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3 years ago

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

Physics

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,

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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a book is sitting still on your desk. are the forces acting on the book balanced or unbalanced? explain.

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<span>They are balanced. If the forces were not balanced the book would move*. In this example, the downward force of gravity on the book is counterbalanced by the upthrust of the desk. </span>

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3 years ago

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WITCHER [35]

The answer to question one is A.
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3 years ago

A 2 kg rubber ball is thrown at a wall horizontally at 3 m/s, and bounces back the way it came at an equal speed. A 2 kg clay ba

Lyrx [107]

Answer:

THE RUBBER BALL

Explanation:

From the question we are told that

The mass of the rubber ball is $m_r = 2 \ kg$

The initial speed of the rubber ball is $u = 3 \ m/s$

The final speed at which it bounces bank $v - 3 \ m/s$

The mass of the clay ball is $m_c = 2 \ kg$

The initial speed of the clay ball is $u = 3 \ m/s$

The final speed of the clay ball is $v = 0 \ m/s$

Generally Impulse is mathematically represented as

$I = \Delta p$

where $\Delta p$ is the change in the linear momentum so

$I = m(v-u)$

For the rubber is

$I_r = 2(-3 -3)$

$I_r = -12\ kg \cdot m/s$

=> $|I_r| = 12\ kg \cdot m/s$

For the clay ball

$I_c = 2(0-3)$

$I_c = -6 \ kg\cdot \ m/s$

=> $| I_c| = 6 \ kg\cdot \ m/s$

So from the above calculation the ball with the a higher magnitude of impulse is the rubber ball

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3 years ago

A 0.10 kg mass is oscillating at a small angle from a light string of length 0.10 m.

cupoosta [38]

Answer:

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Explanation:

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2 years ago

When the current in a toroidal solenoid is changing at a rate of 0.0240 A/s , the magnitude of the induced emf is 12.4 mV . When

Gemiola [76]

Answer:

The number of turns in the solenoid is 230.

Explanation:

Given that,

Rate of change of current, $\dfrac{dI}{dt}=0.0240\ A/s$

Induced emf, $\epsilon=12.4\ mV=12.4\times 10^{-3}\ V$

Current, I = 1.5 A

Magnetic flux, $\phi=0.00338\ Wb$

The induced emf through the solenoid is given by :

$\epsilon=L\dfrac{dI}{dt}$

$L=\dfrac{\epsilon}{(di/dt)}$ ........(1)

The self inductance of the solenoid is given by :

$L=\dfrac{N\phi}{I}$ .........(2)

From equation (1) and (2) we get :

$\dfrac{\epsilon}{(di/dt)}=\dfrac{N\phi}{I}$

N is the number of turns in the solenoid

$N=\dfrac{\epsilon I}{\phi (dI/dt)}$

$N=\dfrac{12.4\times 10^{-3}\times 1.5}{0.00338 \times 0.024}$

N = 229.28 turns

N = 230 turns

So, the number of turns in the solenoid is 230. Hence, this is the required solution.

3 0

2 years ago