What is the metric unit of force? kg kg/cm^3 kg/cm^2 N

Answers

Answer 1
Answer: force is found by multiplying mass (kg) and acceleration (m/s^2), so the metric unit of force is kg*m/s^2 or N (newtons)

Related Questions

A 111 ‑turn circular coil of radius 2.11 cm and negligible resistance is immersed in a uniform magnetic field that is perpendicular to the plane of the coil. The coil is connected to a 14.1 Ω resistor to create a closed circuit. During a time interval of 0.125 s, the magnetic field strength decreases uniformly from 0.669 T to zero. Find the energy, in millijoules, that is dissipated in the resistor during this time interval.
Robert Hook discovered cells when viewing a _____ under a microscope.boxbloodcorkplant
Among types of research, only _______ can establish causality
The temperature at which the air can hold no more water isA. the dew point.B. relative humidity.C. runoff.D. evaporation.
The method of heat transfer by which the sun warms the earth isA) conduction B) convection C) radiation D) solaration

What is single kind of organism that can reproduce on its own?

Answers

A cell will reproduce on its own
A single cell organism can reproduce on its own, and a cell can reproduce on its own

You and a friend ride bicycles to school. Both of you start at the same instant from your house, you riding at 13 m/s and your friend riding at 18 m/s . During the trip your friend has a flat tire that takes him 13 min to fix. He then continues the trip at the same speed of 18 m/s . If the distance to school is 13 km , which of you gets to school first? By comparing the total journey time for you and your friend, this question can be answered.

Answers

V1 = 13 m/s
V2 = 18 m/s
L = 13 km = 13000 m
t0 = 13 min = 780 s

t1 = L/V1 = 13000/13 = 1000 seconds = 16.67 min
t2 = t0 + L/V2 = 780 + 13000/18 = 1500 s = 25 min
You will get to school first

According to Charles's law whenever the temperature of a gas at constant pressure decreases the volume increases

Answers

I believe the answer is false
 

The answer is false.

In terms of density and humidity, which conditions characterize a high-pressure area? A.high density, high humidity


B.high density, low humidity


C.low density, low humidity


D.low density, high humidity

Answers

Answer
A
Source: Meteorologist/Storms spotter & chaser

Answer:

A (high density and high humidity)

Explanation: Hope this helps :)

If an athlete expends 3480. kJ/h, how long does she have to play to work off 1.00 lb of body fat? Note that the nutritional calorie (Calorie) is equivalent to 1 kcal, and one pound of body fat is equivalent to about 4.10 × 103 Calories.

Answers

Answer:

The time required by the Athlete to work off 1.00 lb of body fat = 0.296 minute

Explanation:

1 lb of body fat = 4.1 k cal

1 k cal = 4.184 Kilo joule

1 lb of body fat = 4.1 × 4.184 = 17.1544 Kilo joule

Athlete expends 3480 Kilo joule in one hour

⇒ Time required to expand 3480 Kilo joule  = 60 minute

⇒ Time required to expand 1 Kilo joule = (60)/(3480)(min)/(KJ)

⇒ Time required to expand 17.1544 Kilo joule = (60)/(3480) × 17.1544 = 0.296 min

Therefore the time required by the Athlete to work off 1.00 lb of body fat = 0.296 minute

A small laser emits light at a power of 5.00 mW. The laser beam is focussed until its diameter matches the 1266 nm diameter of a sphere placed in its path. The sphere is perfectly absorbing and has a density of 5.00 x10 3 kg/m 3.a. Calculate the average energy per unit time per unit area (S). b. Find the maximum values of E and B. c. Calculate the radiation pressure on the sphere and the force exerted on the sphere. d. Find the acceleration of the sphere.

Answers

a). The half of the sphere facing the beam has surface area of

                     2 (pi) (R²)

                 = (2 pi) (633 x 10⁻⁹ m)²

                 = (2 pi) (4.01 x 10⁻¹³) m²

                 =      2.52 x 10⁻¹² m²

Average power =  (5 x 10⁻³ W) / (2.52 x 10⁻¹² m²)

                           =  1.984 GW / m²  .

b).
c).
d).  I don't remember how to do these parts of the question.
However, I do sincerely believe that the solution of part-'a' was
a fair return for the penurious and parsimonious award of 5-points.