All categories

2 years ago

A 2.29N force is applied to the plunger of a hypodermicneedle.

Physics

1 answer:

Ede4ka [16]2 years ago

6 0

Answer:

0.04651 N

Explanation:

$F_1$ = Force on plunger = 2.29 N

$d_1$ = Diameter of plunger = 1.27 cm

$d_2$ = Diameter of needle = 0.181 cm

$F_2$ = Force on needle

From Pascal's law we have

$\dfrac{F_1}{A_1}=\dfrac{F_2}{A_2}\\\Rightarrow F_2=\dfrac{F_1\times A_2}{A_1}\\\Rightarrow F_2=\dfrac{2.29\times \pi\dfrac{0.181^2}{4}}{\dfrac{\pi 1.27^2}{4}}\\\Rightarrow F_2=0.04651\ N$

The force with which the fluid leaves the needle is 0.04651 N

You might be interested in

A wire that is 0.65 m long and carrying a current of 8.2 A is at right angles to a uniform magnetic field. The force on the wire

Luda [366]

Answer:

0.075 T

Explanation:

When a current-carrying wire is immersed in a region with magnetic field, the wire experiences a force, given by

$F=ILB sin \theta$

where

I is the current in the wire

L is the length of the wire

B is the strength of the magnetic field

$\theta$ is the angle between the direction of I and B

In this problem we have:

L = 0.65 m is the length of the wire

I = 8.2 A is the current in the wire

F = 0.40 N is the force experienced by the wire

$\theta=90^{\circ}$ since the current is at right angle with the magnetic field

Solving the formula for B, we find the strength of the magnetic field:

$B=\frac{F}{IL sin \theta}=\frac{0.40}{(8.2)(0.65)(sin 90^{\circ})}=0.075 T$

3 0

2 years ago

A stretched string is 2.11 m long and has a mass of 19.5 g. When the string is oscillated at 440 Hz, which is the frequency of t

Katarina [22]

Answer:

efsfefsfsdf

Explanation:

efsfesef

7 0

2 years ago

If all the light waves are reflected in equal amounts off an object then the object will appear

Fantom [35]

The object would appear bright

3 0

3 years ago

Read 2 more answers

A closed-end organ pipe is used to produce a mixture of sounds. The third and fifth harmonics in the mixture have frequencies of

nexus9112 [7]

Answer:

$F_1=366.67Hz$

Explanation:

From the question we are told that:

Frequency of 3rd harmonics $F_3=1100$

Frequency of 5th harmonics $F_3=1833$

Generally the equation for Wavelength at 3rd Harmonics is mathematically given by

$\lambda_3=\frac{4}{3}l$

Therefore

$F_3=\frac{3v}{4l}$

Generally the equation for Wavelength at 1st Harmonics is mathematically given by

$\lambda_1=\frac{4}{1}l$

Therefore

$F_1=\frac{v}{4l}$

Generally the equation for the frequency of the first harmonic is mathematically given by

$F_1=\frac{F_3}{3}$

$F_1=\frac{1100}{3}$

$F_1=366.67Hz$

7 0

2 years ago

A perfectly flexible cable has length L, and initially it is at rest with a length Xo of it hanging over the table edge. Neglect

zaharov [31]

Answer:

$X=X_o+\dfrac{1}{2}gt^2$

Explanation:

Given that

Length = L

At initial over hanging length = Xo

Lets take the length =X after time t

The velocity of length will become V

Now by energy conservation

$\dfrac{1}{2}mV^2=mg(X-X_o)$

$V=\sqrt{2g(X-X_o)}$

We know that

$\dfrac{dX}{dt}=V$

$\dfrac{dX}{dt}=\sqrt{2g(X-X_o)}$

$\sqrt{2g}\ dt=(X-X_o)^{-\frac{1}{2}}dX$

At t= 0 ,X=Xo

So we can say that

$X=X_o+\dfrac{1}{2}gt^2$

So the length of cable after time t

$X=X_o+\dfrac{1}{2}gt^2$

6 0

2 years ago