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Julli [10]
3 years ago
8

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:

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:

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Which describes an object in projectile motion? Check all that apply.
Elodia [21]

Answer:

Gravity acts to pull the object down.

The object’s inertia carries it forward.

The path of the object is curved.

Explanation:

The motion of a projectile consists of two separate motions:

- A uniform motion along the horizontal direction, where the velocity is constant; since there are no forces along this direction, the velocity does not change, and so the object continues its motion for inertia --> so, the statement  "The object’s inertia carries it forward" is true.

- A uniformly accelerated motion along the vertical direction, with a constant downward acceleration (g=9.8 m/s^2, acceleration due to gravity). So, the vertical velocity changes, due to the presence of the gravity that acts to pull the object down.

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5 0
2 years ago
wo slightly out of tune instruments will create beats when the same note is played. This is a pattern of louder and softer sound
Lostsunrise [7]

This interaction is known as <em>constructive interference</em>.  It's a result of linear superposition.

5 0
3 years ago
Part complete How long must a simple pendulum be if it is to make exactly one swing per five seconds?
shtirl [24]

Answer:

L=6.21m

Explanation:

For the simple pendulum problem we need to remember that:

\frac{d^{2}\theta}{dt^{2}}+\frac{g}{L}sin(\theta)=0,

where \theta is the angular position, t is time, g is the gravity, and L is the length of the pendulum. We also need to remember that there is a relationship between the angular frequency and the length of the pendulum:

\omega^{2}=\frac{g}{L},

where \omega is the angular frequency.

There is also an equation that relates the oscillation period and the angular frequeny:

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

where T is the oscillation period. Now, we can easily solve for L:

(\frac{2\pi}{T})^{2}=\frac{g}{L}\\\\L=g(\frac{T}{2\pi})^{2}\\\\L=9.8(\frac{5}{2\pi})^{2}\\\\L=6.21m

3 0
3 years ago
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