Ask question

Ask question

All categories

antoniya [11.8K]

2 years ago

11

What is the change in length of a 1400. m steel, (12x10^-6)/(C0) , pipe for a temperature change of 250.0 degrees Celsius? Remem

ber to include your data, equation, and work when solving this problem.

1 answer:

8090 [49]2 years ago

8 0

Answer:

$\Delta L = 4.2 m$

Explanation:

As per the formula of thermal expansion we know that

$L = L_o(1 + \alpha\Delta T)$

so here we will have

$L_o = 1400 m$

$\alpha = 12 \times 10^{-6} per ^oC$

$\Delta T = 250 degree C$

so here change in the length of the rod is given as

$\Delta L = L - L_o$

$\Delta L = L_o \alpha \Delta T$

$\Delta L = 1400 (12 \times 10^{-6})(250)$

$\Delta L = 4.2 m$

You might be interested in

Stacy measures two quantities: the mass of each washer and the force that the washers exert on the force meter. In general, how

geniusboy [140]

Answer:

When she adds more washers to the meter, the magnitude of force that is shown on the force meter increases.

Explanation:

The force that the washers exert on the force meter is actually the weight of the washers. Weight is actually a force with gravitation acceleration.

F = W = mg

Where g is gravitational acceleration and its value is 9.81 m/s² and m is the mass of any object. As she adds more washers to the meter so the total mass of the washers increases. As the mass of the washers increases, magnitude of the force (Weight) shown on the force meter increases.

6 0

2 years ago

In a semiclassical model of the hydrogen atom, the electron orbits the proton at a distance of 0.053 nm. Part A What is the elec

Bezzdna [24]

Answer with Explanation:

We are given that

$r=0.053 nm=0.053\times 10^{-9} m$

$1 nm=10^{-9} m$

Charge on proton,q= $1.6\times 10^{-19} C$

a.We have to find the electric potential of the proton at the position of the electron.

We know that the electric potential

$V=\frac{kq}{r}$

Where $k=9\times 10^9$

$V=\frac{9\times 10^9\times 1.6\times 10^{-19}}{0.053\times 10^{-9}}$

$V=27.17 V$

B.Potential energy of electron,U= $\frac{kq_e q_p}{r}$

Where

$q_e=-1.6\times 10^{-19} c=$ Charge on electron

$q_p=q=1.6\times 10^{-19} C$ =Charge on proton

Using the formula

$U=\frac{9\times 10^9\times (-1.6\times 10^{-19}\times 1.6\times 10^{-19}}{0.053\times 10^{-9}}$

$U=-4.35\times 10^{-18} J$

8 0

2 years ago

Light is propagated as a transverse wave. For this reason, sunglasses, ski goggles and camera lenses can restrict the vibration

Flura [38]

Polerization is the anwser

6 0

2 years ago

Read 2 more answers

1)After catching the ball, Sarah throws it back to Julie. However, Sarah throws it too hard so it is over Julie's head when it r

DENIUS [597]

Answer:

1)

$v_{oy}=11.29\ m/s$

2)

$y=7.39\ m$

Explanation:

<u>Projectile Motion</u>

When an object is launched near the Earth's surface forming an angle $\theta$ with the horizontal plane, it describes a well-known path called a parabola. The only force acting (neglecting the effects of the wind) is the gravity, which acts on the vertical axis.

The heigh of an object can be computed as

$\displaystyle y=y_o+V_{oy}t-\frac{gt^2}{2}$

Where $y_o$ is the initial height above the ground level, $v_{oy}$ is the vertical component of the initial velocity and t is the time

The y-component of the speed is

$v_y=v_{oy}-gt$

1) We'll find the vertical component of the initial speed since we have not enough data to compute the magnitude of $v_o$

The object will reach the maximum height when $v_y=0$ . It allows us to compute the time to reach that point

$v_{oy}-gt_m=0$

Solving for $t_m$

$\displaystyle t_m=\frac{v_{oy}}{g}$

Thus, the maximum heigh is

$\displaystyle y_m=y_o+\frac{v_{oy}^2}{2g}$

We know this value is 8 meters

$\displaystyle y_o+\frac{v_{oy}^2}{2g}=8$

Solving for $v_{oy}$

$\displaystyle v_{oy}=\sqrt{2g(8-y_o)}$

Replacing the known values

$\displaystyle v_{oy}=\sqrt{2(9.8)(8-1.5)}$

$\displaystyle v_{oy}=11.29\ m/s$

2) We know at t=1.505 sec the ball is above Julie's head, we can compute

$\displaystyle y=y_o+V_{oy}t-\frac{gt^2}{2}$

$\displaystyle y=1.5+(11.29)(1.505)-\frac{9.8(1.505)^2}{2}$

$\displaystyle y=1.5\ m+16,991\ m-11.098\ m$

$y=7.39\ m$

5 0

2 years ago

The cockroach Periplaneta americana can detect a static electric field of magnitude 8.50 kN/C using their long antennae. If the

otez555 [7]

Answer:

0.647 nC

Explanation:

The force experienced by a charge due to the presence of an electric field is given by

$F=qE$

where

q is the charge

E is the magnitude of the electric field

In this problem, each antenna is modelled as it was a single point charge, experiencing a force of

$F=5.50\mu N = 5.50\cdot 10^{-6} N$

Therefore, if the electric field magnitude is

$E=8.50 kN/C = 8500 N/C$

Then the charge on each antenna would be

$q=\frac{F}{E}=\frac{5.50\cdot 10^{-6} N}{8500 N/C}=6.47\cdot 10^{-10} C = 0.647 nC$

8 0

2 years ago