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zavuch27 [327]
3 years ago
10

Assume you have an ice cube and also a small rock that is the same size and shape as the ice cube. Predict what would happen if

you placed the ice cube and rock in a glass of water.
Physics
1 answer:
Step2247 [10]3 years ago
7 0

Answer: The ice cube would float on top of the water and the rock would sink to the bottom.

Explanation: The ice cube has a smaller density than the rock which allows the ice cube to float but makes the rock sink to the bottom of the glass of water.

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Jaime lifts a package weighing 75N. if she lifts it 1.2 m, what work has she done
9966 [12]
Answer:
90 J
Explanation:
W=fd
W=(75)(1.2)
W= 90 J
6 0
3 years ago
Someone please help<br> Just need answers
marishachu [46]

Answer:

As follows,

Explanation:

KE=1/2mv^2

In 1st question,

KE=1/2mv^2=1/2*0.05*12=0.3 J [50g=0.05 kg]

In 2nd question,

KE=1/2mv^2

6.8=1/2*0.046*v^2

v=sqrt(6.8/0.023)

v=17.19

In 3rd question,

KE=1/2mv^2

63/392=m

m=0.16kg=160g

For 4th,

a.

1st case,

KE=1/2mv^2=1/2*28*2.4^2=80.64

2nd case,

KE=1/2mv^2=1/2*28*3.7^2=191.66

Change in KE=191.66-80.64=11.02

b.Speed/velocity gained

6 0
3 years ago
Which of the following is NOT true of all the inner planets?
Artist 52 [7]

Hi there! :)

\large\boxed{\text{B. They are called gas giants.}}

In the solar system, we can categorize the planets as such:

Inner planets: Mercury, Venus, Earth, Mars

Outer planets: Jupiter, Saturn, Uranus, Neptune

The Inner planets are the only planets that consist of solid rock. They also do not have rings and have impact craters due to their early formation.

In comparison, all of the outer planets are primarily made up of gaseous material and contain rings.

Therefore, the correct choice is B.

7 0
3 years ago
In some cases, neither of the two equations in the system will contain a variable with a coefficient of 1, so we must take a fur
Margaret [11]

Answer:

D = -4/7 = - 0.57

C = 17/7 = 2.43

Explanation:

We have the following two equations:

3C + 4D = 5\ --------------- eqn (1)\\2C + 5D = 2\ --------------- eqn (2)

First, we isolate C from equation (2):

2C + 5D = 2\\2C = 2 - 5D\\C = \frac{2 - 5D}{2}\ -------------- eqn(3)

using this value of C from equation (3) in equation (1):

3(\frac{2-5D}{2}) + 4D = 5\\\\\frac{6-15D}{2} + 4D = 5\\\\\frac{6-15D+8D}{2} = 5\\\\6-7D = (5)(2)\\7D = 6-10\\\\D = -\frac{4}{7}

<u>D = - 0.57</u>

Put this value in equation (3), we get:

C = \frac{2-(5)(\frac{-4}{7} )}{2}\\\\C = \frac{\frac{14+20}{7}}{2}\\\\C = \frac{34}{(7)(2)}\\\\C =  \frac{17}{7}\\

<u>C = 2.43</u>

5 0
3 years ago
Can someone help with this question pls
morpeh [17]
I solved it and got 374.2N so i would put 375N

sorry i didn’t get any of the ones on there, i probably made a mistake.
6 0
3 years ago
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