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4 years ago

Is it C?

Physics

2 answers:

DerKrebs [107]4 years ago

8 0

Answer:

Biomass

Explanation:

This is because coal, natural gas, & petroleum are actuall harmful. (LMK)

monitta4 years ago

7 0

No, the correct answer is <span>biomass.</span>

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The "lead" in pencils is a graphite composition with a Young's modulus of about 1 ✕ 10⁹ N/m². Calculate the change in length (in

sweet [91]

Answer:

0.701 mm.

Explanation:

From Hook's law,

Young modulus = Stress/Strain.

Stress = Force/area = F/A

Strain = ΔL/L.

Therefore,

γ = (F/A)/(ΔL/L)

γ = FL/ΔLA ...................... Equation 1

Where γ = Young's modulus, F = Force, L = Length of the pencil, ΔL = change in length of the pencil, A = cross sectional area.

Make ΔL the subject of the equation,

ΔL = FL/γA......................... Equation 2

But,

A = πd²/4......................... Equation 3

Where d = diameter

Substitute equation 3 into equation 2

ΔL = 4FL/γπd²................ Equation 4

Given: F = 4.9 N, L = 55 mm = 0.055 m, d = 0.7 mm = 0.0007 m, γ = 1×10⁹ N/m², π = 3.14.

Substitute into equation 4

ΔL = (4×4.9×0.055)/(1×10⁹×3.14×0.0007²)

ΔL = 1.078/1538.6

ΔL = 7.01×10⁻⁴ m.

ΔL = 0.701 mm.

4 0

3 years ago

Determine the mass of the object below to the correct degree of precision.

OleMash [197]

Answer : The correct degree of precision is 272.94 g.

Explanation :

Precision : It is defined as the closeness of two or more measurements to each other.

For Example: If you weigh a given substance five times and you get 2.7 kg each time. Then the measurement is said to be precise.

Level of precision is determined by the maximum number of decimal places.

In the given figure, the value of mass of object is 280 g. From the given options we conclude that the 272.94 g is more close to 280 g and has maximum number of decimal places. So, 272.94 g measurement is said to be precise.

Therefore, the correct degree of precision is 272.94 g.

6 0

3 years ago

Please help me. Clear answer please and the formula and everything

Diano4ka-milaya [45]

(a) At level A Potential Energy is 2940 J.

(b) At level B Potential Energy is 1960 J.

(d) At level D Potential energy is 0 J.

(e) The change in Potential energy if the object moves from A to B is 980 J.

<u>Explanation</u>:

Given that,

Mass of an object is 5 kg (m) at a height of 60 m (h) above the ground as shown in the given figure.

Potential energy of an object is calculated by using the formula m × g × h. “g” is acceleration due to gravity on earth is 9.8 $m/s^2$ .

(a) At level A:

At height (h) is 60 m above ground.

Potential energy = 5 × 9.8 × 60

Potential energy = 2940 J

(b) At level B:

At height (h) is 40 m above ground.

Potential energy = 5 × 9.8 × 40

Potential energy = 1960 J

(c) At level C:

At height (h) is 30 m above ground.

Potential energy = 5 × 9.8 × 30

Potential energy = 1470 J

(d) At level D:

At height (h) is 0 m above ground (object is on the ground).

Potential energy = 5 × 9.8 × 0

Potential energy = 0 J

(e) Change in “Potential energy” when it moves from “A” to “B”

Change in “Potential energy” in difference in initial “Potential energy” (at level A) and final “Potential energy” (at level B)

= 2940 - 1960

= 980 J

Therefore, change in “Potential energy” is 980 J.

