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AVprozaik [17]
3 years ago
7

(science)pls help, due today and I have alot more to do​

Physics
1 answer:
Effectus [21]3 years ago
8 0

Answer:

I can help

Explanation:

hold on

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Vika [28.1K]

Answer:

make a negative

Explanation:

yep

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A rocket at rest with a mass of 942 kg is acted on by an average net force of 6,731 N upwards for 21 s. What is the final veloci
Galina-37 [17]

Answer:

Explanation:

Givens

m = 942

F = 6731

t = 21 seconds

vi = 0

vf = ?

Formula

F = m * (vf -  vi )  / t

Solution

6731 = 942*(vf - 0)/21          Multiply both sides by 21

6731 * 21 = 942*vf

141351 = 942*vf                   Divide by 942

141351/942 = vf

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3 years ago
HELP URGENT WILL GIVE BRAINLIEST
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The answer for this question is A
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Which of the following will cause a decrease in an object's weight? Increase in the mass of the object Decrease in the mass of t
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A decrease in mass will decrease an objects weight because 
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3 years ago
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One of the harmonics on a string 1.30m long has a frequency of 15.60 Hz. The next higher harmonic has a frequency of 23.40 Hz. F
Alja [10]

Answer:

\large \boxed{\text{(a) 7.800 Hz; (b) 20.3 m/s; 40.6 m/s; 60.8 m/s}}

Explanation:

a) Fundamental frequency

A harmonic is an integral multiple of the fundamental frequency.

\dfrac{\text{23.40 Hz}}{\text{15.60 Hz}} = \dfrac{1.500}{1} \approx \dfrac{3}{2}

f = \dfrac{\text{24.30 Hz}}{3} = \textbf{7.800 Hz}

b) Wave speed

(i) Calculate the wavelength

In a  fundamental vibration, the length of the string is half the wavelength.

\begin{array}{rcl}L & = & \dfrac{\lambda}{2}\\\\\text{1.30 m} & = & \dfrac{\lambda}{2}\\\\\lambda & = & \text{2.60 m}\\\end{array}

(b) Calculate the speed s

\begin{array}{rcl}v_{1}& = & f_{1}\lambda\\& = & \text{7.800 s}^{-1} \times \text{2.60 m}\\& = & \textbf{20.3 m/s}\\\end{array}

\begin{array}{rcl}v_{2}& = & f_{2}\lambda\\& = & \text{15.60 s}^{-1} \times \text{2.60 m}\\& = & \textbf{40.6 m/s}\\\end{array}

\begin{array}{rcl}v_{3}& = & f_{3}\lambda\\& = & \text{23.40 s}^{-1} \times \text{2.60 m}\\& = & \textbf{60.8 m/s}\\\end{array}

4 0
3 years ago
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