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Mars2501 [29]
3 years ago
7

A 4000 kg rocket is launched, shooting 50 kg of burned fuel from its exhaust at a velocity of -625 m/s. What is the velocity of

the rocket after the fuel has burned? (Ignore the effects of gravity as this rocket is in the deepest darkest regions of interstellar space)

Physics
1 answer:
AnnZ [28]3 years ago
8 0
Hopefully this will help you.

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1 A thing ring has a mass of 6kg and a radius of 20cm. calculate the rotational inertia. ​
marissa [1.9K]

Answer:

2400kgm²

Explanation:

Rotational inertia=mass x radius²

7 0
2 years ago
Gold in its pure form is too soft to be used for most jewelry. Therefore, the gold is mixed with other metals to produce an allo
Novay_Z [31]

Answer:

Explanation:18kt alloy contains

i) 75% of gold

rhogold=19.3g/cm^3

=75/100×19.3

=14.475g/cm^3

ii) 16% of silver

rhosilver=10.5g/cm^3

=16/100×10.5

=1.68g/cm^3

iii) 9% of copper

rhocopper =8.90g/cm^3

=9/100×8.9

=0.801g/cm^3

Overall density of 18kt gold

=(0.801+1.68+14.475)g/cm^3

=16.956g/cm^3

=17g/cm^3 to 3s.f

6 0
3 years ago
Technology can best be defined as what?
jasenka [17]
B

Djdjsjdnkajdnsbsjsixhsjbsjsisahjsbxjsjsndnxnsjdd
7 0
2 years ago
Read 2 more answers
Two taut strings of identical mass and length are stretched with their ends fixed, but the tension in one string is 1.10 times g
ollegr [7]

Answer:

The  beat frequency when each string is vibrating at its fundamental frequency is 12.6 Hz

Explanation:

Given;

velocity of wave on the string with lower tension, v₁ = 35.2 m/s

the fundamental frequency of the string, F₁ = 258 Hz

<u>velocity of wave on the string with greater tension;</u>

v_1 = \sqrt{\frac{T_1}{\mu }

where;

v₁ is the velocity of wave on the string with lower tension

T₁ is tension on the string

μ is mass per unit length

v_1 = \sqrt{\frac{T_1}{\mu} } \\\\v_1^2 = \frac{T_1}{\mu} \\\\\mu = \frac{T_1}{v_1^2} \\\\ \frac{T_1}{v_1^2} =  \frac{T_2}{v_2^2}\\\\v_2^2 = \frac{T_2v_1^2}{T_1}

Where;

T₁ lower tension

T₂ greater tension

v₁ velocity of wave in string with lower tension

v₂ velocity of wave in string with greater tension

From the given question;

T₂ = 1.1 T₁

v_2^2 = \frac{T_2v_1^2}{T_1}  \\\\v_2 = \sqrt{\frac{T_2v_1^2}{T_1}} \\\\v_2 = \sqrt{\frac{1.1T_1*(35.2)^2}{T_1}}\\\\v_2 = \sqrt{1.1(35.2)^2} = 36.92 \ m/s

<u>Fundamental frequency of wave on the string with greater tension;</u>

<u />f = \frac{v}{2l} \\\\2l = \frac{v}{f} \\\\thus, \frac{v_1}{f_1}  =\frac{v_2}{f_2} \\\\f_2 = \frac{f_1v_2}{v_1} \\\\f_2 =\frac{258*36.92}{35.2} \\\\f_2 = 270.6 \ Hz<u />

Beat frequency = F₂ - F₁

                          = 270.6 - 258

                          = 12.6 Hz

Therefore, the  beat frequency when each string is vibrating at its fundamental frequency is 12.6 Hz

6 0
3 years ago
A solid sphere of mass 8.6 kg, made of metal whose density is 3,400 kg/, hangs by a cord. When the sphere is immersed in a liqui
creativ13 [48]

Answer:

A.1900 kg/m^3

Explanation:

We are given that

m=8.6 kg

Density,\rho_s=3400 kg/m^3

Tension,T=38 N

We have to find the density of liquid.

T=mg-\rho_l Vg

g=9.8 m/s^2

Volume,V=\frac{m}{\rho_s}

38=8.6\times 9.8-\rho_l\times \frac{8.6}{3400}\times 9.8

\rho_l\times \frac{8.6}{3400}\times 9.8=8.6\times 9.8-38

\rho_l=\frac{(8.6\times 9.8-38)\times 3400}{8.6\times 9.8}

\rho_l=1867kg/m^3\approx 1900 kg/m^3

Option A is true.

4 0
3 years ago
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