Ask question

Ask question

All categories

FromTheMoon [43]

2 years ago

5

Four rods that obey Hooke's law are each put under tension. (a) A rod 50.0 cm50.0 cm long with cross-sectional area 1.00 mm21.00

mm2 and with a 200 N200 N force applied on each end. (b) A rod 25.0 cm25.0 cm long with cross-sectional area 1.00 mm21.00 mm2 and with a 200 N200 N force applied on each end. (c) A rod 20.0 cm20.0 cm long with cross-sectional area 2.00 mm22.00 mm2 and with a 100 N100 N force applied on each end. Order the rods according to the tensile stress on each rod, from smallest to largest.

1 answer:

deff fn [24]2 years ago

8 0

Answer:

c < a = b

Explanation:

The tensile stress = Force applied/(Cross sectional area)

(a) The applied force = 200 N

The cross sectional area = 1.00 mm² = 1 × 10⁻⁶ m²

The tensile stress = 200 N/(1 × 10⁻⁶ m²) = 200,000,000 Pa = 200 MPa

(b) The applied force = 200 N

The cross sectional area = 1.00 mm² = 1 × 10⁻⁶ m²

The tensile stress = 200 N/(1 × 10⁻⁶ m²) = 200,000,000 Pa = 200 MPa

(c) The applied force = 100 N

The cross sectional area = 2.00 mm² = 2 × 10⁻⁶ m²

The tensile stress = 100 N/(2 × 10⁻⁶ m²) = 50,000,000 Pa = 50 MPa

Therefore, the tensile stress from smallest to largest are;

(a) 50 MPa, < (b) 200 MPa = (a) 200 MPa

Therefore, we have;

c < a = b.

You might be interested in

An elementary particle of mass m completely absorbs a photon, after which its mass is 1.01m. (a) what was the energy of the inco

sdas [7]

A.) We use the famous equation proposed by Albert Einstein written below:

E = Δmc²
where
E is the energy of the photon
Δm is the mass defect, or the difference of the mass before and after the reaction
c is the speed of light equal to 3×10⁸ m/s

Substituting the value:

E = (1.01m - m)*(3×10⁸ m/s) = 0.01mc² = 3×10⁶ Joules

b) The actual energy may be even greater than 3×10⁶ Joules because some of the energy may have been dissipated. Not all of the energy will be absorbed by the photon. Some energy would be dissipated to the surroundings.

8 0

2 years ago

23. While sliding a couch across a floor, Andrea and Jennifer exert forces F → A and F → J on the couch. Andrea’s force is due n

PSYCHO15rus [73]

Answer:

a) (95.4 i^ + 282.6 j^) N , b) 298.27 N 71.3º and c) F' = 298.27 N θ = 251.4º

Explanation:

a) Let's use trigonometry to break down Jennifer's strength

sin θ = Fjy / Fj

cos θ = Fjx / Fj

Analyze the angle is 32º east of the north measuring from the positive side of the x-axis would be

T = 90 -32 = 58º

Fjy = Fj sin 58

Fjx = FJ cos 58

Fjx = 180 cos 58 = 95.4 N

Fjy = 180 sin 58 = 152.6 N

Andrea's force is

Fa = 130.0 j ^

We perform the summary of force on each axis

X axis

Fx = Fjx

Fx = 95.4 N

Axis y

Fy = Fjy + Fa

Fy = 152.6 + 130

Fy = 282.6 N

F = (95.4 i ^ + 282.6 j ^) N

b) Let's use the Pythagorean theorem and trigonometry

F² = Fx² + Fy²

F = √ (95.4² + 282.6²)

F = √ (88963)

F = 298.27 N

tan θ = Fy / Fx

θ = tan-1 (282.6 / 95.4)

θ = tan-1 (2,962)

θ = 71.3º

c) To avoid the movement they must apply a force of equal magnitude, but opposite direction

F' = 298.27 N

θ' = 180 + 71.3

θ = 251.4º

4 0

2 years ago

Angular and Linear Quantities: A child is riding a merry-go-round that has an instantaneous angular speed of 1.25 rad/s and an a

serious [3.7K]

To solve this problem we will use the kinematic equations of angular motion in relation to those of linear / tangential motion.

We will proceed to find the centripetal acceleration (From the ratio of the radius and angular velocity to the linear velocity) and the tangential acceleration to finally find the total acceleration of the body.

Our data is given as:

$\omega = 1.25 rad/s \rightarrow$ The angular speed

$\alpha = 0.745 rad/s2 \rightarrow$ The angular acceleration

$r = 4.65 m \rightarrow$ The distance

The relation between the linear velocity and angular velocity is

$v = r\omega$

Where,

r = Radius

$\omega =$ Angular velocity

At the same time we have that the centripetal acceleration is

$a_c = \frac{v^2}{r}$

$a_c = \frac{(r\omega)^2}{r}$

$a_c = \frac{r^2\omega^2}{r}$

$a_c = r \omega^2$

$a_c = (4.65 )(1.25 rad/s)^2$

$a_c = 7.265625 m/s^2$

Now the tangential acceleration is given as,

$a_t = \alpha r$

Here,

$\alpha =$ Angular acceleration

r = Radius

$\alpha = (0.745)(4.65)$

$\alpha = 3.46425 m/s^2$

Finally using the properties of the vectors, we will have that the resulting component of the acceleration would be

$|a| = \sqrt{a_c^2+a_t^2}$

$|a| = \sqrt{(7.265625)^2+(3.46425)^2}$

$|a| = 8.049 m/s^2 \approx 8.05 m/s2$

Therefore the correct answer is C.

7 0

2 years ago

A camera gives a proper exposure when set to a shutter speed of 1/250 s at f-number F8.0. The photographer wants to change the s

Oksana_A [137]

Answer:

F4.0

Explanation:

To obtain a shutter speed of 1/1000 s to avoid any blur motion the f-number should be changed to F4.0 because the light intensity goes up by a factor of 2 when the f-number is decreased by the square root of 2.

5 0

2 years ago

A steel aircraft carrier is 370 m long when moving through the icy North Atlantic at a temperature of 2.0 °C. By how much does t

34kurt

Answer:

The carrier lengthen is 0.08436 m.

Explanation:

Given that,

Length = 370 m

Initial temperature = 2.0°C

Final temperature = 21°C

We need to calculate the change temperature

Using formula of change of temperature

$\Delta T=T_{f}-T_{i}$

$\Delta T=21-2.0$

$\Delta T=19^{\circ}C$

We need to calculate the carrier lengthen

Using formula of length

$\Delta L=\alpha_{steel}\times L_{0}\times\Delta T$

Put the value into the formula

$\Delta L=1.2\times10^{-5}\times370\times19$

$\Delta L=0.08436\ m$

Hence, The carrier lengthen is 0.08436 m.

8 0

2 years ago