Ek = 1/2 mv^2
9 × 10^4 = 1/2 × 800 × v^2
9 × 10^4/400 = 400 v^2 / 400
9 × 10^4/400 = v^2
√225 = v
15 ms⁻¹ = v
That's the only way I know how to work it out
I think in this case velocity and speed would be considered the same because me
s = d/t and v=d/t
one is distance travelled and the other is displacement of a body
Answer:1200
Explanation:
Given data
Upper Temprature
Lower Temprature 
Engine power ouput
Efficiency of carnot cycle is given by
rounding off to two significant figures
Answer:
Following are the responses to this question:
Explanation:
The small current passes thru the capacitor of the strain gauge and the current is generated throughout the resistor. For the very first time, in contrast to what we calculate, its resistance of the multimeter is quite high and therefore the small stream flowing through the bulb would have very little impact on the measure. Thus, as the current flows through the flashbulb, this same calculation is of excellent price, its material is heated and resistance varies with increase. Therefore, when the bulb will be on, sensitivity is greater.
===> Distance fallen from rest in free fall =
(1/2) (acceleration) (time²)
(122.5 m) = (1/2) (9.8 m/s²) (time²)
Divide each side by (4.9 m/s²): (122.5 m / 4.9 m/s²) = time²
(122.5/4.9) s² = time²
Take the square root of each side: 5.0 seconds
===> (Accelerating at 9.8 m/s², he will be dropping at
(9.8 m/s²) x (5.0 s) = 49 m/s
when he goes 'splat'. We'll need this number for the last part.)
===> With no air resistance, the horizontal component of velocity
doesn't change.
Horizontal distance = (10 m/s) x (5.0 s) = 50 meters .
===> Impact velocity = (10 m/s horizontally) + (49 m/s vertically)
= √(10² + 49²) = 50.01 m/s arctan(10/49)
= 50.01 m/s at 11.5° from straight down,
away from the base of the cliff.
It is false. The effect of freezing is almost the exact opposite
