How many cells make up bacteria? 1 , 10, 100, or millions ?

An asteroid that has an orbital period of 3 years will have an orbital with a semi-major axis of about _____ years.

Airida [17]

Answer:

An asteroid that has an orbital period of 3 years will have an orbital with a semi-major axis of about 2 years.

Explanation:

Given;

orbital period of 3 years, P = 3 years

To calculate the years of an orbital with a semi-major axis, we apply Kepler's third law.

Kepler's third law;

P² = a³

where;

P is the orbital period

a is the orbital semi-major axis

(3)² = a³

9 = a³

a = $a = \sqrt[3]{9} \\\\a = 2.08 \ years$

Therefore, An asteroid that has an orbital period of 3 years will have an orbital with a semi-major axis of about 2 years.

5 0

3 years ago

An unknown mass of each substance, initially at 25.0 ∘C, absorbs 1920 J of heat. The final temperature is recorded. Find the mas

zhannawk [14.2K]

Answer: mass for Pyrex glass 84.21g

mass for sand 61.6g

mass for ethanol 41.32g

mass for water 62.07g

Explanation

By definition specific heat is the amount of heat required to change the temperature of 1 kg mas by 1°C

Q=mcΔT is formula for specific heat

Q is heat transfer

m is mass

ΔT is change in temperature

c is specific heat

c of Pyrex glass= 0.75 j/g°C

c of sand = 0.84 j/g°C

c of ethanol= 2.42 j/g°C

c of water = 4.18 j/g°C

now we will make M(mass) the subject, so equation becomes

m=Q/cΔT

for

pyrex glass Tf=55.4°C

m=1920/(55.4-25)*0.75

m=84.21g {after cutting J(joules) and °C we are left with g(grams)}

for

sand Tf=62.1°C

m=1920/(62.1-25)*0.84

m=61.6g {after cutting J(joules) and °C we are left with g(grams)}

for

ethanol Tf=44.2°C

m=1920/(44.2-25)*2.42

m=41.32g {after cutting J(joules) and °C we are left with g(grams)}

for

water Tf=32.4°

m=1920/(32.4-25)*4.18

m=62.07g {after cutting J(joules) and °C we are left with g(grams)}

i hope you understand the solution, thank you.

7 0

3 years ago

Which of the following describes an electric conductor?

IrinaVladis [17]

Yo sup??

the correct answer is option C ie

a material that has a low resistance and allows charges to move freely

this is a basic property shown by conductors

Hope this helps

6 0

3 years ago

Read 2 more answers

Complete the table for the function rule y=3x-2

Orlov [11]

Answer:

3 up 1 across is the answer

Explanation:

Step one: find the Y intercept( Where the line cuts the y-axis)

X = 0

Y = 3(0) (p.s cross out the (o)

y = 2

Step 2: find the gradient (slope of the line)

rise = 3

------ ---

Run 1

5 0

4 years ago

Two metal disks, one with radius R1 = 2.45 cm and mass M1 = 0.900 kg and the other with radius R2 = 5.00 cm and mass M2 = 1.60 k

larisa86 [58]

Answer:

part (a) $a_1\ =\ 2.9\ kg$

Part (b) $a_2\ =\ 6.25\ kg$

Explanation:

Given,

Mass of the larger disk = $M_2\ =\ 1.60\ kg$
Mass of the smaller disk = $M_1\ =\ 0.900\ kg$
Radius of the larger disk = $R_2\ =\ 5.00\ cm\ =\ 0.05\ m$
Radius of the smaller disk = $R_1\ =\ 2.45\ cm\ =\ 0.0245\ m$
Mass of the block = M = 1.60 kg

Both the disks are welded together, therefore total moment of inertia of the both disks are the summation of the individual moment of inertia of the disks.

$\therefore I\ =\ I_1\ +\ I_2\\\Rightarrow I\ =\ \dfrac{1}{2}M_1R_1^2\ +\ \dfrac{1}{2}M_2R_2^2\\\Rightarrow I\ =\ \dfrac{1}{2} (0.9\times 0.0245^2\ +\ 1.60\times 0.05^2)\\\Rightarrow I\ =\ 2.27\times 10^{-3}\ kgm^2$

part (a)

Given that a block of mass m which is hanging with the smaller disk,

Let 'T' be 'a' be the tension in the string and acceleration of the block.

From the free body diagram of the smaller block,

$mg\ -\ T\ =\ ma\\\Rightarrow T\ =\ mg\ -\ ma\,\,\,\,eqn (1)$

From the pulley,

$\sum \tau\ =\ I\alpha\\\Rightarow T\times R_1\ =\ I\alpha\ =\ \dfrac{Ia}{R_1}\\\Rightarrow T\ =\ \dfrac{I\alpha}{R_1^2}\,\,\,eqn(2)$

From the equation (1) and (2),

$mg\ -\ ma\ =\ \dfrac{Ia}{R_1^2}\\\Rightarrow a\ =\ \dfrac{mg}{\dfrac{I}{R_1^2}\ +\ m}\\\Rightarrow a\ =\ \dfrac{1.60\times 9.81}{\dfrac{2.27\times 10^{-3}{0.0245^2}}\ +\ 1.60}\\\Rightarow a\ =\ 2.91\ m/s^2$

part (b)

Above expression for the acceleration of the block is only depended on the radius of the pulley.

Radius of the larger pulley = $R_2\ =\ 0.05\ m$

Let $a_2$ be the acceleration of the block while connecting to the larger pulley. $\therefore a\ =\ \dfrac{mg}{\dfrac{I}{R_2^2}\ +\ m}\\\Rightarrow a\ =\ \dfrac{1.60\times 9.81}{\dfrac{2.27\times 10^{-3}{0.05^2}\ +\ 1.60}}\\\Rightarow a\ =\ 6.25\ m/s^2$

4 0

3 years ago