All categories

3 years ago

Whatis the CGS andSI unit of weight?

Physics

2 answers:

svetlana [45]3 years ago

5 0

Answer:

$\boxed{\mathrm{view \ explanation}}$

Explanation:

Weight is a force that acts on all objects on Earth.

The CGS unit for force is dyne.

The SI unit for force is newtons (N).

Goryan [66]3 years ago

3 0

Weight is force.

The CGS unit of force is the dyne.

The SI unit of force is the Newton.

You might be interested in

PLEASE HELP!!! GIVING BRAINLIEST!! ill also answer questions that you have posted if you answer these correctly!!!! (40pts)

denpristay [2]

Answer:

False, true

Explanation:

3 0

3 years ago

Read 2 more answers

A plane travels down a runway 2750 m before it lifts off at an angle of 37 degrees from the horizontal. The plane has traveled 1

ss7ja [257]

Answer: 4.236km

Explanation:

Let's define the point (x, y) as:

x = horizontal distance moved.

y = vertical distance moved.

If the plane starts in the point (0, 0) then:

"A plane travels down a runway 2750 m before it lifts off..."

At this time, the position will be:

P = (0 + 2750m, 0) = (2750m, 0).

"it lifts off at an angle of 37 degrees from the horizontal. The plane has traveled 1.8 km since its wheels left the ground."

In this case, as the angle is measured from the horizontal, the components will be:

x = 1.8km*cos(37°) = 1.438km

y = 1.8km*sin(37°) = 1.083 km

Then the new position is:

P = (2750m + 1.438 km, 0 + 1.083 km)

Let's write it using the same units for all the quantities:

we know that

1km = 1000m

Then:

2750m = (2750/1000) km = 2.750 km.

Then we can write the new position as:

P = (2.750 km + 1.438km, 1.083km) = (4.188km, 1.083km)

Now, we define the displacement as the distance between the final position and the initial position.

The distance between two points (a, b) and (c, d) is:

D = √( (a c)^2 + (b - d)^2)

In this case the points are:

(0, 0) for the initial position

(4.188km, 1.083km) for the final position.

And the displacement will be:

D = √( (4.188km - 0)^2 + (1.083 - 0)^2) = 4.236km

5 0

2 years ago

Bodies weighing 1 kilogram and 5 kilograms lie on a smooth horizontal surface. If a traction force of 0.6 N acts on another 5 kg

natima [27]

0.6/5,1,5

so calculate it

not so sure though

6 0

3 years ago

What is someones target heart rate at 70% at age 17

baherus [9]

Answer: normal

Explanation:

6 0

2 years ago

I've got an energy and work problem. The premise of the problem is:

Alenkasestr [34]

Refer to the diagram shown below.

μ = the coefficient of dynamic friction between the crate and the ramp.

1. The applied force of F acts over a distance, d.
The work done is F*d.

2. The component of the weight of the crate acting down the ramp is
mg sin(30) = 0.5mg.
The work done by this force is 0.5mgd.

3. The normal force is N = mgcos(30) = 0.866mg.
This force is perpendicular to the ramp, therefore the work done is zero.

4. The frictional force is μN = μmgcos(30) = 0.866μmg.
The work done by the frictional force is 0.866μmgd.

5. The total force acting on the crate up the ramp is
F - mgsin(30) - μmgcos(30) = F - mg(0.5 - 0.866μ)

6. The work done on the crate by the total force is
d*(F - 0.5mg - 0.866μmg)

7 0

3 years ago

Whatis the CGS andSI unit of weight?​

Whatis the CGS andSI unit of weight?