Answer:
Part 1
20 N
Part 2
0.4 m/s²
Part 3
4 m/s
Explanation:
The force which pulls the sled right = 50 N
The friction force exterted towards left by the snow = -30 N
The mass of the sled = 50 kg
Part 1
The sum of the forces on the sled, F = 50 N + (-30) N = 20 N
Part 2
The acceleration of the sled is given as follows;
F = m·a
Where;
m = The mass of the sled
a = The accelertion
a = F/m
∴ a = (20 N)/(50 kg) = 0.4 m/s²
The acceleration of the sled, a = 0.4 m/s²
Part 3
The initial velocity of the sled, u = 2 m/s
The kinematic equation of motion to determine the speed of the sled is v = u + a·t
The speed, <em>v</em>, of the sled after t = 5 seconds is therefore;
v = 2 m/s + 0.4 m/s² × 5 s = 4 m/s.
The relative velocity of the athlete relative to the ground is 5.2 m/s
The given parameters;
constant velocity of the athlete, V = 5.2 m/s
let the velocity of the ground = Vg = 0
The relative velocity concept helps us to determine the velocity of a moving object relative to a stationary observer.
The athlete is the moving object in this question while the ground is stationary.
The relative velocity of the athlete relative to the ground is calculated as follows;

Thus, the relative velocity of the athlete relative to the ground is 5.2 m/s
Learn more here: brainly.com/question/24430414
Answer:
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Explanation:hope this helps
Answer:
The density of the metal is 5200 kg/m³.
Explanation:
Given that,
Weight in air= 0.10400 N
Weight in water = 0.08400 N
We need to calculate the density of metal
Let
be the density of metal and
be the density of water is 1000kg/m³.
V is volume of solid.
The weight of metal in air is



.....(I)
The weight of metal in water is
Using buoyancy force


We know that,
....(I)
Put the value of
in equation (I)

Put the value of Vg in equation (II)



Hence, The density of the metal is 5200 kg/m³.