Torque = r x F
|F| = mg = 60 * 10 N = 600 N ( assuming g ~ 10m/s^2)
distance of fulcrum = torque / Force = 90/600 m = .15 m.
Answer:
207.4 N
Explanation:
The torque
on a body is
where r is the radius vector from the point of rotation to the point at which force F is applied.
The product of r and F is equal to the product of magnitude of r and F multiplied by the sine of angle between both vectors.
Therefore, torque is also given by
Where
is the angle between r and F.
Use the expression of torque.
Substitute L for r in the equation
Where L is the length of the wrench.
Making F the subject
Force required to pull the wrench is given as,
Substitute
for
, 25 cm for L, and 115o for
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Answer:
μ = mg/kx
Explanation:
Since the bock does not slip, the frictional force equals the weight of the block. So, F = mg. Now, the frictional force, F = μN where μ = coefficient of static friction and N = Normal force.
Now, the normal force equals the spring force F' = kx where k = spring constant and x = compression of spring.
N = F' = kx
So, F = μN = μkx
μkx = mg
So, μ = mg/kx
Answer:
3.135 kN/C
Explanation:
The electric field on the axis of a charged ring with radius R and distance z from the axis is E = qz/{4πε₀[√(z² + R²)]³}
Given that R = 58 cm = 0.58 m, z = 116 cm = 1.16m, q = total charge on ring = λl where λ = charge density on ring = 180 nC/m = 180 × 10⁻⁹ C/m and l = length of ring = 2πR. So q = λl = λ2πR = 180 × 10⁻⁹ C/m × 2π(0.58 m) = 208.8π × 10⁻⁹ C and ε₀ = permittivity of free space = 8.854 × 10⁻¹² F/m
So, E = qz/{4πε₀[√(z² + R²)]³}
E = 208.8π × 10⁻⁹ C × 1.16 m/{4π8.854 × 10⁻¹² F/m[√((1.16 m)² + (0.58 m)²)]³}
E = 242.208 × 10⁻⁹ Cm/{35.416 × 10⁻¹² F/m[√(1.3456 m² + 0.3364 m²)]³}
E = 242.208 × 10⁻⁹ Cm/35.416 × 10⁻¹² F/m[√(1.682 m²)]³}
E = 6.839 × 10³ Cm²/[1.297 m]³F
E = 6.839 × 10³ Cm²/2.182 m³F
E = 3.135 × 10³ V/m
E = 3.135 × 10³ N/C
E = 3.135 kN/C