The total distance traveled by the girl will be 40 meters and the displacement will be 30 meters.
Distance = total amount of path followed by any body called distance.
Displacement= The minimum distance between two points irrespective to the path followed by the body is displacement.
It is given that
Distance traveled to north direction = 10 metres
Distance traveled to south direction= 30 metres
So the path followed by the girl in total = North direction + south direction
= 10+30
= 40 meters
And the displacement = min distance between History class and fountain
= 30 meters
total distance traveled by the girl will be 40 meters and the displacement will be 30 meters.
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b. digestion
c. polymerization
d. exchange of ions between two compounds
The box has 3 forces acting on it:
• its own weight (magnitude w, pointing downward)
• the normal force of the incline on the box (mag. n, pointing upward perpendicular to the incline)
• friction (mag. f, opposing the box's slide down the incline and parallel to the incline)
Decompose each force into components acting parallel or perpendicular to the incline. (Consult the attached free body diagram.) The normal and friction forces are ready to be used, so that just leaves the weight. If we take the direction in which the box is sliding to be the positive parallel direction, then by Newton's second law, we have
• net parallel force:
∑ F = -f + w sin(35°) = m a
• net perpendicular force:
∑ F = n - w cos(35°) = 0
Solve the net perpendicular force equation for the normal force:
n = w cos(35°)
n = (15 kg) (9.8 m/s²) cos(35°)
n ≈ 120 N
Solve for the mag. of friction:
f = µn
f = 0.25 (120 N)
f ≈ 30 N
Solve the net parallel force equation for the acceleration:
-30 N + (15 kg) (9.8 m/s²) sin(35°) = (15 kg) a
a ≈ (54.3157 N) / (15 kg)
a ≈ 3.6 m/s²
Now solve for the block's speed v given that it starts at rest, with v₀ = 0, and slides down the incline a distance of ∆x = 3 m:
v² - v₀² = 2 a ∆x
v² = 2 (3.6 m/s²) (3 m)
v = √(21.7263 m²/s²)
v ≈ 4.7 m/s
Answer:
562.5
Explanation:
the answer is 11 km/h ... i just took the test
Answer:
0.276
Explanation: Here we have to consider the kinetic friction force acting on the body. If it is going at a constant speed according to newton's 1 st law. There was'not net force induced at the system. There for kinetic friction foce must be equal to the 50N .
But here we have been asked to get cofficient of static friction . There for you have to get static friction force. It should be 65N.
Mass of the object is 24. Then the reaction between surface and the object would be 24*9.8 = 235.2 N
There for using this,
Static friction force = Cofficient of static friction * Reaction
Cofficient of static friction = 65/235.2
= 0.276
Answer:
Explanation:
It is given that,
Mass of the crate, m = 24 kg
Force acting on the crate when it is at rest,
When the crate is in motion, force acting on the crate,
To find,
The coefficient of static friction between crate and floor
Solution,
When the crate is at rest, the force acting on the crate is given by :
When the crate is in motion, the force acting on the crate is given by :
Hence, this is the required solution.