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3 years ago

Heyy! i’ll give brainliest please help

Physics

2 answers:

Eddi Din [679]3 years ago

7 0

Answer:

5N towards north

Explanation:

Leya [2.2K]3 years ago

4 0

Answer:

I think the answer is B

hope this helps

have a good day :)

Explanation:

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Which of the following is true about ALL energy conversions?

iogann1982 [59]

Answer:

d I think

Explanation:

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3 years ago

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A 1.5 kg ball is thrown upward at 10 m/s. What is its velocity when it is 2 m above the release point?

Anestetic [448]

Answer:

7.75m/s or $2\sqrt{15}m/s$

Explanation:

m = 1.5kg

u= 10m/s

s = 2m

a = $10m/s^{2}$

v = ?

acceleration due to gravity is negative when an object is going up because the force of gravity reduces the speed of the object

using the formula ;

$v^{2} = u^{2} - 2as$ since the object is moving upwards (negative acceleration)

= 100 - 2(10)(2)

100 - 40

$v^{2}$ = 60

v = $\sqrt{60}$

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3 years ago

The only pair of cranial nerves to extend to the thoracic & abdominal cavities is the _____nerves

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4 years ago

A car travels at a constant velocity of 70 mph for one hour. At the second hour, the car’s velocity was 60 mph. At the third hou

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When the velocity increases, then the acceleration will be positive and when the velocity decrease then the acceleration will be negative.

During the first hour, the velocity was 70 mph and during the seconds hour the velocity was 60 mph. Hence, the velocity decrease in the seconds hour. So, the acceleration will be negative during the second hour.

Now, during the third hour the velocity increases as it is 80 mph. Hence, the acceleration will be positive during the third hour.

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4 years ago

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The area of the pond is approximately equal to the area of a circle with radius 297m. Find the mass of the ice. Answer in kilogr

True [87]

Answer:

mass of the ice is 254980463.8T kg

where T is the value of the thickness omitted in the question.

Explanation:

The ice on Walden Pond is .......... thick. The area of the pond is approximately equal to the area of a circle with radius 297 m. Find the mass of the ice. Answer in kg.

The value of the thickness of the ice T is omitted, but I will show the solution, and the real answer can be gotten by multiplying the final calculated answer here by the thickness of the ice omitted.

Given the radius of the equivalent circle of the ice = 297 m'

the area of the ice can be gotten from area A = $\pi r^{2}$ = $3.142*297^{2}$ = 277152.678 m^2

recall that the density of ice p ≅ 920 kg/m^3

also,

density of ice p = (mass of ice, m) ÷ (volume of ice, v)

i.e p = m/v

and,

m = pv

substituting the value of the density of water p into the equation, we have,

mass of the ice, m = 920v ....... equ 1

The volume of the ice above will be = (area of the ice, A) x (thickness of the ice, T)

i.e v = AT

substituting the value of area A into the equation, we have

v = 277152.678T ......equ 2

substitute value of v into equ 1

mass of the ice, m = 920 x (277152.678T)

mass of the ice, m = 254980463.8T kg

where T is the thickness of the ice

NB: To get the mass, multiply this answer with the thickness T given in the question.

7 0

3 years ago