Ask question

Ask question

All categories

2 years ago

5

A 56.6-kg crate rests on a level floor at a shipping dock. The coefficients of static and kinetic friction are 0.517 and 0.260,

respectively. What horizontal pushing force is required to (a) just start the crate moving and (b) slide the crate across the dock at a constant speed

1 answer:

kifflom [539]2 years ago

3 0

Answer:

Explanation:

Magnitude of frictional force = μ mg

μ is either static or kinetic friction.

To start the crate moving , static friction is calculated .

a ) To start crate moving , force required = μ mg where μ is coefficient of static friction .

force required =.517 x 56.6 x 9.8 = 286.76 N .

b ) to slide the crate across the dock at a constant speed , force required

= μ mg where μ is coefficient of kinetic friction , where μ is kinetic friction

= .26 x 56.6 x 9.8 = 144.21 N .

You might be interested in

You drop a steel ball bearing, with a radius of 2.40 mm, into a beaker of honey. Note that honey has a viscosity of 6.00 Pa/s an

Stells [14]

Answer:

The “terminal speed” of the ball bearing is 5.609 m/s

Explanation:

Radius of the steel ball R = 2.40 mm

Viscosity of honey η = 6.0 Pa/s

$\text { Viscosity has Density } \sigma=1360 \mathrm{kg} / \mathrm{m}^{3}$

$\text { Steel has a density } \rho=7800 \mathrm{kg} / \mathrm{m}^{3}$

$\left.\mathrm{g}=9.8 \mathrm{m} / \mathrm{s}^{2} \text { (g is referred to as the acceleration of gravity. Its value is } 9.8 \mathrm{m} / \mathrm{s}^{2} \text { on Earth }\right)$

While calculating the terminal speed in liquids where density is high the stokes law is used for viscous force and buoyant force is taken into consideration for effective weight of the object. So the expression for terminal speed (Vt)

$V_{t}=\frac{2 \mathrm{R}^{2}(\rho-\sigma) \mathrm{g}}{9 \eta}$

Substitute the given values to find "terminal speed"

$\mathrm{V}_{\mathrm{t}}=\frac{2 \times 0.0024^{2}(7800-1360) 9.8}{9 \times 6}$

$\mathrm{V}_{\mathrm{t}}=\frac{0.0048 \times 6440 \times 9.8}{54}$

$\mathrm{V}_{\mathrm{t}}=\frac{302.9376}{54}$

$\mathrm{V}_{\mathrm{t}}=5.609 \mathrm{m} / \mathrm{s}$

The “terminal speed” of the ball bearing is 5.609 m/s

7 0

3 years ago

The mean diameters of Mars and Earth are 6.9 ✕ 103 km and 1.3 ✕ 104 km, respectively. The mass of Mars is 0.11 times Earth's mas

Roman55 [17]

Answer:

(a) Ratio of mean density is 0.735

(b) Value of g on mars 0.920 $m,/sec^2$

(c) Escape velocity on earth is $3.563\times 10^4m/sec$

Explanation:

We have given radius of mars $R_{mars}=6.9\times 10^3km=6.9\times 10^6m$ and radius of earth $R_{E}=1.3\times 10^4km=1.3\times 10^7m$

Mass of earth $M_E=5.972\times 10^{24}kg$

So mass of mars $M_m=5.972\times\times 0.11 \times 10^{24}=0.657\times 10^{24}kg$

Volume of mars $V=\frac{4}{3}\pi R^3=\frac{4}{3}\times 3.14\times (6.9\times 10^6)^3=1375.357\times 10^{18}m^3$

So density of mars $d_{mars}=\frac{mass}{volume}=\frac{0.657\times 10^{24}}{1375.357\times 10^{18}}=477.69kg/m^3$

Volume of earth $V=\frac{4}{3}\pi R^3=\frac{4}{3}\times 3.14\times (1.3\times 10^7)^3=9.198\times 10^{21}m^3$

So density of earth $d_{E}=\frac{mass}{volume}=\frac{5.972\times 10^{24}}{9.198\times 10^{21}}=649.271kg/m^3$

(A) So the ratio of mean density $\frac{d_{mars}}{d_E}=\frac{477.69}{649.27}=0.735$

(B) Value of g on mars

g is given by $g=\frac{GM}{R^2}=\frac{6.67\times 10^{-11}\times0.657\times 10^{24}}{(6.9\times 10^6)^2}=0.920m/sec^2$

(c) Escape velocity is given by

$v=\sqrt{\frac{2GM}{R}}=\sqrt{\frac{2\times 6.67\times 10^{-11}\times 0.657\times 10^{24}}{6.9\times 10^6}}=3.563\times 10^4m/sec$

5 0

3 years ago

Read 2 more answers

Diffusion occurs faster in gases than in liquids because _____.

jolli1 [7]

The correct answer to go in the blank would be A) The particles are moving faster.

5 0

3 years ago

Read 2 more answers

the deepest point in the sea is 100m below sea level .what arr the water pressure at this depth and the total pressure due to wa

fomenos

The water pressure at this depth and the total pressure due to water and atmosphere are 10.3 x 10⁵ Pa and 11.31× 10⁵ Pa.

<h3>What is pressure?</h3>

The pressure is the amount of force applied per unit area.

Atmospheric pressure, Patm =1.01×10⁵ Pa

Density of water, ρ=1030 kg/m³

Depth, h=100 m

Pressure =ρgh

P = 1030×10×100

P = 10.3 x 10⁵ Pa.

Total pressure, P=Po +ρgh

P=1.01×10⁵ + 1030×10×100

P=11.31× 10⁵ Pa

Hence, total pressure is 11.31× 10⁵ Pa.

Learn more about pressure.

brainly.com/question/12971272

#SPJ1

4 0

1 year ago

A migrating robin flies due north with a speed of 12 m/s relative to the air. The air moves due east with a speed of 6.7 m/s rel

mafiozo [28]

Here it is given that speed of migrating Robin is 12 m/s relative to air

so we can say that

$\vec v_{ra} = 12 m/s$ North

so it will be

Let North direction is along Y axis and East direction is along X axis

$\vec v_{ra} = 12\hat j$

also it is given that speed of air is 6.7 m/s relative to ground

$\vec v_a = 6.7 \hat i$

now as we know by the concept of relative motion

$\vec v_{ab} = \vec v_a - \vec v_b$

$\vec v_{ra} = \vec v_r - \vec v_a$

now by rearranging the terms

$\vec v_r = \vec v_{ra} + \vec v_a$

$\vec v_r = 12 \hat j + 6.7 \hat i$

now we need to find the speed of Robin which means we need to find the magnitude of its velocity which we found above

So here we will say

$v_r = \sqrt{12^2 + 6.7^2}$

$v_r = 13.7 m/s$

so the net speed of Robin with respect to ground will be 13.7 m/s

7 0

3 years ago