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3 years ago

Compute the density in g/cm^3 of a piece of metal that had mass of 0.485 kg and a volume of 52cm^3

Physics

1 answer:

steposvetlana [31]3 years ago

8 0

Answer:

9.3 g/cm³

Explanation:

First, convert kg to g:

0.485 kg × (1000 g / kg) = 485 g

Density is mass divided by volume:

D = (485 g) / (52 cm³)

D = 9.33 g/cm³

Rounding to two significant figures, the density is 9.3 g/cm³.

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A disk of a radius 50 cm rotates at a constant rate of 100 rpm. What distance in meters will a point on the outside rim travel d

Dmitry [639]

Answer:

the distance in meters traveled by a point outside the rim is 157.1 m

Explanation:

Given;

radius of the disk, r = 50 cm = 0.5 m

angular speed of the disk, ω = 100 rpm

time of motion, t = 30 s

The distance in meters traveled by a point outside the rim is calculated as follows;

$\theta = \omega t\\\\\theta = (100 \frac{rev}{\min} \times \frac{2\pi \ rad}{1 \ rev} \times \frac{1\min}{60 s} ) \times (30 s)\\\\\theta = 100 \pi \ rad\\\\d = \theta r\\\\d = 100\pi \ \times \ 0.5m\\\\d = 50 \pi \ m = 157.1 \ m$

Therefore, the distance in meters traveled by a point outside the rim is 157.1 m

6 0

3 years ago

A car with a mass of 1380 Kg is traveling at 23 m/s to the north. A truck with a mass of 1625 Kg is traveling at 26 m/s to the s

trasher [3.6K]

Answer: -3.49 m/s (to the south)

Explanation:

This problem can be solved by the Conservation of Momentum principle which establishes the initial momentum $p_{i}$ must be equal to the final momentum $p_{f}$ , and taking into account this is aninelastic collision:

Before the collision:

$p_{i}=mV_{o}+MU_{o}$ (1)

After the collision:

$p_{f}=(m+M)V_{f}$ (2)

Where:

$m=1380 kg$ is the mass of the car

$V_{o}=23 m/s$ is the velocity of the car, directed to the north

$M=1625 kg$ is the mass of the truck

$U_{o}=-26 m/s$ is the velocity of the truck, directed to the south

$V_{f}$ is the final velocity of both the car and the truck

$p_{i}=p_{f}$ (3)

$mV_{o}+MU_{o}=(m+M)V_{f}$ (4)

Isolating $V_{f}$ :

$V_{f}=\frac{mV_{o}+MU_{o}}{m+M}$ (5)

$V_{f}=\frac{(1380 kg)(23 m/s)+(1625 kg)(-26 m/s)}{1380 kg+1625 kg}$ (6)

Finally:

$V_{f}=-3.49 m/s$ The negative sign indicates the direction of the velocity is to the south

8 0

3 years ago

The distance between 4 nodes (3 sections, 2 sections= wavelength) is 15.0cm. The frequency of the source is 10Hz. What's the spe

Natalija [7]

Answer:

the speed of the waves is 150 cm/s

Explanation:

Given;

frequency of the wave, f = 10 Hz = 10

distance between 4 nodes, L = 15.0 cm

The wavelength (λ) of the wave is calculated as follows;

Node to Node = λ/2

L = 2(Node to Node) = (4 Nodes) = 2 (λ/2) = λ

Thus, λ = L = 15.0 cm

The speed (v) of the wave is calculated as follows;

v = fλ

v = 10 Hz x 15.0 cm

v = 150 cm/s

Therefore, the speed of the waves is 150 cm/s

7 0

3 years ago

Air pollutants often cause irritation in the _____ <br> system.

erastova [34]

Respiratory system.

Oversimplified Explanation: they enter the lungs, which is part of the respiratory system.

7 0

3 years ago

A sample of n2 gas occupies a volume of 746 ml at stp. What volume would n2 gas occupy at 155 ◦c at a pressure of 368 torr?

musickatia [10]

Answer:

2.41 L

Explanation:

We can solve the problem by using the ideal gas equation, which can be rewritten as:

$\frac{p_1 V_1}{T_1}=\frac{p_2 V_2}{T_2}$

where we have:

$p_1 = 1.01\cdot 10^5 Pa$ (initial pressure is stp pressure)

$V_1 = 746 mL = 0.746 L = 7.46\cdot 10^{-4}m^3$ is the initial volume

$T_1 = 0^{\circ}=273 K$ is the initial temperature (stp temperature)

$p_2 = 368 torr = 4.9\cdot 10^4 Pa$ is the final pressure

$V_2 = ?$ is the final volume

$T=155^{\circ}=428 K$ is the final temperature

By substituting the numbers inside the formula and solving for V2, we find the final volume:

$V_2 = \frac{p_1 V_1 T_2}{T_1 p_2}=\frac{(1.01\cdot 10^5 Pa)(7.46\cdot 10^{-4} m^3)(428 K)}{(273 K)(4.9\cdot 10^4 Pa)}=2.41\cdot 10^{-3} m^3$

which corresponds to 2.41 L.

7 0

3 years ago