Answer: 0.2 hours
Explanation: In order to solve this question we have to considerer that a recargeable battery can supply 1800 mA in one hour then we have to determine how long could this battery drive current through a long, thin wire of resistance 34 Ω .
Besides, this battery has a voltage of 12 V
so by using the Ohm law we also know that V=R*I,
Fron this we can obtain:
I= V/R= 12 V/ 34 Ω=0.35 A= 350 mA
then considering that this battery can supply 1800 mA in one hour we have this battery can supply 350 mA in x time in the form:
1hour------- 1800 mA
x hour--------350 mA
time= 350/1800= 0.2 hour
The force on each balloon is 2×10^−3 N.
Consider two balloons of diameter 0.200m each with a mass of 1.00g hanging apart with 0.0500m separation on the ends of string making angles of 10.0° with the vertical.

So,

A force is an influence that can change the motion of an object. A force can cause an object with mass to change its velocity (e.g. moving from a state of rest), i.e., to accelerate. Force can also be described intuitively as a push or a pull. A force has both magnitude and direction, making it a vector quantity. It is measured in the SI unit of newton (N).
Learn more about force here:
brainly.com/question/13191643
#SPJ4
Gravity is stronger the closer you get. It is D.
Answer:
Va = 5000 m / 3600 s = 1.39 m/s
(Va - Vb) 60 = 10
Vb = Va - .167 = 1.22 m/s
(Va - Vb) T = 4200 Π where T is time for A to complete 1 more lap
.17 T = 4200 Π
T = 24700 Π time for A to again catch B
N = 1.39 * 24700 Answer:
Va = 5000 m / 3600 s = 1.39 m/s
(Va - Vb) 60 = 10
Vb = Va - .167 = 1.22 m/s
(Va - Vb) T = 4200 Π where T is time for A to complete 1 more lap
.17 T = 4200 Π
T = 24700 Π time for A to again catch B
N = 1.39 * 24700 Π / (4200 Π) = 8.2 laps
A will make 8 but not 9 rounds before catching B
Answer:
D. 21 ml
Explanation:
Since, the cylinder is marked and graduated in the intervals if 1 ml. Therefore, the values between two consecutive ml, such as between 30 ml and 31 ml can not be determined. Because, we do not have any scale in between the ml. So, the least count of this instrument is 1 ml. This graduated cylinder can give the answers to zero decimal places, accurately. And it can not determine any decimal value due to its graduating or the marking limitation. So, all the options given, contain a decimal value, except for the option D. In option D there is no decimal value, hence it is a correct answer.
D. <u>21 ml</u>