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Blababa [14]
2 years ago
8

Wtz the answer to this question? I need to get it right

Physics
2 answers:
Rama09 [41]2 years ago
4 0

Answer: Next time you create a question, add an image or PDF. Because I do not know the question. So, may you please create a new question?

Viktor [21]2 years ago
4 0
Theirs no i picture so can you do the question again or some
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A cube of wood having an edge dimension of 18.0 cm and a density of 651 kg/m3 floats on water.(a) What is the distance from the
astra-53 [7]

Answer:

A. 6.282

B. 2.03kg

Explanation:

A.

We solve using archimedes principle

L³pwood = L²dwater

We make d subject of the formula

d = Lpwood/pester

= 18x651/1000

= 18x0.651

= 11.718cm

Distance from horizontal top to water level

= 18-11.718

= 6.282cm

B.

When we place lead block

WL + L³pwoodg = L³pwaterg

WL = L³g(Pwater-Pwood)

= 0.18³x9.8(1000-651)

= 19.94N

19.94/9.8

= 2.03kg

The mass m is therefore 2.03kg

6 0
2 years ago
Calculate the kinetic energy of a 10kg cart traveling at 4 m/s?
ycow [4]

Answer:

Explanation:

Ke = 1/2mv2 = 1/2.10.(4)2=5.16= 80 J

5 0
3 years ago
Read 2 more answers
Can someone help me on 2 science question,
gavmur [86]
1. I think you should compare diagrams of moon phases from the textbook to diagrams of moon phases online. Because if you pick D it will take to long and C will help you out whith 3 different things to look at. 

2. The moon changes in appearances from the perspective of people on earth because it's revolving around the planet and the earth is revolving around the sun, so  A. Hoped this helped.

8 0
3 years ago
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As the number of resistors in a parallel circuit is increased, what happens to the
aleksandr82 [10.1K]

Answer:

resistance- decreases current-increases

8 0
2 years ago
Consider as a system the Sun with Saturn in a circular orbit around it. Find the magnitude of the change in the velocity of the
Doss [256]

Answer:

v_{su} = 19.44 m/s

Explanation:

m_{su}=5.68x10^{29}kg\\m_{sa}=5.68x10^{26}kg

T=9.29x10^8\\r_{o}=1.43x10^{12}

If the sun considered as x=0 on the axis to put the center of the mass as a:

m_{su}*r_{o}=(m_{sa}+m_{su})*r_{1}

solve to r1

r_1=\frac{m_{sa}*r_{o}}{m_{sa}+m_{su}}=\frac{5.68x10^{26}*1.43x10^{12}}{5.68x10^{26}+5.68x10^{26}}

r_1=1.428x10^9m

Now convert to coordinates centered on the center of mass.  call the new coordinates x' and y' (we won't need y').  Now since in the sun centered coordinates the angular momentum was  

L = \frac{m_{sa}*2*pi*r_1^2}{T}

where T = orbital period

then L'(x',y') = L(x) by conservation of angular momentum.  So that means

L_{sun}=\frac{m_{sa}*2*\pi *( 2r_{o}*r_1 -r_1^2)}{T}

Since

L_{su}= m_{su}*v_{su}*r_1

then

v_{su}=\frac{m_{sa}*2*pi*(2r_{o}*r_{1}-r_{1}^2)}{T*m_{sa}*r_1}

v_{su} = 19.44 m/s

7 0
3 years ago
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