5.972 × 10^24 kg
it is the weight of earth
hope it is helpful to you
Answer:
The resultant velocity is <u>169.71 km/h at angle of 45° measured clockwise with the x-axis</u> or the east-west line.
Explanation:
Considering west direction along negative x-axis and north direction along positive y-axis
Given:
The car travels at a speed of 120 km/h in the west direction.
The car then travels at the same speed in the north direction.
Now, considering the given directions, the velocities are given as:
Velocity in west direction is, 
Velocity in north direction is, 
Now, since 
are perpendicular to each other, their resultant magnitude is given as:
Plug in the given values and solve for the magnitude of the resultant.This gives,
Let the angle made by the resultant be 'x' degree with the east-west line or the x-axis.
So, the direction is given as:
Therefore, the resultant velocity is 169.71 km/h at angle of 45° measured clockwise with the x-axis or the east-west line.
12.00 min = 0.2 hr
8.00 min = 0.15 hr
Total distance:
(10.0 km/hr) (0.2 hr) + (15.0 km/hr) (0.15 hr) + (20.0 km/hr) (0.2 hr)
= 8.25 km
Average speed:
(10.0 km/hr + 15.0 km/hr + 20.0 km/hr) / 3
= 15 km/hr
Change in position:
(10.0 km/hr) (0.2 hr) + (15.0 km/hr) (0.15 hr) - (20.0 km/hr) (0.2 hr)
= 0.25 km
Average velocity:
(10.0 km/hr + 15.0 km/hr - 20.0 km/hr) / 3
≈ 1.67 m/s
The direction of a vector multiplied by a scalar is only affected if the scalar is negative, in which case the vector will now be in the opposite direction. If the scalar is positive, the vector will only change in magnitude
The question is incomplete. I can help you by adding the information missing. They want you to calculate a) the radius of the cyclotron orbit for an electron with speed 1.0 * 10^6 m/s^2 and b) the radius of a cyclotron orbit for a proton with speed 5.0 * 10^4 m/s.
The two tasks involve combining the equations of the magnectic force and the centripetal force in a circular motion.
When you do that, you will obtain an expression to find the radius of the circular motion, which is the radius of the cyclotron that impulses the particles.
a)
Magentic force, F = q*v*B
q is the charge of the electron = 1.6 * 10^ -19 C
v is the speed = 1.0 * 10 ^ 6 m/s
B is the magentic field = 5.0 * 10 ^-5 T
Centripetal force, F = m*Ac = m * v^2 / R
where,
Ac = centripetal acceleration
m = mass of the electron = 9.11 * 10 ^-31 kg
R = the radius of the orbit
Now equal the two forces: q*v*B = m * v^2 / R => R = m*v / (q*B)
=> R = (9.11 * 10^31 kg) (1.0*10^6m/s) / [ (1.6 * 10^-19C)* (5.0 * 10^-5T) ]
=> R = 0.114 m
b) The equations are the same, just now use the speed, charge and mass of the proton instead of those of the electron.
R = m*v / (qB) = (1.66*10^-27 kg)(5.0*10^4 m/s) / [(1.6*10^-19C)(5*10^-5T)]
=> R = 10.4 m
