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3 years ago

7

A 45-gram object that has a volume of 9 mL has a density of

1 answer:

yanalaym [24]3 years ago

7 0

Answer:

<h2> The answer is </h2><h2>a. 5g/mL</h2>

Explanation:

Given data

mass m= 45g

volume v= 9mL

we know that density=m/v

substituting our given data we have

$density=\frac{45}{9} =5g/mL$

What is Density?

The Density of a body can be defined as the ratio of mass to volume,

or

Density, mass of a unit volume of a material substance. The formula for density is $Density= \frac{mass}{volume}$ ,

where d is density,

M is mass, and

V is volume.

Density is commonly expressed in units of grams per cubic centimetre.

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The magnetic dipole moment of Earth has magnitude 8.00 1022 J/T.Assume that this is produced by charges flowing in Earth’s molte

ziro4ka [17]

Answer:

$2.08\cdot 10^9 A$

Explanation:

The magnetic dipole moment of a circular coil with a current is given by

$\mu = IA$

where

I is the current in the coil

$A=\pi r^2$ is the area enclosed by the coil, where

$r$ is the radius of the coil

So the magnetic dipole moment can be rewritten as

$\mu = I\pi r^2$ (1)

Here we can assume that the magnetic dipole moment of Earth is produced by charges flowing in Earth’s molten outer core, so by a current flowing in a circular path of radius

$r=3500 km = 3.5\cdot 10^6 m$

Here we also know that the Earth's magnetic dipole moment is

$\mu = 8.0\cdot 10^{22} J/T$

Therefore, we can re-arrange eq (1) to find the current that the charges produced:

$I=\frac{\mu}{\pi r^2}=\frac{8.00\cdot 10^{22}}{\pi (3.5\cdot 10^6)^2}=2.08\cdot 10^9 A$

3 0

3 years ago

Check the boxes of all the TRUE statements about weight and the acceleration due to gravity.

vlada-n [284]

A is the answer because it is the answer

7 0

3 years ago

List 5 good insulators and 5 good conductors

damaskus [11]

Insulators- fur, plastic, lots of non metals
Conductors- metals, wire, most metals
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3 0

3 years ago

When an object is placed 110 cm from a diverging thin lens, its image is found to be 55 cm from the lens. The lens is removed, a

Kazeer [188]

Explanation:

Given that,

Object distance u= -110 cm

Image distance v= 55 cm

We need to calculate the focal length for diverging lens

Using formula of lens

$\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}$

Put the value into the formula

$\dfrac{1}{-f}=\dfrac{1}{55}-\dfrac{1}{-110}$

$\dfrac{1}{f}=-\dfrac{3}{110}$

$f=-36.6\ cm$

The focal length of the diverging lens is 36.6 cm.

Now given a thin lens with same magnitude of focal length 36.6 cm is replaced.

Here, The object distance is again the same.

We need to calculate the image distance for converging lens

Using formula of lens

$\dfrac{1}{36.6}=\dfrac{1}{v}-\dfrac{1}{-110}$

Here, focal length is positive for converging lens

$\dfrac{1}{v}=\dfrac{1}{36.6}-\dfrac{1}{110}$

$\dfrac{1}{v}=\dfrac{367}{20130}$

$v=54.85\ cm$

The distance of the image is 54.85 cm from converging lens.

Hence, This is the required solution.

5 0

3 years ago

A child whose weight is 287 N slides down a 7.20 m playground slide that makes an angle of 31.0Â° with the horizontal. The coeff

natulia [17]

Answer:

a

$H =212.6 \ J$

b

$v = 7.647 \ m/s$

Explanation:

From the question we are told that

The child's weight is $W_c = 287 \ N$

The length of the sliding surface of the playground is $L = 7.20 \ m$

The coefficient of friction is $\mu = 0.120$

The angle is $\theta = 31.0 ^o$

The initial speed is $u = 0.559 \ m/s$

Generally the normal force acting on the child is mathematically represented as

=> $N = mg * cos \theta$

Note $m * g = W_c$

Generally the frictional force between the slide and the child is

$F_f = \mu * mg * cos \theta$

Generally the resultant force acting on the child due to her weight and the frictional force is mathematically represented as

$F =m* g sin(\theta) - F_f$

Here F is the resultant force and it is represented as $F = ma$

=> $ma = m* g sin(31.0) - \mu * mg * cos (31.0)$

=> $a = g sin(31.0)- \mu * g * cos (31.0)$

=> $a = 9.8 * sin(31.0) - 0.120 * 9.8 * cos (31.0)$

=> $a = 4.039 \ m/s^2$

So

$F_f = 0.120 * 287 * cos (31.0)$

=> $F_f = 29.52 \ N$

Generally the heat energy generated by the frictional force which equivalent tot the workdone by the frictional force is mathematically represented as

$H = F_f * L$

=> $H = 29.52 * 7.2$

=> $H =212.6 \ J$

Generally from kinematic equation we have that

$v^2 = u^2 + 2as$

=> $v^2 = 0.559^2 + 2 * 4.039 * 7.2$

=> $v = \sqrt{0.559^2 + 2 * 4.039 * 7.2}$

=> $v = 7.647 \ m/s$

6 0

3 years ago