PHYSICS COLLEGE

The tensile strength (the maximum tensile stress it can support without breaking) for a certain steel wire is 3000 MN/m2. What is the maximum load that can be applied to a wire with a diameter of 3.0 mm made of this steel without breaking the wire?

Answers

Answer 1

Answer:

Answer:

The correct answer is "21195 N".

Explanation:

The given values are:

Tensile strength,

= 3000 MN/m²

Diameter,

= 3.0 mm

i.e.,

= 3×10⁻³ m

Now,

The maximum load will be:

= $Tensile \ strength* Area$

On substituting the values, we get

= $(3000* 10^6)((\pi)/(4) (3* 10^(-3))^2)$

= $(3000* 10^6)((3.14)/(4) (3* 10^(-3))^2)$

= $21195 \ N$

Answer 2

Answer:

Final answer:

The maximum load that can be applied to a 3.0 mm diameter steel wire with a tensile strength of 3000 MN/m2 without breaking it is 21,200 Newtons.

Explanation:

The subject of this question revolves around the concept of tensile strength in the field of Physics. The maximum load that can be applied to a wire without it breaking depends on the wire's tensile strength and its cross-sectional area. For a steel wire with a tensile strength of 3000 MN/m2 and a diameter of 3.0 mm, we first need to calculate the cross-sectional area, which can be found using the formula for the area of a circle, A = πr^2, where r is the radius of the wire. Given the diameter is 3.0 mm, the radius will be 1.5 mm or 1.5 x 10^-3 m. So, A = π(1.5 x 10^-3 m)^2 ≈ 7.07 x 10^-6 m^2.

We can then use the tensile strength (σ) to find the maximum load (F) using the equation F = σA. Substituting the given values, we get F = 3000 MN/m^2 * 7.07 x 10^-6 m^2 = 21.2 kN, which is equivalent to 21,200 N. Therefore, the maximum load that can be applied to the wire without breaking it is 21,200 Newtons.

Learn more about Tensile Strength here:

brainly.com/question/14293634

#SPJ3

Related Questions

GUYS PLEASE HELP THIS IS DO IN LIKE 30 MIN!!!!!!Question 1 Which of the following color schemes (specifically in jewel tones) is on trend for 2020? a all of the above b complimentary c analogous d monochromaticQuestion 2 Which of the following color pallettes is on trend for 2020? a primary colors b earth tones c cool tones d hues of pinkQuestion 3 Which of the following items could be painted in a dark color to be on trend in 2020? a toilets b doors c ceilings d baseboardsQuestion 4 Which of the following design genres is OUT in 2020? a country chic b traditional c industrial d bohemianQuestion 5 When designing interiors for the elderly, which of the following items can be used to replace traditional door knobs? a clap lights b sliding glass doors c pocket doors d door leversQuestion 6 What design adaptation can be implemented in the bathroom to help with elderly clients? a walk in shower b garden tub c Jacuzzi d sponge bath stationQuestion 7 Which of the following floor types can aid in safe movement throughout the house? a soft area rugs b cork c cement d slick tileQuestion 8 Where should cabinets be place in the kitchen as a safety measure for designing for the elderly? a basic counter tops that are easily accessible b above the refrigerator c over long/extended cabinets d over the stove
Sultan throws a ball horizontally from his window, 12 m above the garden. It reaches the ground afterSelect........seconds.4.05.02.41.6Answer and I will give you brainiliest
A barbell spins around a pivot at its center at A. The barbell consists of two small balls, each with mass 450 grams (0.45 kg), at the ends of a very low mass rod of length d = 20 cm (0.2 m; the radius of rotation is 0.1 m). The barbell spins clockwise with angular speed 120 radians/s.What is the speed of ball 1?
A firefighting crew uses a water cannon that shoots water at 27.0 m/s at a fixed angle of 50.0 ∘ above the horizontal. The firefighters want to direct the water at a blaze that is 12.0 m above ground level. How far from the building should they position their cannon? There are two possibilities (d1Part A:d1=_____mPart B:d2=______m
N5 Swinging a tennis racket against a ball is an example of a third class lever. OT OF 9 Please select the best answer from the choices provided. K

The electric field at the distance of 3.5 meters from an infinite wall of charges is 125 N/C. What is the magnitude of the electric field 1.5 meters from the wall?

Answers

Explanation:

It is given that,

Distance, r = 3.5 m

Electric field due to an infinite wall of charges, E = 125 N/C

We need to find the electric field 1.5 meters from the wall, r' = 1.5 m. Let it is equal to E'. For an infinite wall of charge the electric field is given by :

$E=(\lambda)/(2\pi \epsilon_o r)$

It is clear that the electric field is inversely proportional to the distance. So,

$(E)/(E')=(r')/(r)$

$E'=(Er)/(r')$

$E'=(125* 3.5)/(1.5)$

E' = 291.67 N/C

So, the magnitude of the electric field 1.5 meters from the wall is 291.67 N/C. Hence, this is the required solution.

The atmosphere on Venus consists mostly of CO2. The density of the atmosphere is 65.0 kg/m3 and the bulk modulus is 1.09 x 107 N/m2. A pipe on a lander is 75.0 cm long and closed at one end. When the wind blows across the open end, standing waves are caused in the pipe (like blowing across the top of a bottle). a) What is the speed of sound on Venus? b) What are the first three frequencies of standing waves in the pipe?

