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serg [7]
3 years ago
5

Which is easier for a beginner? (This is for archery) O long bow O recurve bow

Physics
2 answers:
AlexFokin [52]3 years ago
6 0

Answer:

The recurve bows

Explanation:Vote brainliest plz

Zepler [3.9K]3 years ago
4 0
The recurve bow is smaller and easier to handle while the longbow is used more for competitions. So I would say the recurve bow is easier for a beginner
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What are the advantages and disadvantages of renewable and non-renewable energy?
bogdanovich [222]

Renewable energy

<u>Advantages :-</u>

1. Easily regenerate

2. Boost economic growth

3. Easily available

4. Support environment

5. Low maintenance cost

<u>Disadvantages :-</u>

1. Weather dependency

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Non-renewable energy

<u>Advantages :-</u>

1. Concentrated energy source

2. Reliable energy source

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5 0
3 years ago
A 120-kg object and a 420-kg object are separated by 3.00 m At what position (other than an infinitely remote one) can the 51.0-
djverab [1.8K]

Answer:

1.045 m from 120 kg

Explanation:

m1 = 120 kg

m2 = 420 kg

m = 51 kg

d = 3 m

Let m is placed at a distance y from 120 kg so that the net force on 51 kg is zero.

By use of the gravitational force

Force on m due to m1 is equal to the force on m due to m2.

\frac{Gm_{1}m}{y^{2}}=\frac{Gm_{2}m}{\left ( d-y \right )^{2}}

\frac{m_{1}}{y^{2}}=\frac{m_{2}}{\left ( d-y \right )^{2}}

\frac{3-y}{y}=\sqrt{\frac{7}{2}}

3 - y = 1.87 y

3 = 2.87 y

y = 1.045 m

Thus, the net force on 51 kg is zero if it is placed at a distance of 1.045 m from 120 kg.

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Phoenix [80]
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3 years ago
Read 2 more answers
An 10 kg mass has an applied force of magnitude 69N acting on it pushing it to move in the positive x-direction. The mass is on
Oduvanchick [21]

The net force on the mass in the vertical direction is

∑ <em>F</em> = <em>n</em> - <em>w</em> = 0

and the net horizontal force is

∑ <em>F</em> = <em>A</em> - <em>f</em> = <em>ma</em>

where

<em>n</em> = magnitude of the normal force

<em>w</em> = <em>mg</em> = weight of the mass

<em>m</em> = 10 kg, mass of the … uh, mass

<em>g</em> = 9.8 m/s², mag. of acceleration due to gravity

<em>A</em> = 69 N, mag. of the applied force

<em>f</em> = mag. of kinetic friction

From the first equation, we get

<em>n</em> = <em>mg</em> = (10 kg) (9.8 m/s²) = 98 N

Then the friction force has magnitude

<em>f</em> = <em>µ</em> <em>n</em> = 0.5 (98 N) = 49 N

Solve the mass's acceleration in the second equation:

69 N - 49 N = (10 kg) <em>a</em>

<em>a</em> = (20 N) / (10 kg) = 2.0 m/s²

in the positive <em>x</em> direction.

5 0
3 years ago
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