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lana [24]
3 years ago
7

You split wood with an ax. How would you change the ax to make splitting the wood easier? You would increase mechanical advantag

e by___.
Physics
2 answers:
alex41 [277]3 years ago
6 0
I think you forgot to give the choices along with the question. I am answering the question based on my knowledge and research. You would increase mechanical advantage by <span>making the blade longer from the cutting edge. I hope that this is the answer that has actually come to your desired help.</span>
Alecsey [184]3 years ago
5 0

Answer: make a longer wood for the ax

Explanation: torque is the Force that acts on a moment arm, and is used to cause rotational motion.

The motion of the ax is in a rotational form

So using the formula for torque = force ×lenght

We find out that the more the length the more the torque

This increase in length will make the wood splitting easier

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Alex swims at an average speed of 45m/min. how far does he swim in 1 min 24 sec?
saveliy_v [14]

Answer:

63 m

Explanation:

Average speed of swimming = 45 m/min

Time = 1 minute 24 seconds

Converting time into minutes:

Formula:

Putting values, we get.

Thus, Alex swims for 63 m in 1 minute and 24 seconds.

6 0
3 years ago
During a trial run, race car A starts from rest and accelerates uniformly along a straight level track for a particular interval
kondor19780726 [428]
Answer: car B has travelled 4times as far as Car A

d=vi*t+1/2at^2

No initial velocity so equation becomes;

d=1/2at^2 and the acceleration is the same between both only time is different;

Car A d=1/2a(1)^2

Car B d=1/2a(2)^2

Car A d= 1^2=1

Car B d= 2^2=4

Car B d=4*Car A

So car B has travelled 4 times as far as car A
5 0
3 years ago
1. What are three appliances in your house that use a motor that turns electrical energy into mechanical energy?
AleksAgata [21]
A vacuum is an electrical motor and<span> which it  converts electrical energy into mechanical energy.

</span>
6 0
3 years ago
A solid sphere, a solid disk, and a thin hoop are all released from rest at the top of the incline (h0 = 20.0 cm).
Ede4ka [16]

Answer:

a. The object with the smallest rotational inertia, the thin hoop

b. The object with the smallest rotational inertia, the thin hoop

c.  The rotational speed of the sphere is 55.8 rad/s and Its translational speed is 1.67 m/s

Explanation:

a. Without doing any calculations, decide which object would be spinning the fastest when it gets to the bottom. Explain.

Since the thin has the smallest rotational inertia. This is because, since kinetic energy of a rotating object K = 1/2Iω² where I = rotational inertia and ω = angular speed.

ω = √2K/I

ω ∝ 1/√I

since their kinetic energy is the same, so, the thin hoop which has the smallest rotational inertia spins fastest at the bottom.

b. Again, without doing any calculations, decide which object would get to the bottom first.

Since the acceleration of a rolling object a = gsinФ/(1 + I/MR²), and all three objects have the same kinetic energy, the object with the smallest rotational inertia has the largest acceleration.

This is because a ∝ 1/(1 + I/MR²) and the object with the smallest rotational inertia  has the smallest ratio for I/MR² and conversely small 1 + I/MR² and thus largest acceleration.

So, the object with the smallest rotational inertia gets to the bottom first.

c. Assuming all objects are rolling without slipping, have a mass of 2.00 kg and a radius of 3.00 cm, find the rotational and translational speed at the bottom of the incline of any one of these three objects.

We know the kinetic energy of a rolling object K = 1/2Iω²  + 1/2mv² where I = rotational inertia and ω = angular speed, m = mass and v = velocity of center of mass = rω where r = radius of object

The kinetic energy K = potential energy lost = mgh where h = 20.0 cm = 0.20 m and g = acceleration due to gravity = 9.8 m/s²

So, mgh =  1/2Iω²  + 1/2mv² =  1/2Iω²  + 1/2mr²ω²

Let I = moment of inertia of sphere = 2mr²/5 where r = radius of sphere = 3.00 cm = 0.03 m and m = mass of sphere = 2.00 kg

So, mgh = 1/2Iω²  + 1/2mr²ω²

mgh = 1/2(2mr²/5 )ω²  + 1/2mr²ω²

mgh = mr²ω²/5  + 1/2mr²ω²

mgh = 7mr²ω²/10

gh = 7r²ω²/10

ω² = 10gh/7r²

ω = √(10gh/7) ÷ r

substituting the values of the variables, we have

ω = √(10 × 9.8 m/s² × 0.20 m/7) ÷ 0.03 m

= 1.673 m/s ÷ 0.03 m

= 55.77 rad/s

≅ 55.8 rad/s

So, its rotational speed is 55.8 rad/s

Its translational speed v = rω

= 0.03 m × 55.8 rad/s

= 1.67 m/s

So, its rotational speed is of the sphere is 55.8 rad/s and Its translational speed is 1.67 m/s

6 0
3 years ago
The earplug can reduce the sound level to about 18 decibels (dB). What percentage reduction is this intensity?
zlopas [31]

Answer:

1 x 10 -10 whisper at 1m distance.

Explanation:

  • Properly fitted ear plugs an reduce noise form 15-30db. Although they are better for low frequency
8 0
3 years ago
Read 2 more answers
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