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3 years ago

What is the key external force that allows a man to push a wheelbarrow

Physics

1 answer:

ivolga24 [154]3 years ago

8 0

Answer:

Friction

Explanation:

The key and most important external forces that allows a man to push a wheelbarrow is frictional force.

Frictional force is a force that opposes motion and will bring it to a halt.

The reason why motion is possible is due to frictional force.

Without friction, a body will move and continue in motion infinitely. No one will be able to control motion.

Therefore, frictional force is an external force that will allow a man to push a wheelbarrow.

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A single-engine helicopter has two rotors; a main rotor and a tail rotor. The main rotor has a diameter of 14.4 m and rotates at

Ber [7]

Answer:

339.12 m/s, 343.31 m/s

Explanation:

For main rotor

diameter, d = 14.4 m

radius, r = 7.2 m

f = 450 rev/min = 450 / 60 = 7.5 rps

Angular speed, ω = 2 x π x f = 2 x 3.14 x 7.5 = 47.1 rad/s

linear speed, v = r x ω = 7.2 x 47.1 = 339.12 m/s

For tail rotor

diameter, d = 1.6 m

radius, r = 0.8 m

f = 4100 rev/min = 4100 / 60 = 68.33 rps

Angular speed, ω = 2 x π x f = 2 x 3.14 x 68.33 = 429.13 rad/s

linear speed, v = r x ω = 0.8 x 429.13 = 343.31 m/s

6 0

3 years ago

1) When making a digital animation of a person running on a sidewalk in a scene, which parameter would be an initial condition?

lbvjy [14]

Answer:

The position of the person's feet. a running shoe company studying how different surfaces affect the life of a shoe tread

Explanation:

100%

4 0

3 years ago

A 49.6-kg crate rests on a level floor at a shipping dock. The coefficients of static and kinetic friction are 0.584 and 0.399,

NISA [10]

Answer:

(a) Force must be grater than 283.87 N

(B) Force will be equal to 193.945 N

Explanation:

We have given mass of the crate m = 49.6 kg

Acceleration due to gravity $g=9.8m/sec^2$

Coefficient of static friction $\mu _s=0.584$

Coefficient of kinetic friction $\mu _k=0.399$

(a) Static friction force is given by $F_S=\mu _smg=0.584\times 49.6\times 9.8=283.8707N$

So to just start the crate moving we have to apply more force than 283.87 N

(B) This force will be equal to kinetic friction force

We know that kinetic friction force is given by $F_k=\mu _kmg=0.399\times 49.6\times 9.8=193.945N$

3 0

3 years ago

One problem for humans living in outer space is that they are apparently weightless. One way around this problem is to design a

Maurinko [17]

This question is not complete.

The complete question is as follows:

One problem for humans living in outer space is that they are apparently weightless. One way around this problem is to design a space station that spins about its center at a constant rate. This creates “artificial gravity” at the outside rim of the station. (a) If the diameter of the space station is 800 m, how many revolutions per minute are needed for the “artificial gravity” acceleration to be 9.80m/s2?

Explanation:

a. Using the expression;

T = 2π√R/g

where R = radius of the space = diameter/2

R = 800/2 = 400m

g= acceleration due to gravity = 9.8m/s^2

1/T = number of revolutions per second

T = 2π√R/g

T = 2 x 3.14 x √400/9.8

T = 6.28 x 6.39 = 40.13

1/T = 1/40.13 = 0.025 x 60 = 1.5 revolution/minute

6 0

3 years ago

25 points! Please someone help me, this seems like a pretty easy question but I can’t seem to get it right, please list the give

elixir [45]

Answer:

1 my brother say that

Explanation:

i know my brother said it

4 0

3 years ago