Answer:
A.) 12.5 J
B.) 12.5 J
C.) 7.1 m/s
Explanation:
Given that a 0.5 kg object, initially at rest, is pulled to the right along a frictionless horizontal surface by a constant horizontal force of 25 N for a distance of 0.5m.
a. What is the work done by the force?
Work done = force × distance
Work done = 25 × 0.5
Work done = 12.5 J
b. What is the change in the kinetic energy of the block?
Work done = energy
Change in Kinetic energy = work done
Change in kinetic energy = 12.5 J
c. What is the speed of the block after the force is removed?
Kinetic energy = 1/2mV^2
12.5 = 1/2 × 0.5 × V^2
25 = 0.5V^2
V^2 = 25/0.5
V^2 = 50
V = 7.1 m/s
Answer:
Matching of Part A with Part B
Rutherford ------------ discovered proton nucleus
Bohr----------------------postulated the quantum atom
Dalton-------------------Father of atomic theory
Chadwick --------------discovery of neutron
J.J. Thomson----------discovered the electrons
hence , the matching will be
2 -- d
3 -- e
4 -- b
5 -- a
6 -- c
Answer:
Scientists who study the Sun usually divide it up into three main regions: the Sun's interior, the solar atmosphere, and the visible "surface" of the Sun which lies between the interior and the atmosphere. There are three main parts to the Sun's interior: the core, the radiative zone, and the convective zone.
Explanation:
Hopefully this helps :)
The vector c has a magnitude of 24.6m and it is in the negative y direction. Therefore

The vector b is 41.4° up from the x-axis. Therefore
![\vec{b} = b[cos(41.4^{o}) \hat{i} + sin(41.4^{o}) \hat{j} ] =b(0.75\hat{i} + 0.6613 \hat{j})](https://tex.z-dn.net/?f=%5Cvec%7Bb%7D%20%3D%20b%5Bcos%2841.4%5E%7Bo%7D%29%20%5Chat%7Bi%7D%20%2B%20sin%2841.4%5E%7Bo%7D%29%20%5Chat%7Bj%7D%20%5D%20%3Db%280.75%5Chat%7Bi%7D%20%2B%200.6613%20%5Chat%7Bj%7D%29)
The vector a is 27.7° up from the x-axis. Therefore
![\vec{a} = a[cos(22.7^{o})\hat{i} + sin(27.7^{o})\hat{j}] = a(0.8854\hat{i} + 0.4648\hat{j})](https://tex.z-dn.net/?f=%5Cvec%7Ba%7D%20%3D%20a%5Bcos%2822.7%5E%7Bo%7D%29%5Chat%7Bi%7D%20%2B%20sin%2827.7%5E%7Bo%7D%29%5Chat%7Bj%7D%5D%20%3D%20%20a%280.8854%5Chat%7Bi%7D%20%2B%200.4648%5Chat%7Bj%7D%29)
Because

, the sum of the x and y components should be zero. Therefore,
For the x-component,
0.8854a + 0.75b = 0
or
a + 0.847b = 0 (1)
For the y-component,
0.4648a + 0.6613b - 24.6 = 0
or
a + 1.4228b = 52.926 (2)
Subtract (1) from (2).
0.5758b = 52.926
b = 91.917
a = -0.847b = -77.854
Answer:
The magnitude of vector a is -77.85 m
The magnitude of vector b is 91.92 m
Explanation:
Speed: distance/time
Average speed: total distance/total time
Total distance: 400
Total time: 60
Average spead: 400/60= 6.67m/s