Answer:
1. 1 s = 1 x 10⁶ μs
2. 1 g = 0.001 kg
3. 1 km = 1000 m
4. 1 mm = 1 x 10⁻³ m
5. 1 mL = 1 x 10⁻³ L
6. 1 g = 100 dg
7. 1 cm = 1 x 10⁻² m
8. 1 ms = 1 x 10⁻³ s
Explanation:
1.
1 x 10⁻⁶ s = 1 μs
(1 x 10⁻⁶ x 10⁶) s = 1 x 10⁶ μs
<u>1 s = 1 x 10⁶ μs</u>
2.
1000 g = 1 kg
1 g = 1/1000 kg
<u>1 g = 0.001 kg</u>
3.
<u>1 km = 1000 m</u>
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4.
<u>1 mm = 1 x 10⁻³ m</u>
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5.
<u>1 mL = 1 x 10⁻³ L</u>
<u></u>
6.
1 x 10⁻² g = 1 dg
(1 x 10⁻² x 10²) g = 1 x 10² dg
<u>1 g = 100 dg</u>
<u></u>
7.
<u>1 cm = 1 x 10⁻² m</u>
<u></u>
8.
<u>1 ms = 1 x 10⁻³ s</u>
Answer:
a) 
b) the motorcycle travels 155 m
Explanation:
Let 
, then consider the equation of motion for the motorcycle (accelerated) and for the car (non accelerated):
where:
is the speed of the motorcycle at time 2
is the velocity of the car (constant)
is the velocity of the car and the motorcycle at time 1
d is the distance between the car and the motorcycle at time 1
x is the distance traveled by the car between time 1 and time 2
Solving the system of equations:
![\left[\begin{array}{cc}car&motorcycle\\x=v_0\Delta{t}&x+d=(\frac{v_0+v_{m2}}{2}}) \Delta{t}\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Dcar%26motorcycle%5C%5Cx%3Dv_0%5CDelta%7Bt%7D%26x%2Bd%3D%28%5Cfrac%7Bv_0%2Bv_%7Bm2%7D%7D%7B2%7D%7D%29%20%5CDelta%7Bt%7D%5Cend%7Barray%7D%5Cright%5D)
For the second part, we need to calculate x+d, so you can use the equation of the car to calculate x:
Answer:
a) The net force acting on the sleigh is zero because it is moving at a constant speed.
b) 
, then 
.
It's either staying there or is going at the same pace
Answer:
The torque applied by the drill bit is T = 16.2 Nm and the cutting force of the drill bit is F = 33 N.
Explanation:
Given:-
- The diameter of the drill bit, d = 98 cm
- The power at which drill works, P = 5.85 hp
- The rotational speed of drill, N = 1900 rpm
Find:-
What Torque And Force Is Applied To The Drill Bit?
Solution:-
- The amount of torque (T) generated at the periphery of the cutting edges of the drilling bit when it is driven at a power of (P) horsepower at some rotational speed (N).
- The relation between these quantities is given:
T = 5252*P / N
T = 5252*5.85 / 1900
T = 16.171 Nm
- The force (F) applied at the periphery of the drill bit cutting edge at a distance of radius from the center of drill bit can be determined from the definition of Torque (T) being a cross product of the Force (F) and a moment arm (r):
T = F*r
Where, r = d / 2
F = 2T / d
F = 2*16.171 / 0.98
F = 33 N
Answer: The torque applied by the drill bit is T = 16.2 Nm and the cutting force of the drill bit is F = 33 N.
