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3 years ago

In what direction does the induced current flow in the loop?

Physics

1 answer:

bogdanovich [222]3 years ago

6 0

Answer:

Clockwise and counter clockwises, depands.

Explanation:

The direction of current in a loop of wire in a magnatic field depands on the direction in which the loop is moved and the applied magnatic field.

this is determined by what is called right hand rule.

I will give one scenario, let's say that the loop is moved upwards and the applied magnatic field is into the page (if you drew the loop in 2D on a piece of paper), in this case the direction would be clockwise.

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Susan and Hannah are each riding a swing. Susan has a mass of 25 kilograms, and Hannah has a mass of 30 kilograms. Susan’s swing

Sphinxa [80]

Speed has a greater effect than mass

4 0

4 years ago

Read 2 more answers

A playground is on the flat roof of a city school, 6.2m above the street below. The vertical wall of the building is h=7.30m hig

olya-2409 [2.1K]

Answer:

a. 18.13m/s

b. 0.84m

c. 2.4m

Explanation:

a. to find the speed at which the ball was lunched, we use the horizontal component.Since the point distance from the base of the ball is 24m and it takes 2.20 secs to reach the wall,we can say that

t=distance /speed

$speed v_{x}=vcos\alpha=24/2.2=10.9m/s\\V=10.9/cos53^{0}\\V=18.13m/s$

Hence the speed at which ball was lunched is 18.13m/s

b. from the equation

$y=v_{y}t-\frac{1}{2}gt^{2}\\ v_{y}=18.13sin53=14.48m/s\\\\y=(14.48*2.2)-\frac{1}{2}9.8*2.2^{2}\\y=8.14m\\$

the vertical distance at which the ball clears the wall is

y=8.14-7.3=0.84m

c. the time it takes the ball to reach the 6.2m vertically

$6.2=14.47t-4.8t^{2}\\\\14.47t-4.8t^{2}-6.2=0\\\using\\t=-b±\frac{\sqrt{b^{2}-4ac} }{2a}\\ where \\a=-4.8, b=14.47t, c=6.2\\hence t=2.4secs$

the horizontal distance covered at this speed is

$y+24=(18.13cos53)2.4\\y=26.186-24\\y=2.12m$

4 0

3 years ago

Find p1, the gauge pressure at the bottom of tube 1. (Gauge pressure is the pressure in excess of outside atmospheric pressure.)

Taya2010 [7]

Answer:

(a). The gauge pressure at the bottom of tube 1 is $P_{1}=\rho g h_{1}$

(b). The speed of the fluid in the left end of the main pipe $\sqrt{\dfrac{2g(h_{2}-h_{1})}{(1-(\gamma)^2)}}$

Explanation:

Given that,

Gauge pressure at bottom = p₁

Suppose, an arrangement with a horizontal pipe carrying fluid of density p . The fluid rises to heights h1 and h2 in the two open-ended tubes (see figure). The cross-sectional area of the pipe is A1 at the position of tube 1, and A2 at the position of tube 2.

Find the speed of the fluid in the left end of the main pipe.

(a). We need to calculate the gauge pressure at the bottom of tube 1

Using bernoulli equation

$P_{1}=\rho g h_{1}$

(b). We need to calculate the speed of the fluid in the left end of the main pipe

Using bernoulli equation

Pressure for first pipe,

$P_{1}=\rho gh_{1}$ .....(I)

Pressure for second pipe,

$P_{2}=\rho gh_{2}$ .....(II)

From equation (I) and (II)

$P_{2}-P_{1}=\dfrac{1}{2}\rho(v_{1}^2-v_{2}^2)$

Put the value of P₁ and P₂

$\rho g h_{2}-\rho g h_{1}=\dfrac{1}{2}\rho(v_{1}^2-v_{2}^2)$

$gh_{2}-gh_{1}=\dfrac{1}{2}(v_{1}^2-v_{2}^2)$

$2g(h_{2}-h_{1})=v_{1}^2-v_{2}^2$ ....(III)

We know that,

The continuity equation

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

$v_{2}=v_{1}(\dfrac{A_{1}}{A_{2}})$

Put the value of v₂ in equation (III)

$2g(h_{2}-h_{1})=v_{1}^2-(v_{1}(\dfrac{A_{1}}{A_{2}}))^2$

$2g(h_{2}-h_{1})=v_{1}^2(1-(\dfrac{A_{1}}{A_{2}}))^2$

Here, $\dfrac{A_{1}}{A_{2}}=\gamma$

So, $2g(h_{2}-h_{1})=v_{1}^2(1-(\gamma)^2)$

$v_{1}=\sqrt{\dfrac{2g(h_{2}-h_{1})}{(1-(\gamma)^2)}}$

Hence, (a). The gauge pressure at the bottom of tube 1 is $P_{1}=\rho g h_{1}$

(b). The speed of the fluid in the left end of the main pipe $\sqrt{\dfrac{2g(h_{2}-h_{1})}{(1-(\gamma)^2)}}$

8 0

3 years ago

A size-5 soccer ball of diameter 22.6 cm and mass 426 g rolls up a hill without slipping, reaching a maximum height of 5.00 m ab

Drupady [299]

Answer

According the conservation of energy

$\dfrac{1}{2}mv_i^2+\dfrac{1}{2}I\omega_i^2+0 = mgh + 0$

I for ball = $\dfrac{2}{3}mr^2$

$\dfrac{1}{2}mv_i^2+\dfrac{1}{2} \dfrac{2}{3}mr^2\omega_i^2= mgh$

$\omega_i = \dfrac{v_i}{r}$

$v_i^2+\dfrac{2}{3}r^2(\dfrac{v_i}{r})^2 = 2gh$

$v_i^2+\dfrac{2}{3}v_i^2 = 2gh$

$v_i^2+[1+\dfrac{2}{3}]=2gh$

$v_i^2\dfrac{5}{3}=2gh$

$v_i=\sqrt{\dfrac{6gh}{5}}=\sqrt{\dfrac{6\times 9.8\times 5}{5}}$

$v_i = 7.67\ m/s$

a) $\omega_i = \dfrac{v_i}{R}$

$\omega_i = \dfrac{7.67}{0.113}$

$\omega_i =67.87\ rad/s$

b) $K_{rot} = \dfrac{1}{2}\dfrac{2}{3}mR^2\omega_i^2$

$K_{rot} = \dfrac{1}{3}\times 0.426\times 0.112^2\times 67.87^2$

$K_{rot} = 8.205\ J$

4 0

3 years ago

Each statement below represents a feature that has been claimed to exist on Mars. Sort the statements into the correct bin accor

Wittaler [7]

Answer:

Earth’s Moon:

It has no atmosphere due to weak gravity. Venus: It is too hot for liquid water.

Europa:

Is too cold for liquid water Mars: Scientists believe it used have flowing water Europe is a small moon of the planet Jupiter and according to studies, it is confirmed that the moon is too cold to have liquid water.

Venus: Venus is generally known as an extremely hold planet where liquid water would immediately turn into gas.

Mars:

is popularly identified with aliens since several theories talk about a prehistoric alien race that lived on it’s surface.

5 0

3 years ago