What type of friction is using chalk in the summer to draw on the ground in Copley square?

Match each influence to its correct category

Pie

Answer:

healthful I think! Hope that helped

6 0

2 years ago

Read 2 more answers

In a single movable pulley, a load of 500 N is lifted by applying 300 N effort. Calculate MA, VR and efficiency.

galben [10]

Answer:

in pulley there are different kinds.but the most common one are fixed,moveable and compound pulley. in this question we asked about movable pulley.

Explanation:

Given request solutions

L=500N a,M.A=? a,M.A=L/E =5/3

E=300N b,V.R=? b,V.R=2

c,efficiency =? c,£=M.A/V.R=5/6

£=IS NOT THE REAL SYMBOLS OF EFFICIENCY BUT I IS LOOK LIKE THIS

I THINK I HELPED

6 0

3 years ago

Starting from zero, an electric current is established in a circuit made of a battery of emf E, a resistor of resistance R and a

igor_vitrenko [27]

Answer:

time constant will decrease and steady state current will decrease on increasing the resistance

Explanation:

As we know that the EMF of cell is E which is used to connected across a resistor and an inductor.

So we will have

$E - iR - L\frac{di}{dt} = 0$

here we know that

$i = \frac{E}{R}(1 - e^{-Rt/L})$

now here we have

$\tau = \frac{L}{R}$

so if we increase the value of resistance of the wire then the time constant will decrease

and hence it will take less time to reach near the steady state value

also the steady state current will be smaller in that case

8 0

3 years ago

What is the graph of the relationship between the volume of a gas at a constant pressure

Bogdan [553]

We know that P1V1 = P2V2, if there is a constant pressure, then the P1 and P2 can cancel out, so it is V1=V2 that is whats left.

6 0

3 years ago

A 2.00 kg block on a horizontal floor is attached to a horizontal spring that is initially compressed 0.0300 m . The spring has

iogann1982 [59]

Answer:

v = 0.41 m/s

Explanation:

In this case, the change in the mechanical energy, is equal to the work done by the fricition force on the block.
At any point, the total mechanical energy is the sum of the kinetic energy plus the elastic potential energy.
So, we can write the following general equation, taking the initial and final values of the energies:

$\Delta K + \Delta U = W_{ffr} (1)$

Since the block and spring start at rest, the change in the kinetic energy is just the final kinetic energy value, Kf.
⇒ Kf = 1/2*m*vf² (2)
The change in the potential energy, can be written as follows:

$\Delta U = U_{f} - U_{o} = \frac{1}{2} * k * (x_{f} ^{2} - x_{0} ^{2} ) (3)$

where k = force constant = 815 N/m

xf = final displacement of the block = 0.01 m (taking as x=0 the position

for the spring at equilibrium)

x₀ = initial displacement of the block = 0.03 m

Regarding the work done by the force of friction, it can be written as follows:

$W_{ffr} = - \mu_{k}* F_{n} * \Delta x (4)$

where μk = coefficient of kinettic friction, Fn = normal force, and Δx =

horizontal displacement.

Since the surface is horizontal, and no acceleration is present in the vertical direction, the normal force must be equal and opposite to the force due to gravity, Fg:
Fn = Fg= m*g (5)
Replacing (5) in (4), and (3) and (4) in (1), and rearranging, we get:

$\frac{1}{2} * m* v^{2} = W_{ffr} - \Delta U = W_{ffr} - (U_{f} -U_{o}) (6)$

$\frac{1}{2} * m* v^{2} = (- \mu_{k}* m*g* \Delta x) -\frac{1}{2} * k * (x_{f} ^{2} - x_{0} ^{2} ) (7)$

Replacing by the values of m, k, g, xf and x₀, in (7) and solving for v, we finally get:

$\frac{1}{2} * 2.00 kg* v^{2} = (-0.4*2.00 kg*9.8m/s2*0.02m) +( (\frac{1}{2} *815 N/m)* (0.03m)^{2} - (0.01m)^{2}) = -0.1568 J + 0.326 J (8)$

$v =\sqrt{(0.326-0.1568} = 0.41 m/s (9)$

7 0

3 years ago