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3 years ago

3. In Figure 2, which object (A, B, or C) has the highest density? Why?

Physics

1 answer:

Ira Lisetskai [31]3 years ago

6 0

Answer: C Because it has more dots in the rectangle meaning that if it has more dots than others then it has a higher density according to the model

Explanation:

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Two wires are identical, except that one is aluminum and one is tungsten. the aluminum wire has a resistance of 0.22 ω. what is

Margaret [11]

The resistance R of a piece of wire is given by
$R=\rho \frac{L}{A}$
where $\rho$ is the resistivity of the material, L is the length of the wire and A is its cross-sectional area.

Using this formula, and labeling with A the aluminum and with T the tungsten wire, we can write the ratio between $R_T$ (the resistance of the tungsten wire) and $R_A$ (the resistance of the aluminum wire):
$\frac{R_T}{R_A}= \frac{\rho_T \frac{L}{A} }{\rho_A \frac{L}{A} }$

the two wires are identical, so L and A are the same for the two wires and simplify in the ratio, and we get:
$R_T = \frac{\rho_T}{\rho_A} R_A$

By using the resistivity of the aluminum: $\rho_A=2.65 \cdot 10^{-8} \Omega m$ and the resistivity of the tungsten: $\rho_T = 5.6 \cdot 10^{-8} \Omega m$ m we can get the resistance of the tungsten wire:
$R_T = \frac{\rho_T}{\rho_A} R_A = \frac{ 5.6 \cdot 10^{-8} \Omega m}{2.65 \cdot 10^{-8} \Omega m} (0.22 \Omega) = 0.46 \Omega$

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3 years ago

A sled with mass 8.00kg moves in a straight line on a frictionless horizontal surface. At one point in it's path, it's speed is

kati45 [8]

We will use two definitions to solve this problem. The first will be given by the conservation of energy, whereby the change in kinetic energy must be equivalent to work. At the same time, work can be defined as the product between the force by the distance traveled. By matching these two expressions and clearing for the Force we can find the desired variable.

$W = KE_f-KE_i$

$Fd = \frac{1}{2}mv_f^2-\frac{1}{2} mv_i^2$

Thus the force acting on the sled is,

$F = \frac{m}{2s} (v_f^2-v_i^2)$

Replacing,

$F = \frac{8}{2(2.5)}(6^2-4^2)$

$F = 32N$

Therefore the Force acting on the sled is 32N

8 0

3 years ago

A battery-operated car uses a 12.0-V system. Find the charge the batteries must be able to move in order to accelerate the 956kg

DiKsa [7]

<h2>Answer:</h2>

1.99 x 10⁶C

<h2>Explanation:</h2>

The electrical energy ( $E_{E}$ ) stored in the battery will give the car some kinetic energy ( $E_{K}$ ) which will cause the car to move from rest to some other point.

i.e

$E_{E}$ = $E_{K}$ ------------------(i)

$E_{E}$ = $\frac{1}{2}$ x Q x V; -------------------(ii)

Where;

Q = charge on the battery

V = potential difference or voltage of the battery = 12.0V

$E_{K}$ = ( $\frac{1}{2}$ x m x v²) - ( $\frac{1}{2}$ x m x u²) -----------------(iii)

Where;

m = mass of the car = 956kg

v = final velocity of the car = 158m/s

u = initial velocity of the car = 0 [since the car starts from rest]

<em>Substitute equations (ii) and (iii) into equation (i) as follows;</em>

$\frac{1}{2}$ x Q x V = ( $\frac{1}{2}$ x m x v²) - ( $\frac{1}{2}$ x m x u²) -----------------(iv)

<em>Substitute all necessary values into equation (iv) as follows;</em>

$\frac{1}{2}$ x Q x 12.0 = ( $\frac{1}{2}$ x 956 x 158²) - ( $\frac{1}{2}$ x 956 x 0²)

$\frac{1}{2}$ x Q x 12.0 = ( $\frac{1}{2}$ x 956 x 158²) - (0)

$\frac{1}{2}$ x Q x 12.0 = ( $\frac{1}{2}$ x 956 x 158²)

$\frac{1}{2}$ x Q x 12.0 = 11932792

6Q = 11932792

<em>Solve for Q;</em>

Q = 11932792 / 6

Q = 1988798.67 C

Q = 1.99 x 10⁶C

Therefore, the amount of charge the batteries must have is 1.99 x 10⁶C

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4 years ago

_____ varies with the gravitational force. Plz help. Due tomorrow.

MArishka [77]

The weight of anything is determined on how much gravity is pulling on the object. For example a balloon full of helium floats away but iron or concrete would fall to the ground. hope this helps :)

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3 years ago

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A train moving west with an initial velocity of 20 m/s accelerates at 4 m/s2 for 10 seconds. During this tim

dusya [7]

Answer:

400m

Explanation:

4 0

3 years ago