A ) v = v o + a t ( the acceleration will be negative )
9.50 = 16.0 + a * 1.2
a * 1.2 = -16.0 + 9.50
a * 1.2 = - 6.5
a = - 6.5 : 1.2
a = - 5.4167 m/s²
F = m * a = 950 kg * 5.4167 m/s²
F = 5,145.8 N ( the average force exerted on a car during braking )
b ) d = v o - a t² / 2
d = 16.0 * 1.2 - ( 5.4167 * 1.2² / 2 ) =
= 19.20 - 3.90 = 15.30 m
How many times did the original sample lose 50% of its radioactivity ?
-- Start with. . . . . . . . . . . . 12 grams.
-- Lose half of it once. . . . . . 6 grams left.
-- Lose half of it again . . . . . 3 grams left.
-- Lose half of it again . . . . . 1.5 grams left.
-- Lose half of it again . . . . . 0.75 gram left.
-- How many times did it lose half ? 4 times.
-- How long does it take to lose half ? 4.5 days.
(That's why it's called the 'half-life'.)
-- How long did it take to lose half, 4 times ?
(4 x 4.5 days) = 18 days .
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a. <span>FM GmMmr2
</span>= 6.67 x 10-11N.m2kg27 .35 x 1022 kg 70 kg 3.78 x 108 m2
<span>= 2.40 x 10-3 N
b. </span><span>FE GmEmr2
= 6.67 x 10-11 N.m2kg 25 .97 x 1034 kg (70kg) 6.38 x 106 m2
=685 N
FMFE 2.40 x 10-3N685 N= 0.0004%</span>
Answer:
The net change in the internal energy of the gas in the piston is -343J
Explanation:
Because heat and workdone are the only means of energy transfer between the system and the surrounding, change in internal energy is given by;
∆E = q + w
q = heat transfer
w = workdone
Because heat is lost by the system, the heat transfer is negative
q = -413J
Because work is done on the system, workdone is positive
w = +70J
∆E = -413J + 70J
∆E = -343J