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3 years ago

He generation of a magnetic field by an electric current is

Physics

1 answer:

USPshnik [31]3 years ago

6 0

Answer: electromagnetism.

Explanation:The use of coils of wires produces a relationship between electricity and magnetism that gives us another magetism called electromagnetism.

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Water flows into a horizontal, cylindrical pipe at 1.4 m/s. the pipe then narrows until its diameter is halved. what is the pres

inna [77]

According to the Bernoulli's equation,the pressure difference between the wide and narrow ends of the pipe is given by

$\Delta P= \frac{1}{2} \rho ( v^2_{2} - v^2_{1} )$

Here, $v_{1}$ is the velocity of water through wide ends of cylindrical pipe and $v_{2}$ is the velocity of water through narrow ends of cylindrical pipe.

Given, $v_{1} =1.4 m/s$

Now from equation continuity,

$v_{1} A_{1} = v_{2} A_{2}$ .

Here, $A_{1}$ and $A_{2}$ are cross- sectional areas of wide and narrow ends of cylindrical pipe.

As pipe is circular, so

$v_{1} \pi r^2_{1} = v_{2} \pi r^2_{2}$ .

At the second point, the diameter is halved, which means the radius is also halved. Therefore,

$v_{1} r^2_{1} = v_{2}(\frac{1}{2} r_{1})^2 \\\\ v_{2} = 4 v_{1}$

$v_{2} = 4 \times 1.4 = 5.6 m/s$

Substituting these values with the density of water is $1000 \ kg/m^3$ in pressure difference formula we get.

$\Delta P= \frac{1}{2} \rho ( v^2_{2} - v^2_{1} )=\frac{1}{2}\times 1000 kg/m^3(5.6^2-1.4^2)\\\\ \Delta P = 14700\ Pa$

3 0

3 years ago

Read 2 more answers

An object with a mass of 5.0 kg accelerates 2.8 m/s2 towards right when an unknown force is applied to it. What is the amount of

RoseWind [281]

Answer:

Explanation:

5 0

3 years ago

Two circus performers rehearse a trick in which a ball and a dart collide. Horatio stands on a platform 9.8 m above the ground a

xenn [34]

Answer:

Time = 0.55 s

Height = 8.3 m

Explanation:

The ball is dropped and therefore has an initial velocity of 0. Its acceleration, g, is directed downward in the same direction as its displacement, $h_b$ .

The dart is thrown up in which case acceleration, g, acts downward in an opposite direction to its displacement, $h_d$ . Both collide after travelling for a time period, t. Let the height of the dart from the ground at collision be $h_d$ and the distance travelled by the ball measured from the top be $h_b$ .

It follows that $h_d+h_b=9.8$ .

Applying the equation of motion to each body (h = v_0t + 0.5at^2),

Ball:

$h_b=0\times t + 0.5\times 9.8t^2$ (since $v_{b0} =0$ .)

$h_b=4.9t^2$

Dart:

$h_d=17.8\times t - 0.5\times9.8t^2$ (the acceleration is opposite to the displacement, hence the negative sign)

$h_d=17.8\times t - 4.9t^2$

But

$h_b+h_d =9.8$

$17.8\times t - 4.9t^2+4.9t^2 =9.8$

$17.8\times t = 9.8$

$t = 0.55$

The height of the collision is the height of the dart above the ground, $h_d$ .

$h_d=17.8\times t - 4.9t^2$

$h_d=17.8\times 0.55 - 4.9\times(0.55)^2$

$h_d=9.79 - 1.48225$

$h_d=8.3$

8 0

3 years ago

Can anyone help me with questions a and c

Ganezh [65]

Answer:

n=sin i/sin r

n= -0.305/-0.428

n=0.713

sin c=1/n

sin c=1/0.713

sin c= 1.403

c=sin⁻¹(1.403)

c= 40.842°

Explanation:

i hope it will be helpful

plzzz mark as brainliest

5 0

2 years ago

Calculate the ratio of the drag force on a jet flying at 1190 km/h at an altitude of 7.5 km to the drag force on a prop-driven t

Bond [772]

Answer:

$\frac{D_{jet}}{D_{prop}}=2.865$

Explanation:

Given data

Speed of jet Vjet=1190 km/h

Speed of prop driven Vprop=595 km/h

Height of jet 7.5 km

Height of prop driven transport 3.8 km

Density of Air at height 10 km p7.8=0.53 kg/m³

Density of air at height 3.8 km p3.8=0.74 kg/m³

The drag force is given by:

$D=\frac{1}{2}CpAv^2\\$

The ratio between the drag force on the jet to the drag force on prop-driven transport is then given by:

$\frac{D_{jet}}{D_{prop}}=\frac{(1/2)Cp_{7.5}Av_{jet}^2}{1/2)Cp_{3.8}Av_{prop}^2} \\\frac{D_{jet}}{D_{prop}}=\frac{p_{7.5}v_{jet}^2}{p_{3.8}v_{prop}}\\\frac{D_{jet}}{D_{prop}}=\frac{(0.53)(1190)^2}{(0.74)(595)^2}\\ \frac{D_{jet}}{D_{prop}}=2.865$

4 0

3 years ago