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2 years ago

It takes Venus 225 days to orbit the sun. If the Earth-sun distance is 1.5 × 10^11 m, what

Physics

1 answer:

loris [4]2 years ago

6 0

Answer:

Distance between Venus and Sun is 1.0864x10^11 m

Explanation:

Data

T1 (Total time taken by earth to rotate around sun)=365 days

T2 (Total time taken by venus to rotate around sun)=225 days

R1 (Radius b/w Sun and earth)=1.5x10^11m

R2(Distance between Venus and Sun)=?

Solution

Apply Kepler Law

(R2)^3/(R1)^3=(T1)^2/(T2)^2

(R2)^3/(1.5x10^11)^3=(225^2)/(365^2)

R2=1.0864x10^11m

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A zebra runs across a field at a constant speed of 14m/s how far does the zebra go in 8 seconds?

Ratling [72]

Answer:

112m/s

Explanation:

14x8=112 therefore meaning the zebra would run 112m/s

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2 years ago

Read 2 more answers

A curtain hangs straight down in front of an open window. A sudden gust of wind blows past the window; and the curtain is pulled

VikaD [51]

Answer:

option B.

Explanation:

The correct answer is option B.

The phenomenon of the curtains to pull out of the window can be explained using Bernoulli's equation.

According to Bernoulli's Principle when the speed of the moving fluid increases the pressure within the fluid decrease.

When wind flows in the outside window the pressure outside window decreases and pressure inside the room is more so, the curtain moves outside because of low pressure.

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2 years ago

The air in a pipe resonates at 150 Hz and 750 Hz, one of these resonances being the fundamental. If the pipe is open at both end

Xelga [282]

Answer:

Explanation:

Two frequencies with magnitude 150 Hz and 750 Hz are given

For Pipe open at both sides

fundamental frequency is 150 Hz as it is smaller

frequency of pipe is given by

$f=\frac{nv}{2L}$

where L=length of Pipe

v=velocity of sound

$f=150\ Hz$ for n=1

and f=750 is for n=5

thus there are three resonance frequencies for n=2,3 and 4

For Pipe closed at one end

frequency is given by

$f=\frac{(2n+1)}{4L}\cdot v$

for n=0

$f_1=\frac{v}{4L}$

$f_1=150\ Hz$

for n=2

$f_2=\frac{5v}{4L}$

Thus there is one additional resonance corresponding to n=1 , between $f_1$ and $f_2$

8 0

2 years ago

The dial of a scale looks like this: 00.0kg. A physicist placed a spring on it. The dial read 00.6kg. He then placed a metal cha

saveliy_v [14]

Answer:

d. The scale's resolution is too low to read the change in mass

Explanation:

If we want to find the change in energy of the spring, we will have to use the Hooke's Law. Hooke's Law states that:

F = kx

since,

w = Fd

dw = Fdx

integrating and using value of F, we get:

ΔE = (0.5)kx²

where,

ΔE = Energy added to spring

k = spring constant

x = displacement

The spring constant is typically in range of 4900 to 29400 N/m.

So if we take the extreme case of 29400 N/m and lets say we assume an unusually, extreme case of 1 m compression, we get the value of energy added to be:

ΔE = (0.5)(29400 N/m)(1 m)²

ΔE = 1.47 x 10⁴ J

Now, if we convert this energy to mass from Einstein's equation, we get:

ΔE = Δmc²

Δm = ΔE/c²

Δm = (1.47 x 10⁴ J)/(3 x 10⁸ m/s)²

As, you can see from the answer that even for the most extreme cases the value of mass associated with the additional energy is of very low magnitude.

Since, the scale only gives the mass value upto 1 decimal place.

Thus, it can not determine such a small change. So, the correct option is:

<u>d. The scale's resolution is too low to read the change in mass</u>

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2 years ago

7) A heavy dart and a light dart are launched horizontally on a frictionless table by identical ideal springs. Both springs were

Elenna [48]

Answer:

A, B, and E

Explanation:

The springs are identical, and are compressed the same amount, so they have the same initial elastic potential energy. (E is true)

Energy is conserved, so the darts have the same amount of kinetic energy. (A is true, C is false)

The lighter dart has the same energy as the heavier dart. Since it has less mass, it must have a greater speed. (B is true, D is false)

6 0

2 years ago