A magnet attracts more iron dusts at its end

What is quantitative?

7nadin3 [17]

To relate or measure the by the quantity of something, not against the quantity

5 0

4 years ago

The principle of work states that the ratio of work output to work input is always

snow_lady [41]

Answer:

work output is always less than work input - the ratio is less than 1.

Explanation:

This principle comes from the fact that a machine or system cannot produce more work than is supplied to it, because this would violate the energy conservation law (work is a type of mechanical energy).

In theoretical machines called "ideal machines" the input work is the same as the output work, but these machines are only theoretical because in real applications there is always some type of energy loss, either in heat produced by a machine or processes for its operation, for this reason the output work is always less than the input work.

Regarding the ratio work output to work input:

$\frac{WO}{WI} < 1$

because work input WI is always greater than work output WO.

7 0

4 years ago

Three particles lie in the xy plane. Particle 1 has mass m1 = 6.7 kg and lies on the x-axis at x1 = 4.2 m, y1 = 0. Particle 2 ha

krek1111 [17]

Answer:

$F=18.58\times 10^{-11}\ N$

$\theta=30.276^{\circ}$

Explanation:

Given:

mass of first particle, $m_1=6.7\ kg$

mass of second particle, $m_2=5.1\ kg$

mass of third particle, $m_3=3.7\ kg$

coordinate position of first particle in meters, $(x_1,y_1)\equiv(4.2,0)$

coordinate position of second particle in meters, $(x_2,y_2)\equiv(0,2.8)$

coordinate position of third particle in meters, $(x_3,y_3)\equiv(0,0)$

Now, gravitational force on particle 3 due to particle 1:

$F_{31}=G\frac{m_1.m_3}{r_{31}^2}$

$F_{31}=6.67\times 10^{-11} \times \frac{6.7\times 3.7}{4.2^2}$

$F_{31}=9.37\times 10^{-11}\ N$

towards positive Y axis.

gravitational force on particle 3 due to particle 2:

$F_{32}=G\frac{m_2.m_3}{r_{21}^2}$

$F_{32}=6.67\times 10^{-11} \times \frac{5.1\times 3.7}{2.8^2}$

$F_{32}=16.05\times 10^{-11}\ N$

towards positive X axis.

Now the net force

$F=\sqrt{F_{31}\ ^2+F_{32}\ ^2}$

$F=\sqrt{(10^{-11})^2(9.37^2+16.05^2)}$

$F=18.58\times 10^{-11}\ N$

For angle in counterclockwise direction from the +x-axis

$tan\theta=\frac{9.37\times 10^{-11}}{16.05\times 10^{-11}}$

$\theta=30.276^{\circ}$

4 0

4 years ago

This phenomenon is known as the bends. If a scuba diver rises quickly from a depth of 25.0 m in Lake Michigan (which is fresh wa

Sveta_85 [38]

Explanation:

Attachments of the solution are below.

6 0

3 years ago

A certain light truck can go around a flat curve having a radius of 150 m with a maximum speed of 35.5 m/s. a) What is the coeff

postnew [5]

Answer:

The coefficient of friction present between the roadway and the wheels of the truck is 0.833.

Explanation:

Given:

Radius of the curve (R) = 150 m

Maximum speed of truck (v) = 35.5 m/s

Let the coefficient of friction between the roadway and the wheels of the truck be "μ".

As the truck is moving around a circular curve. So, the force acting on it is centripetal force which acts in the radial inward direction towards the center of the circular curve.

The centripetal force acting on the truck is given as:

$F_c=\frac{mv^2}{R}$

Now, the friction between the roadway and the wheels of the truck is responsible for providing the necessary centripetal force. So, frictional force is equal to the centripetal force necessary for circular motion.

Frictional force is given as:

$f=\mu N$