Ask question

Ask question

All categories

3 years ago

9

A 200 kg wood crate sits in the back of a truck. The coefficients of friction between the crate and the truck are ????s = 0.9 an

d ????k = 0.5. The truck starts off up a 20° slope. What is the maximum acceleration the truck can have without the crate slipping out the back?

1 answer:

Xelga [282]3 years ago

4 0

Answer:

Maximum acceleration of the truck is 5.25 m/s^2

Explanation:

To find the maximum acceleration the truck first we need to calculate friction between truck and wood. Because of wood is not moving, coefficient of c_{s} need to be used for it. Then the formula will be:

$F_{s}=c_{s}*200*9.8*cos(20)\\F_{s}=0.9*200*9.8*0.94\\F_{s}=1657.62$

1657.62 Newton is the friction.

So force against friction need to be at most 1657.62 N. Then equation will be:

$F_{s}=F_{a}+F_{m}\\1657.62=200*9.8*sin(20)+200*a*cos(20)\\1657.62=670.36+187.94*a\\987.26=187.94*a\\a=5.25$

Maximum acceleration of the truck need to be 5.25 m/s^2

You might be interested in

If the relative density of a liquid is 0.34 what is the density of the liquid

Elina [12.6K]

Answer:

<h2>3338.98 kg/m³</h2>

Explanation:

The formula for calculating the relative density of a substance is expressed as

Relative density of a liquid = Density of the liquid /density of water

Given relative density of a liquid = 0.34

Density of water 997kg/m³

Substituting into the formula we have;

Density of the liquid = Relative density of a liquid * density of water

Density of the liquid = 0.34 * 997

Density of the liquid = 3338.98 kg/m³

7 0

3 years ago

Red blood cells can be modeled as spheres of 6.53 μm diameter with −2.55×10−12 C excess charge uniformly distributed over the su

yKpoI14uk [10]

Complete Question

Red blood cells can be modeled as spheres of 6.53 μm diameter with −2.55×10−12 C excess charge uniformly distributed over the surface. Find the electric field at the following locations, with radially outward defined as the positive direction and radially inward defined as the negative direction. The permittivity of free space ????0 is 8.85×10−12 C/(V⋅m). What is the electric field

E⃗ 1 inside the cell at a distance of 3.05 μm from the center?

E⃗ 2 Just inside the surface of the cell

E⃗ 3 Just outside the surface of the cell

E⃗ 4 At a point outside the cell 3.05 μm from the surface

Answer:

E⃗ 1

0 V/m

E⃗ 2

0 V/m

E⃗ 3

$E_3 = 2.153 *10^{9} \ V/m$

E⃗ 4

$E_4 = 5.754 *10^ {8} \ V/m$

Explanation:

From the question we are told that

The diameter is $d = 6.53 \mu m = 6.53*10^{-6}\ m$

The charge is $Q = -.2.55 *10^{-12} \ C$

The permittivity of free space is $\epsilon_o = 8.85* 10^{-12}\ C / V.m$

The distance considered is $d = 3.05 \mu m = 3.05 *10^{-6} \ m$

Generally the electric field inside the cell at a distance of 3.05 μm from the center is

0 V/m

This because there is no electric field felt inside the cell according Gauss the cell is taken as a point charge

Generally the electric field just inside the surface of the cell is 0 V/m

This because there is no electric field felt inside the cell according Gauss the cell is taken as a point charge

Generally the electric field just outside the cell is mathematically represented as

$E_3 = \frac{ k * |Q|}{ r^2 }$

Here $k$ is the coulomb constant with value

$k = 9*10^{9}\ kg\cdot m^3\cdot s^{-4} \cdot A^{-2}$

r is the radius of the sphere which is mathematically as

$r = \frac{d}{2} = \frac{6.53*10^{-6}}{2} = 3.265 *10^{-6} \ m$

$E_3 = \frac{ 9*10^{9} * |-2.55 *10^{-12} |}{ [3.265 *10^{-6} ]^2 }$

$E_3 = 2.153 *10^{9} \ V/m$

Generally the electric field at a point outside the cell 3.05 μm from the surface is mathematically represented as

$E_4 = \frac{ k * |Q|}{ R^2 }$

Here R is mathematically represented as

$R = 3.265 *10^{-6} + 3.05 *10^{-6}$

=> $R = 6.315 *10^{-6}$

So

$E_4 = \frac{ 9*10^{9} * |-2.55 *10^{-12} |}{ [ 6.315 *10^{-6} ]^2 }$

$E_4 = 5.754 *10^ {8} \ V/m$

3 0

3 years ago

Answer these questions plz

lozanna [386]

1. The property of a conductor by virtue of which it posses the flow of electric current through it is called resistance.

2. The resistance of a conductor depends on the cross sectional area of the conductor and it's resistivity.

3.This id due to the fact that the resistance of a wire is inversely proportional to the square of its diameter.

4.Due to at high temperatures , the alloy donot oxidize. Alloy doesn't melt readily and get deformed.

8 0

3 years ago

A single conservative force Fx = (6.0x  12) N (x is in m) acts on a particle moving along the x axis. The potential energy asso

jasenka [17]

Answer:

$U(3)=-43J$

Explanation:

Potential energy is minus the integral of Fdx. Doing the integration yields:

$U=\int\limits {6.0x+12}\, dx$

$U=-3x^2-12x+C$

$U(0)=20J$

so

$U(0)=-3(0)^2+12(0)+C=20$

$C=20$

Now for x=3.0m

$U(3)=-3*(3)^2-12(3)+20$

$U(3)=-43J$

6 0

3 years ago

Which statement describes a chemical property related to the particles that make up a material

JulijaS [17]

Answer:

Salt crystals are hard and brittle because they are made up of atoms that combine by ionic bonding.

Explanation:

4 0

2 years ago