetic energy of the box while it is on the conveyor belt
ankoles [38]
Kinwtic energy = (40kg*2*2)/2=40*2=80J
4 0
4 years ago
Read 2 more answers
When a star goes from the main sequence to the red giant phase, it becomes larger rather than smaller. This is true because grav
GuDViN [60]
I believe this is electron degeneracy, because the star is essentially having too many reactions too fast and collapses in on itself eventually.
4 0
3 years ago
Consider two points in an electric field. The potential at point 1, V1, is 33 V. The potential at point 2, V2, is 175 V. An elec
Mnenie [13.5K]

Answer:

ΔU  = e(V₂ - V₁) and its value ΔU = -2.275 × 10⁻²¹ J

Explanation:

Since the electric potential at point 1 is V₁ = 33 V and the electric potential at point 2 is V₂ = 175 V, when the electron is accelerated from point 1 to point 2, there is a change in electric potential ΔV which is given by ΔV = V₂ - V₁.

Substituting the values of the variables into the equation, we have

ΔV = V₂ - V₁.

ΔV = 175 V - 33 V.

ΔV = 142 V

The change in electric potential energy ΔU = eΔV = e(V₂ - V₁) where e = electron charge = -1.602 × 10⁻¹⁹ C and ΔV = electric potential change from point 1 to point 2 = 142 V.

So, substituting the values of the variables into the equation, we have

ΔU = eΔV

ΔU = eΔV

ΔU = -1.602 × 10⁻¹⁹ C × 142 V

ΔU = -227.484 × 10⁻¹⁹ J

ΔU = -2.27484 × 10⁻²¹ J

ΔU ≅ -2.275 × 10⁻²¹ J

So, the required equation for the electric potential energy change is

ΔU  = e(V₂ - V₁) and its value ΔU = -2.275 × 10⁻²¹ J

5 0
3 years ago
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