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ratelena [41]
3 years ago
12

Mechanical advantage worksheet

Physics
1 answer:
noname [10]3 years ago
6 0

Answer:

the ratio of the force produced by a machine to the force applied to it, used in assessing the performance of a machine.

Explanation:

Mechanical advantage is a measure of the force amplification achieved by using a tool, mechanical device or machine system. The device preserves the input power and simply trades off forces against movement to obtain a desired amplification in the output force. The model for this is the law of the lever.

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A car accelerates uniformly from rest tona speed of 30.0 mi/h in 12.0s. Find the distance the car travels during this time? Find
meriva

Answer:

2.5 mi/s^2

Explanation:

please see paper for work!

5 0
2 years ago
RETAKE - A push or a pull that causes an object to change its motion is called!
frozen [14]

Answer: Force

Explanation: Force is defined as a push or pull acting on an object

6 0
3 years ago
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Earl is using his hands to hold a metal pan 10 centimeters above a hot burner. How can this scenario be changed to demonstrate c
Olenka [21]

Answer: touch the pan to the burner

Explanation:

There are three modes of heat transfer:

conduction, convection and radiation.  

For conduction, the heat transfers from a hot object to a cold object when the two are in contact.

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For heat transfer by radiation, medium is not required.

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8 0
3 years ago
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What is the ground-state electron configuration of a neutral atom of titanium?
Georgia [21]

Answer:

\bold {1s^22s^22p^63s^23p^64s^23d^2}

Explanation:

electronic configuration of an atom is the spatial arrangement of the electrons around the nucleus in the energy orbits

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The electronic configuration of Titanium is

1s^22s^22p^63s^23p^64s^23d^2

3 0
3 years ago
The Earth revolves around the Sun once a year at an average distance of 1.50×1011m. Find the orbital radius that corresponds to
DedPeter [7]

Answer:

9.4\cdot 10^{10} m

Explanation:

We can solve the problem by using Kepler's third law, which states that the ratio between the cube of the orbital radius and the square of the orbital period is constant for every object orbiting the Sun. So we can write

\frac{r_a^3}{T_a^2}=\frac{r_e^3}{T_e^2}

where

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T_o=180 d is the orbital period of the object

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T_e=365 d is the orbital period the Earth

Solving the equation for r_o, we find

r_o = \sqrt[3]{\frac{r_e^3}{T_e^2}T_o^2} =\sqrt[3]{\frac{(1.50\cdot 10^{11}m)^3}{(365 d)^2}(180 d)^2}=9.4\cdot 10^{10} m

3 0
3 years ago
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