Mass is a measurement of

You hang a heavy ball with a mass of 10 kg from a gold wire 2.6 m long that is 1.6 mm in diameter. You measure the stretch of th

PolarNik [594]

<u>Answer:</u> The Young's modulus for the wire is $6.378\times 10^{10}N/m^2$

<u>Explanation:</u>

Young's Modulus is defined as the ratio of stress acting on a substance to the amount of strain produced.

The equation representing Young's Modulus is:

$Y=\frac{F/A}{\Delta l/l}=\frac{Fl}{A\Delta l}$

where,

Y = Young's Modulus

F = force exerted by the weight = $m\times g$

m = mass of the ball = 10 kg

g = acceleration due to gravity = $9.81m/s^2$

l = length of wire = 2.6 m

A = area of cross section = $\pi r^2$

r = radius of the wire = $\frac{d}{2}=\frac{1.6mm}{2}=0.8mm=8\times 10^{-4}m$ (Conversion factor: 1 m = 1000 mm)

$\Delta l$ = change in length = 1.99 mm = $1.99\times 10^{-3}m$

Putting values in above equation, we get:

$Y=\frac{10\times 9.81\times 2.6}{(3.14\times (8\times 10^{-4})^2)\times 1.99\times 10^{-3}}\\\\Y=6.378\times 10^{10}N/m^2$

Hence, the Young's modulus for the wire is $6.378\times 10^{10}N/m^2$

3 0

3 years ago

Emily’s vacuum cleaner has a power rating of 200 watts. If the vacuum cleaner does 360,000 joules of work,

gtnhenbr [62]

Divide 360000 by 200 to get 1800 seconds, or half of hour.

7 0

3 years ago

Consider a continuous time signal, x(t) = ∑ Ai 5 i=1 cos(2πFi t + θi), consisting of a sum of sinusoidal signals of frequencies

Wewaii [24]

Answer:

duhhhhh

Explanation:

duhhhhh

4 0

3 years ago

the end of the tracks, 8.8 m lower vertically, is a horizontally situated spring with constant 5 × 105 N/m. The acceleration of

devlian [24]

Answer

Assuming the mass of the car, m = 43000 kg

initial speed u = 0

vertical distance moved, h = 8.8 m

spring constant k = 5 x 10⁵ N / m

acceleration of gravity = 9.8 m/s²

From law of conservation of energy ,

Gravitational potential energy at starting position =potential energy of the spring at maximum compression

$m g h =\dfrac{1}{2}k x^2$

$x = \sqrt{\dfrac{2mgh}{k}}$

$x = \sqrt{\dfrac{2\times 43000\times 9.8\times 8.8}{5\times 10^5}}$

x = 14.83 m

If the mass of the car is equal to 43000 Kg the spring is compressed to 14.83 m

6 0

3 years ago

A horizontal force of magnitude 46.3 n pushes a block of mass 4.14 kg across a floor where the coefficient of kinetic friction i

IrinaVladis [17]

A) Calling F the intensity of the horizontal force and d the displacement of the block across the floor, the work done by the horizontal force is equal to
$W=Fd = (46.3 N)(4.25 m)=196.8 J$

b) The work done by the frictional force against the motion of the block is equal to:
$W_f = -F_f d =- (\mu mg) d =-(0.609)(4.14 kg)(9.81 m/s^2)(4.25 m)=$
$=-105.1 J$
Part of these 105.1 Joules of work becomes increase of thermal energy of the block ( $\Delta E_B$ ), and part of it becomes increase of thermal energy of the floor ( $\Delta E_F$ ). We already know the increase in thermal energy of the block (38.2 J), so we can find the increase in thermal energy of the floor:
$\Delta E_F = 105.1 J - \Delta E_B = 105.1 J-38.2 J=66.9 J$

c) The net work done on the block is the work done by the horizontal force F minus the work done by the frictional force (the frictional force acts against the motion, so we must take it with a negative sign):
$W_{net}=W-W_f=196.8 J-105.1 J=91.7 J$
For the work-energy theorem, the work done on the block is equal to its increase of kinetic energy:
$W_{net} = \Delta K$
So, we have $\Delta K=+91.7 J$

5 0

3 years ago