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2 years ago

Help me on question 4 pls I’m giving out brainlest!!

Physics

1 answer:

Dafna11 [192]2 years ago

7 0

You can not find the density of the box 1 only including the mass. You need the volume as well. Density is mass divided by volume.

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A 65kg person throw a 0.045kg snowball forward with a ground speed of 30m/s. A second person, with a mass of 60kg, catches the s

Kobotan [32]

Well, st first we should find <span>initial momentum for the first person represented in the task which definitely must be :
</span> $(65+0.045)*2.5$
And then we find the final one : $65*x + 0.045*30$
Then equate them together : $x=2.48 m/s$
So we can get the velocity, which is $is 2.48 m/s$
In that way, according to the main rules of <span>conservation of momentum you can easily find the solution for the second person.
Regards!</span>

6 0

3 years ago

Which climate condition is typically found in the tropics due to the interaction of the atmosphere and hydrosphere?

Ilya [14]

The answer is A! Hope I helped!

7 0

3 years ago

Read 2 more answers

Homework help?

bogdanovich [222]

C. unbalanced is the correct answer for newton's first law.

8 0

2 years ago

Read 2 more answers

A 5.45-g combustible sample is burned in a calorimeter. the heat generated changes the temperature of 555 g of water from 20.5°c

Y_Kistochka [10]

Given:
m = 555 g, the mass of water in the calorimeter
ΔT = 39.5 - 20.5 = 19 °C, temperature change
c = 4.18 J/(°C-g), specific heat of water

Assume that all generated heat goes into heating the water.
Then the energy released is
Q = mcΔT
= (555 g)*(4.18 J/(°C-g)*(19 °C)
= 44,078.1 J
= 44,100 J (approximately)

Answer: 44,100 J

3 0

3 years ago

A horizontal beam of light of intensity 25 W/m2 is sent through two polarizing sheets. The polarizing direction of the first mak

Zina [86]

Answer:

option (B)

Explanation:

Intensity of unpolarised light, I = 25 W/m^2

When it passes from first polarisr, the intensity of light becomes

$I'=\frac{I_{0}}{2}=\frac{25}{2}=12.5 W/m^{2}$

Let the intensity of light as it passes from second polariser is I''.

According to the law of Malus

$I'' = I' Cos^{2}\theta$

Where, θ be the angle between the axis first polariser and the second polariser.

$I'' = 12.5\times Cos^{2}15$

I'' = 11.66 W/m^2

I'' = 11.7 W/m^2

7 0

3 years ago