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m_a_m_a [10]
3 years ago
8

Which one is the greatest?

Physics
1 answer:
kipiarov [429]3 years ago
6 0

Answer:

18kg dog walking

Explanation:

the dog walking has more inertia than the boy and girl because its gravity is pulling it down to the earth

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The speed of light in air is 3 x 108 m/s. The speed of light in ice is 2.29 x 108 m/s. What is the refractive index from air to
Studentka2010 [4]

Answer:

η = 1.31

Explanation:

The formula for the refractive index of from air to some other medium is given by the following formula:

\eta = \frac{c}{v}\\

where,

η = refractive index = ?

c = speed of light in air = 3 x 10⁸ m/s

v = speed of light in ice = 2.29 x 10⁸ m/s

Therefore, using these values in the equation we get:

\eta = \frac{3\ x\ 10^8\ m/s}{2.29\ x\ 10^8\ m/s} \\

<u>η = 1.31</u>

4 0
3 years ago
Please Help!!!
algol13

the answer should be C.

4 0
4 years ago
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Machines A, B, and C do the same amount of work, but they each take a different amount of time. Machine A takes more time as com
Amiraneli [1.4K]
So power is considered as the rate of doing work. Base on the problem given, my analysis is that the machine who finish the work faster is machine C. Therefore, in order to finish the same amount of work in a short period of time you are going to expend the most power. My answer is Machine C. 
6 0
3 years ago
Read 2 more answers
A sailboat moves north for a distance of 15.00 km when blown by a wind from the exact southeast with a force of 3.00 x 10^-4 N.
Zolol [24]
These are actually 4 different exercises:

ex 1) The sailboat moves north, while the wind moves from southeast. This means the angle between the direction of the boat and the wind is 45^{\circ}.

Calling F the force of the wind, and d=15~km=15000~m the distance covered by the boat, the work done by the wind is:
W=Fdcos{\theta}=3\cdot10^{-4}~N \cdot 15000~m\cdot cos 45^{\circ}=3.18~J

The total time of the motion is t=1~h=3600~s and therefore the power of the wind is
P= \frac{W}{t} = \frac{3.18~J}{3600~s}=8.8\cdot10^{-4}~W

ex 2) First of all, let's calculate the length of the ramp. Given the two sizes 2.00 m and 6.00 m, we have
d= \sqrt{(2~m)^2+(6~m)^2}= 6.32~m

The mechanical advantage (MA) of the ramp is the ratio between the output load (W) and the input force (F). The output load is the weight of the load, mg, therefore:
MA= \frac{W}{F}= \frac{mg}{F}= \frac{195~Kg\cdot 9.81~m/s^2}{750~N}=2.55

Finally, the efficiency \epsilon of the ramp is the ratio between the output energy and the work done. The output energy is simply the potential energy (Ep) of the load, which is mgh, where h is the height of the ramp. The work done W is the product between the input force, F, and the displacement of the load, which is the length of the ramp: Fd. Therefore:
\epsilon =  \frac{E_p}{W}= \frac{mgh}{Fd}= \frac{195~Kg \cdot 9.81~m/s^2\cdot 2~m}{750~N\cdot6.32~m}=0.81

ex 3) the graph is missing

ex 4) We know that the power is the ratio between the work done W and the time t:
P= \frac{W}{t}
But we can rewrite the work as
W=Fdcos\theta
where F is the force applied, d the displacement of rock and \theta=60^{\circ] is the angle between the direction of the force and the displacement (3 m). 
Therefore we can rewrite the power as
P= \frac{W}{t} = \frac{F d cos\theta}{t}=F v cos\theta
where v=d/t=5~m/s is the velocity, Using the data of the exercise, we can then find the force, F:
F= \frac{P}{v cos\theta} =   \frac{250~W}{5~m/s \cdot cos 60^{\circ}}=100~N

and now we can also calculate the work, which is 
W=Fdcos 60^{\circ}=100~N\cdot 3~m \cos60^{\circ}=150~J
3 0
3 years ago
You are hungry and decide to go to your favorite neighborhood fast-food restaurant. You leave your apartment and take the elevat
Viktor [21]

Answer:

A)  R = (200 i ^ + 100 j ^ + 30k ^) m , B)    L = 223.61 m , C)   R = 225.61 m

Explanation:

Part A

This is a vector summing exercise, let's take a Reference System where the z axis corresponds to the height (flights), the x axis is the East - West and the y axis corresponds to the North - South.

Let's write the displacements

Descending from the apartment

10 flights of 3 m each, the total descent is 30 m

                Z = 30 k ^ m

Offset at street level

            L1 = 0.2 i ^ km

            L2 = 0.1 j ^ km

Let's reduce everything to the SI system

          L1 = 0.2 * 1000 = 200 i ^ m

          L2 = 100 j ^ m

The distance traveled is

          R = (200 i ^ + 100 j ^ + 30k ^) m

Part B

The horizontal distance traveled can be found with the Pythagorean theorem for the coordinates in the plane

                L² = x² + y²

                L = √ (200² + 100²)

                L = 223.61 m

Part C

The magnitude of travel, let's use the Pythagorean theorem for the sum

             R² = x² + y² + z²

              R = √ (30² + 200² + 100²)

             R = 225.61 m

7 0
3 years ago
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