Ask question

Ask question

All categories

3 years ago

15

A scalloped hammerhead shark swims at a steady speed of 1.5 m/s with its 88 cm -cm-wide head perpendicular to the earth's 56 μT

magnetic field. What is the magnitude of the emf induced between the two sides of the shark's head? Express your answer using two significant figures.

1 answer:

Nadya [2.5K]3 years ago

8 0

Answer:

7.4 x 10⁻⁵ T.

Explanation:

Emf induced = BLv

Where B is magnetic field , L is length of rod or any other object moving in magnetic field and v is velocity of moving rod.

Here B = 56 x 10⁻⁶ T

L = 88 X 10⁻² m

v = 1.5 m/s

emf induced = 56 x 10⁻⁶ x 88 x 10⁻² x 1.5

= 7.4 x 10⁻⁵ T.

You might be interested in

At what value of angle between two vectors will the resultant of the two vectors be maximum?

yan [13]

The resultant is maximum when the angle between the two vectors is zero

Explanation:

The formula to calculate the resultant of two vectors A, B is the following:

$R=\sqrt{A^2+B^2+2ABcos \theta}$

where

A is the magnitude of vector A

B is the magnitude of vector B

$\theta$ is the angle between the directions of A and B

This formula can be derived by applying the parallelogram law.

From the formula, we observe that:

- When $\theta=0^{\circ}$ (vectors A and B parallel), $cos \theta = 1$ . so the resultant becomes simply

$R=\sqrt{A^2+B^2 + 2AB}=\sqrt{(A+B)^2}=A+B$

- When $\theta = 90^{\circ}$ (vectors A and B perpendicular), $cos \theta = 0$ , so the resultant becomes

$R=\sqrt{A^2+B^2}$

- When $\theta=180^{\circ}$ (vectors A and B anti-parallel), $cos \theta = -1$ , so the resultant becomes

$R=\sqrt{A^2+B^2 - 2AB}=\sqrt{(A-B)^2}=|A-B|$

By looking at the three cases and at the formula, we see that the maximum value of the resultant is when the angle between the two vectors is zero, since in that case the resultant is simply obtained by adding the magnitudes of the two vectors.

Learn more about vector addition:

brainly.com/question/4945130

brainly.com/question/5892298

#LearnwithBrainly

3 0

3 years ago

Automobile air bags use the decomposition of sodium azide as their sources of gas for rapid inflation, represented in the reacti

NeTakaya

Answer : The mass of $NaN_3$ required is 71.175 grams.

Explanation :

To calculate the moles of nitrogen gas, we use the equation given by ideal gas :

PV = nRT

where,

P = Pressure of nitrogen gas = 763 torr

V = Volume of the nitrogen gas = 40.0 L

n = number of moles of gas = ?

R = Gas constant = $62.364\text{ L.torr }mol^{-1}K^{-1}$

T = Temperature of helium gas = $25^oC=273+25=298K$

Putting values in above equation, we get:

$763torr\times 40.0L=n\times 62.364\text{ L.torr }mol^{-1}K^{-1}\times 298K\\\\n=1.642mol$

Now we have to calculate the moles of $NaN_3$ .

The balanced chemical reaction is:

$2NaN_3(s)\rightarrow 2Na(s)+3N_2(g)$

From the balanced chemical reaction, we conclude that

As, 3 moles of $N_2$ produced from 2 moles $NaN_3$

So, 1.642 moles of $N_2$ produced from $\frac{2}{3}\times 1.642=1.095$ moles $NaN_3$

Now we have to calculate the mass of $NaN_3$ .

Molar mass of $NaN_3$ = 65 g/mol

$\text{Mass of }NaN_3=\text{Moles of }NaN_3\times \text{Molar mass of }NaN_3$

$\text{Mass of }NaN_3=1.095mole\times 65g/mole=71.175g$

Therefore, the mass of $NaN_3$ required is 71.175 grams.

5 0

3 years ago

WILL MARK BRAINLIEST!<br> WORTH A LOT OF POINTS!

Shalnov [3]

Answer:

A is your anwser

Explanation:

4 0

3 years ago

Read 2 more answers

Two copper rods are separated by a small gap at B. Rod AB has a diameter of 200mm and rod BC has a diameter of 150mm. Find the f

gulaghasi [49]

Answer:

A)3.8196 * $10^{9}$ Newtons B) 2.153 * $10^{9}$ Newtons

Explanation:

Force = Pressure * Area.

Pressure is given as 120GPa

Area = Cross Sectional Area of Rod = Area Of A circle = π $R^{2}$ .

Area Expansivity β = $\frac{Change in Area}{Original Area * Temperature Rise}$

Area Expansivity β = 2α

α = 16.9x10-6/ ° C.

Radius of Rod AB = Half 0f Diameter, 200mm = $\frac{0.2m}{2}$ = 0.1 m

Radius of Rod BC = Half Of Diameter, 150mm = $\frac{0.15}{2}$ = 0.075 m.

Note: To convert from millimeter to meter you divide by 1000.

Area of Rod AB = π * $0.1^{2}$ = 0.0314 $m^{2}$

Area of Rod BC = π * $0.075^{2}$ = 0.0177 $m^{2}$ .

Change in Area = 2α * Original Area * Temperature Rise. Therefore for

Rod AB = 2 * 16.9* $10^{-6}$ * 0.0314 * 30 = 3.183 * $10^{-5}$ $m^{2}$ .

For Rod BC we have = 2 * 16.9* $10^{-6}$ * 0.0177 * 30 = 1.794∈-5 $m^{2}$ .

The forces on each rods will be given by.

Force AB = Pressure * Area of Rod AB

= 120GPa * 3.183 * $10^{-5}$ gives 3.8196 * $10^{9}$ Newtons.

Force BC = Pressure * Area of Rod BC

= 120GPa * 1.794 * $10^{-5}$ gives 2.153 * $10^{9}$ Newtons

7 0

3 years ago

The shape of the earth is an _______ ________

Nastasia [14]

The shape of the earth is an <span>oblate spheroid.</span>

3 0

3 years ago