Answer:
723.54 N
Explanation:
Weight of anchor = 833 N
density of iron = 7800 mg/m^3
density of sea water = 1025 kg/m^3
According to the Archimedes principle, if a body is immersed wholly or partly in a liquid it experiences a loss in weight of the body which is equal to the weight of liquid displaced by the body.
The loss in the weight of body is equal to the buoyant force acting on the body.
The formula for the buoyant force acting on the body
= volume of body x density of liquid x acceleration due to gravity
Weight of anchor = mass x acceleration due to gravity
833 = m x 9.8
m = 85 kg
mass of anchor = Volume of anchor x density of iron
85 = V x 7800
V = 0.01089 m^3
Buoyant force on the anchor = Volume of anchor x density of sea water x g
= 0.01089 x 1025 x 9.8 = 109.46 N
So, the tension in the chain = Apparent weight of the anchor
= Weight - Buoyant force
= 833 - 109.46 = 723.54 N
Thus, the tension in the chain is 723.54 N.
Answers:
1) I saw a jogger whose velocity was about 3 m/s going straight North.
2) A car passed me at a velocity of almost 45 mles per hour, straight South.
3) I saw a dog chasing a cat. The cat was runing at about 40 km/h, but the dog barely reached 32 km/h, both 30° North of the East.
Explanation:
- <em>Velocity is a vector</em> and so you must report it using the <em>magnitude and </em>the<em> direction</em>.
- Velocity has <em>units</em> of distance (meter, kilometer, miles, among others) divided by units of time (second, minutes, hours, among others).
- The magnitude of velocity is also called <em>speed</em>, and you calculate speed (average speed) as per the formula <em>Speed = distance / time</em>.
The velocities described above have these meanings:
Velocity Magnitude (speed) units direction
3 m/s going North 3 m/s m/s North
45 m.p.h hour, South 45 miles per hour miles/h South.
40 km/h 30° North of East 40 km/h km/h 30° North of East
40 km/h 30° North of East 35 km/h km/h 30° North of East
Remeber, <em>vectors, like velocity, must be describe by its magnitude and direction.</em>
Answer:
The magnitude of the force exerted by the locomotive on the caboose is 3.08 N.
Explanation:
Given that,
Mass of a toy, m = 2.3 kg
Mass of caboose, m' = 1 kg
The frictional force of the track on the caboose is 0.48 N backward along the track.
The acceleration of train,
The magnitude of the force exerted by the locomotive on the caboose is given by :
So, the magnitude of the force exerted by the locomotive on the caboose is 3.08 N.
<span>An object is acted upon by a force of 22 newtons to the right and a force of 13 newtons to the left.
(1) 22 N to the right
(2) 13 N to the left.
magnitude = 22 - 13
magnitude = 9
Direction would be to the right.
So magnitude is 9N direction to the right.</span>
Answer:
Not between significant digits.
Explanation:
A zero not significant when it's not between significant digits.