3 0

4 years ago

Two solid steel shafts are fitted with flanges that are then connected by bolts as shown. the bolts are slightly undersized and

Lelu [443]

Answer:

The maximum shear stress in shaft AB, $T_{ABmax}$ is 15 MPa

The maximum shear stress in shaft CD, $T_{CDmax}$ is 45.9 MPa

Explanation:

The formula for a shaft polar moment of inertia, J is given by

J = $\pi \times \frac{D^4}{32} =\pi \times \frac{r^4}{2}$

Therefore, we have

$J_{AB} = \pi \times \frac{D_{AB}^4}{32} =\pi \times \frac{r_{AB}^4}{2}$

Where:

D $_{AB}$ = Diameter of shaft AB = 30 mm = 0.03 m

r $_{AB}$ = Radius of shaft AB = 15 mm = 0.015 m

∴ $J_{AB} = \pi \times \frac{0.03^4}{32} =\pi \times \frac{0.015^4}{2}$ = 7.95 × 10⁻⁸ m⁴

and

$J_{CD} = \pi \times \frac{D_{CD}^4}{32} =\pi \times \frac{r_{CD}^4}{2}$

Where:

D $_{CD}$ = Diameter of shaft CD = 36 mm = 0.036 m

r $_{CD}$ = Radius of shaft CD = 18 mm = 0.018 m

Therefore,

$J_{CD} = \pi \times \frac{0.036^4}{32} =\pi \times \frac{0.018^4}{2}$ = 1.65 × 10⁻⁷ m⁴

Given that the shaft AB and CD are rotated 1.58 ° relative to each other, we have;

1.58 °= $1.58 \times \frac{2\pi }{360}$ rad = 2.76 × 10⁻² rad.

That is $\phi_r$ = 2.76 × 10⁻² rad.

However $\phi_r = \phi_{C/D} - \phi_{B/A}$

Where:

$\phi_{B/A} = \frac{T_{AB}\cdot L_{AB}}{J_{AB} \cdot G}$ and

$\phi_{C/D} = \frac{T_{CD}\cdot L_{CD}}{J_{CD} \cdot G}$

$T_{AB}$ and $T_{CD}$ = Torque on shaft AB and CD respectively

$T_{AB}$ = Required

$T_{CD}$ = 500 N·m

$L_{AB}$ and $L_{CD}$ = Length of shafts AB an CD respectively

$L_{AB}$ = 600 mm = 0.6 m

$L_{CD}$ = 900 mm = 0.9 m

G = Shear modulus of the material = 77.2 GPa

Therefore;

$\phi_r = \phi_{C/D} - \phi_{B/A}$ = $\frac{T_{CD}\cdot L_{CD}}{J_{CD} \cdot G} -\frac{T_{AB}\cdot L_{AB}}{J_{AB} \cdot G}$

2.76 × 10⁻² rad = $\frac{T_{CD}\cdot L_{CD}}{J_{CD} \cdot G} -\frac{T_{AB}\cdot L_{AB}}{J_{AB} \cdot G}$

= $\frac{500\cdot 0.9}{1.65 \times 10^{-7} \cdot 77.2\times 10^9} -\frac{T_{AB}\cdot 0.6}{7.95\times 10^{-7} \cdot 77.2\times 10^9}$

Therefore;

T $_{AB}$ = 79.54 N.m

Where T = $T_{AB}$ + T $_{CD}$ =

Therefore T $_{CD total }$ = 500 - 79.54 = 420.46 N·m

τ $_{max}$ = $\frac{T\times R}{J}$

$\tau_{ABmax}$ = $\frac{T_{AB}\times R_{AB}}{J_{AB}}$ = $\frac{79.54\times 0.015}{7.95\times 10^{-8}}$ = 15 MPa

$\tau_{CDmax}$ = $\frac{T_{CD}\times R_{CD}}{J_{CD}}$ = $\frac{420.46\times 0.018}{1.65\times 10^{-7}}$ = 45.9 MPa

7 0

3 years ago

Two forces are acting on an object as shown in Fig. on the right. What is the magnitude of the resultant force?

svetlana [45]

Answer:

185 N

Explanation:

Sum of forces in the x direction:

Fₓ = -(80 N cos 75°) + (120 N cos 60°)

Fₓ = 39.3 N

Sum of forces in the y direction:

Fᵧ = (80 N sin 75°) + (120 N sin 60°)

Fᵧ = 181.2 N

The magnitude of the net force is:

F = √(Fₓ² + Fᵧ²)

F = √((39.3 N)² + (181.2 N)²)

F = 185 N

8 0

3 years ago