During which process do cells use oxygen to release stored energy?

A metal ball has a mass of 2kg and a volume of 6 m3. What is its density

Lena [83]

Answer: The answer is 333.3333 repeating

Explanation:

Divide the mass by the volume.

8 0

3 years ago

Read 2 more answers

A charged box (m=445 g, ????=+2.50 μC) is placed on a frictionless incline plane. Another charged box (????=+75.0 μC) is fixed i

victus00 [196]

The concept required to perform this exercise is given by the coulomb law.

The force expressed according to this law is given by

$F= \frac{kqQ}{r^2}$

Where,

$k = 8.99 * 10^9 N m^2 / C^2.$

q = charges of the objects

r= distance/radius

Our values are previously given, so

$q= 2.5*10^{-6}C\\Q= 75*10^{-6}C\\r=0.59$

Replacing,

$F=\frac{kqQ}{r^2}$

$F= \frac{(8.99 x 10^9)(2.5*10^{-6})(75*10^{-6})}{0.59^2}$

$F= 4.8423N$

The force acting on the block are given by,

$F-mgsin\theta = ma$

$a = \frac{F-mgsin\theta}{m}$

$a = \frac{4.8423-(0.445)(9.8)sin(35)}{0.445}$ $a = 10.31m/s^2$

Therefore the box is accelerated upward.

3 0

3 years ago

o illustrate the work-energy concept, consider the case of a stone falling from xi to xf under the influence of gravity. Using t

Phantasy [73]

Answer: force of gravity on the body due to height difference above the earth's surface

Explanation: as you increase the height of a body above ground, you do work against gravity in moving it from a point on the earth's surface to that point. So a body falling has a stored up gravito-potential energy which acts on it downward due to its mass, accelerating it downwards

Answer b): kinetic energy of the body

Explanation: the downward force produces an acceleration of magnitude 9.81m/s2 downwards which means an increasing velocity. This increasing velocity means the kinetic energy of the body is increasing (kinetic energy is proportional to velocity of the body squared)

5 0

4 years ago

A 75-hp (shaft output) motor that has an efficiency of 91.0 percent is worn out and is to be replaced by a high- efficiency moto

daser333 [38]

Convert the shaft ouput from HP to kW

Shaft output = 75HP = 55.93kW

1st: Finding for the power consumption based on 55.93kW output

Power consumption (Old) = 55.93kW / .910 = 61.46kW

Power consumption (New) = 55.93kW / .954 = 58.63kW

2nd: Total power used in kWh:

Power Used = Power consumption * load factor * Hours:

Power (Old) = 61.46kW * .75 * 4368 = 201343 kWh

Power (New) = 58.63kW * .75 * 4368 = 192072 kWh

Energy saved = 201343 kWh - 192072 kWh = 9,271 kWh

3rd: Calculating for the price:

Price = kW-Hr * $/kWh

Price (Old) = 201343kWh * $0.08/kWh = $16107.44

Price (New) = 192072 kWh * $0.08/kWh = $15365.76

Cost saved = $16107.44 - $15365.76 = $741.68/yr

4th: Setting up the cost equation:

Cost over time, F(t) = Motor_Cost + (Price * Number of Years, t)

Cost (Old) = 5449 + 16107.44*t

Cost (New) = 5520 + 15365.76*t

Equate the two to find for t when they cost equally:

5449 + 16107.44*t = 5520 + 15365.76*t

16107.44*t = 15365.76*t +71

16107.44*t - 15365.76*t = 71

741.68*t = 71

t = 71 / 741.68 = .095 years = 35 days

So the payback period is after 35 days.

6 0

3 years ago

A team of astronauts is on a mission to land on and explore a large asteroid. In addition to collecting samples and performing e

sergij07 [2.7K]

Answer:

9.60 m/s

Explanation:

The escape speed of an object from the surface of a planet/asteroid is given by:

$v=\sqrt{\frac{2GM}{R}}$

where

G is the gravitational constant

M is the mass of the planet/asteroid

R is the radius of the planet/asteroid

In this problem we have

$\rho = 2.02\cdot 10^6 g/m^3$ is the density of the asteroid

$V=3.09\cdot 10^{12}m^3$ is the volume

So the mass of the asteroid is

$M=\rho V=(2.02\cdot 10^6 g/m^3)(3.09\cdot 10^{12} m^3)=6.24\cdot 10^{18} g=6.24\cdot 10^{15} kg$

The asteroid is approximately spherical, so its volume can be written as

$V=\frac{4}{3}\pi R^3$

where R is the radius. Solving for R,

$R=\sqrt[3]{\frac{3V}{4\pi}}=\sqrt[3]{\frac{3(3.09\cdot 10^{12} m^3)}{4\pi}}=9036 m$

Substituting M and R inside the formula of the escape speed, we find:

$v=\sqrt{\frac{2(6.67\cdot 10^{-11})(6.24\cdot 10^{15})}{(9036)}}=9.60 m/s$

7 0

4 years ago