Protons and neutrons grouped in a specific pattern

If a spring requires 20 Newtons to be compressed a distance of 10 centimeters, what is the spring constant in N/m (newton meters

Lady bird [3.3K]

Answer:

20 N / 0.1 meters

Explanation:

10 centimetres is equal to 1 tenth of a meter or 0.1 meters.

6 0

3 years ago

A pilgrim uses a block and tackle to lift a 225 N crate a distance of 16.5 m. She pulls 33.0 m of rope with a force of 125 N. Wh

kenny6666 [7]

Answer:

Ideal mechanical advantage of the machine is 2

Explanation:

Ideal mechanical advantage of the machine is

$MA=\frac{op}{ip} \\$

here we have to consider the distance

$MA=\frac{33}{16.5} \\MA=2$

7 0

3 years ago

Two wires with equal lengths are made of pure copper. The diameter of wire A is three times the diameter of wire B. When 8 kg ma

noname [10]

Answer: c. YA < YB

Explanation:

The formula for Young’s modulus is = Tensile stress / Tensile strain

Tensile stress = Force x Length

Force = mass x acceleration due to gravity

= 8kg x 10m/s

= 80kgm/s

Tensile stress = 80kgm/s x 2m = 160kgm2/s

Tensile strain = Area x change in length

Area = pi x D2 / 4 ; Pi = 3.14

Change in length = L2 – L1 (New length – Initial length)

Given parameters:

Length of wire A = Length of wire B, (let’s use 2meters for the calculation)

For wire A, Diameter = 3 x Wire B diameter

Assuming Diameter of wire B = 1meter

Therefore, diameter of wire A = 1 x 3 = 3meters

It is said that wire B stretches more than wire A when the man of 8kg is placed on both

For wire B, let’s assume new length is = 4m

For wire A let’s assume new length is = 3m.

(i) Tensile strain of wire A =

Area of wire A = 3.14 x (32)/4 = 7.065m2

Change in length = 3m - 2m = 1m.

Therefore, tensile strain = 7.065m2 x 1m = 7.065m3

Young’s modulus for wire A (YA) = 160kgm2/s divided by 7.065m3

= 22.64Pa.

(ii) Tensile strain of wire B =

Area of wire B = 3.14 x (12)/4 = 0.785m2

Change in length = 4m – 2m = 2m

Therefore, tensile strain = 0.785m2 x 2m = 1.57m3

Young’s modulus for wire B (YB) = 160kgm2/s divided by 1.57m3

= 101.91Pa.

From the calculations above, we see that YA is less than YB (YA < YB). This is true given that wire A has a greater diameter than wire B which in turn impacts the Area of the wire since the diameter is directly proportional to area and the area is inversely proportional to the young’s modulus.

5 0

3 years ago

A 67 kg astronaut floating in space throws a 4.2 kg rock at 5.6 m/sec. How fast does the astronaut move backward?

stiks02 [169]

Law of conservation of momentum
67*x=4.2*5.6

5 0

3 years ago

A string of length 1.3 m is oscillating in a standing wave pattern. If the tension in the string is 430 N, the string has a mass

Vlad1618 [11]

To solve this problem we will use the concepts related to the speed of a string which is given by the applied voltage and the linear mass density of it. With the speed value we can find the fundamental frequency that will serve as a step to find the maximum speed through the relation of Amplitude and Angular Speed. So:

$v = \sqrt{\frac{T}{\mu_e}}$