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Damm [24]
3 years ago
11

Ans it fastly i need the ans fastly ​NO SPAM

Physics
1 answer:
Stells [14]3 years ago
3 0

12. C

13. A

14. B

15. D

16. D

17. B

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Four children pull on the same stuffed toy at the same time yet there is no net force on the toy. How is this possible.
Stels [109]

Answer:

Net force is Zero.

Explanation:

If all forces that are equal and opposite are exerted on an object the resulting force will be Zero.

3 0
3 years ago
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A trombone can produce pitches ranging from 85 Hz to 660 Hz approximately. When the trombone is producing a 562 Hz tone, what is
tester [92]

To solve this problem we will apply the concept of wavelength, which warns that this is equivalent to the relationship between the speed of the air (in this case in through the air) and the frequency of that wave. The air is in standard conditions so we have the relation,

Frequency = f = 562Hz

Speed of sound in air = v = 331m/s

The definition of wavelength is,

\lambda = \frac{v}{f}

Here,

v = Velocity

f = Frequency

Replacing,

\lambda = \frac{331m/s}{562Hz}

\lambda = 0.589m

Therefore the wavelength of that tone in air at standard conditions is 0.589m

3 0
3 years ago
You attach a meter stick to an oak tree, such that the top of the meter stick is 2.27 meters above the ground. later, an acorn f
Alexandra [31]

The acorn was at a height of <u>4.15 m</u> from the ground before it drops.

The acorn takes a time t to fall through a distance h₁, which is the length of the scale. When the acorn reaches the top of the scale, its velocity is u.

Calculate the speed of the acorn at the top of the scale, using the equation of motion,

s=ut+ \frac{1}{2} at^2

Since the acorn falls freely under gravity, its acceleration is equal to the acceleration due to gravity g.

Substitute 2.27 m for s (=h₁), 0.301 s for t and 9.8 m/s² for a (=g).

s=ut+ \frac{1}{2} at^2\\ (2.27 m)=u(0.301s)+\frac{1}{2}(9.8m/s^2)(0.301s)^2\\ u=\frac{1.8261m}{0.301s} =6.067m/s

If the acorn starts from rest and reaches a speed of 6.067 m/s at the top of the scale, it would have fallen a distance h₂ to achieve this speed.

Use the equation of motion,

v^2=u^2+2as

Substitute 6.067 m/s for v, 0 m/s for u, 9.8 m/s² for a (=g) and h₂ for s.

v^2=u^2+2as\\ (6.067m/s)^2=(0m/s)^2+2(9.8m/s^2)h_2\\ h_2=\frac{(6.067m/s)^2}{2(9.8m/s^2)} =1.878 m

The height h above the ground at which the acorn was is given by,

h=h_1+h_2=(2.27 m)+(1.878 m)=4.148 m

The acorn was at a height <u>4.15m</u> from the ground before dropping down.

3 0
3 years ago
Hurry help please, number 3
Dovator [93]
C most of the Earths weather occurs in this layer.
6 0
3 years ago
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What is uniform acceleration
kotegsom [21]

Answer:

It means zero acceleration.

Explanation:

When a body moves with constant velocity then it is said to be there is no rate of change of velocity takes place because it is constant. So the acceleration becomes zero or uniform acceleration.

5 0
3 years ago
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