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3 years ago

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(I am doing a lab called, Density of solids and I need help with this question) (I am in HS and I picked Physics because I thoug

ht it was short for Physical Science )
4. Locate the data and observations collected in your lab guide. What are the key results? How would you best summarize the data to relate your findings?

1 answer:

umka2103 [35]3 years ago

4 0

Answer: Location the data and observations collected in your lab guide

Explanation:

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Light travels at 3 × 108 m/s, and it takes about 8 min for light from the sun to travel to Earth. Based on this, the order of ma

N76 [4]

Answer:

The order of magnitude of the distance from the sun to Earth is 10⁸ km.

Explanation:

The order of magnitude of the distance from the sun to Earth can be calculated as follows:

$c = \frac{x}{t}$

Where:

c: is the speed of light = 3x10⁸ m/s

t: is the time = 8 min

Hence, the distance is:

$x = c*t = 3 \cdot 10^{8} m/s*8 min*\frac{60 s}{1 min} = 1.44 \cdot 10^{11} m = 1.44 \cdot 10^{8} km$

Therefore, the order of magnitude of the distance from the sun to Earth is 10⁸ km.

I hope it helps you!

5 0

3 years ago

What happens to a day of light that slows down when it hits a new medium at an angle

Anika [276]

Light travels as transverse waves and faster than sound. It can be reflected, refracted and dispersed. Ray diagrams show what happens to light in mirrors and lenses. Eyes and cameras detect light.

4 0

3 years ago

The barrel of a rifle has a length of 0.89 m. A bullet leaves the muzzle of a rifle with a speed of 620 m/s. What is the acceler

guapka [62]

Answer:

215955.06 m/s^2

Explanation:

length of barrel, s = 0.89 m

initial velocity of the bullet, u = 0 m/s

Final velocity of the bullet, v = 620 m/s

Let a be the acceleration of the bullet in the barrel

Use third equation of motion, we get

$v^{2}=u^{2}+ 2as$

$620^{2}=0^{2}+ 2\times a \times 0.89$

a = 215955.06 m/s^2

Thus, the acceleration of the bullet inside the barrel is 215955.06 m/s^2.

6 0

3 years ago

Consider the nearly circular orbit of Earth around the Sun as seen by a distant observer standing in the plane of the orbit. Wha

ikadub [295]

We have that the spring constant is mathematically given as

$k=2.37*10^{11}N/m$

Generally, the equation for angular velocity is mathematically given by

$\omega=\sqrt{k}{m}$

Where

k=spring constant

And

$\omega =\frac{2\pi}{T}$

Therefore

$\frac{2\pi}{T}=\sqrt{k}{n}$

Hence giving spring constant k

$k=m((\frac{2 \pi}{T})^2$

Generally

Mass of earth $m=5.97*10^{24}$

Period for on complete resolution of Earth around the Sun

$T=365 days$

$T=365*24*3600$

Therefore

$k=(5.97*10^{24})((\frac{2 \pi}{365*24*3600})^2$

$k=2.37*10^{11}N/m$

In conclusion

The effective spring constant of this simple harmonic motion is

$k=2.37*10^{11}N/m$

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3 years ago

Estimate the volume of a typical house (2050 feet squared in size and 10 feet tall) answer in units of meters squared

Aloiza [94]

Answer:

$Volume = 6248.48 m^{3}$

Explanation:

Given:

The area of the house $A = 2050\ ft^{2}$

The height of the house $h=10\ ft$

We need to find the volume of a typical house.

Solution:

We find the volume of the house by multiplying the area of the house and height of the house.

$Volume = Area\times height$

$Volume = A\times h$

Area and height of the house are known, so we substitute these value in above equation.

$Volume = 2050\times 10$

$Volume = 20500\ ft^{3}$

Now we convert the unit from feet to meter.

Divide the volume by 3.2808 for $m^{3}$

$Volume = \frac{20500}{3.2808}$

$Volume = 6248.48\ m^{3}$

Therefore, the volume of the house is $6248.48 m^{3}$

8 0

4 years ago