I think it’s 10 kg since 20/2=10
Answer:
Explanation:
1 ) tire of radius 0.381 m rotating at 12.2 rpm
12.2 rpm = 12.2 /60 rps
n = .20333 rps
angular speed
= 2πn
= 2 x 3.14 x .20333
= 1.277 rad / s
2 ) a bowling ball of radius 12.4 cm rotating at 0.456 rad/s
angular speed = .456 rad/s
3 ) a top with a diameter of 5.09 cm spinning at 18.7∘ per second
18.7° per second = (18.7 / 180) x 3.14 rad/s
= .326 rad/s
4 )
a rock on a string being swung in a circle of radius 0.587 m with
a centripetal acceleration of 4.53 m/s2
centripetal acceleration = ω²R
ω is angular velocity and R is radius
4.53 = ω² x .587
ω = 2.78 rad / s
5 )a square, with sides 0.123 m long, rotating about its center with corners moving at a tangential speed of 0.287 m / s
The radius of the circle in which corner is moving
= .123 x √2
=.174 m
angular velocity = linear velocity / radius
.287 / .174
1.649 rad / s
The perfect order is
4 ) > 5> 1 >2>3.
<span>work = 500*3 = 1500 J
1500J in 5 sec = 300 watts
Hope that helps:)</span>
Answer:
185 N
Explanation:
Sum of forces in the x direction:
Fₓ = -(80 N cos 75°) + (120 N cos 60°)
Fₓ = 39.3 N
Sum of forces in the y direction:
Fᵧ = (80 N sin 75°) + (120 N sin 60°)
Fᵧ = 181.2 N
The magnitude of the net force is:
F = √(Fₓ² + Fᵧ²)
F = √((39.3 N)² + (181.2 N)²)
F = 185 N
The coefficient of static friction is 0.234.
Answer:
Explanation:
Frictional force is equal to the product of coefficient of friction and normal force acting on any object.
So here the mass of the object is given as 2 kg, so the normal force will be acting under the influence of acceleration due to gravity.
Normal force = mass * acceleration due to gravity
Normal force = 2 * 9.8 = 19.6 N.
And the frictional force is given as 4.6 N, then
Coefficient of static friction = 4.6 N / 19.6 N = 0.234
So the coefficient of static friction is 0.234.
