Any clue on this one please help

A train starts from rest and accelerates uniformly, until it has traveled 3.7 km and acquired a velocity of 26 m/s. What is the

bogdanovich [222]

We have:
s=3,7km (3700m)
u=0
v=26m/s
a=
t=
We look up for a formula to solve for a:
v²=u²+2as u²=0 so
v²=2as
(26m/s)²=2a3700m
676/7400=a
a=0,09m/s²

7 0

3 years ago

Which of the following is the best thermal insulator?

sdas [7]

Answer:

you're right it is b....

7 0

3 years ago

An advanced computer sends information to its various parts via infrared light pulses traveling through silicon fibers (n = 3.50

sasho [114]

Answer:

d = 6.43 cm

Explanation:

Given:

- Speed resistance coefficient in silicon n = 3.50

- Memory takes processing time t_p = 0.50 ns

- Information is to be obtained within T = 2.0 ns

Find:

- What is the maximum distance the memory unit can be from the central processing unit?

Solution:

- The amount of time taken for information pulse to travel to memory unit:

t_m = T - t_p

t_m = 2.0 - 0.5 = 1.5 ns

- We will use a basic relationship for distance traveled with respect to speed of light and time:

d = V*t_m

- Where speed of light in silicon medium is given by:

V = c / n

- Hence, d = c*t_m / n

-Evaluate: d = 3*10^8*1.5*10^-9 / 3.50

d = 0.129 m 12.9 cm

- The above is the distance for pulse going to and fro the memory and central unit. So the distance between the two is actually d / 2 = 6.43 cm

8 0

3 years ago

In general, what can be said about metals.

faltersainse [42]

Answer:

They are hard.

Explanation:

that’s all i have to say ngl

6 0

4 years ago

A horizontal disk rotates about a vertical axis through its center. Point P is midway between the center and the rim of the disk

rjkz [21]

<u>If the disk turns with constant angular velocity, the following statements about it are true </u>

The linear acceleration of Q is twice as great as the linear acceleration of P
is moving twice Q as fast as P.

Answer: Options D and E

<u>Explanation: </u>

Let us consider that R is the radius of the circular disc. So as Q is on the rim, so the distance of Q from the centre of the disc is R and as P is the midpoint between centre and rim of the disk, so the distance of P from the centre is R/2.

As we know that the angular velocity of the circular disk will be equal to the ratio of distance covered by that point to the time taken. So the angular velocity at point Q will be

$\text { Angular velocity at point } Q=\frac{\text { Distance covered by point } Q}{\text { Time }}=\frac{2 \pi R}{T}=v$

As R is the distance of point Q from the centre of the disc.

Similarly ,

$\text { Angular velocity at point } P=\frac{\text { Distance covered by point } P}{\text { Time }}=\frac{2 \pi\left(\frac{\mathrm{R}}{2}\right)}{T}=\frac{\pi R}{T}=v^{\prime}$

So if we equate v with v’ we obtain that

$v=2 v^{\prime}$

Therefore, the point Q will be moving twice as fast as P. As the velocity of Q is more than O, the linear acceleration of point Q will also be twice as great as the linear acceleration of P.

This is because acceleration is directly proportional to the rate of change in velocity. So if velocity increases in the factor of 2, the acceleration of point Q will also increase twice with respect to point P.

5 0

4 years ago