<span>Sure, Just change the 2 sec. into hrs. Since 1 hour = 3600 sec. then you can divide 2/3600 = 1/1800 hrs.
Distance in kilometers = (Speed in km/hr * time in hrs)
= 50*(1/1800)*1000 in meters
= 27.77 meters</span>
From the average speed you can fix an equation:
Average speed = distance / time
You know the average speed = 65.1 kg / h, then
65.1 = distance / total time,
where total time is the time traveling plus 22.0 minutes
Call t the time treavelling and pass 22 minutes to hours:
65.1 = distance / [t + 22/60] ==> distance = [t + 22/60]*65.1
From the constant speed, you can fix a second equation
Constant speed = distance / time traveling
94.5 = distance / t ==> distance = 94.5 * t
The distance is the same in both equations, then you have:
[t +22/60] * 65.1 = 94.5 t
Now you can solve for t.
65.1t + 22*65.1/60 = 94.5t
94.5t - 65.1t = 22*65.1/60
29.4t = 23.87
t = 23.87 / 29.4
t = 0.812 hours
distance = 94.5 km/h * 0.812 h = 76.7 km
Answers: 1) 0.81 hours, 2) 76.7 km
Explanation:
It doesn't depends upon other.
It have it's own identity.
It's a lot easier to measure temperature than to measure the motion of component particles.
Answer:
a.
b.1058 N
Explanation:
We are given that
Mass of each dog,M=18.5 kg
Mass of sled with rider,m=250 kg
a.Average force,F=185 N


By Newton's second law



b.By Newton's second law

Substitute the values

Hence, the force in the coupling between the dogs and the sled=1058 N
Answer:
2.23 × 10^6 g of F- must be added to the cylindrical reservoir in order to obtain a drinking water with a concentration of 0.8ppm of F-
Explanation:
Here are the steps of how to arrive at the answer:
The volume of a cylinder = ((pi)D²/4) × H
Where D = diameter of the cylindrical reservoir = 2.02 × 10^2m
H = Height of the reservoir = 87.32m
Therefore volume of cylindrical reservoir = (3.142×202²/4)m² × 87.32m = 2798740.647m³
1ppm = 1g/m³
0.8ppm = 0.8 × 1g/m³
= 0.8g/m³
Therefore to obtain drinking water of concentration 0.8g/m³ in a reservoir of volume 2798740.647m³, F- of mass = 0.8g/m³ × 2798740.647m³ = 2.23 × 10^6 g must be added to the tank.
Thank you for reading.