Answers

Answer:

a. 409.5 m/s b. f₁ = 136.5 Hz, f₂ = 409.5 Hz, f₃ = 682.5 Hz

Explanation:

a. The speed of sound v in a gas is v = √(B/ρ) where B = bulk modulus and ρ = density. Given that on Venus, B = 1.09 × 10⁷ N/m² and ρ = 65.0 kg/m³

So, v = √(B/ρ)

= √(1.09 × 10⁷ N/m²/65.0 kg/m³)

= √(0.01677 × 10⁷ Nm/kg)

= √(0.1677 × 10⁶ Nm/kg)

= 0.4095 × 10³ m/s

= 409.5 m/s

b. For a pipe open at one end, the frequency f = nv/4L where n = mode of wave = 1,3,5,.., v = speed of wave = 409.5 m/s and L = length of pipe = 75.0 cm = 0.75 m

Now, for the first mode or frequency, n = 1

f₁ = v/4L

= 409.5 m/s ÷ (4 × 0.75 m)

= 409,5 m/s ÷ 3 m

= 136.5 Hz

Now, for the second mode or frequency, n = 2

f₂ = 3v/4L

= 3 ×409.5 m/s ÷ (4 × 0.75 m)

= 3 × 409,5 m/s ÷ 3 m

= 3 × 136.5 Hz

= 409.5 Hz

Now, for the third mode or frequency, n = 5

f₃ = 5v/4L

= 5 × 409.5 m/s ÷ (4 × 0.75 m)

= 5 × 409,5 m/s ÷ 3 m

= 682.5 Hz

Proper design of automobile braking systems must account for heat buildup under heavy braking. Part A Calculate the thermal energy dissipated from brakes in a 1600 kg car that descends a 15 ∘ hill. The car begins braking when its speed is 95 km/h and slows to a speed of 40 km/h in a distance of 0.34 km measured along the road.

Answers

Answer:

1838216 J

Explanation:

95 km/h = 26.39 m/s

40 km/h = 11.11 m/s

Initial kinetic energy

= .5 x 1600 x(26.39)²

= 557145.67 J

Final kinetic energy

= .5 x 1600 x ( 11.11)²

= 98745.68 J

Loss of kinetic energy

= 458400 J

Loss of potential energy

= mg x loss of height

= 1600 x 9.8 x 340 sin 15

= 1379816 J

Sum of Loss of potential energy and Loss of kinetic energy

= 1379816 + 458400

= 1838216 J

This is the work done by the friction . So this is heat generated.

Final answer:

To calculate the thermal energy dissipated from the brakes of a car, use the equation Q = Mgh/10, where Q is the energy transferred to the brakes, M is the mass of the car, g is the acceleration due to gravity, and h is the height of the hill. The temperature change of the brakes can then be calculated using the equation Q = mc∆T, where m is the mass of the brakes and c is its specific heat capacity.

Explanation:

The thermal energy dissipated from the brakes of a car can be calculated by converting the gravitational potential energy lost by the car into internal energy of the brakes. By using the equation Q = Mgh/10, where Q is the energy transferred to the brakes, M is the mass of the car, g is the acceleration due to gravity, and h is the height of the hill, we can calculate the thermal energy dissipated. From there, the temperature change of the brakes can be calculated using the equation Q = mc∆T, where m is the mass of the brakes and c is its specific heat capacity.

Learn more about Thermal energy dissipation here:

brainly.com/question/16043710

#SPJ11

Which of the following expressions is equivalent to the expression 2(x + 5)? x + 7 x + 10 2x + 5 2x + 10

Answers

Answer:

d) 2x+10

Explanation:

$2(x+5)\n=2x+10$

Question:-

Which of the following expressions is equivalent to the expression 2(x + 5)?

x + 7
x + 10
2x + 5
2x + 10

Answer:-

2x+10

Explanation:-

take 2 as common from both the terms :-

$: \implies \sf 2x + 10$

$: \implies \sf 2(x + 5)$

Hope u got what u were looking for

(a) How fast and in what direction must galaxy A be moving if an absorption line found at 550 nm (green) for a stationary galaxy is shifted to 450 nm (blue) for A? (b) How fast and in what direction is galaxy B moving if it shows the same line shifted to 700 nm (red)?

Answers

Explanation:

For Part (a)

Since the apparent wavelength decreases hence galaxy moving towards the stationary observer.

Δλ/λ=v/c

$=(v)/(c)\n v=(550-450)/(550)*3*10^(8)\n v=5.4545*10^(7)m/s$

For Part (b)

Since the apparent wavelength increases hence galaxy moving towards the stationary observer.

Δλ/λ=v/c

$=(v)/(c)\n v=(700-550)/(550)*3*10^(8)\n v=8.1818*10^(7)m/s$

At the instant the traffic light turns green, a car that has been waiting at an intersection starts ahead with a constant acceleration of 3.60 m/s^2 . At the same instant a truck, traveling with a constant speed of 23.5 m/s , overtakes and passes the car. a. How far beyond its starting point does the car overtake the truck?b. How fast is the car traveling when it overtakes the truck?