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3 years ago

How long will it take the ball to reach the height of 35m?

1 answer:

7 0

Ahhh... hello! I’ve done a science experiment that has asked this sort of question. Not as easy as it sounds... I know. I’ll tell you what I learned from that experiment. The answer will depend on the materials the ball is made of and then also your drop length. After these two things are determined, you can get a pretty close time and measurement. I hope this information helps. Have a great night!
~Brooke❤️

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Given that average speed is distance traveled divided by time, determine the values of m and n when the time it takes a beam of

schepotkina [342]

If speed = distance/time , then time = speed/distance.

So...

Speed of light = 3*10^8(m/s)
Average distance from Earth to Sun = 149.6*10^9(m)

Therefore, t=(3*10^8(m/s))/(149.6*10^9(m))

I hope this was a helpful explanation, please reply if you have further questions about the problem.

Good luck!

5 0

2 years ago

A man has a power of 90 W and mass 60 kg runs up a staircase in 40 s. If each staircase is 20cm high find the number of steps? (

34kurt

I showed my working in the images above. if you have any questions please feel free to ask

4 0

2 years ago

A string with a mass density of 3 * 10^-3 kg/m is under a tension of 380 N and is fixed at both ends. One of its resonance frequ

Delvig [45]

Answer:

(a) the fundamental frequency of this string is 65 Hz

(b) the harmonics of the given frequencies are third and fourth respectively.

Explanation:

Given;

mass density of the string, μ = 3 x 10⁻³ kg/m

tension of the string, T = 380 N

resonating frequencies, 195 Hz and 260 N

For the given resonant frequencies;

$195 = \frac{n}{2l} \sqrt{\frac{T}{\mu} } ---(1)\\\\260 = \frac{n+1}{2l} \sqrt{\frac{T}{\mu} } ---(2)\\\\divide \ (2) \ by (1)\\\\\frac{260}{195} = \frac{n+1 }{n} \\\\260n = 195(n+1)\\\\260 n = 195 n + 195\\\\260n - 195n = 195\\\\65n = 195\\\\n = \frac{195}{65} \\\\n = 3$

$195 = \frac{n}{2l} \sqrt{\frac{T}{\mu} } \\\\195 = \frac{3}{2l}\sqrt{\frac{380}{3\times 10^{-3}} } \\\\l = \frac{3}{2\times 195}\sqrt{\frac{380}{3\times 10^{-3}} }\\\\l = 2.74 \ m$

(a) the fundamental frequency is calculated as;

$f_o = \frac{1}{2l} \sqrt{\frac{T}{\mu} } \\\\f_o = \frac{1}{2\times 2.74} \sqrt{\frac{380}{3\times 10^{-3} } }\\\\f_o = 65 \ Hz$

(b) harmonics of the given frequencies;

the first harmonic (n = 1) = f₀ = 65 Hz

the second harmonic (n = 2) = 2f₀ = 130 Hz

the third harmonic (n = 3) = 3f₀ = 195 Hz

the fourth harmonic (n = 4) = 4f₀ = 260 Hz

Thus, the harmonics of the given frequencies are third and fourth respectively.

7 0

2 years ago

A fisherman is fishing from a bridge and is using a "42.0-N test line." In other words, the line will sustain a maximum force of

lara31 [8.8K]

Answer:

(a) 42 N

(b)36.7 N

Explanation:

Nomenclature

F= force test line (N)

W : fish weight (N)

Problem development

(a) Calculating of weight of the heaviest fish that can be pulled up vertically, when the line is reeled in at constant speed

We apply Newton's first law of equlibrio because the system moves at constant speed:

∑Fy =0

F-W= 0

42N -W =0

W = 42N

(b) Calculating of weight of the heaviest fish that can be pulled up vertically, when the line is reeled with an acceleration whose magnitude is 1.41 m/s²

We apply Newton's second law because the system moves at constant acceleration:

m= W/g , m= W/9.8 , m:fish mass , W: fish weight g:acceleration due to gravity

∑Fy =m*a

m= W/g , m= W/9.8 , m:fish mass , W: fish weight g:acceleration due to gravity

F-W= ( W/9.8 )*a

42-W= ( W/9.8 )*1.41

42= W+0.1439W

42=1.1439W

W= 42/1.1439

W= 36.7 N

8 0

3 years ago

The diagram shows the forces on an object.

jok3333 [9.3K]

Number 2 because it is

6 0

3 years ago

Read 2 more answers