What is the mass of 1 l of water? what is its weight in newtons?

A box rests on top of a flat bed truck. The box has a mass of m = 16.0 kg. The coefficient of static friction between the box an

3241004551 [841]

Answer:

1) 1.31 m/s2

2) 20.92 N

3) 8.53 m/s2

4) 1.76 m/s2

5) -8.53 m/s2

Explanation:

1) As the box does not slide, the acceleration of the box (relative to ground) is the same as acceleration of the truck, which goes from 0 to 17m/s in 13 s

$a = \frac{\Delta v}{\Delta t} = \frac{17 - 0}{13} = 1.31 m/s2$

2)According to Newton 2nd law, the static frictional force that acting on the box (so it goes along with the truck), is the product of its mass and acceleration

$F_s = am = 1.31*16 = 20.92 N$

3) Let g = 9.81 m/s2. The maximum static friction that can hold the box is the product of its static coefficient and the normal force.

$F_{\mu_s} = \mu_sN = mg\mu_s = 16*9.81*0.87 = 136.6N$

So the maximum acceleration on the block is

$a_{max} = F_{\mu_s} / m = 136.6 / 16 = 8.53 m/s^2$

4)As the box slides, it is now subjected to kinetic friction, which is

$F_{\mu_s} = mg\mu_k = 16*9.81*0.69 = 108.3 N$

So if the acceleration of the truck it at the point where the box starts to slide, the force that acting on it must be at 136.6 N too. So the horizontal net force would be 136.6 - 108.3 = 28.25N. And the acceleration is

28.25 / 16 = 1.76 m/s2

5) Same as number 3), the maximum deceleration the truck can have without the box sliding is -8.53 m/s2

3 0

3 years ago

All of the following apply to a business plan EXCEPT:

PtichkaEL [24]

- - D. It includes payroll records.

This is because it does not apple to a business plan, when all other choices do.

6 0

3 years ago

In an experiment, a rectangular block with height h is allowed to float in two separate liquids. In the first liquid, which is w

Amiraneli [1.4K]

Answer:

The relative density of the second liquid is 7.

Explanation:

From archimede's principle we know that the force that a liquid exerts on a object equals to the weight of the liquid that the object displaces.

Let us assume that the volume of the object is 'V'

Thus for the liquid in which the block is completely submerged

The buoyant force should be equal to weight of liquid

Mathematically

$F_{buoyant}=Weight\\\\\rho _{1}\times V\times g=m\times g\\\\\therefore \rho _{1}=\frac{m}{V}...............(i)$

Thus for the liquid in which the block is 1/7 submerged

The buoyant force should be equal to weight of liquid

Mathematically

$F'_{buoyant}=Weight\\\\\rho _{2}\times \frac{V}{7}\times g=m\times g\\\\\therefore \rho _{2}=\frac{7m}{V}.................(ii)$

Comparing equation 'i' and 'ii' we see that

$\rho_{2}=7\times \rho _{1}$

Since the first liquid is water thus $\rho _{1}=1gm/cm^3$

Thus the relative density of the second liquid is 7.

6 0

3 years ago

Read 2 more answers

Who is Tim peak where did he come from

Paul [167]

I believe you mean Tim Peake ? .Major Timothy Nigel "Tim" Peake CMG is a British Army Air Corps officer, European Space Agency astronaut and International Space Station crew member.

Born: April 7, 1972 (age 44), Chichester, United Kingdom

Space missions: Expedition 46, Expedition 47, Soyuz TMA-19M

Nationality: British

8 0

3 years ago

All of the following are factors affecting flow rate except what?

faust18 [17]

Answer:

c. Vessel side holes

Explanation:

The "Poiseuille formula" which is given by $\\\begin{aligned} \small Q& = \small \frac{\pi r^4}{8 \eta}.\frac{\Delta P}{\Delta L}\\\end{aligned}$ describes the volumetric flow rate ( $\small Q$ ) through tubular sections.
Here, $\Delta P,\,\, \Delta L,\,\, r,\,\, \eta$ represent the injection pressure difference, the length of the section, the radius of the section and the viscosity index of the fluid that flows through the section respectively.
With this, one can confirm that all the factors except the vessel side holes affect the flow rate.
Side holes, however, are a factor that could give a measure of how much volume would flow to a particular location. In such a situation the flow rate remains unchanged and one location would receive a lower volume (not the whole) as some volume would spill out at the side holes.

#SPJ4

5 0

1 year ago