A sled of mass 10 kg slides along the ice. it has an initial speed of 2 m/s but stops because of friction. How much work is done
2 answers:
Answer: The correct answer is option B.
Explanation:
Mass of the sled = 10 kg
Initial speed of the sled = 2 m/s
Kinetic energy of the sled =
Work done by the sled = 20 joules
The work done by the friction will be in opposite direction and equal to the magnitude of the work done of the sled that - 20 J.
Hence, correct answer is option B.
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Answer:
Find the attachments for complete solution
Answer:
2.13 x 10^-19 J or 0.53 eV
Explanation:
cut off wavelength, λo = 700 nm = 700 x 10^-9 m
λ = 400 nm = 400 x 10^-9 m
Use the energy equation
Where, K be the work function
K = 2.13 x 10^-19 J
K = 0.53 eV
Answer:
tension wire 104 N, horizontal force hinge 104 N, vertical force hinge 63.7 N
Explanation:
The system described is in static equilibrium under the action of several forces, which are shown in the attached diagram, where T is the tension of the wire, W is the weight of the bar, Fx is the horizontal reaction of the wall and Fy It is the vertical reaction of the wall.
The directions of the forces are indicated by the arrows and are marked intuitively, but when solving the problem if one gives a negative value this indicates that the direction is wrong, but does not alter the results of the problem
For the resolution we use Newton's second law, both in translational and rotational equilibrium, if necessary.
We must establish a reference system to assign the positive meaning, we place it with the origin in the hinge, and the positive directions to the right and up. The location of the coordinate system allows us to eliminate the reaction of the hinge by having zero distance to origin. We write the equilibrium equations for each axis
∑ Fx = 0
Fx -T =0
∑ Fy =0
Fy -W =O W = m g
∑τ =0
-T dy + W dx =0
For rotational equilibrium we take the positive direction as counterclockwise rotation and distance is the perpendicular distance of the force to the axis of the coordinate system
dy = L sin 17
dx = (L/2) cos 17
Where L is the length of the bar and the weight is applied at the center of it, we write and simplify the equation
-T L sin 17 + mg (L / 2) cos 17 = 0
- T sin 17 + mg/2 Cos 17 =0
We write the three equations together
Fx- T = 0
Fy -W = 0
-T sin 17 + (m g / 2) cos 17 = 0
a) With the third equation we can find the wire tension
T = m g / 2 cos 17 / sin17
T = 6.5 9.8/2 cotan 17
T = 104 N
b) We use the first equation to find Fx
Fx- T =0
Fx = T = 104 N
c) We use the second equation to find Fy
Fy = W = m g
Fy = 6.5 9.8
Fy = 63.7 N
