un fir de cauciuc de lungimea 1 m si sectiunea 10mm patrati se alungeste acu 1 cm ce marime are forta de deformare

A piece of wire length 30cm and cross sectional area of 0.5mm^2 has a resistance of 5ohms at 20°c. It is then heated to a temper

Grace [21]

Answer:

8.333*10^-6 ohms

Explanation:

Resistivity of a material is expressed as;

p = RA/l

R is the resistance of the material

A is the cross sectional area

l is the length of the material

Given

R = 5 ohms

A = 0.5mm^2

A = 5 * 10^-7m^2

l = 30cm = 0.3m

Substitute into the formula;

p = (5 * 5 * 10^-7m^2)/0.3

p = 25 * 10^-7/0.3

p = 0.0000025/0.3

p = 8.333*10^-6

Hence its resistivity at 20 degrees Celsius is 8.333*10^-6 ohms

7 0

3 years ago

A plane designed for vertical takeoff has a mass of 8.0 × 10³ kg. Find the net work done by all forces on the plane as it accele

artcher [175]

Answer:

the net work done after starting from rest is = 2.4 × 10⁵ J

Explanation:

Work: Work can be defined as the product of force and distance. The fundamental unit of work is Joules (J), The unit of Energy is Joules (J), as such Energy and work are interchangeable during calculation, This is illustrated below

E = W = 1/2mv².......................... Equation 1

Where m = mass of the plane, v = velocity of the plane, E = Energy, W = work done.

v² = u² + 2as ................................. Equation 2.

Where v = final velocity of the plane, u = initial velocity of the plane, a = acceleration of the plane, distance of the plane.

Given: a = 1.0 m/s², s = 30 m, u = 0 m/s (at rest)

Substituting these values into equation 2

v² = 0² +2×1×30

v² = 60

v = √60

v = 7.75 m/s

Also given: m = 8.0 × 10³ kg, and v = 7.75 m/s

Substituting these values into equation 1,

W = 1/2(8.0×10³)(7.75)²

W = (4.0×10³)(60)

W = 240 × 10³ J

W = 2.4 × 10⁵ J

Therefore the net work done after starting from rest is = 2.4 × 10⁵ J

4 0

3 years ago

PLEASE HELP! I'LL GIVE BRAINLEST

erma4kov [3.2K]

Answer:

8.162

Explanation:

W=mg

= 2.2 x 3.71

= 8.162 N

5 0

3 years ago

A glass tube 1 mm in diameter is dipped into glycerin. The density of the glycerin is 1260 kg/m3, surface tension is 6.3x10-2 N/

Sunny_sXe [5.5K]

Answer:

The capillary rise of the glycerin is most nearly $y = 0.0204 \ m$

Explanation:

From the question we are told that

The diameter of the glass tube is $d = 1 \ mm = 0.001 \ m$

The density of glycerin is $\rho = 1260 \ kg /m^3$

The surface tension of the glycerin is $\sigma = 6.3 *10^{-2} \ N /m$

The capillary rise of the glycerin is mathematically represented as

$y = \frac{4 * \sigma * cos (\theta )}{ \rho * g * d}$

substituting value

$y = \frac{4 * 6.3 *10^{-2} * cos (0 )}{ 1260 * 9.8 * 0.001}$

$y = 0.0204 \ m$

Therefore the height of the glass tube the glycerin was able to cover is

$y = 0.0204 \ m$

4 0

3 years ago

Kai swims for the school swim team. He specializes in a backstroke event where he has to swim the 50 m length of the pool three

Westkost [7]

Answer:

Displacement is 50 m

Explanation:

Distance is simply the measurement of the sum of all paths travelled.

Thus, since he swims the 50m length pool 3 times, then the total distance = 50 × 3 = 150 m

Whereas, displacement is the measurement of length of the shortest path from initial point to final point.

In this case initial point to final point is 50m. Thus, the displacement is 50 m

4 0

3 years ago

Read 2 more answers