How does a parallel circuit differ from a series circuit?

8) A racecar accelerates uniformly from 18.5 m/s to 46.1 m/s in 2.47 seconds. Determine the

Vlada [557]

8)Acceleration of the car is 11.17 m/s².

9) Acceleration of skater is 0.57 m/s².

10) The deceleration is 1.8 m/s².

11) The average acceleration of the sprinter is 2 m/s².

12) The velocity of the stroller after 4.75 minutes is 171 m/s.

13) The skier will be moving as 11 m/s speed.

14) 0.56 seconds is required by the rattle snake to reach the speed of 28 m/s from rest.

Answer:

Explanation:

8) Acceleration of the car is determined by finding the ratio of change in velocity to the time taken for the change in velocity. As here the initial velocity is 18.5 m/s and the final velocity is 46.1 m/s, then in 2.47 seconds, the acceleration is

Acceleration = (46.1-18.5)/2.47 = 11.17 m/s².

Acceleration of the car is 11.17 m/s².

9)Similarly, the acceleration of the skater with initial velocity zero to final velocity as 5.7 m/s in 10 seconds is

Acceleration = (5.7-0)/10=0.57 m/s².

So acceleration of skater is 0.57 m/s².

10) In this case, the shuttle bus is stopping so its speed is decreasing from high value to low value. This kind of acceleration related to the decrease in velocity is termed as deceleration as the value of acceleration will be coming as negative.

Acceleration = (0-9)/5=-1.8 m/s².

So the deceleration is 1.8 m/s².

11) Average acceleration of sprinter = (Final velocity-Initial velocity)/Time

Average acceleration = (7.5-5)/1.25=2 m/s².

So the average acceleration of the sprinter is 2 m/s².

12)Since the acceleration of the stroller is given as 0.6 m/s². And the initial velocity is given as zero. So for the time of 4.75 minutes, the velocity will be equal to the final velocity.

As per equations of motion,

v = u +at

As u =0, a = 0.6 m/s² and t = 4.75 ×60 s = 285 s

So v = 0.6×285 = 171 m/s

Thus, the velocity of the stroller after 4.75 minutes is 171 m/s.

13) Similarly, in this case, u = 0, a = 2.2 m/s² and t = 5 s

Then v = u+at=0+(2.2×5)=11 m/s

So the skier will be moving as 11 m/s speed.

14) Here a = 50 m/s² and v = 28 m/s with u = 0 so time taken to reach the speed of 28 m/s is

v = u +at = 0+ (50 t)

28 = 50 t

t = 28/50 = 0.56 s

So 0.56 seconds is required by the rattle snake to reach the speed of 28 m/s from rest.

4 0

3 years ago

To practice Problem-Solving Strategy 27.1: Magnetic Forces. A particle with mass 1.81×10−3 kgkg and a charge of 1.22×10−8 CC has

Roman55 [17]

Answer:

The magnitude of the acceleration is $a = 0.33 m/s^2$

The direction is $- \r k$ i.e the negative direction of the z-axis

Explanation:

From the question we are that

The mass of the particle $m = 1.8*10^{-3} kg$

The charge on the particle is $q = 1.22*10^{-8}C$

The velocity is $\= v = (3.0*10^4 m/s ) j$

The the magnetic field is $\= B = (1.63T)\r i + (0.980T) \r j$

The charge experienced a force which is mathematically represented as

$F = q (\= v * \= B)$

Substituting value

$F = 1.22*10^{-8} (( 3*10^4 ) \r j \ \ X \ \ ( 1.63 \r i + 0.980 \r j )T)$

$= 1.22 *10^{-8} ((3*10^4 * 1.63)(\r j \ \ X \ \ \r i) + (3*10^4 * 0.980) (\r j \ \ X \ \ \r j))$

$= (-5.966*10^4 N) \r k$

Note :

$i \ \ X \ \ j = k \\\\j \ \ X \ \ k = i\\\\k \ \ X \ \ i = j\\\\j \ \ X \ \ i = -k \\\\k \ \ X \ \ j = -i\\\\i \ \ X \ \ k = - j\\$

Now force is also mathematically represented as

$F = ma$

Making a the subject

$a = \frac{F}{m}$

Substituting values

$a =\frac{(-5.966*10^4) \r k}{1.81*10^{-3}}$

$= (-0.33m/s^2)\r k$

$= 0.33m/s^2 * (- \r k)$

6 0

4 years ago

HELP When the forces are applied in the same direction, how do you determine net force?

aleksklad [387]

Answer:

Explanation:

If two forces act on an object in the same direction, the net force is equal to the sum of the two forces.

5 0

3 years ago

Two application of heat energy

densk [106]

Answer:

Heat is very important in our daily life in warming the house, cooking, heating the water, and drying the washed clothes. The heat has many usages in the industry as making and processing the food and manufacture of the glass, the paper, the textile, and etc.

Explanation:

4 0

3 years ago

What is the gauge pressure of the water right at the point p, where the needle meets the wider chamber of the syringe? neglect t

Helen [10]

Missing details: figure of the problem is attached.

We can solve the exercise by using Poiseuille's law. It says that, for a fluid in laminar flow inside a closed pipe,

$\Delta P = \frac{8 \mu L Q}{\pi r^4}$

where:

$\Delta P$ is the pressure difference between the two ends

$\mu$ is viscosity of the fluid

L is the length of the pipe

$Q=Av$ is the volumetric flow rate, with $A=\pi r^2$ being the section of the tube and $v$ the velocity of the fluid

r is the radius of the pipe.

We can apply this law to the needle, and then calculating the pressure difference between point P and the end of the needle. For our problem, we have:

$\mu=0.001 Pa/s$ is the dynamic water viscosity at $20^{\circ}$

L=4.0 cm=0.04 m

$Q=Av=\pi r^2 v= \pi (1 \cdot 10^{-3}m)^2 \cdot 10 m/s =3.14 \cdot 10^{-5} m^3/s$

and r=1 mm=0.001 m

Using these data in the formula, we get:

$\Delta P = 3200 Pa$

However, this is the pressure difference between point P and the end of the needle. But the end of the needle is at atmosphere pressure, and therefore the gauge pressure (which has zero-reference against atmosphere pressure) at point P is exactly 3200 Pa.

8 0

3 years ago