Ask question

Ask question

All categories

sergiy2304 [10]

3 years ago

11

The truck travels in a circular path having a radius of 50 m at a speed of v = 4 m>s. For a short distance from s = 0, its sp

eed is increased by v # = (0.05s) m>s 2, where s is in meters. Determine its speed and the magnitude of its acceleration when it has moved s = 10 m.

1 answer:

miss Akunina [59]3 years ago

6 0

Answer:

Magnitude of velocity when s is 10 m= 4.58m/s

Magnitude of acceleration when s is 10 m = 0.653 m/s²

Explanation:

Radius of circular path = 50m

Speed = 4m/s

increase in acceleration = (0.05s)m/s²

speed and acceleration when s is 10 m = ?

Magnitude of Velocity:

From third equation of motion

$v_{f}^{2}-v_{i}^{2}=2aS\\\\v_{f}=\sqrt{2aS+v_{i}^{2}} \\\\v_{f}=\sqrt{2(0.05)(10)+(4)^{2}} \\\\v_{f}=4.58\,m/s$

Magnitude of Acceleration:

$a=\sqrt{a_{t}^{2}+a_{n}^{2}}--(1)$

The tangential component of acceleration is

$a_{t}=(0.05)(10)\\a_{t}=0.5\,m/s^{2}$

The normal component of acceleration is

$a_{n}=\frac{v^{2}}{R}\\\\a_{n}=\frac{16}{50}\\\\a_{n}=0.42\,m/s^{2}$

Substituting tangential and normal component in (1)

$a=\sqrt{(0.5)^{2} +(0.42)^{2} }$

a=0.653 m/s²

You might be interested in

A particular electromagnetic wave travelling in vacuum is detected to have a frequency of 3 × 10 12 Hz. How much time will it ta

irina1246 [14]

3×10^-12 seconds

Explanation:

T=1/f

7 0

2 years ago

A composite wall is composed of 20 cm of concrete block with k = 0.5 W/m-K and 5 cm of foam insulation with k = 0.03 W/m-K. The

wariber [46]

Answer:

4.8°C

Explanation:

The rate of heat transfer through the wall is given by:

$q=\frac{Ak}{L}dT$

$\frac{q}{A}=\frac{k}{L}dT$

Assumptions:

1) the system is at equilibrium

2) the heat transfer from foam side to interface and interface to block side is equal. There is no heat retention at any point

3) the external surface of the wall (concrete block side) is large enough that all heat is dissipated and there is no increase in temperature of the air on that side

${k_{fi}= 0.03 W/m.K$

${L_{fi}= 5 cm = 0.05 m$

${T_{fi}= 25 \°C$

${k_{cb} = 0.5 W/m.K$

${L_{cb}= 20 cm = 0.20 m$

${T_{cb}= 0 \°C$

${T_{m}= ? \°C =$ temperature at the interface

Solving for ${T_{m}$ will give the temperature at the interface:

$\frac{q}{A}=\frac{k_{fi} }{L_{fi} }(T_{fi} -T_{m})=\frac{k_{cb} }{L_{cb} }(T_{m} -T_{cb})$

$\frac{0.03}{0.05 }(25 -T_{m})=\frac{0.5}{0.2}(T_{m} -0})$

$15 -0.6T_{m}=2.5T_{m}$

$3.1T_{m}=15$

$T_{m}=4.8$

3 0

3 years ago

Two well-known NP-complete problems are 3-SAT and TSP, the traveling salesman problem. The 2-SAT problem is a SAT variant in whi

Hitman42 [59]

3-SAT ≤p TSP

If P ¹ NP, then no NP-complete problem can be solved in polynomial time.

both the statements are true.

<u>Explanation:</u>

3-SAT ≤p TSP due to any complete problem of NP to other problem by exits of reductions.
If P ¹ NP, then 3-SAT ≤p 2-SAT are the polynomial time algorithm are not for 3-SAT. In P, 2-SAT is found, 3- SAT polynomial time algorithm implies the exit of reductions. 3 SAT does not have polynomial time algorithm when P≠NP.
If P ¹ NP, then no NP-complete problem can be solved in polynomial time. because for the NP complete problem individually gets the polynomial time algorithm for the others. It may be in P for all the problems, the implication of latter is P≠NP.

7 0

3 years ago

3. Low-voltage conductors rarely cause<br> injuries.

hram777 [196]

True
the answer to this would be true

8 0

2 years ago

British standered institution

Scorpion4ik [409]

Answer:

The British Standards Institution, is the national standards body of the United Kingdom. BSI produces technical standards on a wide range of products and services and also supplies certification and standards-related services to businesses

6 0

3 years ago