The law of conservation of energy applies:

An engineer in a locomotive sees a car stuck on the track at a railroad crossing in front of the train. When the engineer first

GarryVolchara [31]

Answer:

The right answer is "1.369 m/s²".

Explanation:

The given values are:

Distance (s)

= 260 m

Initial speed (u)

= 26 m/s

Reaction time (t')

= 0.51 s

During reaction time, the distance travelled by locomotive will be:

⇒ $s'=ut'$

$=26\times 0.51$

$=13.26 \ m$

Remained distance between locomotive and car:

⇒ $x=s-s'$

$=260-13.26$

$=246.74 \ m$

Now,

The final velocity to avoid collection is, V = 0 m/s

From third equation of motion:

⇒ $V^2=u^2+2ax$

On putting the estimated values, we get

⇒ $0=(26)^2+2\times a\times 246.74$

⇒ $0=676+493.48a$

⇒ $493.48a=-676$

⇒ $a=-\frac{676}{493.48}$

⇒ $a=1.369 \ m/s^2$

3 0

3 years ago

A wheel with radius 36 cm is rotating at a rate of 19 rev/s.(a) What is the angular speed in radians per second? rad/s(b) In a t

Sedaia [141]

(a) 119.3 rad/s

The angular speed of the wheel is

$\omega= 19 rev/s$

we need to convert it into radiands per second. We know that

$1 rev = 2 \pi rad$

Therefore, we just need to multiply the angular speed of the wheel by this factor, to get the angular speed in rad/s:

$\omega = 19 rev/s \cdot (2\pi rad/rev))=119.3 rad/s$

(b) 596.5 rad

The angular displacement of the wheel in a time interval t is given by

$\theta= \omega t$

where

$\omega=119.3 rad$

and

t = 5 s is the time interval

Substituting numbers into the equation, we find

$\theta=(119.3 rad/s)(5 s)=596.5 rad$

(c) 127.3 rad/s

At t=10 s, the angular speed begins to increase with an angular acceleration of

$\alpha = 1.6 rad/s^2$

So the final angular speed will be given by

$\omega_f = \omega_i + \alpha \Delta t$

where

$\omega_i = 119.3 rad/s$ is the initial angular speed

$\alpha = 1.6 rad/s^2$ is the angular acceleration

$\Delta t = 15 s - 10 s = 5 s$ is the time interval

Solving the equation,

$\omega_f = (119.3 rad/s) + (1.6 rad/s^2)(5 s)=127.3 rad/s$

(d) 616.5 rad

The angle through which the wheel has rotated during this time interval is given by

$\theta = \omega_i \Delta t + \frac{1}{2} \alpha (\Delta t)^2$

Substituting the numbers into the equation, we find

$\theta = (119.3 rad/s)(5 s) + \frac{1}{2} (1.6 rad/s^2) (5 s)^2=616.5 rad$

(e) 222 m

The instantaneous speed of the center of the wheel is given by

$v_{CM} = \omega R$ (1)

where

$\omega$ is the average angular velocity of the wheel during the time t=10 s and t=15 s, and it is given by

$\omega=\frac{\omega_i + \omega_f}{2}=\frac{127.3 rad/s+119.3 rad/s}{2}=123.3 rad/s$

and

R = 36 cm = 0.36 m is the radius of the wheel

Substituting into (1),

$v_{CM}=(123.3 rad/s)(0.36 m)=44.4 m/s$

And so the displacement of the center of the wheel will be

$d=v_{CM} t = (44.4 m/s)(5 s)=222 m$

8 0

3 years ago

True or False: <br><br> Solids always<br> have a higher density than<br> liquids and gases.

Hunter-Best [27]

Answer:

<h3>False</h3>

<h3> I hoped this helpful for you....</h3>

Thank you ☺️☺️

7 0

3 years ago

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Find the density of seawater at a depth where the pressure is 680 atm if the density at the surface is 1030 kg/m3. Seawater has

Pie

Answer:

$1060.41kg/m^3$

Explanation:

Bulk modulus is defined as the relative change in the volume of a body produced by a unit of compressive acting uniformly over its surface:

$B=\rho _o \frac{\bigtriangleup P }{\bigtriangleup \rho}$

Hence the density of the seawater at a depth of 680atm is calculated as:-

$\rho=\rho_o +\bigtriangleup \rho=\rho_o(1+\frac{\bigtriangleup P}{B})\\\\=1030 \times (1+ \frac{(680-1)\times10^5}{2.3\times 10^9})\\=1060.41kg/m^3$

4 0

3 years ago

Goal posts at the ends of football fields are padded as a safety measure for players who might run into them. How does thick pad

zepelin [54]

The answer given by E2020 is

"The padding around the goal post increases the time of the collision between the player and the post, which decreases the force exerted to bring the player to a stop"

7 0

3 years ago

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