Can enter through any cavity lined with mucus membranes

Consider the system shown in fig. 6-26. the rope and pulley have negligible mass, and the pulley is frictionless. the coefficien

Anestetic [448]

Need and answer choice if you have one

4 0

3 years ago

A 95kg clock initially at rest on a horizontal floor requires a 560N horizontal force to set it in motion. After the clock is in

miv72 [106K]

Complete Question

A 95 kg clock initially at rest on a horizontal floor requires a 650 N horizontal force to set it in motion. After the clock is in motion, a horizontal force of 560 N keeps it moving with a constant velocity. Find the coefficient of static friction and the coefficient of kinetic friction.

Answer:

The value for static friction is $\mu_s = 0.60$

The value for static friction is $\mu_k = 0.70$

Explanation:

From the question we are told that

The mass of the clock is $m = 95 \ kg$

The first horizontal force is $F _s = 560 \ N$

The second horizontal force is $F _k = 650 \ N$

Generally the static frictional force is equal to the first horizontal force

So

$F _s = m * g * \mu_s$

=> $560 = 95 * 9.8 * \mu_s$

=> $\mu_s = 0.60$

Generally the kinetic frictional force is equal to the second horizontal force

So

$F _k = m * g * \mu_k$

$650 = 95 * 9.8 * \mu_k$

$\mu_k = 0.70$

3 0

3 years ago

A ball is dropped off the side of a bridge.<br> After 1.55 s, how far has it fallen?

KengaRu [80]

Answer:11.7 meters

Explanation: Gravitational acceleration (g)

9.8 m/s²

Initial velocity (v₀)

0 ft/s

Height (h)

11.77225 m

Time of fall (t)

1.55 sec

Velocity (v)

15.19 m/s

6 0

3 years ago

Three beads are placed along a thin rod. The first bead, of mass m1 = 23 g, is placed a distance d1 = 1.1 cm from the left end o

Mila [183]

Answer:

a) $x=\frac{m_{1}d_{1}+m_{2}(d_{1}+d_{2})+m_{3}(d_{1}+d_{2}+d_{3} ) }{m_{1}+m_{2}+m_{3} }$

b) x = 4.47 cm

c) $x=\frac{m_{1}d_{2}+m_{2}(0)+m_{3}d_{3} }{m_{1}+m_{2}+m_{3} }$

d) x = 1.48 cm

Explanation:

a) The center of mass is equal to:

$x=\frac{m_{1}x_{1}+m_{2}x_{2}+m_{3}x_{3} }{m_{1}+m_{2} +m_{3}}$

Where m is the mass of beads and x is the distances, if x₁ = d₁, x₂ = d₂ and x₃ = d₃

$x=\frac{m_{1}d_{1}+m_{2}(d_{1}+d_{2})+m_{3}(d_{1}+d_{2}+d_{3} ) }{m_{1}+m_{2}+m_{3} }$

b) If

m₁ = 23g

m₂ = 15 g

m₃ = 58 g

d₁ = 1.1 cm

d₂ = 1.9 cm

d₃ = 3.2 cm

$x=\frac{23*1.1+15*(1.1+1.9)+58(1.1+1.9+3.2) }{23+15+58 } =4.47cm$

c) The center of the mass of the beads realtive to the center of bead is:

$x=\frac{m_{1}d_{2}+m_{2}(0)+m_{3}d_{3} }{m_{1}+m_{2}+m_{3} }$

d) $x=\frac{23*(-1.9)+(15*0)+(58*3.2) }{23+15+58 } =1.48cm$

6 0

3 years ago

Read 2 more answers

Compute the velocity of an electron that has been accelerated through a difference of potential of 100 volts. express your answe

Elodia [21]

The velocity of an electron that has been accelerated through a difference of potential of 100 volts will be 5.93 * $10^{6}$ m/s

Electrons move because they get pushed by some external force. There are several energy sources that can force electrons to move. Voltage is the amount of push or pressure that is being applied to the electrons.

By conservation of energy, the kinetic energy has to equal the change in potential energy, so KE=q*V. The energy of the electron in electron-volts is numerically the same as the voltage between the plates.

given

charge of electron = 1.6 × $10^{-19}$ C