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m = 80 kg
R = 6.4 x 10^6 m
Earth completes one rotation in 24 hours
Thus, T = 24 x 3600 s
Centripetral acceleration is given by
ac = w^2R
ac = (2 pie/T)^2R
ac = (2 pie/T)^2R = 4 pie^2/(24 x 3600)^2 x 6.4 x 10^6
ac = 0.034 m/s^2
Solution
distance travelled by Chris
\Delta t=\frac{1}{3600}hr.
X_{c}= [(\frac{21+0}{2})+(\frac{33+21}{2})+(\frac{55+47}{2})+(\frac{63+55}{2})+(\frac{70+63}{2})+(\frac{76+70}{2})+(\frac{82+76}{2})+(\frac{87+82}{2})+(\frac{91+87}{2})]\times\frac{1}{3600}
=\frac{579.5}{3600}=0.161miles
Kelly,
\Delta t=\frac{1}{3600}hr.
X_{k}=[(\frac{24+0}{2})+(\frac{3+24}{2})+(\frac{55+39}{2})+(\frac{62+55}{2})+(\frac{71+62}{2})+(\frac{79+71}{2})+(\frac{85+79}{2})+(\frac{85+92}{2})+(\frac{99+92}{2})+(\frac{103+99}{2})]\times\frac{1}{3600}
=\frac{657.5}{3600}
\Delta X=X_{k}-X_{C}=0.021miles
Answer:
24m/s²
Explanation:
Given
Distance S = 3m
Time of fall = 0.5sec
Required
Acceleration due to gravity
Using the equation of motion
S = ut+1/2gt²
Substitute the given values
3 = 0+1/2g(0.5)²
3 = 1/2(0.25)g
3 = 0.125g
g = 3/0.125
g = 24
Hence the value for the acceleration of gravity on this new planet is 24m/s²
The cyclist accelerates from 0 m/s to 9 m/s in 3 seconds with an acceleration of 3 m/s².
Answer:
Explanation:
Acceleration exerted by an object is the measure of change in speed or velocity of that object with respect to time. So the initial and final velocities play a major role in determining the acceleration of the cyclist. As here the initial velocity of the cyclist is the speed at rest and that is given as 0 m/s. Then after 3 seconds, the velocity of the cyclist changes to 9 m/s.
Then acceleration = change in velocity/Time.
Acceleration = (9-0)/3=9/3=3 m/s².
So the cyclist accelerates from 0 m/s to 9 m/s in 3 seconds with an acceleration of 3 m/s².
Answer:
Hold on Ill answer it..When do u need it by
Explanation:
