Most of the nutrients in the rainforest ecosystem are in the ____

Two wires of the same material and having the same volume, are fixed

Setler79 [48]

Answer:

48 kg

Explanation:

Given that the two wires are of same material, so their value of young's modulus will be same

Assuming that the wires are cylindrical in shape

As radius of the first wire is half that of the second wire and therefore the area of cross-section of the first wire will be one-fourth of the second wire( ∵ wire is cylindrical, the cross-sectional part will be circle and the area of the circle = π × r² )

As the volume is same for both wires

∴ π × ( $r_{1}$ )² × $l_{1}$ = π × ( $r_{2}$ )² × $l_{2}$

Here

$r_{1}$ is the radius of the first wire

$r_{2}$ is the radius of the second wire

$l_{1}$ is the length of the first wire

$l_{2}$ is the length of the second wire

⇒ π × (( $r_{2}$ )² ÷ 4) × $l_{1}$ = π × ( $r_{2}$ )² × $l_{2}$ (∵ radius of first wire is half that of the second wire)

By cancelling the same terms on both sides

we get

$l_{1}$ = 4 × $l_{2}$

⇒ Length of first wire will be four times of the length of second wire

<h3>Strain is defined as the elongation per unit length</h3>

Strain in first wire = ΔL ÷ $l_{1}$ = ΔL ÷ (4 × $l_{2}$ )

where ΔL is the elongation of the wire which in this case is same in both wires

Strain in second wire = ΔL ÷ $l_{2}$

∴ Strain in second wire is four times of strain in first wire

<h3>Stress = F ÷ A</h3>

where F is the force perpendicular to the cross-sectional area

A is the area of cross-section

Force in first wire = $m_{1}$ × g

where $m_{1}$ is the mass hanged to the first wire

g is the acceleration due to gravity

Force in second wire = $m_{2}$ × g

where $m_{2}$ is the mass hanged to the second wire

g is the acceleration due to gravity

Let $A_{1}$ be the cross-sectional area of first wire

$A_{2}$ be the cross-sectional area of second wire

$A_{2}$ = 4 × $A_{1}$ (∵ cross=sectional area of the wire = π × (radius of the wire)² )

Stress in first wire = ( $m_{1}$ × g) ÷ ( $A_{1}$ )

Stress in second wire = ( $m_{2}$ × g) ÷ ( $A_{2}$ ) = ( $m_{2}$ × g) ÷ (4 × $A_{1}$ )

<h3>Young's modulus is defined as Stress per unit strain</h3>

As Young's modulus is same for both wires, Stress per unit strain must be same for both wires

Stress per unit strain of first wire = (( $m_{1}$ × g) ÷ ( $A_{1}$ )) ÷ (ΔL ÷ (4 × $l_{2}$ ))

Stress per unit strain of second wire = (( $m_{2}$ × g) ÷ (4 × $A_{1}$ )) ÷ (ΔL ÷ $l_{2}$ )

By equating them we get

$m_{2}$ = 16 × $m_{1}$

⇒ $m_{2}$ = 16 × 3 = 48 kg

∴ $m_{2}$ = 48 kg

5 0

3 years ago

A ball with an initial velocity of 2 m/s rolls for a period of 3 seconds. If the ball is uniformly accelerating at a rate of 3 m

ikadub [295]

Answer: 11 m/s

vinitial=2 m/s

time=3 s

acceleration = 3 m/s^2

vfinal = ?

The key here is that it is a constant acceleration, so we can use the constant acceleration equations. The easiest one to use would be:

vfinal=vinitial + a*t

We need vfinal, so algebraically we are ready to put in numbers into the equation:

vfinal=vinitial + a*t = 2 m/s + (3 m/s^2)*(3 s ) = 11 m/s is the final velocity

7 0

3 years ago

You are driving your car on a very cold late Fall day. You clear a turn and see a couple of pedestrians standing at the cross wa

Diano4ka-milaya [45]

Answer:

t_pass = 2.34 m

t_stop = 4.68 s

Thus, for the car passing at constant speed the pedestrian will have to wait less.

Explanation:

If the car is moving with constant speed, then the time taken by it will be given as:

$t_{pass} = \frac{D}{v}$

where,

t_pass = time taken = ?

D = Distance covered = 23 m

v = constant speed = (22 mi/h)(1609.34 m/1 mi)(1 h/3600 s) = 9.84 m/s

Therefore,

$t_{pass} = \frac{23\ m}{9.84\ m/s} \\$

<u></u>

Now, for the time to stop the car, we will use third equation of motion to get the acceleration first:

$2as = v_{f}^{2} - v_{i}^2\\a = \frac{v_{f}^{2} - v_{i}^2}{2D}\\\\a = \frac{(0\ m/s)^{2}-(9.84\ m/s)^2}{(2)(23\ m)}\\\\a = -2.1\ m/s^2$

Now, for the passing time we use first equation of motion:

$v_{f} = v_{i} + at_{stop}\\t_{stop} = \frac{v_{f}-v_{i}}{a}\\\\t_{stop} = \frac{0\ m/s - 9.84\ m/s}{-2.1\ m/s^2}$

7 0

3 years ago

Calculate the effective radiation dosage in sieverts for a 57 57 ‑kg person who is exposed to 4.2 × 10 9 4.2×109 particles of al

Rus_ich [418]

Answer:

Dose = 5.4 10⁻⁴ Sv

Explanation:

The radiation dose absorbed by the body is 81% of the incident radiation, which is the number of particles per energy of each particle

E = 0.81 n₀ E₀

E = 0.81 4.2 10⁹ 0.81 6.1 10⁻¹³

E = 2.075 10⁻³ J

Effective radiation dose

Dose1 = n x RBE

Dose1 = 2,075 10⁻³ 15

Dose1 = 3.1 10⁻² REM

The effective radiation dose in sievert

Dose = dose1 / m

Dose = 3.1 10⁻² / 57

Dose = 5.4 10⁻⁴ [j / kg]

Dose = 5.4 10⁻⁴ Sv

6 0

4 years ago

A carpet is to be installed in a room of length 9.72 meters and width 17.30 of the room retaining the proper number of significa

Usimov [2.4K]

Answer:

168 m^2, 380 m^2

Explanation:

length of the room, l = 9.72 m

width of the room, b = 17.30 m

Area of teh rectangle is given by

A = length x width

So, A = 9.72 x 17.30 = 168.156 m^2

the significant digits should be 3 in the final answer

So, A = 168 m^2

Now length = 72 m

width = 17.39 feet = 5.3 m

Area, A = 72 x 5.3 = 381.6 m^2

There should be two significant digits in the answer so, by rounding off

A = 380 m^2

3 0

3 years ago

Most of the nutrients in the rainforest ecosystem are in the _____.