it means you send it out to everybody that is in your contact list
why do you text in a work sheet
each work is discussed separately.
To calculate the work done by a 1 kW motor in one minute, we can use the formula: [ \text{Work} = \text{Power} \times \text{Time} ] Given that power is 1 kW (or 1000 watts) and time is 1 minute (or 60 seconds), we can substitute the values: [ \text{Work} = 1000 , \text{W} \times 60 , \text{s} = 60000 , \text{J} ] Thus, a 1 kW motor does 60,000 joules of work in one minute.
What area codes work for text now on fire tablets
To find the time it takes for a 500 W electric motor to do (1.50 \times 10^5) J of work, we can use the formula: [ \text{Power} = \frac{\text{Work}}{\text{Time}}. ] Rearranging gives us: [ \text{Time} = \frac{\text{Work}}{\text{Power}} = \frac{1.50 \times 10^5 \text{ J}}{500 \text{ W}} = 300 \text{ seconds}. ] Thus, it would take 300 seconds, or 5 minutes, for the motor to perform that amount of work.
It means that the text is too short to be able to work and/or fit.
a text area is the area where we write our text. and a workspace is the place where we can work anything.
a text area is the area where we write our text. and a workspace is the place where we can work anything.
To find the force exerted, we can use the formula for work: ( \text{Work} = \text{Force} \times \text{Distance} ). Rearranging gives us ( \text{Force} = \frac{\text{Work}}{\text{Distance}} = \frac{1470 , \text{joules}}{20 , \text{meters}} = 73.5 , \text{newtons} ). To convert this force to pounds, we can use the conversion factor ( 1 , \text{newton} \approx 0.2248 , \text{pounds} ), resulting in a weight of approximately 16.56 pounds for the bowling ball.
It means that the text is too short to be able to work and/or fit.
Lifting a 50 newton weight 3 meters straight upward requires more work than lifting a 30 newton weight the same distance. Work is calculated using the formula ( \text{Work} = \text{Force} \times \text{Distance} ). For the 50 newton weight, the work done is ( 50 , \text{N} \times 3 , \text{m} = 150 , \text{J} ), while for the 30 newton weight, it is ( 30 , \text{N} \times 3 , \text{m} = 90 , \text{J} ). Thus, lifting the 50 newton weight involves more work.
The work done in lifting the crate is equal to its change in potential energy: ( \text{Work} = \text{Force} \times \text{distance} = m \cdot g \cdot h ). The power required is the work done divided by the time taken: ( \text{Power} = \frac{\text{Work}}{\text{time}} ). Plug in the values to calculate the power required.