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Sergeeva-Olga [200]

1 year ago

10

Experimental tests have shown that hammerhead sharks can detect magnetic fields. In one such test, 100 turns of wire were wrappe

d around a 7.0-m-diameter cylindrical shark tank. A magnetic field was created inside the tank when this coil of wire carried a current of 1.5 A. Sharks trained by getting a food reward when the field was present would later unambiguously respond when the field was turned on.Part A)What was the magnetic field strength in the center of the tank due to the current in the coil?Part B)Is the strength of the coil's field at the center of the tank larger or smaller than that of the earth?- larger- smaller

1 answer:

Triss [41]1 year ago

7 0

B = µo*N*I/2r

So B = 4πx10^-7*150*1.6/2*3.5 = 4.31x10^-5T

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A radioactive substance decays exponentially. A scientist begins with 200 milligrams of a radioactive substance. After 17 hours,

lianna [129]

75.17 mg of the radioactive substance will remain after 24 hours.

Answer:

Explanation:

Any radioactive substance will obey the exponential decay behavior. So according to this behavior, any radioactive substance will be decaying in terms of exponential form of disintegration constant and Time.

Disintegration constant is the rate of decay of radioactive elements. It can be measured using the half life time of the radioactive element .While half life time is the time taken by any radioactive element to decay half of its concentration. Like in this case, at first the scientist took 200 mg then after 17 hours, it got reduced to 100 g. So the half life time of this element is 17 hours.

Then Disintegration constant = 0.6932/Half Life time

Disintegration constant = 0.6932/17=0.041

Then as per the law of disintegration constant:

$N = N_{0}e^{-xt}$

Here N is the amount of radioactive element remaining at time t and $N_{0}$ is the initial amount of sample, x is the disintegration constant.

So here, $N_{0}$ = 200 mg, x = 0.041 and t = 24 hrs.

N = 200 × $e^{-24*0.041}$ =75.17 mg.

So 75.17 mg of the radioactive substance will remain after 24 hours.

3 0

1 year ago

Suppose you push a hockey puck of mass m across frictionless ice for a time 1.0 s, starting from rest, giving the puck speed v a

EleoNora [17]

Newton's second law ...Force = momentum change/time.momentum change = Forcextme.also, F=ma -> a=F/m - the more familiar form of Newton's second law
using one of the kinematic equations for m ... V=u+at; u=0; a=F/m -> V=(F/m)xt.-> t=mV/F using one of the kinematic equations for 2m ... V=u+at; u=0; a=F/2m -> V=(F/2m)xt. -> t=2mV/F (twice as long, maybe ?)
I think I've made a mistake somewhere below, but I think that the principle is right ...using one of the kinematic equations for m ... s=ut + (1/2)at^2); s=d;u=0;a=F/m; t=1; -> d=(1/2)(F/m)=F/2musing one of the kinematic equations for 2m ... s=ut + (1/2)at^2); s=d;u=0;a=F/2m; t=1; -> d=(1/2)(F/2m)=F/4m (half as far ????? WHAT ???)

3 0

1 year ago

Read 2 more answers

Assume the radius of an atom, which can be represented as a hard sphere, is r = 1.95 Å. The atom is placed in a (a) simple cubic

Nuetrik [128]

Answer:

(a) A = $3.90 \AA$

(b) $A = 4.50 \AA$

(c) $A = 5.51 \AA$

(d) $A = 9.02 \AA$

Solution:

As per the question:

Radius of atom, r = 1.95 $\AA = 1.95\times 10^{- 10} m$

Now,

(a) For a simple cubic lattice, lattice constant A:

A = 2r

A = $2\times 1.95 = 3.90 \AA$

(b) For body centered cubic lattice:

$A = \frac{4}{\sqrt{3}}r$

$A = \frac{4}{\sqrt{3}}\times 1.95 = 4.50 \AA$

(c) For face centered cubic lattice:

$A = 2{\sqrt{2}}r$

$A = 2{\sqrt{2}}\times 1.95 = 5.51 \AA$

(d) For diamond lattice:

$A = 2\times \frac{4}{\sqrt{3}}r$

$A = 2\times \frac{4}{\sqrt{3}}\times 1.95 = 9.02 \AA$

6 0

1 year ago

5.16 An insulated container, filled with 10 kg of liquid water at 20 C, is fitted with a stirrer. The stirrer is made to turn by

Anna007 [38]

Answer:

a) W=2.425kJ

b) $\Delta E=2.425kJ$

c) $T_f=20.06^{o}C$

d) Q=-2.425kJ

Explanation:

a)

First of all, we need to do a drawing of what the system looks like, this will help us visualize the problem better and take the best possible approach. (see attached picture)

The problem states that this will be an ideal system. This is, there will be no friction loss and all the work done by the object is transferred to the water. Therefore, we need to calculate the work done by the object when falling those 10m. Work done is calculated by using the following formula:

$W=Fd$

Where:

W=work done [J]

F= force applied [N]

d= distance [m]

In this case since it will be a vertical movement, the force is calculated like this:

F=mg

and the distance will be the height

d=h

so the formula gets the following shape:

$W=mgh$

so now e can substitute:

$W=(25kg)(9.7 m/s^{2})(10m)$

which yields:

W=2.425kJ

b) Since all the work is tansferred to the water, then the increase in internal energy will be the same as the work done by the object, so:

$\Delta E=2.425kJ$

c) In order to find the final temperature of the water after all the energy has been transferred we can make use of the following formula:

$\Delta Q=mC_{p}(T_{f}-T_{0})$

Where:

Q= heat transferred

m=mass

$C_{p}$ =specific heat

$T_{f}$ = Final temperature.

$T_{0}$ = initial temperature.

So we can solve the forula for the final temperature so we get:

$T_{f}=\frac{\Delta Q}{mC_{p}}+T_{0}$

So now we can substitute the data we know:

$T_{f}=\frac{2 425J}{(10000g)(4.1813\frac{J}{g-C})}+20^{o}C$

Which yields:

$T_{f}=20.06^{o}C$

d)

For part d, we know that the amount of heat to be removed for the water to reach its original temperature is the same amount of energy you inputed with the difference that since the energy is being removed this means that it will be negative.

$\Delta Q=-2.425kJ$

3 0

1 year ago

A quarterback throws a football at 40km/hr to a receiver 50yd away. How much time does it take the ball to reach the receiver

Akimi4 [234]

Given:

Distance = 50 yard = 45.72 meter

Speed = 40 km/hr = 11.11 m/s

To find:

Time required by ball to reach the receiver = ?

Formula used:

speed = $\frac{distance}{time}$

Solution:

The speed of the ball is given by,

speed = $\frac{distance}{time}$

Thus,

Time = $\frac{distance}{speed}$

Distance = 50 yard = 45.72 meter

Speed = 40 km/hr = 11.11 m/s

Time = 4.12 second

Hence, ball reaches the receiver in 4.12 second.

3 0

1 year ago