All categories

4 years ago

What are the cons of using a cup made of plastic?

Physics

1 answer:

mafiozo [28]4 years ago

3 0

-Most are not biodegradable or recyclable
-can contain harmful materials in long-term use

You might be interested in

Atoms are very small. Which measurement is the approximate diameter of a helium atom?

stich3 [128]

Atoms diameter are on the order of 60 to 600pm (picometers). This is around 6*10^-11.

3 0

3 years ago

Assuming a roughly spherical shape and a density of 4000 kg/m3, estimate the diameter of an asteroid having the average mass of

vredina [299]

Explanation:

It is given that,

Density of asteroid, $\rho=4000\ kg/m^3$

Mass of asteroid, $m=10^{17}\ kg$

We need to find the diameter of the asteroid. The formula of density is given by:

$\rho=\dfrac{m}{V}$

V is the volume of spherical shaped asteroid, $V=\dfrac{4}{3}\pi r^3$

$r^3=\dfrac{3M}{4\pi \rho}$

$r^3=\dfrac{3\times 10^{17}\ kg}{4\pi \times 4000\ kg/m^3}$

$r^3=\sqrt{5.96\times 10^{12}}$

r = 2441311.12 m

Diameter = 2 × radius

d = 4882622.24 m

$d=4.88\times 10^6\ m$

Hence, this is the required solution.

8 0

3 years ago

A wave has a wavelength of 3m and a frequency of 10Hz. How fast is the wave traveling

andreev551 [17]

Firstly, do you understand the concept of a wavelength?

7 0

3 years ago

The magnitude of the momentum of an object is the product of its mass m and speed v. If m = 3 kg and v = 1.5 m/s, what is the ma

Oliga [24]

Answer:

Momentum, p = 5 kg-m/s

Explanation:

The magnitude of the momentum of an object is the product of its mass m and speed v i.e.

p = m v

Mass, m = 3 kg

Velocity, v = 1.5 m/s

So, momentum of this object is given by :

$p=3\ kg\times 1.5\ m/s$

p = 4.5 kg-m/s

p = 5 kg-m/s

So, the magnitude of momentum is 5 kg-m/s. Hence, this is the required solution.

5 0

3 years ago

When an airplane is flying 200 mph at 5000-ft altitude in a standard atmosphere, the air velocity at a certain point on the wing

Sedaia [141]

Answer:

P1 = 0 gage

P2 = 87.9 lb/ft³

Explanation:

Given data

Airplane flying = 200 mph = 293.33 ft/s

altitude height = 5000-ft

air velocity relative to the airplane = 273 mph = 400.4 ft/s

Solution

we know density at height 5000-ft is 2.04 × $10^{-3}$ slug/ft³

so here P1 + $\frac{\rho v1^2}{2}$ = P2 + $\frac{\rho v2^2}{2}$

and here

P1 = 0 gage

because P1 = atmospheric pressure

and so here put here value and we get

P1 + $\frac{\rho v1^2}{2}$ = P2 + $\frac{\rho v2^2}{2}$

0 + $\frac{2.048 \times 10^{-3} \times 293.33^2}{2}$ $= P2 + \frac{2.048 \times 10^{-3} \times 400.4^2}{2}$

solve it we get

P2 = 87.9 lb/ft³

5 0

3 years ago