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gregori [183]
4 years ago
12

Use newton's third law to explain why a blown up balloon but untied balloon will fly around the room when you let it go? really

on a time crunch please please help!
Physics
1 answer:
sergij07 [2.7K]4 years ago
7 0
Newton's third law states that for every action, there is an equal and opposite reaction. When you let go of the ballon, you are letting the force out but the force also pushes the balloon back.
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What would happen if earth was further from the sun
Colt1911 [192]
Earth would have different climates and would be more colder. 
3 0
3 years ago
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A street musician sounds the A string of his violin, producing a tone of 440 Hz, What frequency does a bicyclist hear as he a) a
DanielleElmas [232]

Answer:

a) f_o=454.11Hz

b)f_o=425.89Hz

Explanation:

Let´s use Doppler effect, in order to calculate the observed frequency by the byciclist. The Doppler effect equation for a general case is given by:

f_o=\frac{v\pm v_o}{v\pm v_s} *f_s

where:

f_o=Observed\hspace{3}frequency

f_s=Actual\hspace{3}frequency

v=Speed\hspace{3}of\hspace{3}the\hspace{3}sound\hspace{3}waves

v_s=Velocity\hspace{3}of\hspace{3}the\hspace{3}source

v_o=Velocity\hspace{3}of\hspace{3}the\hspace{3}observer

Now let's consider the next cases:

+v_o\hspace{3}is\hspace{3}used\hspace{3}when\hspace{3}the\hspace{3}observer\hspace{3}moves\hspace{3}towards\hspace{3}the\hspace{3}source

-v_o\hspace{3}is\hspace{3}used\hspace{3}when\hspace{3}the\hspace{3}observer\hspace{3}moves\hspace{3}away\hspace{3}from\hspace{3}the\hspace{3}source

-v_s\hspace{3}is\hspace{3}used\hspace{3}when\hspace{3}the\hspace{3}source\hspace{3}moves\hspace{3}towards\hspace{3}the\hspace{3}observer

+v_s\hspace{3}is\hspace{3}used\hspace{3}when\hspace{3}the\hspace{3}source\hspace{3}moves\hspace{3}away\hspace{3}from\hspace{3}the\hspace{3}observer

The data provided by the problem is:

f_s=440Hz\\v_o=11m/s

The problem don't give us aditional information about the medium, so let's assume the medium is the air, so the speed of sound in air is:

v=343m/s

Now, in the first case the observer alone is in motion towards to the source, hence:

f_o=\frac{v+v_o}{v}*f_s=\frac{343+11}{343} *440=454.1107872Hz

Finally, in the second case the observer alone is in motion away from the source, so:

f_o=\frac{v-v_o}{v}*f_s=\frac{343-11}{343} *425.8892128Hz

6 0
4 years ago
WILL GIVE BRAINLIEST TO THE CORRECT ANSWER
LenKa [72]

Answer:

See the answers below.

Explanation:

We can solve both problems using Newton's second law, which tells us that the sum of forces on a body is equal to the product of mass by acceleration.

∑F =m*a

where:

F = force [N] (units of newtons)

m = mass = 1000 [kg]

a = acceleration = 3 [m/s²]

F = 1000*3\\F=3000[N]

And the weight of any body can be calculated by means of the mass product by gravitational acceleration.

W=m*g\\W=1000*9.81\\W=9810 [N]

4 0
3 years ago
Which muscles help you sit up from laying down?​
rosijanka [135]

Answer: abdominal muscles.

Explanation:

3 0
3 years ago
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The driver or a sports car traveling at 10 m/s steps down hard on the accelerator for 5 seconds. If they accelerate at a rate of
ivanzaharov [21]

Answer:

53.125m

Explanation:

The displacement of the car, denoted by S, can be calculated using the formula:

S = ut + 1/2at²

Where;

u = initial velocity/speed (m/s)

t = time (s)

a = acceleration (m/s²)

According to the information provided in this question, u = 10m/s, t = 5s, a = 0.25m/s², S = ?

S = ut + 1/2at²

S = (10 × 5) + 1/2 (0.25 × 5²)

S = 50 + 1/2 (0.25 × 25)

S = 50 + 1/2(6.25)

S = 50 + 3.125

S = 53.125m

4 0
3 years ago
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