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4 years ago

A solid object has a mass of 104 kg and a volume of 1,278 m3. What is its density?

1 answer:

4 0

The density is 81.4 g/m3. Before you start plugging numbers into the density formula (D=M/V), you should convert 104 kg to grams, which ends up being 104,000 grams. Then you can plug in the 104,000 grams and 1,278 m3 into the formula. When you divide the mass by the volume, you get a really long decimal, which you can round to 81.4 g/m3, or whatever place your teacher wants you to round to.

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500km is equal to how many millimeters

Sindrei [870]

Answer:

500000000

if you can give me brainliest that would be great

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4 years ago

Explain the role that the ant and the acacia tree play in their symbiotic relationships

MAXImum [283]

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3 years ago

21. A toy car starts from rest and begins to

Bond [772]

Answer:

v = 66 m/s

Explanation:

Given that,

The initial velocity of a car, u = 0

Acceleration of the car, a = 11 m/s²

We need to find the final velocity of the toy after 6 seconds.

Let v is the final velocity. It can be calculated using first equation of motion. It is given by :

v = u +at

v = 0 + 11 m/s² × 6 s

v = 66 m/s

So, the final velocity of the car is 66 m/s.

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3 years ago

Which of the following is a category of mechanical wave?

givi [52]

Answer:

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3 years ago

Read 2 more answers

Frank’s automobile engine runs at 100∘C. One day, when the outside temperature is 15∘C, he turns off the ignition and notes that

lapo4ka [179]

Answer:

the engine cool to 40 $^{o}C$ at 14.07 minutes

Explanation:

Given information

T(5) = 70 $^{o}C$

$T_{0}$ = 100 $^{o}C$

C = 15 $^{o}C$

Newton's law of cooling :

T(t) = C + ( $T_{0}$ - C) $e^{-kt}$

where

T(t) = temperature at any given time

C = surrounding temperature

$T_{0}$ = initial temperature of heated object

k = cooling constant

to find the the time when the engine will be cooled down to 40 $^{o}C$ , we first need to find the cooling constant, k

when t = 5, T(5) = 70 $^{o}C$

so,

T(t) = C + ( $T_{0}$ - C) $e^{-kt}$

T(5) = 15 + (100 - 15) $e^{-5k}$

70 = 15 + (85) $e^{-5k}$

$e^{-5k}$ = (70 - 15) / 85

-5k = ln (55/85)

k = - ln (55/85) / 5

k = 0.087

thus, we have the eqaution

T(t) = 15 + (85) $e^{-0.087t}$

now we can determine the time when T(t) = 40 $^{o}C$

40 = 15 + (85) $e^{-0.087t}$

$e^{-0.087t}$ = (40-15)/85

-0.0087t = ln (25/85)

t = - ln (25/85)/0.087

t = 14.07 minutes

8 0

3 years ago