Lower Secondary Science

[sunnymoon]

Can someone help me with this question?

A basket is being carried by a boy at a constant height above the ground. Is work being done?
Thanks for your help

[superkiasume]

Hi, can someone help my son with the following questions:


(a) The engine of a car of mass 800kg can develop a constant 80kW of power. If its maximum speed on a flat road is 100km/h, determine the resistance to its motion.

(b) With the engine working at the same rate, find the acceleration of the car when is travelling at a speed of 50km/h, if the resistance is constant.

Thank you.

[tutor76]

spicy:

Hi

I'm a stranger to Physics :scratchhead: , do appreciate help with this question pl. Thanks :please:

http://i60.tinypic.com/xp90k3.jpg\">

Charge in coulomb = current x time
Curremt in xy is a series circuit (using conventional current flow )

Therefore current = volt/resistance

Current =1.2ampere

Therefore q=it


Therefore 12v/

1.3333x 10^-19 seconds


I is current 1.2a

Time is charge divide by current =


Formule ,voltage = work done /charge

8v x 1.6x10^-19 coulomb


1.28x 10^-19 J

[tutor76]

[quote="superkiasume"]Hi, can someone help my son with the following questions:


(a) The engine of a car of mass 800kg can develop a constant 80kW of power. If its maximum speed on a flat road is 100km/h, determine the resistance to its motion.

(b) With the engine working at the same rate, find the acceleration of the car when is travelling at a speed of 50km/h, if the resistance is constant.

Thank you.[/quo
There is too many assumption to be made in these two question,is the teacher trying so much to impress the principal with these question ?

Quite unsolvable …

[tutor76]

45has:

http://i60.tinypic.com/23sjh95.jpg\">


1. Calculate current flowing through XY.
2. Calculate voltage across XY.
3. Calculate work done.

I used Kirchoff law to calculate current faster.

Secondary school does not require kirchoff law ,from my experience teaching secondary student science .cheers

[tutor76]

http://i60.tinypic.com/2zg5eo0.jpg\"> Something for everyone to try !

[Dr.033430Daniel]

superkiasume:

Hi, can someone help my son with the following questions:


(a) The engine of a car of mass 800kg can develop a constant 80kW of power. If its maximum speed on a flat road is 100km/h, determine the resistance to its motion.

(b) With the engine working at the same rate, find the acceleration of the car when is travelling at a speed of 50km/h, if the resistance is constant.

Thank you.

For Part a, you can use the formula that at constant velocity

Power = Force X velocity

100km/hr converts to 27.8 m/s

80,000 W = Force X 27.8m/s

So Force = 2880 Newtons.

The driving force pushing the car forward is equal to the force of friction because it is moving at constant velocity.

Usually students remember that Power = Energy / time
At constant velocity the work done to push the car is the force times the distance. Then if we divide by time we get power. So if the car travels 27.8 meters in one second, then force X 27.8m is the work - then divide by 1 second of time. So this is equivalent to saying that Power = Force X velocity when moving at constant velocity.

part (b) is a bit confusing to me. If the engine is working at the same rate, that means it is delivering the same power. But the statement that the resistance is constant is confusing. Most of the resistance to a car is air resistance which depends on speed. So it is a bit hard to decipher part (b).

[Dr.033430Daniel]

spicy:

Hi

I'm a stranger to Physics :scratchhead: , do appreciate help with this question pl. Thanks :please:

http://i60.tinypic.com/xp90k3.jpg\">

The best place to start is

Volts = Work/charge

I've got the charge - they gave me just the charge of 1 electron, so calculating current in the circuit is not needed.

The key skill is to be able to look at that circuit diagram and realize that the 10 Ohm resistor has 8 Volts across it.

The thought process to arrive at that goes like this:

The two 10 Ohm resistors in parallel have an equivalent resistance of 5 Ohms. So this means there is a 5 Ohm and 10 Ohm resistor in series. For resistors in series the voltage across each resistor is determined by the proportion of the total resistance. This means that the 12 Volts of the battery is divided up into three parts. 2/3 of the voltage is dropped across the 10 Ohm resistor and 1/3 across the 5 Ohm resistor since they are in series. If you can \"see\" that, you just write

8Volts = Work/ 1.6E-19 C and you are done. The answer is B.

Being able to recognize how voltages are distributed across resistors in series is an important skill for O'Level pure physics and beyond. If you had 12 volts across two 10 Ohm resistors in series, each has 6 Volts. If you had 12 volts across a 1 Ohm and 11 Ohm resistor in series, the 1 Ohm resistor has 1 Volt across it and the 11 Ohm resistor has 11 Volts. The proportion of the total resistance is the proportion of the voltage. This is true for resistors in series, but not in parallel.

[Dr.033430Daniel]

sunnymoon:

Can someone help me with this question?

A basket is being carried by a boy at a constant height above the ground. Is work being done?
Thanks for your help

The work done is force times distance traveled in the same direction as the force. The boy will have to exert an upward force on the basket to hold it up, but if it is at constant height there is no distance traveled in the direction of the upward force, so therefore no work done.

In the horizontal direction, the general assumption in these kinds of questions is that there is no horizontal force needed if a person is walking along at constant velocity and air resistance to the basket is negligible (since it is at a slow speed). So the assumption is that no work is needed to move it horizontally at constant velocity with no friction.

[spicy]

:thankyou:


Your help is much appreciated :lovesite: