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3 years ago

What is the force required to produce a 1.4 Nm torque when applied to a door at a 60.0º angle and 0.40 m from the hinge?

Physics

1 answer:

Marysya12 [62]3 years ago

5 0

Refer to the diagram shown below.

The component of the applied force perpendicular to the door is
F * sin(60°) = 0.866F N

Because the moment arm is 0.40 m, the torque is
(0.866F N)*(0.4 m) = 0.3464F N-m

This torque is equal to 1.4 N-m, therefore
0.3464F = 1.4
F = 4.04 N

Answer: 4.04 N

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The national high magnetic field laboratory holds the world record for creating the strongest magnetic field. for brief periods

max2010maxim [7]

Newton’s 2nd law states that Force is equal to the product of mass (m) and acceleration (a):

F = m a ---> 1

While in magnetic forces, force can also be expressed as:

F = q v B ---> 2

where,

q = total charge

v = velocity = 45 cm / s = 0.45 m / s

B = the magnetic field = 85 T

First we solve for the total charge, q:

q = 3.8 × 10^-23 g (1 mol / 23 g) (6.022 × 10^23 electrons / mol) (1.602 × 10^-19 C / electron)

q = 1.594 × 10^-19 C

We equate equations 1 and 2 then solve for acceleration a:

m a = q v B

a = q v B / m

a = [1.594 × 10^-19 C * 0.45 m / s * 85 T] / 3.8 × 10-26 kg

a = 160,437,862.2 m/s^2

Therefore the maximum acceleration of Na ions is about 160 × 10^6 m/s^2.

5 0

3 years ago

Read 2 more answers

Two Earth satellites, A and B, each of mass m, are to be launched into circular orbits about Earth's center. Satellite A is to o

Pachacha [2.7K]

(a) 0.448

The gravitational potential energy of a satellite in orbit is given by:

$U=-\frac{GMm}{r}$

where

G is the gravitational constant

M is the Earth's mass

m is the satellite's mass

r is the distance of the satellite from the Earth's centre, which is sum of the Earth's radius (R) and the altitude of the satellite (h):

r = R + h

We can therefore write the ratio between the potentially energy of satellite B to that of satellite A as

$\frac{U_B}{U_A}=\frac{-\frac{GMm}{R+h_B}}{-\frac{GMm}{R+h_A}}=\frac{R+h_A}{R+h_B}$

and so, substituting:

$R=6370 km\\h_A = 5970 km\\h_B = 21200 km$

We find

$\frac{U_B}{U_A}=\frac{6370 km+5970 km}{6370 km+21200 km}=0.448$

(b) 0.448

The kinetic energy of a satellite in orbit around the Earth is given by

$K=\frac{1}{2}\frac{GMm}{r}$

So, the ratio between the two kinetic energies is

$\frac{K_B}{K_A}=\frac{\frac{1}{2}\frac{GMm}{R+h_B}}{\frac{1}{2}\frac{GMm}{R+h_A}}=\frac{R+h_A}{R+h_B}$

Which is exactly identical to the ratio of the potential energies. Therefore, this ratio is also equal to 0.448.

(c) B

The total energy of a satellite is given by the sum of the potential energy and the kinetic energy:

$E=U+K=-\frac{GMm}{R+h}+\frac{1}{2}\frac{GMm}{R+h}=-\frac{1}{2}\frac{GMm}{R+h}$

For satellite A, we have

$E_A=-\frac{1}{2}\frac{GMm}{R+h_A}=-\frac{1}{2}\frac{(6.67\cdot 10^{-11})(5.98\cdot 10^{24}kg)(28.8 kg)}{6.37\cdot 10^6 m+5.97\cdot 10^6 m}=-4.65\cdot 10^8 J$

For satellite B, we have

$E_B=-\frac{1}{2}\frac{GMm}{R+h_B}=-\frac{1}{2}\frac{(6.67\cdot 10^{-11})(5.98\cdot 10^{24}kg)(28.8 kg)}{6.37\cdot 10^6 m+21.2\cdot 10^6 m}=-2.08\cdot 10^8 J$

So, satellite B has the greater total energy (since the energy is negative).

(d) $-2.57\cdot 10^8 J$

The difference between the energy of the two satellites is:

$E_B-E_A=-2.08\cdot 10^8 J-(-4.65\cdot 10^8 J)=-2.57\cdot 10^8 J$

4 0

3 years ago

On Earth, 1 kg = 9.8 N = 2.2 lbs. On the Moon, 1 kg = 1.6 N = 0.37 lbs. Use these relationships to answer the following question

romanna [79]

Answer:

(a) 490 N on earth

(b) 80 N on earth

(d) 270.27 kg on moon

Explanation:

We have given 1 kg = 9.8 N = 2.2 lbs on earth

And 1 kg = 1.6 N = 0.37 lbs on moon

(a) We have given mass of the person m = 50 kg

As it is given that 1 kg = 9.8 N

So 50 kg = 50×9.8 =490 N

(b) Mass of the person on moon = 50 kg

As it is given that on moon 1 kg = 1.6 N

So 50 kg = 50×1.6 = 80 N

As it is given that 1 kg = 2.2 lbs on earth

So 100 lbs = 45.4545 kg

(d) We have given weight of the person on moon = 100 lbs

As it is given that 1 kg = 0.37 lbs

So 100 lbs $\frac{100}{0.37}=270.27kg$

8 0

3 years ago

If a substance is found to be reactive flammable soluble and explosive what observation is also a physical property

Ainat [17]

Answer:

soluble

Explanation:

Chemical properties only manifest when a chemical reaction occurs. Being reactive, flammable and explosive are chemical properties, because they involve chemical reactions: the substances are changed; the chemical bonds of some substances, called reactants, are broken, and the chemical bonds are created, forming other substances, called products.

Solubility is a physical property because during dissolution no new substances are formed. You can prove it when the solvent evaporates leaving behind the same original substance.

The the observation that the substance is soluble is describing a physical property.

3 0

4 years ago

you should begin viewing a bacteria specimen with what objective lens? view available hint(s)for part g you should begin viewing

Sergio039 [100]

Some people view bacteria specimens with a 100x objective lens in order to see the smallest details.
Others may use a 10x objective lens for more general purposes, such as examining stained slides or pictures.
And still others may use a 40x objective lens to gain maximum resolution when viewing images of thick samples.

It is important to choose the appropriate magnification for your needs so that you can properly examine the specimen under study.

<h3>Why is the 100x objective lens necessary to see bacteria?</h3>

Bacteria must, of course, be viewed at the maximum magnification and resolution possible because to their small size.
Due to optical restrictions, this is approximately 1000x in a light microscope.
To improve resolution, the oil immersion method is performed. This calls for a unique 100x objective.

To learn more about bacterial specimen, visit:

brainly.com/question/1412064

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5 0

2 years ago