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Dahasolnce [82]
3 years ago
5

Your cousin is moving into an apartment in San Francisco. This apartment is on a street that is angled at 25∘ above the horizont

al. He places a box of mass 15kg on the street. The coefficients of friction between the box and the street are μs=0.45 and μk=0.25.
A: He releases the box after he places it on the ground. Will it start to slide down the sloped road?
Yes or No?
B: As the box slides down the slope, what is its acceleration?
C: What is the magnitude of the minimum force he would have to apply to the box to keep it from sliding down the hill? Assume that he pushes parallel to the road and wants to hold it in place.
D: What is the magnitude of the minimum force he would have to apply to start pushing the box up the hill? Again assume that he pushes parallel to the road.
Physics
1 answer:
Vesnalui [34]3 years ago
5 0

Answer:

70

Explanation:

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The position of a particle moving along the x-axis depends on the time according to the equation x = ct2 - bt3, where x is in me
Sav [38]

Answer:

(a):  \rm meter/ second^2.

(b):  \rm meter/ second^3.

(c):  \rm 2ct-3bt^2.

(d):  \rm 2c-6bt.

(e):  \rm t=\dfrac{2c}{3b}.

Explanation:

Given, the position of the particle along the x axis is

\rm x=ct^2-bt^3.

The units of terms \rm ct^2 and \rm bt^3 should also be same as that of x, i.e., meters.

The unit of t is seconds.

(a):

Unit of \rm ct^2=meter

Therefore, unit of \rm c= meter/ second^2.

(b):

Unit of \rm bt^3=meter

Therefore, unit of \rm b= meter/ second^3.

(c):

The velocity v and the position x of a particle are related as

\rm v=\dfrac{dx}{dt}\\=\dfrac{d}{dx}(ct^2-bt^3)\\=2ct-3bt^2.

(d):

The acceleration a and the velocity v of the particle is related as

\rm a = \dfrac{dv}{dt}\\=\dfrac{d}{dt}(2ct-3bt^2)\\=2c-6bt.

(e):

The particle attains maximum x at, let's say, \rm t_o, when the following two conditions are fulfilled:

  1. \rm \left (\dfrac{dx}{dt}\right )_{t=t_o}=0.
  2. \rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}

Applying both these conditions,

\rm \left ( \dfrac{dx}{dt}\right )_{t=t_o}=0\\2ct_o-3bt_o^2=0\\t_o(2c-3bt_o)=0\\t_o=0\ \ \ \ \ or\ \ \ \ \ 2c=3bt_o\Rightarrow t_o = \dfrac{2c}{3b}.

For \rm t_o = 0,

\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}=2c-6bt_o = 2c-6\cdot 0=2c

Since, c is a positive constant therefore, for \rm t_o = 0,

\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}>0

Thus, particle does not reach its maximum value at \rm t = 0\ s.

For \rm t_o = \dfrac{2c}{3b},

\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}=2c-6bt_o = 2c-6b\cdot \dfrac{2c}{3b}=2c-4c=-2c.

Here,

\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}

Thus, the particle reach its maximum x value at time \rm t_o = \dfrac{2c}{3b}.

7 0
3 years ago
If each of the three rotor helicopter blades is 3.50 m long and has a mass of 120 kg , calculate the moment of inertia of the th
devlian [24]

Answer:

1470kgm²

Explanation:

The formula for expressing the moment of inertial is expressed as;

I = 1/3mr²

m is the mass of the body

r is the radius

Since there are three rotor blades, the moment of inertia will be;

I = 3(1/3mr²)

I = mr²

Given

m = 120kg

r = 3.50m

Required

Moment of inertia

Substitute the given values and get I

I = 120(3.50)²

I = 120(12.25)

I = 1470kgm²

Hence the moment of inertial of the three rotor blades about the axis of rotation is 1470kgm²

7 0
3 years ago
1. The illuminance on a surface is 6 lux and the surface is 4 meters from the light source. What is the intensity of the source?
Crank

<u>Answer</u>

1) A. 96 Candelas

2) A. Both of these types of lenses have the ability to produce upright images.

3) C. 5 meters


<u>Explanation</u>

Q1

The formula for calculation the luminous intensity is;

Luminous intensity = illuminance × square radius

Lv = Ev × r²

= 6 × 4²

= 6 × 16

= 96 Candelabra

Q2

For converging lenses, an upright image is formed when the object is between the lens and the principal focus while a diverging lens always forms and upright image.

A. Both of these types of lenses have the ability to produce upright images.

Q3

Luminous intensity = illuminance × square radius

square radius = Luminous intensity/ illuminance

r² = 100/4

= 25

r = √25

= 5 m




5 0
4 years ago
Read 2 more answers
a swimmer experiences a total (absolute) pressure of 117,500 pa in a pool. how far below the surface are they located?
marta [7]

Answer:

Explanation:

We know that the pressure can be calculated in the following way:

p = d·g·h

with d being the density of the water, g the gravitational acceleration and h the depth.

Also d of the water = 1000 kg/m^3 circa and g = 9.8 m/s^2 circa

117,500 Pa = 1000kg/m³ · 9.8m/s² · h

Therefore h = 11,9 m

4 0
2 years ago
An object of length 3.00 cm is inside a plastic block with index of refraction 1.40. If the object is viewed from directly above
Lilit [14]
  • The length of its image = -3cm

Given,

length of object, h_o = 3 cm

We know, for flat refracting surface,  

Image distance = object distance

So,

magnification is = -1

length of the image,

h_i = magnification * h_o\\\\h_i = -1 * 3 \\\\h_i = -3cm

Here, negative sign means inverted image.

For more information on refraction, visit

brainly.com/question/14760207?referrer=searchResults

3 0
3 years ago
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