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3 years ago

11

A 4kg block slides freely across a rough surface such that the block slows down with an acceleration of −1.25 m/s2 . What is the

coefficient of kinetic friction between the block and the surface?

1 answer:

Sophie [7]3 years ago

6 0

Answer:

The coefficient of kinetic friction between the block and the surface is 0.127.

Explanation:

Given that,

The mass of a block, m = 4 kg

The acceleration of the block, a = -1.25 m/s²

We need to find the coefficient of kinetic friction between the block and the surface. The force of friction is given by :

$F=\mu mg\\\\ma=\mu mg\\\\\mu=\dfrac{a}{g}\\\\\mu=\dfrac{1.25}{9.8}\\\\\mu=0.127$

So, the coefficient of kinetic friction between the block and the surface is 0.127.

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Two wheels roll side-by-side without sliding, at the same speed. The radius of wheel 2 is one-half (1/2) the radius of wheel 1.

Lemur [1.5K]

Answer:

w'=(1/2)w

Explanation:

In order to calculate the angular velocity of the second wheel, you use the following formula:

$\omega=\frac{v}{r}$ (1)

v: speed of the wheel 1 = speed of the wheel 2

r: radius of the wheel 1

For the second wheel you have:

r'=2r

You replace this value of r' in the following equation:

$\omega'=\frac{v}{r'}=\frac{v}{2r}=\frac{1}{2}\frac{v}{r}=\frac{1}{2}\omega\\\\\omega'=\frac{1}{2}\omega$

The angular velocity of the second wheel is one half of the angular velocity of the first wheel

6 0

3 years ago

A hydraulic car jack needs to be designed so it can lift a 2903.57 lb car assuming that a person can exert a force of 24.41 lbs.

marishachu [46]

Answer:

Diameter of the piston would be 0.71 m (71.1 cm)

Explanation:

From the principle of pressure;

$\frac{F_{1} }{A_{1} }$ = $\frac{F_{2} }{A_{2} }$

Let $F_{1}$ = 2903.57 lb, $F_{2}$ = 24.41 lbs, $r_{2}$ = 3.26 cm = 0.0326 m.

$A_{2}$ = $\pi r^{2}$

= $\frac{22}{7}$ x $(0.0326)^{2}$

= 0.00334 $m^{2}$

So that:

$\frac{2903.57}{A_{1} }$ = $\frac{24.41}{0.00334}$

$A_{1}$ = $\frac{2903.57*0.00334}{24.41}$

= 0.3973

$A_{1}$ = 0.4 $m^{2}$

The radius of the piston can be determined by:

$A_{1}$ = $\pi r^{2}$

0.3973 = $\frac{22}{7}$ x $r^{2}$

$r^{2}$ = $\frac{0.3973*7}{22}$

= 0.1264

r = $\sqrt{0.1264}$

= 0.3555

r = 0.36 m

Diameter of the piston = 2 x r

= 2 x 0.3555

= 0.711

Diameter of the piston would be 0.71 m (71.1 cm).

4 0

3 years ago

U. A hockey player takes a slap shot at a puck at rest on

faltersainse [42]

Answer:

a) 7200 ft/s²

b) 140 ft

c) 3.7 s

Explanation:

(a) Average acceleration is the change in velocity over change in time.

a_avg = Δv / Δt

We need to find what velocity the puck reached after it was hit by the hockey player.

We know it reached 40 ft/s after traveling 90 feet over rough ice at an acceleration of -20 ft/s². Therefore:

v² = v₀² + 2a(x − x₀)

(40 ft/s)² = v₀² + 2(-20 ft/s²)(100 ft − 10 ft)

v₀² = 5200 ft²/s²

v₀ = 20√13 ft/s

So the average acceleration impacted to the puck as it is struck is:

a_avg = (20√13 ft/s − 0 ft/s) / (0.01 s)

a_avg = 2000√13 ft/s²

a_avg ≈ 7200 ft/s²

(b) The distance the puck travels before stopping is:

v² = v₀² + 2a(x − x₀)

(0 ft/s)² = (5200 ft²/s²) + 2(-20 ft/s²)(x − 10 ft)

x = 140 ft

(c) The time the puck takes to travel 10 ft without friction is:

t = (10 ft) / (20√13 ft/s)

t = (√13)/26 s

The time the puck travels over the rough ice is:

v = at + v₀

(0 ft/s) = (-20 ft/s²) t + (20√13 ft/s)

t = √13 s

So the total time is:

t = (√13)/26 s + √13 s

t = (27√13)/26 s

t ≈ 3.7 s

8 0

3 years ago

Which is an outer planet?

saw5 [17]

Answer:

pluto

Explanation:

Dwarf because it is very minut

8 0

3 years ago

Read 2 more answers

An automobile with a tangential speed of 54.1 km/h follows a circular road that has a radius of 41.6 m. The automobile has a mas

Viefleur [7K]

1) Available force of friction: 6174 N

2) No

Explanation:

1)

The magnitude of the frictional force between the car's tires and the pavement of the road is given by

$F_f=\mu mg$

where

$\mu$ is the coefficient of friction

m is the mass of the car

g is the acceleration of gravity

For the car in this problem, we have:

$\mu=0.500$ (coefficient of friction)

m = 1260 kg (mass of the car)

$g=9.8 m/s^2$

Therefore, the force of friction is

$F_f=(0.500)(1260)(9.8)=6174 N$

2)

In order to mantain the car in circular motion, the force of friction must be at least equal to the centripetal force.

The centripetal force is given by

$F=m\frac{v^2}{r}$

where

m is the mass of the car

v is the tangential speed

r is the radius of the curve

In this problem, we have

m = 1260 kg

$v=54.1 km/h =15.0 m/s$ is the tangential speed

r = 41.6 m is the radius of the curve

Therefore, the centripetal force is

$F=(1260)\frac{15.0^2}{41.6}=6814 N$

Therefore, the force of friction is not enough to keep the car in the curve, since $F_f$

4 0

4 years ago