What is the difference between mass and weight?

As a new electrical technician, you are designing a large solenoid to produce a uniform 0.170 T magnetic field near the center o

MrRa [10]

Answer:

18.6012339739 A

Explanation:

$\mu_0$ = Vacuum permeability = $4\pi \times 10^{-7}\ H/m$

L = Length of wire = 55 cm

N = Number of turns = 4000

I = Current

Magnetic field is given by

$B=\dfrac{\mu_0NI}{L}\\\Rightarrow I=\dfrac{BL}{\mu_0N}\\\Rightarrow I=\dfrac{0.17\times 0.55}{4\pi \times 10^{-7}\times 4000}\\\Rightarrow I=18.6012339739\ A$

The current necessary to produce this field is 18.6012339739 A

7 0

3 years ago

An ice cube at 0c was dropped into 30.0 g of water in a cup at 45.0c. at the instant that all of the ice was melted, the tempera

Ede4ka [16]

The amount of heat given by the water to the block of ice can be calculated by using
$Q=m_w C_{sw} \Delta T_w$
where
$m_w = 30 g$ is the mass of the water
$C_{sw}=4.18 J/(g ^{\circ}C)$ is the specific heat capacity of water
$\Delta T_w = 45.0^{\circ}-19.5^{\circ}C = 20.5^{\circ}C$ is the variation of temperature of the water.

Using these numbers, we find
$Q=(30 g)(4.18 J/(g^{\circ}C))(20.5^{\circ}C)=2571 J$

This is the amount of heat released by the water, but this is exactly equal to the amount of heat absorbed by the ice, used to melt it into water according to the formula:
$Q = m_i L_f$
where $m_i$ is the mass of the ice while $L_f =334 J/g$ is the specific latent heat of fusion of the ice.
Re-arranging this formula and using the heat Q that we found previously, we can calculate the mass of the ice:
$m_i = \frac{Q}{L_f}= \frac{2571 J}{334 J/g} =7.7 g$

3 0

2 years ago

A cylindrical container with a cross-sectional area of 66.2 cm2 holds a fluid of density 856 kg/m3 . At the bottom of the contai

Ugo [173]

Answer:

A. h = 2.15 m

B. Pb' = 122 KPa

Explanation:

The computation is shown below:

a) Let us assume the depth be h

As we know that

$Pb - Pat = d \times g \times h \\\\ ( 119 - 101) \times 10^3 = 856 \times 9.8 \times h$

After solving this,

h = 2.15 m

Therefore the depth of the fluid is 2.15 m

b)

Given that

height of the extra fluid is

$h' = \frac{2.35 \times 10^{-3}}{ area} \\\\ h' = \frac{2.35 \times 10^{-3}} { 66.2 \times 10^{-4}}$

h' = 0.355 m

Now let us assume the pressure at the bottom is Pb'

so, the equation would be

$Pb' - Pat = d \times g \times (h + h')\\\\Pb' = 856 \times 9.8 \times ( 2.15 + 0.355) + 101000$

Pb' = 122 KPa

3 0

3 years ago

Which describes a positive or negative net force acting on an obiect? Check all that apply.

Murrr4er [49]

Answer:

It can cause an object to accelerate.

It can cause an object to stop moving.

It can cause an object to start moving.

It can cause an object to change directions.

Explanation:

When the velocity of an object is increased in the same direction, the object is said to have positive acceleration. If it increases its velocity in a direction that is opposite to the original direction, it is negative acceleration.

When an object that's already moving is made to stop, it is said to have decelerated. Deceleration is negative acceleration.

When an object at rest is made to move by applying a force, it is said to have accelerated to some final velocity, during its motion for some duration.

An object at rest will remain at rest is said to have no net force acting on it.

4 0

2 years ago

Show that Ns is the same as kgmls.<br>

Minchanka [31]

1 N = 1kgm/s²

then,

1Ns = 1kgm/s² * s

1Ns = 1kgm/s

Hence shown.

3 0

2 years ago