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2 years ago

The density of water is 1.00 g/cm3. What is its density in kg/m3?

1 answer:

3 0

1 g = 1 ÷ 1000 kg
= 0.001 kg

1 cm³ = 1 ÷ 100 ÷ 100 ÷ 100 m³
= 0.000001 m³

1 g/cm³ = 1 g / 1 cm³
= 0.001 kg / 0.000001 m³
= 1000 kg/m³

The density is 1000 kg/m³.

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A spherical ball is dropped through a liquid, explain why it reaches terminal velocity.

Alekssandra [29.7K]

Probably because of the drag coefficient and the density of the liquid.

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2 years ago

An automobile starter motor has an equivalent resistance of 0.0510 Ω and is supplied by a 12.0 V battery with a 0.0090 Ω interna

Alisiya [41]

Answer:

a) 200A

b) 10.2V

c) 2.04kW

I=80A

V=4.08V

P=0.326kW

Explanation:

Here we have a circuit of one power source and two resistors in series, the first question is asking for the current, so according to Ohm's Law:

$I=\frac{V}{R}$

Where R is the equivalent resistance of the resistors in series

$R=0.0510+0.0090=0.0600[ohm]$

$I=\frac{12.0}{0.0600}=200A$

To calculate the voltage dropped by the motor we have to apply the voltage divider rule:

$V_m=V*\frac{R_m}{R_m+R_s}\\V_m=12.0*\frac{0.0510}{0.0600}\\V_m=10.2V$

The power dissipated supplied to the motor is given by:

$P=I^2*R_m\\P=(200)^2*0.0510=2.04kW$

now solving adding a 0.0900 ohm resistor:

$I=\frac{12.0}{0.15}=80A$

$V_m=12.0*\frac{0.0510}{0.15}\\V_m=4.08V$

$P=I^2*R_m\\P=(200)^2*0.0510=0.326kW$

5 0

2 years ago

A spring on Earth has a 0.500 kg mass suspended from one end and the mass is displaced by 0.3 m. What will the displacement of t

julia-pushkina [17]

To solve this problem we will apply the concepts related to the Force of gravity given by Newton's second law (which defines the weight of an object) and at the same time we will apply the Hooke relation that talks about the strength of a body in a system with spring.

The extension of the spring due to the weight of the object on Earth is 0.3m, then

$F_k = F_{W,E}$

$kx_1 = mg$

The extension of the spring due to the weight of the object on Moon is a value of $x_2$ , then

$kx_2 = mg_m$

Recall that gravity on the moon is a sixth of Earth's gravity.

$kx_2 = m\frac{g}{6}$

$kx_2 = \frac{1}{6} mg$

$kx_2 = \frac{1}{6} kx_1$

$x_2 = \frac{1}{6} x_1$

We have that the displacement at the earth was $x_1 = 0.3m$ , then

$x_2 = \frac{1}{6} 0.3$

$x_2 = 0.05m$

Therefore the displacement of the mass on the spring on Moon is 0.05m

6 0

2 years ago

a car initially at 65.00 m/s accelerates at 22.39 m/s^2. How far has the car traveled before it comes to a stop

Gennadij [26K]

Answer:

94.35 meter

Step by step explanation

7 0

2 years ago

Joe first focuses his attention (and his eyes) on the tree. The focal length of the cornea-lens system in his eye must be ______

kotykmax [81]

Answer: The focal length of the cornea-lens system in his eye must be LESS THAN the distance between the front and back of his eye.

Explanation:

The human eye the front part of the eye is the CORNEA. This is the tough white transparent part of the eye that helps in the refraction of light rays. While the backside of the eye is the RETINA. This is the part of the eye when images are focused.

When a normal eye is at rest, parallel rays from a distant object are focused on the retina. The ability of the eye - lens to focus points at different distances on the retina is known as accomodation. The adjustment of the eye lens to focus objects of varying distances is brought about by the ciliary muscles. The have the ability to change the shape of the eye which leads to change in focal length.

When a person with normal vision looks at a distant object at infinity, the lens brings parallel rays to focus on the retina. Thus, the furthest point which the eye can see distinctly is called the far point of the eye and it's infinity for a normal eye. But Joe was able to focus his eye on the tree, meaning that the tree was within his near point. This is the nearest point at which an object is clearly seen. Therefore, when the effective focal length of the cornea-lens system changes, it changes the location of the image of any object in one's field of view.

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2 years ago