Power = (work done) / (time to do the work)
Work done = (force to lift the object) x (distance lifted)
In this question, the force is the (weight of the basket)+(your weight).
Work done = (weight of basket+you) x (3 meters)
Time to do the work = 6 seconds.
Power = (weights x 3 meters) / (6 seconds)
<em>Power = (1/2)·(weight of the basket+you, in Newtons) watts</em>
Answer:
M2 = 278.06 kg
Explanation:
We calculate the weight of M1
W=m*g
Where
m: mass (kg)
g: acceleration due to gravity (m/s²)
W₁=288* 9.8= 2822.4 N
Look at the attached graphic
We calculate the x-y components of the weight :
W₁x= 2822.4*sin41° N =1851.66 N
W₁y= 2822.4 *cos41° N = 2130.09 N
We apply Newton's first law for the balance in M1:
Σ Fy=0
Fn-W₁y=0 , Fn: normal force
Fn=W₁y=2130.09N
Friction Force = Ff=μs *Fn = 0.41*2130.09 =873.34 N
Σ Fx=0
T- W₁x- Ff=0
T= 1851.66 + 873.34
T= 1851.66 + 873.34
T=2725 N
We apply Newton's first law for the balance in M2:
Σ Fy=0
T- W₂ =0
W₂ = T = 2725 N
W₂ = M2*g
M2 = W₂/g
M2 = 2725/9.8
M2 = 278.06 kg
The solution for this problem is computed by through this formula, F = kQq / d²Plugging in the given values above, we can now compute for the answer.
F = 8.98755e9N·m²/C² * -(7e-6C)² / (0.03m)² = -489N, the negative sign denotes attraction.
The force between two parallel wires carrying currents in the same direction is attractive. It is repulsive if the currents are in opposite directions.
The position of the object at time t =2.0 s is <u>6.4 m.</u>
Velocity vₓ of a body is the rate at which the position x of the object changes with time.
Therefore,

Write an equation for x.

Substitute the equation for vₓ =2t² in the integral.

Here, the constant of integration is C and it is determined by applying initial conditions.
When t =0, x = 1. 1m

Substitute 2.0s for t.

The position of the particle at t =2.0 s is <u>6.4m</u>