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3 years ago

What are 2 examples of levers

Physics

2 answers:

GaryK [48]3 years ago

4 0

Answer:

Some examples of levers include more than one class, such as a nut cracker, a stapler, nail clippers, ice tongs and tweezers. Other levers, called single class levers include the claw end of a hammer.

Explanation:

EastWind [94]3 years ago

4 0

Answer:

Explanation:

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How are mass and inertia related?

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The greater the mass the greater is inertia.

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3 years ago

Review Conceptual Example 6 as background for this problem. A car is traveling to the left, which is the negative direction. The

DiKsa [7]

Answer:

(a) 1.21 m/s² (b) 1.75 m/s²

Explanation:

The initial speed of the car, u = 17.8 m/s

Case 1.

Final speed of the car, v = 23.5 m/s

Time, t = 4.68-s

Acceleration = rate of change of velocity

$a=\dfrac{23.5 -17.8 }{4.68}\\\\a=1.21\ m/s^2$

Case 2.

Final speed of the car, v = 15.3 m/s

$a=\dfrac{23.5 -15.3}{4.68}\\\\a=1.75\ m/s^2$

Hence, this is the required solution.

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2 years ago

10. Explain the principle of electric motor. Write its uses.

konstantin123 [22]

Explanation:

The principle of an electric motor is based on the current carrying conductor which produces magnetic field around it. A current carrying conductor is placed perpendicular to the magnetic field so that it experiences a force.

The largest electric motors are used for ship propulsion, pipeline compression and pumped-storage applications with ratings reaching 100 megawatts. Electric motors are found in industrial fans, blowers and pumps, machine tools, household appliances, power tools and disk drives.

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2 years ago

a 42.3 kg girl and a 7.93 kg sled are on the surface of a frozen lake, 15.0m apart and linked by a rope, but not moving yet. the

ycow [4]

Answer:

they meet from the girl's original position at: 2.37 (meters)

Explanation:

We need to use the Newton's law, exactly the second law that relate force, mass and acceleration as: $F=m*a$ with this we can get both accelerations; solving for acceleration $a=\frac{F}{m}$ . Now $a_{girl}=\frac{5.76}{42.3}=0.14 (m/s^{2})$ and $a_{sled}=\frac{5.76}{7.93}=0.73(m/s^{2})$ . Then knowing that they both travel at the same time and assuming that the distance among the girl and the sled is: 15.0-x, so, $x=\frac{1}{2}*a_{girl}*t^{2}$ and $15.0-x=\frac{1}{2}*a_{sled}*t^{2}$ , solving for the time we get: $t=\sqrt{\frac{2x}{a_{girl} } }$ and $t=\sqrt{\frac{2*(15.0-x)}{a_{sled} } }$ with this equations we solving for the x that is the distance between the girl and the sled after the apply the force and we get: $\sqrt{\frac{2x}{a_{girl}}} = \sqrt{\frac{2*(15.0-x)}{a_{sled} }$ . Finally we get: $\frac{x}{a_{girl} }=\frac{(15.0-x)}{a_{sled} }$ and replacing the values we have got: $\frac{x}{0.14} =\frac{(15.0-x)}{0.73}$ so $5.33*x=15-x$ so x=2.37 (meters).

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3 years ago

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