Answer: a) the greater speed for the ball is getting with the large radius of the circle. b) 1.68* 10 ^3 m/s^2 c) 1.25*10^3 m/s^2
Explanation: In order to solve this problem firstly we have to consider that speed in a of the circular movement is directly the angular rotation multiply the radius of the circle so by this we found that the second radius get large speed.
Secondly to calculate the centripetal acceleration for the ball we have to considerer the relationship given by:
acceleration in a circular movement= ω^2*r
so
a1= (8.44 *2*π)^2*r1=1.68 *10^3 m/s^2
a2= (5.95*2*π)^2*r2=1.25*10^3 m/s^2
Answer:
Explanation:
a )
Reaction force of the ground
R = mg
= 160 N
Maximum friction force possible
= μ x R
= μ x 160
= .4 x 160
= 64 N .
b )
160 N will act at middle point . 740N will act at distance of 3 / 5 m from the wall ,
Taking moment about top point of ladder
160 x 1.5 + 740 x 3/5 + f x 4 = 900 x 3
240 + 444 + 4f = 2700
f = 504 N
c )
Let x be the required distance.
Taking moment about top point of ladder
160 x 1.5 + 740 x 3 x / 5 + .4 x 900 x 4 = 900 x 3 ( .4 x 900 is the maximum friction possible )
240 + 444 x + 1440 = 2700
x = 2.3 m
so man can go upto 2.3 at which maximum friction acts .
Answer:
The shortest distance in which you can stop the automobile by locking the brakes is 53.64 m
Explanation:
Given;
coefficient of kinetic friction, μ = 0.84
speed of the automobile, u = 29.0 m/s
To determine the the shortest distance in which you can stop an automobile by locking the brakes, we apply the following equation;
v² = u² + 2ax
where;
v is the final velocity
u is the initial velocity
a is the acceleration
x is the shortest distance
First we determine a;
From Newton's second law of motion
∑F = ma
F is the kinetic friction that opposes the motion of the car
-Fk = ma
but, -Fk = -μN
-μN = ma
-μmg = ma
-μg = a
- 0.8 x 9.8 = a
-7.84 m/s² = a
Now, substitute in the value of a in the equation above
v² = u² + 2ax
when the automobile stops, the final velocity, v = 0
0 = 29² + 2(-7.84)x
0 = 841 - 15.68x
15.68x = 841
x = 841 / 15.68
x = 53.64 m
Thus, the shortest distance in which you can stop the automobile by locking the brakes is 53.64 m
The mass of the astronaut is still 65 kilograms. Mass is constant or doesn't change no matter where you are.
Answer:
1.t=-1.96sec
2.H=4.8m
3.T=1.96sec
4.R=19.2m
Explanation:
u=9.8,t=?,sin theta=1
using formula t=2usintheta/g
t=2x9.8x1=19.6/10
t=1.96seconds
using formula H=u(squared)sin(squared)theta/2g
H=9.8(squared)x1(squared)/2x10
H=96x1/20
H=96/20
H=4.8m
using formula T=2usintheta/g
T=2x9.8x1/10
T=19.6/10
T=1.96sec
using the formula R=u(squared)sin2theta/g
R=9.8(squared)x2/10
R=96x2/10
R=192.08/10
R=19.2
