What is required foe electricity to flow

Calculate the speed of body moving a distance of 320km in 4h

ololo11 [35]

Convert km to m. -> 320km=320000 m
Convert hours to seconds -> 4 hrs=14400 s
Divide the distance by time because v=d/t
320000/14400=22.22 m/s

6 0

2 years ago

Steel train rails are laid in 12.0-m-long segments placed end to end. The rails are laid on a winter day when their temperature

Yanka [14]

Answer:

a) Space = $6.05 x 10^{-3} m = 0.605 cm$

b) Stress= $-100.8 x 10^{6} Pa$

Explanation:

<em>1) Data Given</em>

$L = 12 m , T_i = -9 C \degree, T_f = 33 C \degree$

<em>2) Calculate the space using Linear thermal expansion formula</em>

We need to use Linear thermal expansion formula since the space created would be a change on 1 dimension, the increase of the temperature will increase the length of the steel. The formula is given by:

$\Delta L = L_i \alpha_{steel} \Delta T$

We have everything except the $\alpha_{steel}$ , so we look for this on a book and we find that $\alpha_{steel} = 1.2 x 10^{-5} C^{-1}$ , so we can replace.

$\Delta L = 12 m (1.2 x 10^{-5} C^{-1}) (33 C \degree -(-9 C \degree)) = 12 m (1.2 x 10^{-5} C^{-1}) 42 C \degree =6.048 x 10^{-3} m = 0.6048cm$

<em>3) Calculate the stress of the steel </em>

The Stress is the ratio of applied force F to a cross section area - defined as

$\sigma = \frac{F_n}{A}$

Since we don't have the force and the Area, we need to look for another way to find the stress.

For this we can use the concept called Young's Modulus, defined as : "the mechanical property that measures the stiffness of a solid material", and the formula for this is given by:

$Y =\frac{F L}{A \Delta L}$ (1)

Solving $\frac{F}{A}$ from the previous formula we have this:

$\frac{F}{A}$ = (Y Δ L)/L (2)

From the <em>Linear thermal expansion formula</em> we can solve like this

$\frac{\Delta L}{L}$ = α ΔT (3)

And replacing equation (3) into equation (2) we have:

$\frac{F}{A}$ = Y α ΔT (4)

We have that the Young's Modulus for the steel is 20x10^{10} Pa, so replacing into equation (4)

$\frac{F}{A}$ = $20x10^{10}$ Pa (1.2x10^-5 C^-1) (42C) = $100.8 *10^{6}$ Pa

That represent the absolute value for the Stress, the sign on this case would be negative since there is a compression.

3 0

3 years ago

In 2000, NASA placed a satellite in orbit around an asteroid. Consider a spherical asteroid with a mass of 5.00�1015kg and a rad

irakobra [83]

Answer:

a). $v_{1}=13.49 \frac{m}{s}$

b). $v_{2}=17.54\frac{m}{s}$

Explanation:

Using the conservation of energy and the kinetic energy of the satellite around the asteroid can model the motion in

$\frac{1}{2}*m*v^2=\frac{G*M*m}{a_{o}}$

$v^2=\frac{2*G*M*m}{a_{o}}$

$G=6.673x10^{-11}\frac{m^3}{kg*s^2}$

$M=1.20x10^{16}kg$

a).

$a_{o}=8.8km*\frac{1000m}{1km}=8800m$

$v=\sqrt{\frac{2*6.673x10^{-11}\frac{m^3}{kg*s^2}*1.2x10^{10}kg}{8800m}}$

$v_{1}=13.49 \frac{m}{s}$

b).

$a_{f}=5.2km*\frac{1000m}{1km}=5200m$

$v=\sqrt{\frac{2*6.673x10^{-11}\frac{m^3}{kg*s^2}*1.2x10^{10}kg}{5200m}}$

$v_{2}=17.54\frac{m}{s}$

3 0

3 years ago

Read 2 more answers

An object is thrown straight up with a velocity, in ft/s, given by v(t)= -32t + 83, where t is in seconds, from a height of 46 f

____ [38]

<h2>a) Initial velocity = 83 ft/s</h2><h2>b) Object's maximum speed = 99.4 ft/s</h2><h2>c) Object's maximum displacement = 153.64 ft</h2><h2>d) Maximum displacement occur at t = 2.59 seconds.</h2><h2>e) The displacement is zero when t = 5.70 seconds</h2><h2>f) Object's maximum height = 153.64 ft</h2>

Explanation:

We have velocity

v(t)= -32t + 83

Integrating

s(t) = -16t²+83t+C

At t = 0 displacement is 46 feet

46 = -16 x 0²+83 x 0+C

C = 46 feet

So displacement is

s(t) = -16t²+83t+46

a) Initial velocity is

v(0)= -32 x 0 + 83 = 83 ft/s

Initial velocity = 83 ft/s

b) Maximum velocity is when the object reaches ground, that is s(t) = 0 ft

Substituting

0 = -16t²+83t+46

t = 5.70 seconds

Substituting in velocity equation

v(t)= -32 x 5.70 + 83 = -99.4 ft/s

Object's maximum speed = 99.4 ft/s

c) Maximum displacement is when the velocity is zero

That is

-32t + 83 = 0

t = 2.59 s

Substituting in displacement equation

s(2.59) = -16 x 2.59²+83 x 2.59+46 = 153.64 ft

Object's maximum displacement = 153.64 ft

d) Maximum displacement occur at t = 2.59 seconds.

e) Refer part b

The displacement is zero when t = 5.70 seconds

f) Same as option d

Object's maximum height = 153.64 ft

5 0

3 years ago

A baseball is thrown a distance of 20 meters what is its speed if it takes 0.5 seconds to cover the distance?

anastassius [24]

Speed=distance/time
=20m/0.5s
=40ms^-1

6 0

3 years ago