Ask question

Ask question

All categories

2 years ago

14

A 50 wt% ni–50 wt% cu alloy is slowly cooled from 1400°c to 1200°c:

1 answer:

g100num [7]2 years ago

6 0

This composition from Ni and Cu is called cupronickel.It is high in copper and silver in colour and highly resistant to corrosion, particularly seawater. Cooper nickel is used in the industry, in marine engineering . Melting point is 2254,73 F, which is 1234.85 Celsius. In this range of tempereature the cupronickel melts.

You might be interested in

A pendulum is released from rest at point A. If the horizontal line represents the reference point, the pendulum has energy at p

Rzqust [24]

Only kinetic

______________

Okay?

6 0

2 years ago

Read 2 more answers

A person on a cruise ship is doing laps on the promenade deck. On one portion of the track the person is moving north with a spe

Aleksandr-060686 [28]

T<u>he direction of motion</u> of the person relative to the water is <u>16.7° north of east.</u>

Why?

We can solve the problem by applying the Pitagorean Theorem, where the first speed (to the north) and the second speed (to the east) corresponds to two legs of the right triangle formed with them. (north and east directions are perpendicular each other)

We can calculate the angle that give the direction using the following formula:

$Tan(\alpha)=\frac{NorthSpeed}{EastSpeed}\\\\Tan(\alpha)^{-1}=Tan(\frac{NorthSpeed}{EastSpeed})^{-1}\\\\\alpha=Tan(\frac{NorthSpeed}{EastSpeed})^{-1}$

Now, substituting the given information we have:

$alpha=Tan(\frac{NorthSpeed}{EastSpeed})^{-1}$

$\alpha =Tan(\frac{3.6\frac{m}{s} }{12\frac{m}{s} })^{-1}\\\\\alpha =Tan(0.3)^{-1}=16.69\°(North-East)=16.7\°(North-East)$

Hence, we have that <u>the direction of motion</u> of the person relative to the water is 16.7° north of east.

Have a nice day!

7 0

2 years ago

Gamma rays are waves of energy that have no charge. <br> True<br> False

Andre45 [30]

Gamma rays are waves of energy that have no charge. TRUE.

3 0

2 years ago

Read 2 more answers

"What is the magnifying power of an astronomical telescope using a reflecting mirror whose radius of curvature is 5.9 m and an e

notka56 [123]

Answer:

The Magnifying power of a telescope is $M = 109.26$

Explanation:

Radius of curvature R = 5.9 m = 590 cm

focal length of objective $f_{objective}$ = $\frac{R}{2}$

⇒ $f_{objective}$ = $\frac{590}{2}$

⇒ $f_{objective}$ = 295 cm

Focal length of eyepiece $f_{eyepiece}$ = 2.7 cm

Magnifying power of a telescope is given by,

$M = \frac{f_{objective} }{f_{eyepiece} }$

$M = \frac{295}{2.7}$

$M = 109.26$

therefore the Magnifying power of a telescope is $M = 109.26$

4 0

2 years ago

A 3.00 kg object is moving in the XY plane, with its x and y coordinates given by x = 5t³ !1 and y = 3t ² + 2, where x and y are

Hatshy [7]

Answer:

The net force acting on this object is 180.89 N.

Explanation:

Given that,

Mass = 3.00 kg

Coordinate of position of $x= 5t^3+1$

Coordinate of position of $y=3t^2+2$

Time = 2.00 s

We need to calculate the acceleration

$a = \dfrac{d^2x}{dt^2}$

For x coordinates

$x=5t^3+1$

On differentiate w.r.to t

$\dfrac{dx}{dt}=15t^2+0$

On differentiate again w.r.to t

$\dfrac{d^2x}{dt^2}=30t$

The acceleration in x axis at 2 sec

$a = 60i$

For y coordinates

$y=3t^2+2$

On differentiate w.r.to t

$\dfrac{dy}{dt}=6t+0$

On differentiate again w.r.to t

$\dfrac{d^2y}{dt^2}=6$

The acceleration in y axis at 2 sec

$a = 6j$

The acceleration is

$a=60i+6j$

We need to calculate the net force

$F = ma$

$F = 3.00\times(60i+6j)$

$F=180i+18j$

The magnitude of the force

$|F|=\sqrt{(180)^2+(18)^2}$

$|F|=180.89\ N$

Hence, The net force acting on this object is 180.89 N.

3 0

3 years ago