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3 years ago

Some machines do not multiply the force that is applied to them

Physics

1 answer:

Katyanochek1 [597]3 years ago

5 0

Answer:

Machine Efficiency

Explanation:

Efficiency is the percent of work put into a machine by the user (input work) that becomes work done by the machine (output work). The output work is always less than the input work because some of the input work is used to overcome friction. Therefore, efficiency is always less than 100 percent

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A car speeds up as it rolls down a hill. which is this an example of? positive acceleration negative acceleration relative veloc

alex41 [277]

Positive acceleration

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4 years ago

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A river 500 ft wide flows with a speed of 8 ft/s with respect to the earth. A woman swims with a speed of 4 ft/s with respect to

White raven [17]

Answer:

1) $\Delta s=1000\ ft$

2) $\Delta s'=998.11\ ft.s^{-1}$

3) $t\approx125\ s$

$t'\approx463.733\ s$

Explanation:

Given:

width of river, $w=500\ ft$

speed of stream with respect to the ground, $v_s=8\ ft.s^{-1}$

speed of the swimmer with respect to water, $v=4\ ft.s^{-1}$

Now the resultant of the two velocities perpendicular to each other:

$v_r=\sqrt{v^2+v_s^2}$

$v_r=\sqrt{4^2+8^2}$

$v_r=8.9442\ ft.s^{-1}$

Now the angle of the resultant velocity form the vertical:

$\tan\beta=\frac{v_s}{v}$

$\tan\beta=\frac{8}{4}$

$\beta=63.43^{\circ}$

Now the distance swam by the swimmer in this direction be d.

so,

$d.\cos\beta=w$

$d\times \cos\ 63.43=500$

$d=1118.034\ ft$

Now the distance swept downward:

$\Delta s=\sqrt{d^2-w^2}$

$\Delta s=\sqrt{1118.034^2-500^2}$

$\Delta s=1000\ ft$

On swimming 37° upstream:

The velocity component of stream cancelled by the swimmer:

$v'=v.\cos37$

$v'=4\times \cos37$

$v'=3.1945\ ft.s^{-1}$

Now the net effective speed of stream sweeping the swimmer:

$v_n=v_s-v'$

$v_n=8-3.1945$

$v_n=4.8055\ ft.s^{-1}$

The component of swimmer's velocity heading directly towards the opposite bank:

$v'_r=v.\sin37$

$v'_r=4\sin37$

$v'_r=2.4073\ ft.s^{-1}$

Now the angle of the resultant velocity of the swimmer from the normal to the stream:

$\tan\phi=\frac{v_n}{v'_r}$

$\tan\phi=\frac{4.8055}{2.4073}$

$\phi=63.39^{\circ}$

Now let the distance swam in this direction be d'.

$d'\times \cos\phi=w$

$d'=\frac{500}{\cos63.39}$

$d'=1116.344\ ft$

Now the distance swept downstream:

$\Delta s'=\sqrt{d'^2-w^2}$

$\Delta s'=\sqrt{1116.344^2-500^2}$

$\Delta s'=998.11\ ft.s^{-1}$

Time taken in crossing the rive in case 1:

$t=\frac{d}{v_r}$

$t=\frac{1118.034}{8.9442}$

$t\approx125\ s$

Time taken in crossing the rive in case 2:

$t'=\frac{d'}{v'_r}$

$t'=\frac{1116.344}{2.4073}$

$t'\approx463.733\ s$

7 0

4 years ago

The force of air resistance acts to oppose the motion of an object moving through the air. A ball is thrown upward and eventuall

ozzi

Answer:

For a (1) net force will be greater than the weight of the ball

For b (2) net force will be lesser than the weight of the ball

Explanation:

For (a):

For a linear motion of a system, one must have to understand, according to Newtons first law of motion, which is also known as law of inertia, a body which is at motion will continue to move or a body at rest will continue to rest until an external force is applied to it. In the given case, when ball goes upward, one thing is for sure, the net force is greater than the weight of the ball, because three forces are applied during upward motion:

gravity or weight which is pulling the ball downward,

air resistance, which is also acting downward as it is creating friction between ball and air molecules, so creating hindrance in upward motion

External force to throw ball upward

Net Force = Upward force - Air friction - Weight

Since ball is going upward, so net force is greater than both weight and air friction which are pulling ball downward.

For (b):

For a linear motion of a system, one must have to understand, according to Newtons first law of motion, which is also known as law of inertia, a body which is at motion will continue to move or a body at rest will continue to rest until an external force is applied to it. In the given case, when ball goes downward, one thing is for sure, the net force is lesser than the weight of the ball, because two forces are applied during downward motion:

gravity or weight which is pulling the ball downward,

air resistance, which is acting upward as it is creating friction between ball and air molecules, so creating hindrance in downward motion

Net Force = Weight - Air friction

Since ball is going downward, so weight is greater than net force which is in this case is air friction which is pulling ball upward.

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4 years ago

Compared to the size and density of Earth, the Moon has a A. smaller diameter and lower density B. smaller diameter and higher d

MrRa [10]

Answer:

(a) Moon has smaller diameter and lower density than earth

Explanation:

Moon has smaller diameter than earth earth has a diameter of 12742 km whereas moon has a diameter of 3474.21 km

So for density density is given by $density=\frac{mass}{volume}$

From the relation we can see that density is dependent on mass and volume directly proportional to mass and inversely proportional to diameter

But in case of earth and moon mass is more dominating than volume so density is less for moon than earth

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3 years ago

What is the magnitude and direction of the electric field atradiaConsider a coaxial conducting cable consisting of a conductingr

Alexxx [7]

Answer:

E = 9.4 10⁶ N / C , The field goes from the inner cylinder to the outside

Explanation:

The best way to work this problem is with Gauss's law

Ф = E. dA = qint / ε₀

We must define a Gaussian surface, which takes advantage of the symmetry of the problem. We select a cylinder with the faces perpendicular to the coaxial.

The flow on the faces is zero, since the field goes in the radial direction of the cylinders.

The area of the cylinder is the length of the circle along the length of the cable

dA = 2π dr L

A = 2π r L

They indicate that the distance at which we must calculate the field is

r = 5 R₁

r = 5 1.3

r = 6.5 mm

The radius of the outer shell is

r₂ = 10 R₁

r₂ = 10 1.3

r₂ = 13 mm

r₂ > r

When comparing these two values we see that the field must be calculated between the two housings.

Gauss's law states that the charge is on the outside of the Gaussian surface does not contribute to the field, the charged on the inside of the surface is

λ = q / L

Qint = λ L

Let's replace

E 2π r L = λ L /ε₀

E = 1 / 2piε₀ λ / r

Let's calculate

E = 1 / 2pi 8.85 10⁻¹² 3.4 10-12 / 6.5 10-3

E = 9.4 10⁶ N / C

The field goes from the inner cylinder to the outside

5 0

3 years ago